\sloppy

1 Information measures and Typicality

1.1 Entropy

Mrs Gerber’s Lemma (MGL).
Let H − 1:[0, 1] → [0, 1 ⁄ 2] be the inverse of the binary entropy function, i.e., H(H − 1(v)) = v.
Scalar MGL
Let X be the binary random variable and let U be an arbitrary random variable. If Z ~ Bern(p) is independent of (X, U) and Y = XZ, then
\mathchoiceH(Y|U) ≥ H(H − 1(H(X|U)*p))H(Y|U) ≥ H(H − 1(H(X|U)*p))H(Y|U) ≥ H(H − 1(H(X|U)*p))H(Y|U) ≥ H(H − 1(H(X|U)*p))
Vector MGL
Let Xn be a binary random vector and U be an arbitrary random variable. If Zn is a vector of independent and identically distributed Bern(p) random variables independent of (Xn, U) and Yn = XnZn, then
(H(Yn|U))/(n) ≥ HH − 1(H(Xn|U))/(n)*p
a*b = a⋅(1 − b) + b(1 − a)
H − 1-инверзна функција
The proof of the lemma follows by the convexity of the function H(H − 1(v)*p) in v and using induction; see Problem 2.5
Binary analogue of Mrs Gerber’s Lemma to the EPI
We will introduce an analogue between the EPI (Entropy Power Inequality) and the Mrs Gerber Lemma. The EPI states that for any independent X ~ f(x) and Z ~ f(z)
22h(X + Y) ≥ 22h(X) + 22h(Y)
if we define η = (1)/(2)log(2πex) then,
η(η − 1(x)) = x = η − 1(η(x))
x = (1)/(2πe)⋅22η = η(η − 1(x))
\mathchoiceη − 1(x) = (1)/(2πe)22xη − 1(x) = (1)/(2πe)22xη − 1(x) = (1)/(2πe)22xη − 1(x) = (1)/(2πe)22x → η(η − 1(x)) = (1)/(2)log(2πeη − 1(x)) = (1)/(2)log(22x) = x
(1) 22h(X + Y) ≥ 22h(X) + 22h(Y) ⇒ η − 1(h(X + Y)) ≥ η − 1(h(X)) + η − 1(h(Y))
H(p) =  − p⋅logp − (1 − p)log(1 − p) =  − logpp(1 − p)(1 − p) pp(1 − p)(1 − p) = 2 − H(p)
H(H − 1) = p
Now let X and Y be independent r.v. with finite alphabet and entropy H(X) < 1,  H(Y) < 1 respectively. We define σ(x) = h − 1(H(x))
H(X, Y) = H(X) + H(Y|X) ≤ H(X) + H(Y)
X ~ B(q) Y ~ B(p)
X, Y|XY 0 1 00 1 0 01 0 1 10 0 1 11 1 0 X, Y|XY 0 1 00 pq 0 01 0 q(1 − p) 10 0 (1 − q)p 11 (1 − p)(1 − q) 0 p(XY) 1 − p(1 − q) + (1 − p)q p(1 − q) + (1 − p)q 1 − p*q \mathchoicep*qp*qp*qp*q
H(XY) = H(p*q)
H − 1(H(p*q)) = p*q
σ(p*q) = p*q
σ(XY) ≥ σ(X)*σ(Y) = σ(X)(1 − σ(Y)) + σ(Y)(1 − σ(X)) = σ(X) − 2σ(X)σ(Y) + σ(Y)
(2) h − 1(H(XY)) ≥ h − 1(H(X))*h − 1(H(Y))
Notice the similarity of 1↑ and 2↑. Now we set Y ~ B(p)(X, U) thus we obtain
h − 1(H(XY)|U) = h − 1(H(X|U))*h − 1(H(Y|U)) = h − 1(H(X|U))*h − 1(H(Y))
h − 1(H(XY)|U) = h − 1(H(X|U))*p
Taking h() on both sides (which is possible since h is monotonically increasing), we obtain the Mrs. Gerber’s Lemma.
H(Y|U) ≥ H(H − 1(H(X|U)*p))
Binary Broadcast Channel (Lecture 9 from Ben-Gurion University) Ова е илустрација за користење на Mrs Gerber’s Lemma.
We define the Binary Broadcast Channel depicted in 1↑
figure Figure 1.1.png
Figure 1 Binary Broadcast channel
излгеда на сликава p2 = 2.
We set the following distributions
Z1 ~ B(p1) Z2 ~ B(p2) 2 ~ B(2) Z1Z2 Z2 = Z12
thus p2 = p1*p2̃ = p1(1 − 2) + (1 − p1)2. Additionally, we define
Y1 = XZ1 Y2 = Y12 = XZ12 = XZ2
The capacity region of the degraded broadcast channel is the set of all pair rates (R1, R2) that satisfies
(3) R1 < I(X;Y1|U) R2 < I(U;Y2)
for some joint distribution p(u)p(x|u)p(y1y2|x).
We state the capacity region of the Binary Broadcast Channel is the of all pairs (R1R2) that satisfies
R1 < h(αp1) − h(p1) R2 < 1 − h(αp2)
where α ∈ [0, 1].
Proof:
Achievability: Let us define two r.v. \mathchoiceU ~ B(1)/(2),  V ~ B(α)U ~ B(1)/(2),  V ~ B(α)U ~ B(1)/(2),  V ~ B(α)U ~ B(1)/(2),  V ~ B(α) and change X = UV. We now apply this new definitions to 3↑.
R2 = I(U;Y2) = H(Y2) − H(Y2|U) = 1 − H(αp2)
Furthermore
R1 = I(X;Y1|U) = I(UV;UVZ1|U) = I(V;VZ1) = h(VZ1) − h(VZ1|V) = h(\mathchoicea*p1a*p1a*p1a*p1) − h(p1)
H(UV|U) = H(U)
V, Z1|VZ1 0 1 00 1 0 01 0 1 10 0 1 11 1 0 V, Z1|VZ1 0 1 00 αp1 0 01 0 α(1 − p1) 10 0 (1 − α)p1 11 (1 − α)(1 − p1) 0 p(VZ1) 1 − α(1 − p1) + (1 − α)p1 α(1 − p1) + (1 − α)p1 1 − αp1 \mathchoice\alphap1\alphap1\alphap1\alphap1
Expand[αp1 + (1 − α)(1 − p1)] = 2αp1 − α − p1 + 1
Expand[1 − α(1 − p1) − (1 − α)p1] = 2αp1 − α − p1 + 1
Истиoв резултати се добиваат и во Example 15.6.5 од EIT но овде одат чекор подалеку со тоа што го опишуваат каналот со mod2 и воведуваат уште една променилва V.
Thus we obtained the desired region and proved the achievability part. In order to prove the converse we use 1.1↑.
Converse:
I(Y2;U) = h(Y2) − h(Y2|U) ≤ 1 − h(Y2|U)
we bound h(Y2|U) from both sides
U → X → Y2
I(U;X) ≥ I(U;Y2) → H(U) − H(U|X) ≥ H(U) − H(U|Y2) → H(U|X) ≤ H(U|Y2)
I(U;X) ≥ I(U;Y2) → H(X) − H(X|U) ≥ H(Y2) − H(Y2|U) → H(U|X) ≤ H(U|Y)
I(U;Y2) ≤ I(X;Y2) → H(Y2) − H(Y2|U) ≥ H(Y2) − H(Y2|X) → H(Y2|U) ≤ H(Y2|X)
1 = h(Y2|U) ≥ h(Y2|X) = h(p2)
by the Markov chain Y2XU. Therefore exist an α s.t.
h(Y2|U) = h(αp2)
thus
R2 ≤ 1 − h(αp2)
Now for R1,
I(X;Y1|U) = h(Y1|U) − h(Y1|U, X) = \mathchoiceh(Y1|U)h(Y1|U)h(Y1|U)h(Y1|U) − h(p1)
By setting X = Y1,  Y = Y2 in the Mrs. Gerber Lemma we obtain
Let H − 1:[0, 1] → [0, 1 ⁄ 2] be the inverse of the binary entropy function, i.e., H(H − 1(v)) = v.
Let X be the binary random variable and let U be an arbitrary random variable. If Z ~ Bern(p) is independent of (X, U) and Y = XZ, then
H(Y|U) ≥ H(H − 1(H(X|U)*p))
\mathchoiceh(Y2|U)h(Y2|U)h(Y2|U)h(Y2|U) ≥ h(h − 1(h(Y1|U))*2)
h(αp2) ≥ h(h − 1(h(Y1|U))*2)
since h is monotonically increasing function
αp1*2 ≥ h − 1(h(Y1|U))*2 αp1(1 − \cancel2) + (1 − \cancelαp1)2 ≥ h − 1(h(Y1|U))(1 − \cancel2) + (1 − \cancelh − 1(h(Y1|U)))2 αp1 + \cancel2 ≥ h − 1(h(Y1|U)) + \cancel2
αp1 ≥ h − 1(h(Y1|U)) → h(αp1) = h(Y1|U)
therefore
R1 ≤ h(αp1) − h(p1)
which concludes the proof.
И без толку да комплицираш очигледно е дека h(Y1|U) = h(αp1) ако не комплицираш и ја користиш сликата и дефиницијата на X како во примерот 15.6.5 од EIT.
Не е тоа поентата. Добар си е доказот.

2 Point-to-point information theory

2.1 Packing-lemma

Let (U, X, Y) ~ p(u, x, y) . Let (n, n) ~ p(n, n) be a pair of arbitrarily distributed random sequences, not necessarily distributed according to ni = 1pU, Y(i, i). Let Xn(m),  m ∈ A, where |A| ≤ 2nR, be random sequences, each distributed according ot ni = 1pX|U(xi|i). Further assume that Xn(m), m ∈ A is pairwise conditionally independent of n given n, but is arbitrarily dependent on other Xn(m) sequences. Then, there exist δ(ϵ) that tends to zero as ϵ → 0 such that
limn → ∞P{(n, Xn(m), n) ∈ A(n)ϵfor some m ∈ A} = 0
if R < I(X;Y|U) − δ(ϵ).

3 Relay Channels

In this chapter, we begin our discussion of communication over general multihop networks. We study the three node relay channel, which is a model for point-to-point com- munication with the help of a relay, such as communication between two base stations through both a terrestrial link and a satellite, or between two nodes in a mesh network with an intermediate node acting as a relay. The capacity of the relay channel is not known in general. We establish a cutset upper bound on the capacity and discuss several coding schemes that are optimal in some special cases. We first discuss the following two extreme schemes.
- Direct transmission: In this simple scheme, the relay is not actively used in the communication.
- Decode–forward: In this multihop scheme, the relay plays a central role in the communication. It decodes for the message and coherently cooperates with the sender to communicate it to the receiver. This scheme involves the new techniques of block Markov coding, backward decoding, and the use of binning in channel coding. We observe that direct transmission can outperform decode–forward when the channel from the sender to the relay is weaker than that to the receiver. This motivates the devel- opment of the following two schemes.
- Partial decode–forward: Here the relay recovers only part of the message and the rest of the message is recovered only by the receiver. We show that this scheme is optimal for a class of semideterministic relay channels and for relay channels with orthogonal sender components.
- Compress–forward: In this scheme, the relay does not attempt to recover the message. Instead, it uses Wyner–Ziv coding with the receiver’s sequence acting as side informa- tion, and forwards the bin index. The receiver then decodes for the bin index, finds the corresponding reconstruction of the relay received sequence, and uses it together with its own sequence to recover the message. Compress–forward is shown to be op- timal for a class of deterministic relay channels and for a modulo 2 sum relay channel example whose capacity turns out to be strictly lower than the cutset bound.
Motivated by wireless networks, we study the following three Gaussian relay channel models.
- Full-duplex Gaussian RC: The capacity for this model is not known for any set of nonzero channel parameter values. We evaluate and compare the cutset upper bound and the decode–forward and compress–forward lower bounds. We show that the partial decode–forward lower bound reduces to the largest of the rates for direct trans- mission and decode–forward.
- Half-duplex Gaussian RC with sender frequency division: In contrast to the full- duplex RC, we show that partial decode–forward is optimal for this model.
- Half-duplex Gaussian RC with receiver frequency division: We show that the cutset bound coincides with the decode–forward bound for a range of channel parameter values. We then present the amplify–forward coding scheme in which the relay sends a scaled version of its previously received signal. We generalize amplify–forward to linear relaying functions that are weighted sums of past received signals and establish a single-letter characterization of the capacity with linear relaying.
In the last section of this chapter, we study the effect of relay lookahead on capacity. In the relay channel setup, we assume that the relaying functions depend only on past re- ceived relay symbols; hence they are strictly causal. Here we allow the relaying functions to depend with some lookahead on the relay received sequence. We study two extreme lookahead models—the noncausal relay channel and the causal relay channel. We present upper and lower bounds on the capacity for these two models that are tight in some cases. In particular, we show that the cutset bound for the strictly causal relay channel does not hold for the causal relay channel. We further show that simple instantaneous relaying can be optimal and achieves higher rates than the cutset bound for the strictly causal relay channel. We then extend these results to the (full-duplex) Gaussian case. For the non- causal Gaussian RC, we show that capacity is achieved via noncausal decode–forwardthe channel from the sender to the relay is sufficiently strong, while for the causal Gaussian RC, we show that capacity is achieved via instantaneous amplify–forward if the channel from the sender to the relay is sufficiently weaker than the other two channels. These re- sults are in sharp contrast to the strictly causal case for which capacity is not known for any nonzero channel parameter values.

3.1 Discrete Memoryless Relay Channel

Consider the 3-node point-to-point communication system with a relay depicted in 2↓. The sender (node 1) wishes to communicate a message M to the receiver (node 3) with the help of the relay (node 2). We first consider the discrete memoryless relay channel (DM-RC) model (X1×X2, p(y2, y3|x1, x2), Y2×Y3) that consists of four finite sets X1, X2, Y2, Y3 , and a collection of conditional pmfs p(y2, y3|x1, x2) on Y2×Y3 .
figure Fig16.1 Point-to-point Communication System.png
Figure 2 Poit-to-Poitn Communication system with a relay
A (2nR, n) code for the DM-RC consists of
- a message set [1:2nR] ,
- an encoder that assigns a codeword xn1(m) to each message m∈[1:2nR],
- a relay encoder that assigns a symbol x2i(y2i−1) to each past received sequence yi2 ∈ Yi−12 for each time i∈[1:n], and
- a decoder that assigns an estimatê or an error message e to each received sequence yn3 ∈Yn3 .
The channel is memoryless in the sense that the current received symbols (Yi2, Yi3) and the message and past symbols (m, Xi−11, Xi−12, Yi−12, Yi−13) are conditionally independent given the current transmitted symbols (Xi1, Xi2). We assume that the message M is uniformly distributed over the message set. The average probability of error is defined as P(n)e = P( ≠ M). A rate R is said to be achievable for the DM-RC if there exists a sequence of (2nR, n) codes such that limn→∞P(n)e = 0. The capacity C of the DM-RC is the supremum of all achievable rates. The capacity of the DM-RC is not known in general. We discuss upper and lower bounds on the capacity that are tight for some special classes of relay channels.

3.2 Cutset upper bound on the capacity

The following upper bound is motivated by the cutset bounds for graphical networks we discused in Chapter 15
Cutset bound for DMRC
\mathchoiceC ≤ maxp(x1x2)min{I(X1, X2;Y3), I(X1;Y2, Y3|X2)}C ≤ maxp(x1x2)min{I(X1, X2;Y3), I(X1;Y2, Y3|X2)}C ≤ maxp(x1x2)min{I(X1, X2;Y3), I(X1;Y2, Y3|X2)}C ≤ maxp(x1x2)min{I(X1, X2;Y3), I(X1;Y2, Y3|X2)}
The cutset bound is tight for many classes of DMRC with known capacity. However, it is not tight in general as shown via an example in Section 16.7.3.
first term:
ni = 1I(M;Y3i|Yi − 13)
I(M;Y3i|Yi − 13) ≤ I(Yi − 13;Yi) + I(M;Y3i|Yi − 13) = I(M, Yi − 13;Y3i) ≤ I(X1iX2iM, Yi − 13;Yi)
I(X1iX2iM, Yi − 13;Yi) = H(Y3i) − H(Y3i|X1iX2iM, Yi − 13) ≥ H(Y3i) − H(Y3i|M, Yi − 13) conditioning reduces entropy
I(X1iX2iM, Yi − 13;Yi) = H(Y3i) − H(Y3i|X1iX2iM, Yi − 13) = H(Y3i) − H(Y3i|X1iX2i) = I(X1iX2i;Y3i)
Последново следи од 3.1↑.
\mathchoiceI(M;Yn3) ≤ ni = 1I(X1iX2i;Y3i)I(M;Yn3) ≤ ni = 1I(X1iX2i;Y3i)I(M;Yn3) ≤ ni = 1I(X1iX2i;Y3i)I(M;Yn3) ≤ ni = 1I(X1iX2i;Y3i)
second term:
I(M;Yn3) ≤ I(M;Yn3Yn2) = ni = 1I(M;Y2iY3i|Yi − 12Yi − 13)\overset(a) = ni = 1\mathchoiceI(M;Y2iY3i|Yi − 12Yi − 13X2i)I(M;Y2iY3i|Yi − 12Yi − 13X2i)I(M;Y2iY3i|Yi − 12Yi − 13X2i)I(M;Y2iY3i|Yi − 12Yi − 13X2i) = 
\overset(b) ≤ ni = 1\mathchoiceI(X1i, M, Yi − 12Yi − 13;Y2iY3i|X2i)I(X1i, M, Yi − 12Yi − 13;Y2iY3i|X2i)I(X1i, M, Yi − 12Yi − 13;Y2iY3i|X2i)I(X1i, M, Yi − 12Yi − 13;Y2iY3i|X2i)\overset(c) = ni = 1I(X1i;Y2iY1i|X2i)
\mathchoiceI(M;Yn3) ≤ ni = 1I(X1i;Y2iY1i|X2i)I(M;Yn3) ≤ ni = 1I(X1i;Y2iY1i|X2i)I(M;Yn3) ≤ ni = 1I(X1i;Y2iY1i|X2i)I(M;Yn3) ≤ ni = 1I(X1i;Y2iY1i|X2i)
——————————————————————————––
(a) X2i is function of Yi − 12
(b)
I(M;Y2iY3i|Yi − 12Yi − 13X2i) + I(Yi − 12Yi13;Y2iY3i|X2i) = I(Yi − 12Yi13, M;Y2iY3i|X2i) ⇒ \mathchoiceI(Yi − 12Yi13, M;Y2iY3i|X2i)I(Yi − 12Yi13, M;Y2iY3i|X2i)I(Yi − 12Yi13, M;Y2iY3i|X2i)I(Yi − 12Yi13, M;Y2iY3i|X2i) ≥ \mathchoiceI(M;Y2iY3i|Yi − 12Yi − 13X2i)I(M;Y2iY3i|Yi − 12Yi − 13X2i)I(M;Y2iY3i|Yi − 12Yi − 13X2i)I(M;Y2iY3i|Yi − 12Yi − 13X2i)
I(X1iYi − 12Yi13, M;Y2iY3i|X2i) = I(Yi − 12Yi13, M;Y2iY3i|X2i) + I(X1i;Y2iY3i|M, X2iYi − 12Yi13) ⇒ 
\mathchoiceI(X1iYi − 12Yi13, M;Y2iY3i|X2i)I(X1iYi − 12Yi13, M;Y2iY3i|X2i)I(X1iYi − 12Yi13, M;Y2iY3i|X2i)I(X1iYi − 12Yi13, M;Y2iY3i|X2i) ≥ \mathchoiceI(Yi − 12Yi13, M;Y2iY3i|X2i)I(Yi − 12Yi13, M;Y2iY3i|X2i)I(Yi − 12Yi13, M;Y2iY3i|X2i)I(Yi − 12Yi13, M;Y2iY3i|X2i) ≥ I(M;Y2iY3i|Yi − 12Yi − 13X2i)
(c)
I(X1i, M, Yi − 12Yi − 13;Y2iY3i|X2i) = I(M;Y2iY3i|X2i) + I(M;Y2iY3i|X2iX1i, Yi − 12Yi − 13) = (*)
(*) = I(M;Y2iY3i|X2i) + I(M;Y2iY3i|X2iX1i) = I(M;Y2iY3i|X2i) + I(M;Y2iY3i|X2iX1i, Yi − 12Yi − 13) ≥ I(M;Y2iY3i|X2iX1i, Yi − 12Yi − 13)
——————————————————————————–———–
I(X1i, M, Yi − 12Yi − 13;Y2iY3i|X2i) = H(Y2iY3i|X2i) − H(Y2iY3i|X2iX1i, M, Yi − 12Yi − 13) = H(Y2iY3i|X2i) − H(Y2iY3i|X2iX1i) = I(X1i;Y2iY1i|X2i)
—————————————————————
I(M;Yn3) ≤ ni = 1I(X1iX2i;Y3i) = nI(X1X2;Y3|Q) ≤ nI(X1X2;Y3)
I(M;Yn3) ≤ ni = 1I(X1i;Y2iY1i|X2i) = nI(X1;Y2Y1|X2Q) ≤ nI(X1;Y2Y1|X2)
R = min{I(X1X2;Y3), I(X1;Y2Y1|X2)}

3.3 Direct Tranmission Lower Bound

One simple coding scheme for the relay channel is to fix the relay transmission at the most favorable symbol to the channel from the sender to the receiver and to communicate the message directly using optimal point-to-point channel coding. The capacity of the relay channel is thus lower bounded by the capacity of the resulting DMC as
(4) \mathchoiceCmaxp(x1), x2I(X1;Y3|X2 = x2)Cmaxp(x1), x2I(X1;Y3|X2 = x2)Cmaxp(x1), x2I(X1;Y3|X2 = x2)Cmaxp(x1), x2I(X1;Y3|X2 = x2)
Bound is tight for reversely degraded DMRC
p(y2, y3|x1, x2) = p(y3|x1, x2)p(y2|y3, x2)

3.4 Decode–Forward Lower Bound

At the other extreme of direct transmission, the decode–forward coding scheme relies heavily on the relay to help communicate the message to the receiver. We develop this scheme in three steps.
In the first two steps, we use a multihop relaying scheme in which the receiver treats the transmission from the sender as noise. The decode–forward scheme improves upon this multihop scheme by having the receiver decode also for the information sent directly by the sender.

3.4.1 Multihop Lower Bound

In the multihop relaying scheme, the relay recovers the message received from the sender in each block and retransmits it in the following block. This gives the lower bound on the capacity of the DMRC
(5) \mathchoiceC ≥ maxp(x1)p(x2)min{I(X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1)p(x2)min{I(X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1)p(x2)min{I(X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1)p(x2)min{I(X2;Y3), I(X1;Y2|X2)}
It is not difficult to show that this lower bound is tight when the DM-RC consists of a cascade of two DMCs, i.e., p(y2, y3|x1, x2) = p(y2|x1)p(y3|x2). In this case the capacity expression simplifies to:
C = maxp(x1)p(x2)min{I(X2;Y3), I(X1;Y2|X2)} = maxp(x1)p(x2)min{I(X2;Y3), I(X1;Y2)} = min{maxp(x2){I(X2;Y3)}, maxp(x1){I(X1;Y2)}}
Achievability of the multihop lower bound uses b transmission blocks, each consisting of n transmissions, as illustrated in 3↓. A sequence of (b−1) messages M j , j ∈ [1:b − 1] , each selected independently and uniformly over [1:2nR], is sent over these b blocks, We assume mb = 1 by convention. Note that the average rate over the b blocks is R⋅(b−1) ⁄ b, which can be made as close to R as desired.
figure Figure 16.3.png
Figure 3 Multiple transmission blocks used in multiphop scheme
Codebook generation:
Fix the product pmf [A]  [A] Probability Mass Function p(x1)p(x2) that attains the multihop lower bound in 5↑. Randomly and independently generate a codebook for each block. For each j ∈ [1:b], randomly and independently generate 2nR sequences xn1(mj), mj ∈ [1:2nR], each according to ni = 1pX1(x1i). Similarly, generate 2nR sequences xn2(mj−1), mj−1 ∈ [1:2nR], each according to ni = 1pX2(x2i). This defines the codebook:
Cj = {(xn1(mj), xm2(mj − 1)):mj − 1, mj ∈ [1:2nR]},  j ∈ [1, ..b].
Глупава е оваа индексирање. Претпоставувам сака да каже дека xn1(m1), xm2(m1) треба да бидат здружено типични. можеби сака да каже ако mj = m1 тогаш и mj − 1 = m1 т.е. тоа ќе се случи во наредниот блок.
The codebooks are revealed to all parties.
Encoding:
Let mj ∈ [1:2nR] be the new message to be sent in block j. The encoder transmits xn1(mj) from codebook Cj.
Relay encoding:
By convention, let 0 = 1. At the end of block j, the relay finds the unique message j such (xn1(mj̃), xn2(j − 1), yn2(j)) ∈ \mathnormalA(n)ϵ. In block j + 1, it transmits xn2(j) from the codebook Cj + 1.
Сам ми текна на ова после гледам во следната глава и они истото го прават.
Јас каде што имам ставено за X2 = 0 они имаат ставено x2(1).
Они поинаку го толкуваат y не го третираат како сигнал на влез од релето туку како декодирана верзија од примениот сигнал. Вау!!! Ако е така ова секаде го имам грешено.
S——>R——->D
j 1 2 3 4 ... j ... b X1 x1(m1) x1(m2) x1(m3) x1(m4) x1(mj) 1 Y2 y2(m1) y2(m2) y2(x3) y2(m4) y2(mj) 1 X2 0 x2(m1) x2(m2) x2(m3) x2(mj − 1) x2(mb − 1) Y3 y(1) y(1) y(3) y(4) y(5) y(b − 1)
j = 3
At the end of block j = 3, the relay finds the unique message 3 such (xn1(m3̃), xn2(2), yn2(3)) ∈ A(n)ϵ. In block j + 1 = 4, it transmits xn2(3) from the codebook C4
Since the relay codeword transmitted in a block depends statistically on the message transmited in the previous block, we reffer to this scheme as block Markov coding (BMC)
Decoding:
figure Miscellaneous/Fig. Linear Relaying for FD-AWGN.png
Figure 4 Слика за AF од S. Zahedi
At the end of block j + 1, the receiver finds the unique message j such that (xn2(j), yn3(j + 1)) ∈ A(n)ϵ. Бранче е релето, а капа е дестианцијата.
Analysis of the probability of error. We analyze the probability of decoding error for the message Mj averaged over codebooks. Assume without loss of generality that Mj = 1. Let j be the relay message estimate at the end of block j. Since
{j ≠ 1} ⊆ {j ≠ 1}{j ≠ j}
the decoder makes an error only if one of more of following events occur:
(j) = {(Xn1(1), Xn2(j − 1), Yn2(j)) ≠ A(n)ϵ}
2(j) = {(Xn1(mj), Xn2(j − 1), Yn2(j)) ∈ A(n)ϵfor some mj ≠ 1}
E1(j) = {Xn2(j), Yn3(j + 1) ≠ A(n)ϵ}
E2(j) = {(Xn2(mj), Yn3(j + 1)) ∈ A(n)ϵ for some mj ≠ j}
Thus, the probability of erro is upper boundd as:
P(E(j) = P{j ≠ 1}) ≤ P(1(j)2(j)E1(j)E2(j)) ≤ P(1(j)) + P(2(j)) + P(E1(j)) + P(E2(j))
where the first two terms upper bound P{j ≠ 1} and the last two terms upper bound P{j ≠ j}.
Now, by the independence of the codebooks, the relay message estimate j − 1 , which is a function of Yn2(j−1) and codebook Cj−1 , is independent of the codewords Xn1(mj), Xn2(mj−1), mj, mj−1 ∈ [1:2nR], from codebook Cj . Hence, by the LLN [B]  [B] LLN - Law of Large Numbers P((j))tends to zero as n → ∞, and by the packing lemma [1] Или eq6.27 од докторската. , P(2(j)) tends to zero as n→∞ if \mathchoiceR < I(X1;Y2|X2) − δ(є)R < I(X1;Y2|X2) − δ(є)R < I(X1;Y2|X2) − δ(є)R < I(X1;Y2|X2) − δ(є). Similarly, by the independence of the codebooks and the LLN, P(E1(j)) tends to zero as n → ∞, and by the same independence and the packing lemma, P(E2(j)) tends to zero as n→∞ if \mathchoiceR < I(X2;Y3) − δ(є)R < I(X2;Y3) − δ(є)R < I(X2;Y3) − δ(є)R < I(X2;Y3) − δ(є). Thus we have shown that under the given constraints on the rate, {j ≠ j} tends to zero as n→∞ for each j∈[1:b−1]. This completes the proof of the multihop lower bound.
Од Capacity Theorem
Лема 2:
Да земеме дека (S1, S2, S3) ~ ni = 1p(s1i, s2i, s3i) и (S1, S2, S3) ~ ni = 1p(s1i|s3i)p(s2i|s3i)p(s3i). Тогаш, за такво n за кое P{Aϵ(S1, S2, S3)} > 1 − ϵ,
(6) (1 − ϵ)2 − n(I(S1;S2|S3 + 7⋅ϵ) ≤ P{(S1, S2, S3) ∈ Aϵ(S1, S2, S3)} ≤ 2 − n(I(S1;S2|S3) − 7ϵ)

3.4.2 Coherent Multihop Lower Bound

The multihop scheme discussed in the previous subsection can be improved by having the sender and the relay coherently cooperate in transmitting their codewords. With this improvement, we obtain the lower bound on the capacity of the DM-RC
(7) \mathchoiceC ≥ maxp(x1, x2)min{I(X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1, x2)min{I(X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1, x2)min{I(X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1, x2)min{I(X2;Y3), I(X1;Y2|X2)}
Again we use a block Markov coding scheme in which a sequence of (b−1) i.i.d. messages Mj, j∈[1:b−1], is sent over b blocks each consisting of n transmissions.
Codebook generation:
Fix the pmf p(x1, x2) that attains the lower bound in 7↑. For j∈[1:b], randomly and independently generate 2nR sequences xn2(mj−1), mj−1 ∈ [1:2nR], each according to ni = 1pX2(x2i). For each mj−1 ∈ [1:2nR], randomly and conditionally independently generate 2nR sequences xn1(mj|mj−1), mj ∈ [1:2nR], each according to ni = 1pX1|X2(x1i|x2i(mj−1)). This defines the codebook
(8) Cj = {(xn1(mj|mj − 1), xn2(mj − 1)):mj − 1, mj ∈ [1:2nR]},  j ∈ [1:b]
The codebooks are revealed to all parties.
Encoding and decoding are expalined with the help of 1↓.
Block b X1 xn1(m1|x) xn1(m2|m1) xn1(m3|m2) ... xn1(mb − 1|mb − 2) xn1(mb|mb − 1) Y2 1 2 3 ... b − 1 0 X2 xn1(m1|x) xn1(m2|m1) xn1(m3|m2) ... xn1(mb − 1|mb − 2) xn1(mb|mb − 1) Y3 0 1 2 ... b − 2 b − 1
Block 1 2 3 4 b − 1 b
X1 xn1(m1|1) xn1(m2|m1) xn1(m3|m2) ... xn1(mb − 1|mb − 2) xn1(mb|mb − 1)
Y2 1 2 3 ... b − 1 0
X2 xn2(1) xn2(1) xn2(2) ... xn1(b − 2) xn2(b − 1)
Y3 0 1 2 b − 2 b − 1
Table 1 Encoding and decoding for the coherent multihop lower bound
Encoding:
Let mj ∈ [1:2nR] message to be send in lock j. Te encoder transmits x1(mj|mj − 1) from codebook Cj, where m0 = mb = 1 by convention.
Relay encoding:
By convention, let 0 = 1. At the end of block j, the relay finds the unique message j such that (xn1(j|j − 1), xn2(j − 1), yn2(j)) ∈ A(n)ϵ. In block j + 1, it transmits xn2(mj) from codebook Cj + 1.
Decoding:
At the end of block j + 1, the receiver finds the unique message j such that (xn2(j), yn2(j + 1)) ∈ A(n)ϵ.
Analysis of the probability of error:
We analyze the probability of decoding error for Mj average over codebooks. Assume without loss of generality that Mj − 1 = Mj = 1 . Let j be the relay message estimate at the end of block j. As before, the decoder makes an error only if one or more of the following events occur:
(j) = {j ≠ 1}
E1(j) = {Xn2(j), Yn3(j + 1) ≠ A(n)ϵ}
E2(j) = {(Xn2(mj), Yn3(j + 1)) ∈ A(n)ϵ for some mj ≠ j}
Thus, the probability of erro is upper bounded as:
P(E(j)) = P{j ≠ 1} ≤ P((j)E1(j)E2(j)) ≤ P(1(j)) + P(E1(j)) + P(E2(j))
Following the same steps to the analysis of the probability of error for the (noncoherent) multihop scheme, the last two terms, P(E1(j)) and P(E2(j)), tend to zero as n→∞ if R < I(X2;Y3)−є. To upper bound the first term P(1(j)), define:
1(j) = {(Xn1(1|j − 1), Xn2(j − 1), Yn2(j))A(n)ϵ}
2(j) = {(Xn1(mj|j − 1), Xn2(j − 1), Yn2(j)) ∈ A(n)ϵfor some mj ≠ 1}
Then
P((j)) ≤ P((j − 1)1(j)2(j)) ≤ P((j − 1)) + P(1(j)c(j − 1)) + P(2(j))
Consider the second term
P(1(j)c(j − 1)) = P{(Xn1(1|j − 1), Xn2(j − 1), Yn2(j))A(n)ϵ, j − 1 = 1} ≤ P{(Xn1(1|1), Xn2(1), Yn2(j))A(n)ϵ|j − 1 = 1}
which, by the independence of the codebooks and the LLN, tends to zero as n→∞. By the packing lemma, P(2(j)) tends to zero as n→∞ if R < I(X1;Y2|X2)−δ(є). Note that 0 = 1 by definition. Hence, by induction, \strikeout off\uuline off\uwave offP((j))\uuline default\uwave default tends to zero as n→∞ for every j∈[1:b−1]. Thus we have shown that under the given constraints on the rate, P{j ≠ j} tends to zero as n→∞ for every j ∈ [1:b−1]. This completes the proof of achievability of the coherent multihop lower bound in 7↑.

3.4.3 Decode-Froward Lower Bound

The coherent multihop scheme can be further improved by having the receiver decode simultaneously for the messages sent by the sender and the relay. This leads to the following.
(Decode-Forward Lower Bound)
The capacity of the DMRC is lower bounded as
\mathchoiceC ≥ maxp(x1x2)min{I(X1X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1x2)min{I(X1X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1x2)min{I(X1X2;Y3), I(X1;Y2|X2)}C ≥ maxp(x1x2)min{I(X1X2;Y3), I(X1;Y2|X2)}
Note that the main difference between this bound an the cutset bound is that the latter includes Y3 in the second mutual information term.
This lower bound is tight when the DMRC is degraded, i.e.,
p(y2y3|x1x2) = p(y2|x1x2)p(y3|x2y2)
The proof of the converse for this case follows by the cutset upper bound in Theorem 3.2↑ since the degradedness of the channel implies that I(X1;Y2, Y3|X2) = I(X1;Y2|X2). We illustrate this capacity result in the following.
(Sato Relay Channel)
Consider degraded DMRC with X1 = Y2 = Y3 = {0, 1, 2}, X2 = {0, 1},   and Y2 = X1 as depicted in 5↓
figure Figure 16.4 Sato relay channel.png
Figure 5 Sato relay channel
With direct transmission, R0 = 1 bits/transmission can be achieved by setting X2 = 0 or X2 = 1. By comparison using the optimal first-order Markov relay function x2(y2, i − 1) yields R1 = 1.0437, and using the optimal second-order Markov relay function x2i(y2, i − 1, y2, i − 2) yields R2 = 1.0549. Since the channel is degraded, the capacity coincides with the decode-forward lower bound in 3.4.3↑. Evaluating this bound yields C = 1.1619. Оваа вредност се добива за дистрибуцијата p(x1x2) дадена во чланакот Capacity Theorem.mw. Со извод јас јa добивам дистрибуцијата [7 ⁄ 8, 1 ⁄ 8, 1 ⁄ 8, 1 ⁄ 8, 1 ⁄ 8, 7 ⁄ 8] како што добиваат во Lecture Notes во Relay with Unlimited Lookahead.
I(X1;Y3) = H(Y3) − H(Y3|X1) = 
H(Y3|X1) = p(X1 = 0)\cancelto0H(Y3|X1 = 0) + p(X1 = 1)\cancelto1H(Y3|X1 = 1) + p(X1 = 2)\cancelto1H(Y3|X1 = 2) = p(X1 = 1) + p(X1 = 2) = p1 + p2
 − H(Y3) =  − H(p(Y3 = 0), p(Y3 = 1), p(Y3 = 2)) =  − Hp(X1 = 0), (p(X1 = 1) + p(X1 = 2))/(2), (p(X1 = 1) + p(X1 = 2))/(2) = 
 = p(X1 = 0)logp(X1 = 0) + (p(X1 = 1) + p(X1 = 2))/(2)log(p(X1 = 1) + p(X1 = 2))/(2) + (p(X1 = 1) + p(X1 = 2))/(2)logp(p(X1 = 1) + p(X1 = 2))/(2) = 
 = p(X1 = 0)logp(X1 = 0) + (p(X1 = 1) + p(X1 = 2))log(p(X1 = 1) + p(X1 = 2))/(2) = 
p0logp0 + (p1 + p2)log(p1 + p2)/(2) = (1 − p1 − p2)log(1 − p1 − p2) + (p1 + p2)log(p1 + p2)/(2)
I(X1;Y3) =  − (1 − p1 − p2)log(1 − p1 − p2) − (p1 + p2)log(p1 + p2)/(2) + p1 + p2
y = log3x → x = 3y
y = log2x → x = 2y
log3x = log32y → y = (log3(x))/(log3(2)) = log2(x)
p = p1 + p2
I(X1;Y3) =  − (1 − p)log(1 − p) − plog(p)/(2) + p =  − H(p) + plog2 + p
Не треба вака!!! Треба како во Three Terminal Communication т.е. (1, 2) третирај ги како еден смбол или пренесувај само (0, 1) или само (0, 2). Тие дефинитивно ќе ги пренесеш со капацитет од 1 бит/трансмисија.
30.06.2014
во табелата горе е дадена условна веројатност!!!!
H(Y3|X1) = p(X1 = 0)H(Y2|X1 = 0) + (p(X1 = 1) + p(X2 = 2))H(Y2|X1 = 1, 2) = 0
I(X1;Y3) = H(Y3) − 0 = H(1)/(2) = 1
H(Y3) = H(p(Y3 = 0), p(Y3 = 1) + p(Y3 = 2)) = plogp + (1 − p)log(1 − p) = 1бит/трансмисија
p = p1 1 − p = 1 − p2 − p3
I(X1;Y3) = 1
Y3|X1 0 1 2 0 1 0 0 1 0 1 ⁄ 2 1 ⁄ 2 2 0 1 ⁄ 2 1 ⁄ 2
Решението е во Capacity Theorem.lyx

3.4.4 Proof via Backward Decoding

Again we consider b transmission blocks, each consisting of n transmissions, and use a block Markov coding scheme. A sequence of (b − 1) i.i.d message Mj ∈ [1:2nR],  j ∈ [1:b − 1] is to be sent over the channel in nb transmissions.
Codebook generation. Fix the pmf p(x1x2)that attains the lower bound. As in the coherent multihop scheme, we randomly and independently generate codebooks
Cj = {(xn1(mj|mj − 1), xn2(mj − 1)):mj − 1, mj ∈ [1:2nR]},  j ∈ [1:b]
Encoding and backward decoding are explained with the help of the 3.4.3↑.
Block 1 2 3 4 b − 1 b
X1 xn1(m1|1) xn1(m2|m1) xn1(m3|m2) ... xn1(mb − 1|mb − 2) xn1(mb|mb − 1)
Y2 1 →  2 →  3 →  ... b − 1 0
X2 xn2(1) xn2(1) xn2(2) ... xn1(b − 2) xn2(b − 1)
Y3 0  ← 1  ← 2  ← b − 2  ← b − 1
Table 2 Encoding and decoding for the decode-forward lower bound
Relay encoding.
Relay encoding is also the same as in the coherent multihop scheme. By convention, let 0 = 1. At the end of block j, the relay finds the unique message j such that (xn1(j|j − 1), xn2(j − 1), yn2(j)) ∈ A(n)ϵ. In block j + 1 it transmits xn2(j) from codebook Cj + 1.
Backward decoding.
Decoding at the receiver is done backwards after all b blocks are received. For j = b − 1, b − 2, ...1 the receiver finds the unique message mj such that (xn1(j + 1), xn2(mj), yn3(j + 1)) ∈ A(n)ϵ, successively with the initial condition b = 1.
Analysis of the probability of error.
We analyze the probability of decoding error for the message Mj averaged over codebooks. Assume without loss of generality that Mj = Mj + 1 = 1. The decoder makes an error only if one or more of the following events occur:
(j) = {j ≠ 1}
E(j + 1) = {j + 1 ≠ 1}
E1(j) = {(Xn1(j + 1|j), Xn2(j), Yn3(j + 1))A(n)ϵ}
E2(j) = {(Xn1(j + 1|\mathchoicemjmjmjmj), Xn2(mj), Yn3(j)) ∈ A(n)ϵfor some mj ≠ j}
Thus the probability of error is upper bounded as:
P(E(j)) = P(j ≠ 1) ≤ P((j)E(j + 1)E1(j)E2(j)) ≤ P((j)) + P(E(j + 1)) + P(E1(j)c(j)Ec(j + 1)) + P(E2(j)).
Following the same steps as in the analysis of the probability of error for the coherent multihop scheme, the first term P((j)) tends to zero as n → ∞ if \mathchoiceR < I(X1;Y2|X2) − δ(ϵ)R < I(X1;Y2|X2) − δ(ϵ)R < I(X1;Y2|X2) − δ(ϵ)R < I(X1;Y2|X2) − δ(ϵ). The third term is upper bounded as:
P(E1(j)c(j)Ec(j + 1)) = P(E1(j){j + 1 = 1}{j = 1}) = P{(X1(1|1), Xn2(1), Yn3(j + 1))A(n)ϵ, j + 1 = 1, j = 1}
 ≤ P{(X1(1|1), Xn2(1), Yn3(j + 1))A(n)ϵ|j = 1}
which by the independence of the codebooks and the LLN, tends to zero as n → ∞. By the same independence and the packing lemma, the fourth term P(E2(j)) tens to zero as n → ∞ if \mathchoiceR < I(X1, X2;Y3)R < I(X1, X2;Y3)R < I(X1, X2;Y3)R < I(X1, X2;Y3). Finally fort the second term P(E(j + 1)), note that b = Mb = 1. Hence, by induction, P{j ≠ Mj} tends to zero as n → ∞ for every j ∈ [1:b − 1] if the given constraints on the rate are satisfied. This competes the proof of decode-forward lower bound.

3.4.5 Proof via Binning

The excessive delay of backward decoding can be alleviated by using binning or sliding window decoding. We describe the binning scheme here. The proof using sliding window decoding will be given in Chapter 18.
In the binning scheme, the sender and the relay cooperatively send the bin index Lj of the message Mj (instead of the message itself) in block j + 1 to help the receiver recover Mj.
Codedbook generation.
Fix the pmf p(x1x2) that attains the lower bound. Let 0 ≤ R2 ≤ R. For each j ∈ [1:b], randomly and independently generate 2nR2 sequences xn2(lj − 1), lj − 1 ∈ [1:2nR2], each according to ni = 1px2(x2i). For each lj − 1 ∈ [1:2nR], randomly and conditionally independently generate 2nR sequences xn1(mj|lj − 1). Ова ја дефинира кодната книга
Cj = {(xn1(mj|lj − 1), xn2(lj − 1)):mj ∈ [1:2nR], lj − 1 ∈ [1:2nR2]},  j ∈ [1:b]
Partition the set of messages into 2nR2 equal size bins B(l) = [(l − 1)2n(R − R2) + 1:l⋅2n(R − R2)],  l ∈ [1:2nR2]. The codebooks and bin assignments are revelaled to all parties.
Encoding and decoding are explainded with the help of 3↓
Block 1 2 3 ... j j + 1 ... b − 1 b
X1 xn1(m1|1) xn1(m2|l1) xn1(m3|l2) ... xn1(mj|lj − 1) xn1(mj + 1|lj) ... xn1(mb − 1|lb − 2) xn1(1|lb − 1)
Y2 , 1 2, 2 3, 3 ... j, j j + 1, j + 1 ... b − 1, b − 1 0
X2 xn2(1) xn2(1) xn2(2) ... xn2(j − 1) xn2(j) ... xn1(b − 2) xn2(b − 1)
Y3 0 11 22 ... j − 1j − 1 jj ... b − 1b − 2 b − 1b − 1
Table 3 Encoding and decoding of the binning scheme for decode-forward
Encoding.
Let mj ∈ [1:2nR] be the new message to be sent in block j and assume that mj − 1 ∈ B(lj − 1). The encoder transmits xn1(mj|lj − 1) from codebook Cj, where l0 = mb = 1 by convention.
Relay encoding. At the end of block j, the relay finds the unique message j such that (xn1(mj|lj − 1), xn2(j − 1), yn2(j)) ∈ A(n)ϵ. If j ∈ B(j), the relay transmits x2(j) from codebook Cj + 1 in block j + 1, where by convention 0 = 1.
Decoding.
At the end of block j + 1 the receiver finds the unique index j such that (xn2(j), yn3(j + 1)) ∈ A(n)ϵ. It then finds the unique message j such that (x1(j|j − 1), xn2(j − 1), yn3(j)) ∈ A(n)ϵ and j ∈ B(j).
Малку е чудно зошто прво го поминав чланакто Capacity Theorem па сега ме буни. Сепак ОК е и овој пристап. Дури можеби е и поправилен зошто индексот за време е секогаш егзактен. Анализата малку те враќа еден чекор во минатото и не ги користи во целост податоците од последнито чекор
Analysis of the probability of error.
We analyze the probability of decoding error for the message Mj averaged over codebooks. Assume without loss of generality that Mj = Lj − 1 = Lj = 1 and let j be the relay estimate of Lj. Then the decoder makes an error only if one or more of the following events occur:
(j − 1) = {Lj − 1 ≠ 1}
E1(j − 1) = {j − 1 ≠ 1}
E1(j) = {j ≠ 1}, 
E2(j) = {(Xn1(1|j − 1), Xn2(j − 1), Yn3(j))A(n)ϵ}
E3(j) = {(Xn1(mj|j − 1), Xn2(j − 1), Yn3(j)) ∈ A(n)ϵfor some mj ≠ 1, mj ∈ B(j)}
Thus the probability of error is upper bounded as
P(E(j)) = P{j ≠ 1} ≤ P((j − 1)E1(j − 1)E1(j)E2(j)E3(j)) ≤ 
P((j − 1)) + P(E1(j)) + P(E1(j − 1)) + P(E2(j)c(j − 1)Ec1(j − 1)Ec1(j)) + P(E3(j)c(j − 1)Ec1(j − 1)Ec1(j)).
Потсетување за унија.
P(AB) = P(A) + P(B) − P(AB)
P(3∪ odd) = P(3) + P(odd) − P(3∩ odd)
 = (1)/(6) + (1)/(2) − (1)/(6) = (1)/(2)
Одземањето е затоа што два пати се јавува тројката и во непарни и сама сама
P(3∪ even) = P(3) + P(even) − P(3∩even) = (1)/(6) + (1)/(2) − 0 = (4)/(6) = (2)/(3)
пикова дама
P(дамаслика) = (1)/(52) + (20)/(52) − (1)/(52) = (20)/(52)
Ова е затоа што два пати се јавува дамата и во слика и сама дама
Following similar steps to the analysis of the error probability for coherent multihop scheme, with j − 1 replacing j − 1, the first term P((j − 1)) tends to zero as n → ∞ if \mathchoiceR2 < I(X1;Y2|X2) − δ(ϵ)R2 < I(X1;Y2|X2) − δ(ϵ)R2 < I(X1;Y2|X2) − δ(ϵ)R2 < I(X1;Y2|X2) − δ(ϵ). Again following similar steps to the analysis of the error probability for the coherent multihop scheme, the second and third terms P(E1(j)) and P(E1(j − 1)) tend to zero as n → ∞ if R2 < I(X2;Y3) − δ(ϵ). The fourth term is upper bounded as:
P(E2(j)c(j − 1)Ec1(j − 1)Ec1(j)) = P(E2(j){j − 1 = 1}{j − 1 = 1}{j = 1}) ≤ 
 ≤ P{(Xn1(1|1), Xn2(1), Yn3(j)) ∈ A(n)ϵ|j − 1 = 1}
Ова го разбирам на следниов начин: P(X, Y, Z, W) = P(Z)P(X, Y, W|Z) ≤ P(X, Y, W|Z). Ова го мапираш со изразот за E2(j).
which, by the independence of the codebooks and the LLN, tends to zero as n → ∞. The last term is upper bounded as:
P(E3(j)c(j − 1)Ec1(j − 1)Ec1(j)) = P(E3(j){j − 1 = 1}{j − 1 = 1}{j = 1})
 ≤ P{(Xn1(1|1), Xn2(1), Yn3(j)) ∈ A(n)ϵ for some mj ≠ 1, mj ∈ B(1)|j − 1 = 1}
which, by the same independence and the packing lemma, tends to zero as n → ∞ if R − R2 ≤ I(X1;Y3|X2) − δ(ϵ). Combining the bounds and eliminating R2, we have shown that P(j ≠ Mj) tends to zero as n → ∞ for each j ∈ [1:b − 1] if R < I(X1;Y2|X2) − δ(ϵ) and:
R ≤ I(X1;Y3|X2) + I(X2;Y3) − 2δ(ϵ) = I(X1X2;Y3) − 2δ(ϵ)
This completes the proof of the decode-forward lower bound using binning.
R − R2 ≤ I(X1;Y3|X2) − δ(ϵ) → R ≤ I(X1;Y3|X2) + R2 − δ(ϵ); R2 ≤ I(X2;Y3) − δ(ϵ) → 
R ≤ I(X1;Y3|X2) + I(X2;Y3) − 2δ(ϵ) → R ≤ I(X1, X2;Y3) − 2δ(ϵ)

3.5 Gaussian Relay Channel

Оваде дават резултати за општ релен канал а во Capacity theorem велат дека работат со деградиран релеен канал. Башка ги земаат во предвид коефициентите на каналот т.е. федингот.
Consider the Gaussian relay channel depicted on 6↓, which is a simple model for wireless point-to-point communication with a relay. The channel outputs corresponding to the inputs X1 and X2 are:
Y2 = g21X1 + Z2
Y3 = g31X1 + g32X2 + Z3, 
where g21, g31,  and g32 are channel gains, and Z2N(0, 1) and Z3N(0, 1) are independent noise components. Assume average power constraints P on each of X1 and X2. Since the relay can both send X2 and receive Y2 at the same time, this model is sometimes referred to as the full-duplex Gaussian RC, compared to the half duplex models we discuss in section 16.6.3 and 16.8.
figure Figure 16.5.png
Figure 6 Gaussian relay channel
We denote the SNR of the direct channel by \mathchoiceS31 = g231PS31 = g231PS31 = g231PS31 = g231P, the SNR of the channel form the sender to the relay receiver by \mathchoiceS21 = g221PS21 = g221PS21 = g221PS21 = g221P, and the SNR of the channel from the relay to the receiver by \mathchoiceS32 = g232PS32 = g232PS32 = g232PS32 = g232P. Note that under this model, the RC cannot be degraded or reversely degraded. In fact, the capacity is not known for any S21, S31, S32 > 0.

3.5.1 Upper bound and Lower bounds on the Capacity of the Gaussian RC

We evaluate the upper and lower bounds we discussed in the previous sections.
Cutset upper bound.
The proof of the cutset bound in Theorem 3.2↑ applies to arbitrary alphabets. By optimizing the bound subject to power constraints, we can show that it is attained by jointly Gaussian(X1, X2) (See Appendix 16A)
While increasing ρ would increase the mutual information term I(X1, X2;Y) by helping the transmission in multiple access cut of relay channel, it is limiting the information transfer rate in broadcast cut of the channel.
Clearly, introducing correlation between the channel input and relay signal increases the information transfer rate in the multiple access cut of the relay channel. However, it has its own drawback, i.e., it means less information transfer in the broadcast cut of the relay channel which can be interpreted as having prior knowledge of some part of the transmitted message from the source at the relay node (because of the correlation).
(9) \mathchoiceC ≤ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)(S31 + S21))} = C ≤ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)(S31 + S21))} = C ≤ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)(S31 + S21))} = C ≤ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)(S31 + S21))} = 
(10)  =  C(((S21S32) + (S31(S31 + S21 − S32)))2 ⁄ (S31 + S21))  if S21 ≥ S32       C(S31 + S21)  otherwise
The cutset bound for the Gaussian RC is given by:
C ≤ supF(x1x2):E(X21) ≤ P, E(X22) ≤ Pmin{I(X1X2;Y3)I(X1;Y2, Y3|X2)}

We perform the maximization by first establishing an upper bound on the right hand side of this expression and then showing that it is attained by jointly Gaussian(X1, X2).
We begin with the first mutual information term. Assume without loss of generality that E(X1) = E(X2) = 0. Consider
Y3 = g31X1 + g32X2 + Z3,  Y2 = g21X1 + Z2
\mathchoiceI(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3) = h(Y3) − h(Y3|X1X2) = h(Y3) − (1)/(2)log(2πe) ≤ (1)/(2)log(E(Y23)) ≤ (1)/(2)log(1 + g231E[X21] + g232E[X22] + 2g31g32E[X1X2])
 ≤ (1)/(2)⋅log(1 + S31 + S32 + 2ρ(S31S32)) = C(S32 + S32 + 2ρ(S31S32))
where
ρ = E(X1X2) ⁄ (E(X21)E(X22))
is correlation coefficient на здружената дистрибуција p(X1X2) .
2ρ(S31S32) = 2(E(X1X2))/((E(X21)E(X22)))(g231g232P2) = 2(E(X1X2))/((P2))g31g32(P2) = 2g31g32E(X1X2)

Next we consider the second mutual information term in the cutset bound
\mathchoiceI(X1;Y2, Y3|X2)I(X1;Y2, Y3|X2)I(X1;Y2, Y3|X2)I(X1;Y2, Y3|X2) = h(Y2, Y3|X2) − h(Y2Y3|X1X2) = h(Y2Y3|X2) − h(Z2Z3) = \mathchoiceh(Y3|X2)h(Y3|X2)h(Y3|X2)h(Y3|X2) + \mathchoiceh(Y2|X2Y3)h(Y2|X2Y3)h(Y2|X2Y3)h(Y2|X2Y3) − h(Z1) − h(Z2) = 
h(g31X1 + g32X2 + Z3|X2) − h(Y2|X2Y3) − \undersetlog(2πe)h(Z1) − h(Z2) ≤ (1)/(2)logE(Var(Y3|X2)) + (1)/(2)log(E(Var(Y2|Y3X2)))\overset(a) ≤ 
(1)/(2)⋅log(2πe)(g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22] + 1) + (1)/(2)log(1 + (g221 + g231)(E(X21) − E2(X1X2) ⁄ E(X22)))/(1 + g231(E(X21) − E2(X1X2) ⁄ E(X22)))
 = (1)/(2)log(1 + (g221 + g231)(E(X21) − E2(X1X2) ⁄ E(X22))) = (1)/(2)log(1 + (g221 + g231)E(X21)(1 − E2(X1X2) ⁄ E(X21)E(X22))) ≤ 
 ≤ (1)/(2)log(1 + (1 − ρ2)(S21 + S31)) = C((1 − ρ2)(S21 + S31))
(a)-Gaussian maximizes differential entropy. Првиот член се добива во 19↓. Вториот го добив но со одредени претпоставки
(g221 + g231)(E(X21) − E2(X1X2) ⁄ E(X22)) = (g221 + g231)E(X21)(1 − E2(X1X2) ⁄ E(X21)E(X22)) = (g221\oversetPE(X21) + g231E(X21))(1 − \oversetρ2E2(X1X2) ⁄ E(X21)E(X22)) = (S21 + S31)(1 − ρ2)
Од Capacity theorem со користење на Шварцовото неравенство \mathchoice|E(XY)|2 ≤ E(X2)E(Y2)|E(XY)|2 ≤ E(X2)E(Y2)|E(XY)|2 ≤ E(X2)E(Y2)|E(XY)|2 ≤ E(X2)E(Y2)
EX1X2(X1X2) = EX2[X2\oversetYEX1[X1|X2]] = E[X2E[X1|X2]] ≤ (E[X22]E{E2[X1|X2]})
(E2X1X2(X1X2))/(E[X22]) ≤ E{E2[X1|X2]}
——————————————————————————–———————
First term in the cutset bound
\mathchoiceh(Y3|X2)h(Y3|X2)h(Y3|X2)h(Y3|X2) = h(g31X1 + g32X2 + Z3|X2) ≤ (1)/(2)log(2πe)E(Var(Y3|X2)) ≤ (1)/(2)⋅log(2πe)Ex2(var(g31X1 + g32X2 + Z3|X2)
Ex2(Var(g31X1 + g32X2 + Z3|x2) = Ex2{var[g31X1|X2]} + 1 = 
 = |σ2 = ξ2 − ξ2| = Ex2[Ex1[(g31X1)2|X2] − E2x1[g31X1|X2]] + 1 = Ex1[g231X21] − g31\mathchoiceEx2E2x1[X1|X2]Ex2E2x1[X1|X2]Ex2E2x1[X1|X2]Ex2E2x1[X1|X2] + 1 = (*)
\mathchoiceEx2E2x1[g31X1|X2]Ex2E2x1[g31X1|X2]Ex2E2x1[g31X1|X2]Ex2E2x1[g31X1|X2] ≥ (E(X1X2))2 ⁄ E(X22)
[EX1X2(X1X2)]2 = [EX2[X2EX1[X1|X2]]]2 ≤ E[X22]E[E2[X1|X2]] → \mathchoiceEx2[E2x1[X1|X2]] = [EX1X2(X1X2)]2 ⁄ E[X22]Ex2[E2x1[X1|X2]] = [EX1X2(X1X2)]2 ⁄ E[X22]Ex2[E2x1[X1|X2]] = [EX1X2(X1X2)]2 ⁄ E[X22]Ex2[E2x1[X1|X2]] = [EX1X2(X1X2)]2 ⁄ E[X22]
(*) ≤ g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22] + 1
\mathchoiceEx2{var[X1|X2]} = Ex1[X21] − [EX1X2(X1X2)]2 ⁄ E[X22]Ex2{var[X1|X2]} = Ex1[X21] − [EX1X2(X1X2)]2 ⁄ E[X22]Ex2{var[X1|X2]} = Ex1[X21] − [EX1X2(X1X2)]2 ⁄ E[X22]Ex2{var[X1|X2]} = Ex1[X21] − [EX1X2(X1X2)]2 ⁄ E[X22]
\mathchoiceh(Y3|X2)h(Y3|X2)h(Y3|X2)h(Y3|X2) = (1)/(2)⋅log(2πe)(1 + g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22]) Q.E.D!!!
\mathchoiceE(var(Y3|X2)) = 1 + g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22]E(var(Y3|X2)) = 1 + g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22]E(var(Y3|X2)) = 1 + g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22]E(var(Y3|X2)) = 1 + g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22]
Second term in cutset bound (Сеуште не го имам докажано)
I(X1;Y2, Y3|X2) = (1)/(2)⋅log(g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22] + 1) + \mathchoice(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])
\mathchoiceE[var(Y2|X2Y3)]E[var(Y2|X2Y3)]E[var(Y2|X2Y3)]E[var(Y2|X2Y3)] = Ex2y3[var(Y2|X2Y3)] = Ex2y3[E(Y22|X2Y3) − E2[Y2|X2Y3]] = E[Y22] − \mathchoiceEx2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}
\mathchoiceY2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, 
E[Y22] = g221E[X21] + 1
\mathchoiceEx2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]} = Ex2y3{E2[g21X1 + Z2|X2Y3]} = g221\mathchoiceEx2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]} + 1
Ex2y3{E2[X1|X2Y3]} = ((E(X1X2Y3))2)/(E[(Y3X2)2])
E[(Y3X2)2] = E[(g31X1 + g32X2 + Z3)2X22] = g31E(X1X2) + g32E(X22)
E[(g31X1 + g32X2 + Z3)2X22] = E[X22g312X12 + 2 X23g31X1g32 + 2 X22g31X1Z3 + X24g322 + 2 X23g32Z3 + X22Z32] = 
 = g31E[X21X22]
\mathchoiceEx2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]} = ((\mathchoiceE(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2))2)/(Ex2[Ey3[Y23|X2]])
\mathchoiceE(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2) = E((g31X1 + g32X2 + Z3)X1|X2) = E(g31X21 + g32X1X2 + Z3X1|X2) = g31E(X21|X2) + g32E[X1X2|X2] = 
 = g31E(X21|X2) + g32E[X1|X2] = g31E(X21) + g32E[X1|X2] = g31E(X21) + g32E[X1|X2]
x1x2p(x1x2|x2)dx1dx2 = x1x2p(x2|x2)p(x1|x2x2)dx1dx2 = x1x2p(x1|x2)dx1dx2 = E[X1|X2]
E2[X1, X2] = E[X21]E[E2[X1|X2]]
E[var(Y2|X2Y3)] = g221E[X21] + 1 − ((g31E(X21) + g32E[X1|X2])2)/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22])
g221E[X21] + 1 − ((g31E(X21) + g32E[X1X2])2)/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]) = ||E2[X1|X2] = (E2[X1X2])/(E[X22])|| = 
 = (g221E[X21] + \cancelg231E2[X1] − g221g231E[X21][E(X1X2)]2 ⁄ E[X22] + 1 + g231Ex1[X21] − g231E2(X1X2) ⁄ E[X22] − \cancelg231E2(X21) − g31g32E(X21)E[X1|X2] − g232E2[X1|X2])/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]) = 
(1 + (g221 + g231)E[X21] − g221g231E[X21][E(X1X2)]2 ⁄ E[X22] − (g231 + g232)E2(X1X2) ⁄ E[X22] − g31g32E(X21)E[X1|X2])/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]) ???
Second term in cutset bound (Сеуште не го имам докажано)
I(X1;Y2, Y3|X2) = h(Y3|X2) + h(Y2|X2Y3) − log(2πe) = (1)/(2)⋅log(g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22] + 1) + \mathchoice(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])
\mathchoiceE[var(Y2|X2Y3)]E[var(Y2|X2Y3)]E[var(Y2|X2Y3)]E[var(Y2|X2Y3)] = Ex2y3[var(Y2|X2Y3)] = Ex2y3[E(Y22|X2Y3) − E2[Y2|X2Y3]] = E[Y22] − \mathchoiceEx2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}
\mathchoiceY2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, 
E[Y22] = g221E[X21] + 1
\mathchoiceEx2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]}Ex2y3{E2[Y2|X2Y3]} = Ex2y3{E2[g21X1 + Z2|X2Y3]} = g221\mathchoiceEx2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]} + 1
\mathchoiceEx2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]} = ((\mathchoiceE(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2))2)/(Ex2[Ey3[Y23|X2]])
\mathchoiceE(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2) = E((g31X1 + g32X2 + Z3)X1|X2) = E(g31X21 + g32X1X2 + Z3X1|X2) = g31E(X21|X2) + g32Ex1x2[X1X2|X2] = 
 = g31E(X21|X2) + g32E[X1] = g31(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]) + g32\cancelto0E[X1] = g31(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])
x1x2p(x1x2|x2)dx1dx2 = x1x2p(x2|x2)p(x1|x2x2)dx1dx2 = x1x2p(x1|x2)dx1dx2 = E[X1|X2]
E2[X1, X2] = E[X21]E[E2[X1|X2]]
E[var(Y2|X2Y3)] = g221E[X21] + \cancel1 − g221(g231(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])2)/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]) − \cancel1 = g221E[X21] − g221(g231(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])2)/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22])
 = (g221E[X21] + \cancelg221g231E2[X1] − \cancelg221g231E[X21][E(X1X2)]2 ⁄ E[X22] − \cancelg231g221E2x1[X21] + \cancel2⋅g231g221Ex1[X21][E(X1X2)]2 ⁄ E[X22] − g231g221[E(X1X2)]4 ⁄ E4[X22])/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]) = 
 = (g221E[X21] + g231g221E[X21][E(X1X2)]2 ⁄ E[X22] − g231g221[E(X1X2)]4 ⁄ E4[X22])/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]) = (1 + g221E[X21](1 + g231[E(X1X2)]2 ⁄ E[X22]) − 1 − g231g221[E(X1X2)]4 ⁄ E2[X22]2)/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22])
(g221E[X21] + g231g221[E(X1X2)]2 ⁄ E[X22](E[X21] − [E(X1X2)]2 ⁄ E2[X22]))/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22])
(g221 + g231)(E(X21) − E2(X1X2) ⁄ E(X22)) = g221E(X21) − g221E2(X1X2) ⁄ E(X22) + g231E(X21) − g231E2(X1X2) ⁄ E(X22)
Докажано но со претпоставките (a) и (b)
I(X1;Y2, Y3|X2) = h(Y3|X2) + \mathchoiceh(Y2|X2Y3)h(Y2|X2Y3)h(Y2|X2Y3)h(Y2|X2Y3) − log(2πe) = (1)/(2)⋅log(g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22] + 1) + \mathchoice(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])(1)/(2)log(E[var(Y2|X2Y3)])
\mathchoiceY2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, Y2 = g21X1 + Z2 Y3 = g31X1 + g32X2 + Z3, 
\mathchoiceE[var(Y2|X2Y3)]E[var(Y2|X2Y3)]E[var(Y2|X2Y3)]E[var(Y2|X2Y3)] = Ex2y3[var(g21X1 + Z2|X2Y3)] = Ex2y3[var(g21X1|X2Y3)] + 1 = 1 + g221Ex2y3[E(X21|X2Y3)] − g221\mathchoiceEx2y3E2[X1|X2Y3]Ex2y3E2[X1|X2Y3]Ex2y3E2[X1|X2Y3]Ex2y3E2[X1|X2Y3]
\mathchoiceEx2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]} = ((\mathchoiceE(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2))2)/(Ex2[Ey3[Y23|X2]])
Ex2[Ey3[Y23|X2]] = E(var(Y3|X2)) = 1 + g231Ex1[X21] − g231[EX1X2(X1X2)]2 ⁄ E[X22] (a) Не е баш најпрецизно зошто ја испуштам статичната компонента од варијансата!!!
\mathchoiceE(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2)E(Y3X1|X2) = E((g31X1 + g32X2 + Z3)X1|X2) = E(g31X21 + g32X1X2 + \overset non correlatedZ3X1|X2) = g31E(X21|X2) + g32Ex1x2[X1X2|X2] = 
 = g31E(X21|X2) + g32E[X1] = g31(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]) + g32\cancelto0E[X1] = g31(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])
\mathchoiceEx2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]}Ex2y3{E2[X1|X2Y3]} = (g231(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])2)/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]);
Ex2y3[E(X21|X2Y3)] = E[var(X1|X2)] = (Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]) (b) Ова не е баш најпрецизно!!!
\mathchoiceE[var(Y2|X2Y3)]E[var(Y2|X2Y3)]E[var(Y2|X2Y3)]E[var(Y2|X2Y3)] = 1 + g221(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]) − g221(g231(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])2)/(1 + g231(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])) = 
 = (1 + g231(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]) + g221(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]) + \cancelg231g221(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])2 − \cancelg221g231(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22])2)/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]) = 
 = (1 + g231(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]) + g221(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]))/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22]) = (1 + (g221 + g231)(Ex1[X21] − [E(X1X2)]2 ⁄ E[X22]))/(1 + g231Ex1[X21] − g231[E(X1X2)]2 ⁄ E[X22])  Q.E.D
Q.E.D!!!
Ex2y3{E2[X1|X2Y3]} ≥ (E2(Y3X1|X2))/(Ex2[Ey3[Y23|X2]])
E[X1X2Y3] = Ex2[X2Ex1y3[X1Y3|X2]] ≤ Ex2[X22]Ex2[E2x1y3[X1Y3|X2]]
E[X1X2Y3] = Ex2y3[X2Y3Ex1[X1|X2Y3]] ≤ Ex2y3[(X2Y3)2]Ex2y3[E2x1[X1|X2Y3]] Ex2y3[E2x1[X1|X2Y3]] ≥ E2[X1X2Y3] ⁄ Ex2[(X2Y3)2]
-Проба да го добијам изразот од Khojastepour со Theorem 6
\mathchoiceR*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2) = supI(X1;Y2, Y3, |X2)
I(X2;Y3) ≥ I(Y2;2|X2, Y3)
I(Y2;2|X2, Y3) = 0
Y3 = g32X2 + g31X1 + Z3 Y2 = g21X2 + Z2
I(X2;Y3) = H(Y3) − H(Y3|X2) = (1)/(2)log(2πe)(1 + S32 + S31 + g31g32E(X1X2)) − (1)/(2)log(2πe)(Ex2[Var(g31X1|X2)] + 1) = 
g31g32E(X1X2) = (S31S32)ρ = (S31S32)(E[X1X2])/((E[X21]E[X22])) = (g231Pg232P)(E[X1X2])/((PP)) = g31g32E[X1X2]
(1)/(2)log(2πe)(S32 + S31 + 1 + ρ(S31S32)) − (1)/(2)log(2πe)g231E(X21) − (g231E2(X1X2))/(E[X22]) + 1 = (*)
g231E(X21) − (E2(X1X2))/(E[X22]) + 1 = S31 − g231E[X21](E2(X1X2))/(E[X22]E[X21]) = S31 − S31ρ2 = S31(1 − ρ2)
(*) = (1)/(2)log(2πe)((S32 + S31 + 1 + ρ(S31S32)))/(S31(1 − ρ2))
((S32 + S31 + 1 + ρ(S31S32)))/(S31(1 − ρ2)) ≥ 0
S32 + S31 + 1 + ρ(S31S32) ≥ 0 ρ ≥  − (S32 + S31 + 1)/((S31S32))
I(X1;Y2, Y3, |X2) = (1)/(2)log(1 + (1 − ρ2)(S32 + S21)) = (1)/(2)log1 + 1 − (S32 + S31 + 1)/((S31S32))2(S32 + S21)
figure /home/jovan/Dropbox/Books Read/Notebooks/LyxNotebooks/El Gamal, Network Information Theory/Minimum Of Two Functions.png
C ≤ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)(S31 + S21))}

На сликата горе се прикажани првиот и вториот член од минимумот за S31 = 10 S32 = 5 S21 = 15. Јасно е дека до пресекот помал е првиот член, а после пресекот е помал вториот член. Видливо е дека пресекот е максимум на резултантаната функција која е минимум од двете функции.
C ≤ max0 ≤ ρ ≤ 1min{log(1 + S31 + S32 + 2ρ(S31S32)), log(1 + (1 − ρ2)(S31 + S21))}

Solve[log(1 + S31 + S32 + 2ρ(S31S32)) − log(1 + (1 − ρ2)(S31 + S21)) =  = 0, ρ]
Од Maple се добива:
ρ = ( − (S31S32) + (S31S21 − S21S32 + S212))/(S31 + S21)

f(ρ) = 1 + S31 + S32 + 2ρ(S31S32)

f(ρ) = 1 + S31 + S32 + 2 (( − (S31S32) + (S21S31 − S32S21 + S221))(S31S32))/(S31 + S21) = \mathchoice1 + S31 + S32 + 2⋅(( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32))/(S31 + S21)1 + S31 + S32 + 2⋅(( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32))/(S31 + S21)1 + S31 + S32 + 2⋅(( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32))/(S31 + S21)1 + S31 + S32 + 2⋅(( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32))/(S31 + S21)

 = 1 + ((S31 + S32)(S31 + S21) + 2⋅( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32))/(S31 + S21) = 

1 + ((S31 + S32)(S31 + S21) + 2⋅( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32))/(S31 + S21) = 1 + (S231 + S21S31 + S32S31 + S21S32 − 2⋅S31S32 + 2⋅(S21S31S32(S31 − S32 + S21)))/(S31 + S21) = (*)

Expand[(S31 + S32)(S31 + S21)] = S231 + S21S31 + S32S31 + S21S32

————————————
((S21S32) + (S31(S31 + S21 − S32)))2 = S21S32 + 2(S31S21S32(S31 + S21 − S32)) + S31(S31 + S21 − S32) = S21S32 + 2(S31S21S32(S31 + S21 − S32)) + S231 + S21S31 − S32S31

 = \mathchoiceS231 + S21S31 − S32S31 + S21S32 + 2(S31S21S32(S31 + S21 − S32))S231 + S21S31 − S32S31 + S21S32 + 2(S31S21S32(S31 + S21 − S32))S231 + S21S31 − S32S31 + S21S32 + 2(S31S21S32(S31 + S21 − S32))S231 + S21S31 − S32S31 + S21S32 + 2(S31S21S32(S31 + S21 − S32))

(*) = 1 + (\mathchoiceS231 + S21S31 + S21S32 − S31S32 + 2⋅(S21S31S32(S31 − S32 + S21))S231 + S21S31 + S21S32 − S31S32 + 2⋅(S21S31S32(S31 − S32 + S21))S231 + S21S31 + S21S32 − S31S32 + 2⋅(S21S31S32(S31 − S32 + S21))S231 + S21S31 + S21S32 − S31S32 + 2⋅(S21S31S32(S31 − S32 + S21)))/(S31 + S21) = 1 + (((S21S32) + (S31(S31 + S21 − S32)))2)/(S31 + S21)

За да ги добијам условите ќе го гледам изразот во зелено
1 + S31 + S32 + 2(( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32))/(S31 + S21)
\mathchoiceS21 ≥ S32S21 ≥ S32S21 ≥ S32S21 ≥ S32
( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32) = ( − S31S32 + (S21S31S32(S31 + S21 − S32)))  ≥ ( − S31S32 + (S21S31S32(S31 + S21 − S21))) = 
( − S31S32 + S31(S21S32)) ≥ ( − S31S32 + S31(S32S32)) = ( − S31S32 + S31S32) ≥ 0
\mathchoiceS21 < S32S21 < S32S21 < S32S21 < S32
( − (S31S32) + (S21(S31 − S32 + S21)))(S31S32) = ( − S31S32 + (S21S31S32(S31 + S21 − S32))) <( − S31S32 + (S21S31S32(S31 + S21 − S21))) = 
( − S31S32 + S31(S21S32)) < ( − S31S32 + S31(S32S32)) = ( − S31S32 + S31S32) < 0
Заради тоа капацитетот е:
C ≤  C(((S21S32) + (S31(S31 + S21 − S32)))2)/(S31 + S21)  if S21 ≥ S32       C(S31 + S21)  otherwise
Q.E.D!!!
Најубави докази со користење на естимации и тоталната естимација и условни естимации има понатаму. Овде добри се ознаките ама на пример не се користи Jensen неравенството што мислам дека е круциално за разбирање на изразите со варијанса.
Direct-transmission lower bound.
It is straightforward to see that the lower bound in 4↑ yields the lower bound
\mathchoiceC ≥ C(S31)C ≥ C(S31)C ≥ C(S31)C ≥ C(S31)
Cmaxp(x1), x2I(X1;Y3|X2 = x2)
Y3 = g31X1 + g21X2 + Z3
I(X1;Y3|X2 = x2) = H(Y3|X2 = x2) − H(Y3|X1X2 = x2) = H(g31X1 + Z3) − H(Z3) ≤ (1)/(2)log2(2πe)(\oversetS31g231P + 1) − (1)/(2)log(2πe) = (1)/(2)log2(S31 + 1) = C(S31)
Q.E.D!
Multihop lower bound.
Consider the multihop lower bound in 5↑ subject to the power constraint. The distributions on the inputs X1 and X2 that optimize the bound are not known in general. Assuming X1 and X2 to be Gaussian,we obtain the lower bound
(11) \mathchoiceC ≥ min{C(S21), C(S32 ⁄ (S31 + 1))}C ≥ min{C(S21), C(S32 ⁄ (S31 + 1))}C ≥ min{C(S21), C(S32 ⁄ (S31 + 1))}C ≥ min{C(S21), C(S32 ⁄ (S31 + 1))}
To prove achievability of this bound, we extend the multiphop achievability to the case with input costs and use the discretization procedure in Section 3.4.
C ≥ maxp(x1)p(x2)min{I(X2;Y3), I(X1;Y2|X2)}
Y3 = g32X2 + g31X1 + Z3
Y2 = g21X1 + Z2
S32 = g32P
I(X2;Y3) = H(Y3) − H(Y3|X2) = (1)/(2)log(2πe)(g232P + g231P + 1) − (1)/(2)log(2πe)(g231P + 1) = (1)/(2)log((g232P + g231P + 1))/((g231P + 1)) = (1)/(2)log((S32 + S31 + 1))/((S31 + 1)) = (1)/(2)log1 + (S32)/((S31 + 1)) = \mathchoiceC(S32)/(S31 + 1)C(S32)/(S31 + 1)C(S32)/(S31 + 1)C(S32)/(S31 + 1)
I(X1;Y2|X2) = H(Y2|X2) − H(Y2|X1X2) = (1)/(2)log(2πe)(E(Var(X1|X2)) + 1) − (1)/(2)log(2πe)E[(g21X1 + Z1)2|X1] = (1)/(2)log(Ex2[Ex1[g221X21|X2]] + 1) = 
 = (1)/(2)log(g221Ex2[Ex1[X21|X2]] + 1) = (1)/(2)log(g221E[X21] + 1) = (1)/(2)log(g221P + 1) =  = (1)/(2)log(S21 + 1) = \mathchoiceC(S21)C(S21)C(S21)C(S21)
До ова одма доажаќ ако претпоставиш дека X1 и X2 се независни.
Q.E.D!!!
Decode-forward lower bound.
Maximizing the decode-forward lower bound in Theorem 3.4.3↑ subject to constraints yields.
\mathchoiceC ≥ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)S21)}C ≥ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)S21)}C ≥ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)S21)}C ≥ max0 ≤ ρ ≤ 1min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)S21)} = 
(12)  =  C(((S31(S21 − S32)) + (S32(S21 − S31)))2 ⁄ S21)  if S21 ≥ S31 + S32       C(S21)  otherwise
C ≥ maxp(x1x2)min{I(X1X2;Y3), I(X1;Y2|X2)}
Y3 = g31X1 + g32X2 + Z3,  Y2 = g21X1 + Z2
I(X1;Y2|X2) = H(Y2|X2) − H(Y2|X1X2) ≤ (1)/(2)log(2πe)E[Var(Y2|X2)] + (1)/(2)log(2πe) = (1)/(2)log(2πe)(E[Var(g21X1|X2)] + 1) − (1)/(2)log(2πe) = (*)
E[Var(g21X1|X2)] = Ex2[Ex1(g221X21|X2)] − g221Ex2[E2x1[X1|X2]] ≤ g221Ex1(X21) − (g221E2x2[X1X2])/(E[X22]) = g221Ex1(X21) − (g221E[X21]E2x2[X1X2])/(E[X22]E[X21]) = S21 − S21ρ2 = S21(1 − ρ2)
E[XY] = Ey[YEx[X|Y]] ≤ (Ey[Y2]Ey[E2x[X2|Y]]) (E2[XY])/(E[Y2]) ≤ EyE2x[X2|Y]
(*) = (1)/(2)log(2πe)(1 + S21(1 − ρ2)) = C(S21(1 − ρ2))
Q.E.D!!!
f(ρ0) = 1 + S31 + S32 − 2 (((S31S32) + (S31S32 − S31S21 − S21S32 + S221))(S31S32))/(S21)
ρ0: =  − ((S31S32) − (S31S32 − S31S21 − S21S32 + S221))/(S21)
1 + S31 + S32 − (2(S31S32 + (S31S32(S31S32 − S31S21 − S21S32 + S221))))/(S21) = ((1 + S31 + S32)S21 − 2(S31S32 − (S31S32(S31S32 − S31S21 − S21S32 + S221))))/(S21)

(S21 + S31S21 + S32S21 − 2⋅S31S32 + 2⋅(S31S32(S31S32 − S31S21 − S21S32 + S221)))/(S21) = 1 + (\mathchoiceS31S21 + S32S21 − 2⋅S31S32 + 2⋅(S31S32(S21 − S31)(S21 − S32))S31S21 + S32S21 − 2⋅S31S32 + 2⋅(S31S32(S21 − S31)(S21 − S32))S31S21 + S32S21 − 2⋅S31S32 + 2⋅(S31S32(S21 − S31)(S21 − S32))S31S21 + S32S21 − 2⋅S31S32 + 2⋅(S31S32(S21 − S31)(S21 − S32)))/(S21)

S31S32(S31S32 − S31S21 − S21S32 + S221) = S31S32(S31(S32 − S21) + S21(S21 − S32)) = S31S32( − S31(S21 − S32) + S21(S21 − S32)) = S31S32(S21 − S31)(S21 − S32)

((S31(S21 − S32)) + (S32(S21 − S31)))2 = S31(S21 − S32) + S32(S21 − S31) + 2(S31(S21 − S32)S32(S21 − S31)) = 
S31S21 − S31S32 + S32S21 − S32S31 + 2(S31(S21 − S32)S32(S21 − S31)) = \mathchoiceS31S21 + S32S21 − 2⋅S32S31 + 2(S31S32(S21 − S32)(S21 − S31))S31S21 + S32S21 − 2⋅S32S31 + 2(S31S32(S21 − S32)(S21 − S31))S31S21 + S32S21 − 2⋅S32S31 + 2(S31S32(S21 − S32)(S21 − S31))S31S21 + S32S21 − 2⋅S32S31 + 2(S31S32(S21 − S32)(S21 − S31))

\mathchoiceS21 ≥ S31 + S32S21 ≥ S31 + S32S21 ≥ S31 + S32S21 ≥ S31 + S32
S32(S21 − 2⋅S31) − 2⋅(S31S32(S21 − S31)(S21 − S32)) = S32(S21 − 2⋅S31) − 2⋅(S31S32(S21 − S31)(S21 − S32)) ≥ 
S32(S31 + S32 − 2⋅S31) − 2⋅(S31S32(S31 + S32 − S31)(S31 + S32 − S32)) = S32(S32 − S31) − 2⋅(S31S32S32S31) = S32(S32 − S31) − 2⋅S32S31 = S32S32 − S32S31 − 2⋅S32S31 = S232 − 3⋅S32S31
(S31S32(S21 − S31)(S21 − S32)) ≥ (S31S32(S31 + S32 − S31)(S31 + S32 − S32)) ≥ (S31S32S32S31) = S31S32
S31S21 + S32S21 − 2⋅S32S31 + 2(S31S32(S21 − S32)(S21 − S31)) ≥ S31S21 + S32S21 − 2⋅S32S31 + 2S32S31 = S31S21 + S32S21 > S21(S31 + S32) > (S31 + S32)2
\mathchoicef > 1 + ((S31 + S32)2)/(S21)f > 1 + ((S31 + S32)2)/(S21)f > 1 + ((S31 + S32)2)/(S21)f > 1 + ((S31 + S32)2)/(S21)

\mathchoiceS21 < S31 + S32S21 < S31 + S32S21 < S31 + S32S21 < S31 + S32 S31 > S21 − S32 S32 > S21 − S31
(S31S32(S21 − S31)(S21 − S32)) > ((S21 − S32)(S21 − S31)(S21 − S31)(S21 − S32)) = (S21 − S31)(S21 − S32)
S31S21 + S32S21 − 2⋅S32S31 + 2(S31S32(S21 − S32)(S21 − S31)) > S31S21 + S32S21 − 2⋅S32S31 + 2(S21 − S31)(S21 − S32) = 2S221 − S31S21 − S32S21
 = S21(2S21 − S31 − S32) > S21(2(S31 + S32) − S31 − S32) = S21(S31 + S32) > S21S21 = S221
Expand[S31S21 + S32S21 − 2⋅S32S31 + 2(S21 − S31)(S21 − S32)] = 2S221 − S31S21 − S32S21
\mathchoicef > 1 + (S221)/(S21) = 1 + S21f > 1 + (S221)/(S21) = 1 + S21f > 1 + (S221)/(S21) = 1 + S21f > 1 + (S221)/(S21) = 1 + S21
Не е добар овде знакот треба помало да биде!!!
Q.E.D!!!
Achievability follows by setting X2 ~ N(0, P) and X1 = ρX2 + X1 where X1’ ~  N(0, (1 − ρ2)P) is independent of X2 and carries the new message to be recovered first by the relay. Note that when S21 ≤ S31, the decode-forward rate becomes lower that the direct transmission rate C(S31).
C = (1)/(2)log(1 + S31 + S32 + 2ρ(S31S32))
(d)/(d)C(ρ) = (1)/(2)(2(S31S32))/(1 + S31 + S32 + 2ρ(S31S32)) = 0 → ρ → ∞ ⇒ C(ρ) = Cmax
ρ = E(X1X2) ⁄ (E(X21)E(X22))
C = (1)/(2)log(1 + (1 − ρ2)(S32 + S21))
(d)/(d)C(ρ) = (1)/(2)( − 2(S32 + S31)ρ)/(1 + (1 − ρ2)(S32 + S31)) = 0 ⇒ ρ → 0 ⇒ C(ρ) = Cmax
S32 = g32P ≤ P
ρ = E(X1X2) ⁄ (E(X21)E(X22))

min{C(S31 + S32 + 2ρ(S31S32)), C((1 − ρ2)(S32 + S21))} = (log(1 + S31 + S32 + 2ρ(S31S32)) + log((1 − ρ2)(S32 + S21)))/(2) − ||(log(1 + S31 + S32 + 2ρ(S31S32)) − log(1 + (1 − ρ2)(S32 + S21)))/(2)||Solve(log(1 + S31 + S32 + 2ρ(S31S32)) + log((1 − ρ2)(S32 + S21)))/(2) − ||(log(1 + S31 + S32 + 2ρ(S31S32)) − log(1 + (1 − ρ2)(S32 + S21)))/(2)|| =  = 0, ρ
1 + S31 + S32 + 2ρ(S31S32) ≥ 1 + (1 − ρ2)(S32 + S21)
S31 + S32 + 2ρ(S31S32) ≥ (1 − ρ2)(S32 + S21) S31 + S32 + 2ρ(S31S32) ≥ S32 + S21 − ρ2(S32 + S21) S31 + 2ρ(S31S32) ≥ S21 − ρ2(S32 + S21)
Noncoherent decode-forward lower bound.
Since implementing coherent communication is difficult in wireless systems, one may consider a noncoherent decode-forward coding scheme, where X1 and X2 are independent. This gives the lower bound
(13) \mathchoiceC ≥ min{C(S31 + S32), C(S21)}C ≥ min{C(S31 + S32), C(S21)}C ≥ min{C(S31 + S32), C(S21)}C ≥ min{C(S31 + S32), C(S21)}
This scheme uses the same codebook generation and encoding steps as the (noncoherent) multihop scheme, but achieves a higher rate by performing simultaneous decoding.
Y3 = g31X1 + g32X2 + Z3, 
I(X1, X2;Y3) = h(Y3) − h(Y3|X1X2) = h(Y3) − (1)/(2)log(2πe) ≤ (1)/(2)log(E(Y23)) ≤ (1)/(2)log(1 + g231E[X21] + g232E[X22] + 2g31g32E[X1X2])
 ≤ (1)/(2)⋅log(1 + S31 + S32 + 2ρ\cancelto0E(X1)\cancelto0E(X2)) = (1)/(2)⋅log(1 + S31 + S32) = C(S31 + S32)
S31 = g31P = g31E(X21)
E[X1X2] = x1x2p(x1x2)dx1dx2 = x1x2p(x1)p(x2)dx1dx2 = E(X1)E(X2)
Y2 = g21X1 + Z2
I(X1;Y2|X2) = H(Y2|X2) − H(Y2|X1X2) = H(g21X1 + Z2|X2) − H(g21X1 + Z2|X1, X2) = (1)/(2)log(2πe)(g221P + 1) − (1)/(2)log(2πe) = (1)/(2)log(1 + S21) = C(S21)
Q.E.D!!!

3.6 Partial Decode-Forward Lower Bound

In decode-forward, the relay fully recovers the message, which is optimal for the degraded RC because the relay receives a strictly better version of X1 than receiver. But n some cases (e.g. the Gaussian RC with S21 < S31), the channel to the relay can be a bottleneck and decode-forward can be strictly worse than direct transmission. In partial decode-forward, the relay recovers only part of the message. This yields a tighter lower bound on the capacity than both decode-forward and direct transmission.
(Partial Decode-Forward Lower Bound)
The capacity of DMRC is lower bounded as
\mathchoiceC ≥ maxp(u, x1, x2)min{I(X1, X2;Y3), I(U;Y2|X2) + I(X1;Y3|X2U)}C ≥ maxp(u, x1, x2)min{I(X1, X2;Y3), I(U;Y2|X2) + I(X1;Y3|X2U)}C ≥ maxp(u, x1, x2)min{I(X1, X2;Y3), I(U;Y2|X2) + I(X1;Y3|X2U)}C ≥ maxp(u, x1, x2)min{I(X1, X2;Y3), I(U;Y2|X2) + I(X1;Y3|X2U)}
where |U| ≤ |X1||X2|.
Note that if we set U = X1, this lower bound reduces to the decode-forward lower bound, and if we set U = 0, it reduces to direct transmission bound.
I(X1;Y2|X2) + I(X1;Y3|X2X1)| ако се прептостави I(X1;Y3|X2X1) = I(X1;Y3|X2Y2)| = I(X1;Y2|X2) + I(X1;Y3|X2Y2) = I(X1;Y2Y3|X2)
Proof outline
We use block Markov coding and backward decoding. Divide the message Mj, j ∈ [1:b − 1] into two independent messages Mj at rate R and Mj’’ at rate R’’. Hence R = R’ + R’’. Fix the pmf p(u, x1x2)that attains the lower bound and randomly generate an independent codebook
Cj = {(un(mj’|mj − 1), xn1(mj’, mj’’|mj − 1), xn2(mj − 1)):mj − 1’, mj’ ∈ [1:2nR], mj’’ ∈ [1:2nR’’]}
for each blok j ∈ [1:b]. The sender and the relay cooperate to communicate mj to the receive. The relay recovers mj at the end of block j using joint typical decoding (with un(mj’|mj − 1) replacing xn1(mj|mj − 1) in decode-foward). The probability of error for this step tends to zero as n → ∞ if \mathchoiceR’ < I(U;Y2|X2) − δ(ϵ)R’ < I(U;Y2|X2) − δ(ϵ)R’ < I(U;Y2|X2) − δ(ϵ)R’ < I(U;Y2|X2) − δ(ϵ). After receiving all the blocks, the messages mj’, j ∈ [1:b − 1], are first recovered at the receiver using backward decoding (with (un(mj + 1’|mj), xn2(mj)) replacing (xn1(mj + 1|mj)xn2(mj)) in decode-forward). Probability of error for this step tends to zero as n → ∞ if \mathchoiceR’ < I(U, X2;Y3) − δ(ϵ)R’ < I(U, X2;Y3) − δ(ϵ)R’ < I(U, X2;Y3) − δ(ϵ)R’ < I(U, X2;Y3) − δ(ϵ). The receiver then finds the unique message mj’’, j ∈ [1:b − 1], such that
(un(mj’|mj − 1), xn1(mj’, mj’’|mj − 1), xn2(mj − 1)) ∈ A(n)ϵ
The probability of error for this step tends to zero as n → ∞ if \mathchoiceR’’ < I(X1;Y3|U, X2) − δ(ϵ)R’’ < I(X1;Y3|U, X2) − δ(ϵ)R’’ < I(X1;Y3|U, X2) − δ(ϵ)R’’ < I(X1;Y3|U, X2) − δ(ϵ). Eliminating R and R’’ from the rate constraints establishes the parital delcode-forward lower bound in Theorem 3.6↑.
The partial decode-forward scheme is optimal in some special cases.

3.6.1 Semideterminsitic DMRC

Suppose that Y2 is a function of (X1X2) i.e. Y2 = y2(X1X2). Then, the capacity of this semideterministic DMRC
C = maxp(x1x2)min{I(X1, X2;Y3), H(Y2|X2) + I(X1;Y3|X1X2)}
Achievability follows by setting U = Y2 in the partial decode-forward lower bound in 3.6↑ which is feasible since Y2 is function of (X1X2). The converse follows by the cutset bound in 3.2↑.
I(Y2;Y2|X2) = H(Y2|X2) − H(Y2|Y2, X2) = H(Y2|X2)
C ≥ maxp(u, x1, x2)min{I(X1, X2;Y3), I(Y2;Y2|X2) + I(X1;Y3|X2Y2)} = maxp(x1x2)min{I(X1, X2;Y3), H(Y2|X2) + I(X1;Y3|X1X2)}

3.6.2 Relay Channel with Orthogonal Sender Components

The relay channel with orthogonal components is motivated by the fact that in many wireless communication systems the relay cannot send and receive in hte same time slot or in the same frequency band. The relay channel model can be specialized to accommodate this constraint by assuming orthogonal sender or receiver components. Here we consider hte DMRC with orthogonal sender components depicted in 7↓, where X1 = (X1’, X1’’) and p(y2y3|x1x2) = p(y3|x1’, x2)p(y2|x1’’, x2). The relay channel with orthogonal receiver components is discused in Section 16.7.3. It turns out that the capacity is known fo this case.
The capacity of DMRC wiht orhogonal sender components
C = maxp(x2)p(x1’|x2)p(x1’’|x2)min{I(X1’, X2;Y3), I(X1’’;Y2|X2) + I(X1’;Y3|X2)}
p(x2x1x1’’) = p(x2)p(x1’|x2)p(x’’|x2x1) = p(x2)p(x1’|x2)p(x’’|x2)
C ≥ maxp(u, x1, x2)min{I(X1, X2;Y3), I(U;Y2|X2) + I(X1;Y3|X2U)}
C ≥ maxp(u, x1, x2)min{I(X1’, X2;Y3), I(X1’’;Y2|X2) + I(X1’;Y3|X2\mathchoiceX’’X’’X’’X’’)} = maxp(u, x1, x2)min{I(X1’, X2;Y3), I(X1’’;Y2|X2) + I(X1’;Y3|X2)}
под претпоставка дека X1’ → X2 → X1’’
I(X1’;Y3|X2X’’) = H(X1’|X2X’’) − H(X1’|Y3X2X’’) = H(X1’|X2) − H(X1’|Y3X2) = I(X1’;Y3|X2)
The proof of achievability uses partial decode-forward with U = X1’’ . The proof of the converse follows by the cutset bound.
figure Figure 16.6.png
Figure 7 Relay channel with orthogonal sender components

3.6.3 SFD Gaussian Relay Channel

We consider the Gaussian counterpart of the relay channel with orthogonal sender components depicted in 8↓, which we refer to as the sender frequency-divison (SFD) Gaussian RC. In this half-duplex model, the channel from the sender to the relay uses separate frequency band. More specifically, in this model X1 = (X1’, X1’’) and
Y2 = g21X1’’ + Z2
Y3 = g31X1’ + g32X2 + Z3
where g21, g31 and g32 are channel gains, and Z2 ~  N(0, 1) and Z3 ~  N(0, 1) are independent noise components.
figure Figure 16.7.png
Figure 8 Sender frequency-division Gaussian relay channel
Assume average power constraint P on each of X1 = (X1’, X1’’) and X2. The capacity of hte SFD Gaussian RC is:
(14) \mathchoiceC ≤ max0 ≤ ρ ≤ 1min{C(αS31 + S32 + 2ρ(αS31S32)), C(αS21) + C(α(1 − ρ2)S31)}C ≤ max0 ≤ ρ ≤ 1min{C(αS31 + S32 + 2ρ(αS31S32)), C(αS21) + C(α(1 − ρ2)S31)}C ≤ max0 ≤ ρ ≤ 1min{C(αS31 + S32 + 2ρ(αS31S32)), C(αS21) + C(α(1 − ρ2)S31)}C ≤ max0 ≤ ρ ≤ 1min{C(αS31 + S32 + 2ρ(αS31S32)), C(αS21) + C(α(1 − ρ2)S31)}
C ≥ maxp(u, x1, x2)min{I(X1, X2;Y3), I(U;Y2|X2) + I(X1;Y3|X2U)}
Achievability is proved by extending the partial decode-forward lower bound in 3.6↑ to the case with input cost constraints and using the discretization procedure in section 3.4 with U = X1’’ ~ N(0, αP),  X1’ ~ N(0, αP),  X2 ~ N(0, P) where (X1’, X2) is jointly Gaussian with correlation coefficient ρ and is independent of X1’’. The converse follows by showing that the capacity in Proposition 3.6.2↑ is attained by the same choice of (X1’, X1’’, X2).
It can be readily verified that direct transmission achieves C(S31) Ова се добива од изразот 14↑ доколку се замени α = 1 и ρ = 0 (Corresponding to α = 1 in 14↑, while decode-forward achieves min{C(S21), C(S32)} Не постои директна компонента т.е. S32 = 0 затоа што α = 0 a X1’ ~ N(0, αP) corresponding to α = 0 in 14↑. Both of these coding schemes are strictly sub-optimal in general.
By contrast it can be shown that the partial decode-forward lower bound for the (full-duplex) Gaussian RC in Section 3.5↑ is equal to the maximum of the direct-transmission and decode-forward lower bounds; see Appendix 16B. As such, partial decode-forward does not offer any rate improvement over these simpler schemes for the full-duplex case.
- Доказ на првиот член од 16.9
X1’’ ~ N(0, αP),  X1’ ~ N(0, αP),  X2 ~ N(0, P)
Y2 = g21X1’’ + Z2 Y3 = g31X1’ + g32X2 + Z3
I(X1’, X2;Y3) = H(Y3) − H(Y3|X1X2) ≤ (1)/(2)log(2πe)(g231αP + g32P + 2⋅E(X1X2)) − (1)/(2)log(2πe) = 
 = (1)/(2)⋅log(g231P + g232P + g31g32\oversetρ(E(X1X2))/((E(X1’2)E(X22)))(E(X12)E(X22))) = (1)/(2)⋅log(g231P + g232P + 2⋅ρ(αg231Pg232P)) = 
 = (1)/(2)⋅log(S31 + S32 + 2⋅ρ(αS31S32))
- Доказ на вториот член во 16.9
U = X1’’ ~ N(0, αP),  X1’ ~ N(0, αP),  X2 ~ N(0, P)
Y2 = g21X1’’ + Z2 Y3 = g31X1’ + g32X2 + Z3
C = maxp(x2)p(x1’|x2)p(x1’’|x2)min{I(X1’, X2;Y3), I(X1’’;Y2|X2) + I(X1’;Y3|X2)}
I(X1’’;Y2|X2) + I(X1’;Y3|X2) = h(Y2|X2) − h(Y2|X2, X1’’) + h(Y3|X2) − h(Y3|X1’, X2) = 
X1’’ е независен од X2
 = (1)/(2)log(2πe)E[Var(g21X1’’ + Z2|X2)] − (1)/(2)log(2πe) + (1)/(2)log(2πe)E[Var(g31X1’ + g32X2 + Z2|X2)] − (1)/(2)log(2πe) = 
 = (1)/(2)log(2πe)E[Var(g21X1’’ + Z2|X2)] − (1)/(2)log(2πe) + (1)/(2)log(2πe)E[Var(g31X1’ + g32X2 + Z2|X2)] − (1)/(2)log(2πe) = 
 = (1)/(2)log(2πe)E[Var(g21X1’’ + Z2|X2)] + (1)/(2)log(2πe)E[Var(g31X1’ + g32X2 + Z2|X2)] = 
 = (1)/(2)logE[Var(g21X1’’ + Z2|X2)] + (1)/(2)logE[Var(g31X1’ + g32X2 + Z2|X2)] = 
 = (1)/(2)logE[Var(g21X1’’ + Z2)] + (1)/(2)logE[Var(g31X1’ + Z2|X2)] = 
 = (1)/(2)log(g21αP + 1) + (1)/(2)log(E[Var(g31X1’|X2)] + 1) = 
 = (1)/(2)log(αS21 + 1) + (1)/(2)log(1 + g231(E[X’21] − E2[X1X2] ⁄ E[X22])) = 
 = (1)/(2)log(αS21) + (1)/(2)log(1 + g231E[X’21](1 − \oversetρ2(E2[X1X2] ⁄ E[X’21]E[X22]))) = 
 = (1)/(2)log(αS21 + 1) + (1)/(2)log(1 + αS31(1 − ρ2)) = C(αS21) + C(S31(1 − ρ2))
Appendix 16B
——————————————————
\mathchoiceI(U;Y2|X2) + I(X1;Y3|X2, U)I(U;Y2|X2) + I(X1;Y3|X2, U)I(U;Y2|X2) + I(X1;Y3|X2, U)I(U;Y2|X2) + I(X1;Y3|X2, U) = h(Y2|X2) − h(Y2|X2, U) + h(Y3|X2, U) − h(Y3|X1, X2, U) = 
 ≤ (1)/(2)log(2πeE[Var(Y2|X2)]) − h(Y2|X2, U) + h(Y3|X2, U) − (1)/(2)⋅log(2πe) ≤ 
E[Var(Y2|X2)] = E[Var(g21X1 + Z2|X2)] = E[Var(g21X1|X2)] + 1 = Ex2Ex1(g221X21|X2) − Ex2E2x1(g21X1|X2) ≤ (*)
Coshy Shwartz
E2(X1, X2) = E2x2x1(X1X2) = E2x2(X2Ex1(X1|X2)) = Ex2(X22)Ex2E2x1(X1|X2) → \mathchoiceEx2E2x1(X1|X2) = (E2(X1X2))/(Ex2(X22))Ex2E2x1(X1|X2) = (E2(X1X2))/(Ex2(X22))Ex2E2x1(X1|X2) = (E2(X1X2))/(Ex2(X22))Ex2E2x1(X1|X2) = (E2(X1X2))/(Ex2(X22))
(*) ≤ g221E(X21) − g221(E2(X1X2))/(E(X22)) = g221E(X21)(1 − ρ2)
 ≤ (1)/(2)log(1 + g221(E[X21] − E(X1X2)2 ⁄ E[X22])) − h(Y2|X2, U) + h(Y3|X2, U) = (1)/(2)log(1 + g221E[X21](1 − \oversetρ2(E(X1X2)2 ⁄ E[X21]E[X22]))) − h(Y2|X2, U) + h(Y3|X2, U) = \mathchoiceC((1 − ρ2)S21) − \mathchoiceh(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)C((1 − ρ2)S21) − \mathchoiceh(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)C((1 − ρ2)S21) − \mathchoiceh(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)C((1 − ρ2)S21) − \mathchoiceh(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)h(Y2|X2U) + h(Y3|X2U)
We now upper bound (h(Y3|X2, U) − h(Y2|X2, U)). First consider the case \mathchoiceS21 > S31S21 > S31S21 > S31S21 > S31 i.e., \mathchoice|g21| > |g31||g21| > |g31||g21| > |g31||g21| > |g31|. In this case
h(Y2|X2, U) = h(g21X1 + Z2|X2, U) = \mathchoiceh(g21X1 + Z3|X2, U) ≥ h(g31X1 + Z3|X2, U)h(g21X1 + Z3|X2, U) ≥ h(g31X1 + Z3|X2, U)h(g21X1 + Z3|X2, U) ≥ h(g31X1 + Z3|X2, U)h(g21X1 + Z3|X2, U) ≥ h(g31X1 + Z3|X2, U) = h(Y3|X1, U) → h(Y2|X2U) ≥ h(Y3|X1, U)
Hence,
C((1 − ρ2)S21) − h(Y2|X2U) + h(Y3|X2U) ≥ C((1 − ρ2)S21) − \cancelh(Y3|X2U) + \cancelh(Y3|X2U) = C((1 − ρ2)S21)
\mathchoiceI(U;Y2|X2) + I(X1;Y3|X2, U) ≤ C((1 − ρ2)S21)I(U;Y2|X2) + I(X1;Y3|X2, U) ≤ C((1 − ρ2)S21)I(U;Y2|X2) + I(X1;Y3|X2, U) ≤ C((1 − ρ2)S21)I(U;Y2|X2) + I(X1;Y3|X2, U) ≤ C((1 − ρ2)S21)
——————————————————————————–—-
and the rate of partial decode-forward reduces to decode-forward.
Next consider the case \mathchoiceS21 ≤ S31S21 ≤ S31S21 ≤ S31S21 ≤ S31. Since
\mathchoice(1)/(2)log(2πe) ≤ h(Y2|X2, U) ≤ h(Y2) = (1)/(2)log(2πe)(1 + S21)(1)/(2)log(2πe) ≤ h(Y2|X2, U) ≤ h(Y2) = (1)/(2)log(2πe)(1 + S21)(1)/(2)log(2πe) ≤ h(Y2|X2, U) ≤ h(Y2) = (1)/(2)log(2πe)(1 + S21)(1)/(2)log(2πe) ≤ h(Y2|X2, U) ≤ h(Y2) = (1)/(2)log(2πe)(1 + S21), првата нееднаквост следи од таму што RHS сигурно е поголем Y2 = g21X1 + Z2 зошто X1 ≠ 0 !!!
there exist a constan β ∈ [0, 1] such that
h(Y2|X2, U) = (1)/(2)log(2πe)(1 + βS21)
Now consider
\mathchoiceh(g21X1 + Z2|X2U)h(g21X1 + Z2|X2U)h(g21X1 + Z2|X2U)h(g21X1 + Z2|X2U) = h(g21)/(g31)g31X1 + (g31)/(g21)Z2|X2, U = \oversetEIT Theorem8.6.4hg31X1 + (g31)/(g21)Z2|X2 + log||(g21)/(g31)||\overset(a) = h(g31X1 + Z3’ + Z3’’|X2, U) + log||(g21)/(g31)||
\overset(b) ≥ (1)/(2)log(22h(g31X1 + Z3’|X2U) + 22h(Z3’’|X2U)) + log||(g21)/(g31)|| = (1)/(2)log(22h(g31X1 + Z3’|X2U) + 2πe(g231 ⁄ g221 − 1)) + log||(g21)/(g31)|| = 
\mathchoice = (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)) + (1)/(2)log(S21)/(S31) = (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)) + (1)/(2)log(S21)/(S31) = (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)) + (1)/(2)log(S21)/(S31) = (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)) + (1)/(2)log(S21)/(S31)
where in (a) Z3’ ~ N(0, 1) and \mathchoiceZ3’’ ~ N0, (g231)/(g221) − 1Z3’’ ~ N0, (g231)/(g221) − 1Z3’’ ~ N0, (g231)/(g221) − 1Z3’’ ~ N0, (g231)/(g221) − 1 are independent, and (b) follows by the entropy power inequality.
entropy power inequality EIT 9.181
2(2)/(n)(h(X + Y)) ≥ 2(2)/(n)H(X) + 2(2)/(n)H(Y)
Since
h(g21X1 + Z2|X2U) = (1)/(2)log(2πe)(1 + βS21)
we obtain
2πe(S31 ⁄ S21 + βS31) ≥ 22h(Y3|X2, U) + 2πe(S31 ⁄ S21 − 1)
h(g21X1 + Z2|X2U) ≥ (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)) + (1)/(2)log(S21)/(S31)
(1)/(2)log(2πe)(1 + βS21) ≥ (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)) + (1)/(2)log(S21)/(S31)
(1)/(2)log(2πe)(1 + βS21) − (1)/(2)log(S21)/(S31) ≥ (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)) (1)/(2)log(2πe)(1 + βS21)(S31)/(S21) ≥ (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1))
(1)/(2)log(2πe)(1 + βS21)(S31)/(S21) ≥ (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)) (1)/(2)log(2πe)(1 + βS21)(S31)/(S21) ≥ (1)/(2)log(22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1))
2πe(S31)/(S21) + βS31 ≥ 22h(Y3|X2U) + 2πe(S31 ⁄ S21 − 1)⋅ ⁄ log(...)
log((S31)/(S21) + βS31)/((S31 ⁄ S21 − 1)) ≥ 2h(Y3|X2U) → h(Y3|X2U) ≤ (1)/(2)log((S31)/(S21) + βS31)/((S31 ⁄ S21 − 1)) = (1)/(2)log((S31)/(S21)(1 + βS21))/((S31)/(S21)1 − (S21)/(S31)) = (1)/(2)log((1 + βS21))/((1 − \underset ≤ 1(S21)/(S31))) ≤ (1)/(2)log(1 + βS21)
Thus h(Y3|X2, U) ≤ (1)/(2)log(2πe)(1 + βS31) and
h(Y3|X2U) − h(Y2|X2, U) ≤ (1)/(2)log(1 + βS31)/(1 + βS21)\overset(a) ≤ (1)/(2)log(1 + S31)/(1 + S21)
where (a) follows since if S21 ≤ S31, (1 + βS31) ⁄ (1 + βS21) is strictly increasing function of β and attains its maximum when β = 1. Substituting we obtain
I(U;Y2|X2) + I(X1;Y3|X2U) ≤ C((1 − ρ2)S21) − h(Y2|X2U) + h(Y3|X2U) ≤ C((1 − ρ2)S21) + (1)/(2)log(1 + S31)/(1 + S21) = (1)/(2)⋅log(1 + (1 − ρ2)S21) + (1)/(2)log(1 + S31)/(1 + S21) = 
 ≤ (1)/(2)⋅log(1 + S21) + (1)/(2)log(1 + S31)/(1 + S21) ≤ (1)/(2)⋅log(1 + S31) = \mathchoiceC(S31)C(S31)C(S31)C(S31)
Which is the capacity of direct channel. Thus the rate of partial decode-forward reduces to that of direct transmission.

3.7 Compress-Forward Lower Bound

In the (partial) decode-forward coding scheme, the relay recovers the entire message (or part of it). If the channel from the sender to the relay is weaker than the direct channel to the receiver, this requirement can reduce the rate below that for direct transmission in which the relay is not used at all. In the compress-forward coding scheme the relay helps communication by sending a description ова директно ме потсеќа на Distortion Rate. Се работи за реконструкција т.е примерок. of its received sequence to the receiver. Because this description is correlated with the received sequence, Wyner-Ziv coding is used to reduce the rate needed to communicate it to the receiver. This scheme achieves the following lower bound.
Compress Forward Lower Bound
The capacity of the DMRC is lower bounded as:
\mathchoiceC ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
where the maximum is over all conditional pmfs \mathchoicep(x1)p(x2)p(2|x2y2)p(x1)p(x2)p(2|x2y2)p(x1)p(x2)p(2|x2y2)p(x1)p(x2)p(2|x2y2) with |Ŷ2| ≤ | X2|| Y2| + 1.
Compared to the cutset bound in 3.2↑, the first term in the minimum is the multiple access bound without coherent cooperation (X1 and X2 are independent) and with a subtracted term, and the second term resembles the broadcast bound but with Y2 replaced by the description 2.
The compress-forward lower bound can be equivalently characterized as
(15) C ≥ maxI(X1;2, Y3|X2)
where the maximum is over all conditional pmfs p(x1)p(x2)p(2|x2y2) such that
I(X2;Y3) ≥ I(Y2;2|X2Y3)
We establish this equivalence in Appendix 16C.
The bound (before maximization) is not in general convex in p(x1)p(x2)p(2|x2, y2) and hence the compress-forward scheme can be improved using coded time sharing to yield the lower bound
C ≥ maxmin{I(X1, X2;Y3|Q) − I(Y2;2|X1, X2, Y3, Q), I(X1;2, Y3|X2, Q)}
where the maximum is over all conditional pmfs p(q)p(x1|q)p(x2|q)p(2|x2, y2, q) with |Ŷ2| ≤ |X2||Y2| + 1 and |Q| ≤ 2.
Appendix 16.CDenote the compress-forward lower bound in 3.7↑ by
R’ ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
where the maximum is over all conditional pmfs p(x1)p(x2)p(2|x2y2), and denote the alternative characterization in 15↑ by:
R’’ = maxI(X1;2, Y3|X2)

where the maximum is over all conditional pmfs p(x1)p(x2)p(2|x2y2) that satisfy the constraint
(16) \mathchoiceI(X2;Y2)I(X2;Y2)I(X2;Y2)I(X2;Y2) ≥ I(Y2;2|X2Y3)

we first show that R’’ ≤ R. For the conditional pmf that attains R’’ we have
(17) R’’ = I(X1;Y3, 2|X2) = I(X1;Y3|X2) + I(X1;2|X2, Y3) = I(X1, X2;Y3) − \mathchoiceI(X2;Y3)I(X2;Y3)I(X2;Y3)I(X2;Y3) + I(X1;2|X2, X3)

\overset(a) ≤ I(X1, X2;Y3) − I(Y2;2|X2Y3) + I(X1;2|X2, X3) = I(X1, X2;Y3) − I(X1, Y2;2|X2Y3) + I(X1;2|X2, X3) = 

 = I(X1, X2;Y3) − I(X1;2|X2Y3) − I(Y2;2|X1X2Y3) + I(X1;2|X2, X3) = I(X1, X2;Y3) − I(Y2;2|X1X2Y3) → 

R’’ = I(X1;Y3, 2|X2) ≤ I(X1, X2;Y3) − I(Y2;2|X1X2Y3)
p(x1x2, y22) = p(x1)p(x2|x1)p(y2|x1x2)p(2|y2x1x2) = p(x1)p(x2)\overset = 1  assumptionp(y2|x1x2)p(2|y2x2) = p(x1)p(x2)p(2|x2y2)
Oчигледно се работи за марков ланец: x1 → (x2y2) → 2
I(X1, Y2;2|X2Y3) = I(Y2;2|X2Y3) + I(X1;2|X2Y3Y2) = I(Y2;2|X2Y3) + H(2|X2Y3Y2) − H(2|X2Y3Y2, X1) = 
I(Y2;2|X2Y3) + H(2|X2Y3Y2) − H(2|X2Y3Y2) = I(Y2;2|X2Y3)
where (a) follows from 16↑
To show that R’ ≤ R’’, note that this is the case if I(X1;Y3, 2|X2) ≤ I(X1X2;Y3) − I(Y2;2|X1X2Y3) for the conditional pmf that attains R. Now assume that at the optimum conditional pmf, I(X1;Y3, 2|X2) > I(X1X2;Y3) − I(Y2;2|X1X2Y3). We show that a higher rate can be achieved. Fix a product pmf p(x1)p(x2) and let 2’ = 2 with probability p and 2’ = 0 with probability (1 − p). Then \mathchoice2’ → 2 → (Y2, X2)2’ → 2 → (Y2, X2)2’ → 2 → (Y2, X2)2’ → 2 → (Y2, X2) form a markov chain. Note that the two mutual information terms are continuous in p and that as p increases, the first term decreases and the second increases. Thus there exist p* such
R’ ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
I(X1;2’, Y3|X2) = 
H(2, 2) = H(2) + \overset = 0H(2|2) = H(2) + H(2|2) = H(2 = 2, 2 = 0) + p(2 = 2)H(2|2 = 2) + p(2 = 0)H(2|2 = 0) = 
 = H(p) + p⋅0 + (1 − p)H(2|2’ = 0) =  = H(p) + p⋅0 + (1 − p)H(2|2’ = 0) = H(p) + (1 − p)H(2) → \mathchoiceH(2) = H(p) + (1 − p)H(2)H(2) = H(p) + (1 − p)H(2)H(2) = H(p) + (1 − p)H(2)H(2) = H(p) + (1 − p)H(2)
Мене во Notebook 17p6 ми излегува дека H(2) = H(p) + pH(2) + (1 − p) но тоа не менува многу во илустрацијата.
I(X1;2’, Y3|X2) = I(X1;Y3|X2) + I(X1;2’|X2, Y3) = I(X1;Y3|X2) + H(2’|X2Y3) − H(2’|X1X2Y3) = 
I(X1;Y3|X2) + H(2’|X2Y3) − H(2’|X1X2Y3) = 
 = I(X1;Y3|X2) + H(p) + (1 − p)H(2|X2Y3) − H(p) − (1 − p)H(2|X1X2Y3) = I(X1;Y3|X2) + (1 − p)H(2|X2Y3) − (1 − p)H(2|X1X2Y3) = 
 = \mathchoiceI(X1;Y3|X2) + (1 − p)I(X1;Y2|X2Y3)I(X1;Y3|X2) + (1 − p)I(X1;Y2|X2Y3)I(X1;Y3|X2) + (1 − p)I(X1;Y2|X2Y3)I(X1;Y3|X2) + (1 − p)I(X1;Y2|X2Y3)
——————————————————————————–—–
I(X1, X2;Y3) − I(Y2;2’|X1X2Y3) = ?
I(Y2;2’|X1X2Y3) = H(2’|X1X2Y3) − H(2’|Y2X1X2Y3) = (1 − p)H(2|X1X2Y3) − (1 − p)H(2|Y2X1X2Y3)
 = I(X1, X2;Y3) − I(Y2;2’|X1X2Y3) = I(X1, X2;Y3) − (1 − p)H(2|X1X2Y3) + (1 − p)H(2|Y2X1X2Y3) = 
 = \mathchoiceI(X1, X2;Y3) − (1 − p)I(Y2;2|X1X2Y3)I(X1, X2;Y3) − (1 − p)I(Y2;2|X1X2Y3)I(X1, X2;Y3) − (1 − p)I(Y2;2|X1X2Y3)I(X1, X2;Y3) − (1 − p)I(Y2;2|X1X2Y3)
Reacall:
X1, X2
X =  X1 P = α       X2 P = 1 − α
θ = f(X) =  1  X = X1       2  X = X2
H(X, θ) = H(θ) + H(X|θ) = H(X) + H(θ|X) = H(X)
H(X) = H(θ) + H(X|θ) = H(α) + p(θ = 1)H(X1) + p(θ = 2)H(X2) = H(α) + αH(X1) + (1 − α)H(X2)
I(X1;Y3, 2’|X2) = I(X1X2;Y3) − I(Y2;2|X1, X2, Y3)

and the rate using p(2|y2x2) is larger than that using p(2|y2x2). By the above argument, at the optimum conditional pmf,
(18) I(X1X2;Y3) − I(Y2;2|X1, X2, Y3) = I(X1;Y3, 2|X2)
Thus (from 18↑)
(19) I(X1X2;Y3) = I(Y2;2|X1, X2, Y3) + I(X1;Y3, 2|X2)
I(X1X2;Y3) = I(X1;Y3, 2|X2) + I(Y2;2|X1, X2, Y3) = I(X1;Y3, 2|X2) + I(Y2;2|X1, X2, Y3)I(X1X2;Y3) = I(X2;Y3) + I(X1;Y3|X2) → I(X2;Y3) = I(X1X2;Y3) − I(X1;Y3|X2) Ако се замени 19↑ се добива:
I(X2;Y3) = I(X1;Y3, 2|X2) + I(Y2;2|X1X2, Y3) − I(X1;Y3|X2) = \cancelI(X1;Y3|X2) + I(X1;2|X2Y3) + I(Y2;2|X1X2, Y3) − \cancelI(X1;Y3|X2) = I(X1Y2;2|X2Y3) = H(2|X2Y3) − H(2|X2Y3X1Y2) = | assumption: x1 → (x2y2) → 2| = H(2|X2Y3) − H(2|X2Y3Y2) = I(Y2;2|X2Y3)
I(X2;Y3) = I(Y2;2|X2Y3)
These completes the poor of the equivalence.

3.7.1 Proof of the Compress-Forward Lower Bound

Again a block Markov coding scheme is used to communicate (b − 1) i.i.d. messages in b blocks. At the end of block j, a reconstruction sequence n2(j) of yn2(j) conditioned on xn2(j) (which is known to both the relay and the receiver) is chosen by the relay. Since the receiver has side information yn3(j) about y2(j), we use binning as in Wyner-Ziv coding to reduce hte rate necessary to send n2(j). The bin index is sent to the receiver in block j + 1 via xn2(j + 1). At the end of block j + 1, the receiver decodes for xn2(j + 1). It then uses yn3(j) and xn2(j) to decode yn2(j) and xn1 simultaneously. We now give the details of the proof.
Codebook generation. Fix the conditional pmf p(x1)p(x2)p(2|y2, x2) that attains the lower bound. Again, we randomly generate an independent codebook for each block. For j ∈ [1:b], randomly and independently generate 2nR sequences xn1(mj), mj ∈ [1:2nR2] each according to ni = 1pX1(x1i). Randomly and independently generate 2nR2 sequences xn2(lj − 1), lj − 1 ∈ [1:2nR2], each according to ni = 1pX2(x2i). For each lj − 1 ∈ [1:2nR2] , randomly and conditionally independently generate 2n2 sequences n2(kj|lj − 1),  kj ∈ [1:2n2], each according to ni = 1p2|X2(2i|x2i(li − 1)). This defines the codebook
Cj = {(xn1(mj), xn2(lj − 1), n2(kj|lj − 1)):mj ∈ [1:2nR], lj − 1 ∈ [1:2nR2], kj ∈ [1:2n2]}
Partition the set [1:2n2] into 2nR2 equal size bins B(lj),  lj ∈ [1:2nR2]. The codebook and bin assignment are revealed to all parties.
Block 1 2 3 ... j ... b − 1 b
X1 xn1(m1) xn1(m2) xn1(m3) ... xn1(mj) ... xn1(mb − 1) xn1(1)
Y2 2(k1|1), l1 2(k2|l1), l2 n2(k3|l2), l3 ... n2(kj|lj − 1), lj ... n2(kb − 1|lb − 2), lb − 1 0
X2 xn2(1) xn2(l1) xn2(l2) ... xn2(lj − 1) ... xn1(lb − 2) xn2(lb − 1)
Y3 0 11 22 ... j − 1j − 1 ... b − 1b − 2 b − 1b − 1
1 2 j − 1 b − 2 b − 1
Table 4 Encoding and decoding for the compress-forward lower bound
Encoding
Let mj ∈ [1:2nR] be the message to be sent in block j. The encoder transmits xn1(mj) from codebook Cj, where mb = 1 by convention.
Relay encoding
By convention, let l0 = 1. At the end of block j, the relay finds the index kj such that (y2(j), 2(kj|lj − 1), xn2(lj − 1)) ∈ A(n)ϵ. If there is more than one such index, it select one of them uniformly at random. If there is no such index, it selects an index form [1:2n2] uniformly at random. In block j + 1 the relay transmits xn2(lj), where lj is bin index of kj.
Decoding
Let ϵ > ϵ. At the end of block j + 1, the receiver finds the unique index j, such that (xn2(j), yn3(j + 1)) ∈ A(n)ϵ. It then finds unique message j such that
(xn1(j), xn2(j − 1), n2(j|j − 1), yn3(j)) ∈ A(n)ϵ for some j ∈ B(j).
Analysis of the probability of error
We analyze the probability of decoding error for the message Mj averaged over codebooks. Assume without loss of generality that Mj = 1 and let Lj − 1, Lj, Kj denote the indices chosen by the relay in block j. The decoder makes an error only if one or more of the following events occur:
(j) = {(Xn2(j − 1), n2(kj|j − 1), Yn2(j)) ∈ A(n)ϵfor all kj ∈ [1:2nR2̃]}
E1(j − 1) = {j − 1 ≠ Lj − 1}
E1(j) = {j ≠ Lj}, 
E2(j) = {(Xn1(1), Xn2(j − 1), n2(Kj|j − 1)Yn3(j))A(n)ϵ}
E3(j) = {(Xn1(mj), Xn2(j − 1), n2(Kj|j − 1)Yn3(j)) ∈ A(n)ϵfor some mj ≠ 1}
E4(j) = {(Xn1(mj), Xn2(j − 1), n2(j|j − 1), Yn3(j)) ∈ A(n)ϵfor some j ∈ B(j),  j ≠ Kj,  mj ≠ 1}
Thus the probability of error is upper bounded as
P(E(j)) = P{j ≠ 1} ≤ 
P((j)) + P(E1(j − 1)) + P(E1(j)) + P(E2(j)c(j)Ec1(j − 1)) + E3(j) + P(E4(j)Ec1(j − 1)Ec1(j)).
By independence of the codebooks and the covering lemma, the first term P((j)) tends to zero as n → ∞ if \mathchoice2 > I(2;Y2|X2) + δ(ϵ)2 > I(2;Y2|X2) + δ(ϵ)2 > I(2;Y2|X2) + δ(ϵ)2 > I(2;Y2|X2) + δ(ϵ). Following the analysis of the error probability in the multihop coding scheme, the next two terms P(E1(j − 1)) = P{j − 1 ≠ Lj − 1} and P(E1(j) = P{j ≠ Lj}) tends to zero as n → ∞ if \mathchoiceR2 < I(X2;Y3) − δ(ϵ)R2 < I(X2;Y3) − δ(ϵ)R2 < I(X2;Y3) − δ(ϵ)R2 < I(X2;Y3) − δ(ϵ). The fourth term is upper bounded as
P(E2(j)c(j)Ec1(j − 1)) ≤ P{(Xn1(1), Xn2(Lj − 1), n2(Kj|Lj − 1), Yn3(j) ≠ A(n)ϵ|c(j))}
which, by the independence of ht codebooks and the conditional typicality lemma tends to zero as n → ∞. By the same independence and the packing lemma, P(Ej(j)) tends to zero as n → ∞ if \mathchoiceR ≤ I(X1;X2, Y2, Y3) + δ(ϵ) = I(X1;2, Y3|X2) + δ(ϵ)R ≤ I(X1;X2, Y2, Y3) + δ(ϵ) = I(X1;2, Y3|X2) + δ(ϵ)R ≤ I(X1;X2, Y2, Y3) + δ(ϵ) = I(X1;2, Y3|X2) + δ(ϵ)R ≤ I(X1;X2, Y2, Y3) + δ(ϵ) = I(X1;2, Y3|X2) + δ(ϵ). finally, following similar steps as in Lemma 11.1 the last term is upper bounded as
P(E4(j)Ec1(j − 1)Ec1(j)) ≤ P{Xn1(mj), Xn2(Lj − 1), n2(j|Lj − 1), Yn3(j) ∈ A(n)ϵfor some j ∈ B(j),  j ≠ Kj,  mj ≠ 1} ≤ 
 ≤ P{Xn1(mj), Xn2(Lj − 1), n2(j|Lj − 1), Yn3(j) ∈ A(n)ϵfor some j ∈ B(1),  j ≠ Kj,  mj ≠ 1}
which, y the independence of the codebooks, the joint typicality lemma (twice), and the union of events bound, tends to zero as n → ∞ if \mathchoiceR + 2 − R2 < I(X1;Y3|X2) + I(2;X1, Y3|X2) − δ(ϵ)R + 2 − R2 < I(X1;Y3|X2) + I(2;X1, Y3|X2) − δ(ϵ)R + 2 − R2 < I(X1;Y3|X2) + I(2;X1, Y3|X2) − δ(ϵ)R + 2 − R2 < I(X1;Y3|X2) + I(2;X1, Y3|X2) − δ(ϵ). Combining the bounds and eliminating R2 and 2, we have shown that P(j ≠ Mj) tends to zero as n → ∞ for every j ∈ [1:b − 1] if
R ≤ R2 − 2 + I(X1;Y3|X2) + I(2;X1, Y3|X2) − δ(ϵ) ≤ I(X2;Y3) − δ(ϵ) − I(2;Y2|X2) − δ(ϵ) + I(X1;Y3|X2) + I(2;X1, Y3|X2) − δ(ϵ) = 
 = I(X2;Y3) − I(2;Y2|X2) + I(X1;Y3|X2) + I(2;X1, Y3|X2) − 2δ(ϵ) − δ(ϵ) = I(X1X2;Y3) − I(2;Y2|X2) + I(2;X1, Y3|X2)
R < I(X1, X2;Y3) + I(2;X1, Y3|X2) − I(2;Y2|X2) − 2δ(ϵ) − δ(ϵ)\overset(a) = I(X1, X2;Y3) + I(2;X1Y3|X2) − I(2;X1Y2Y3|X2) − δ(ϵ) = I(X1X2;Y3) − I(2;Y2|X1X2Y3) − δ(ϵ)
I(2;X1Y2Y3|X2) = H(2|X2) − H(2|X2Y2) = I(2;Y2|X2)
I(2;X1Y3|X2) − I(2;X1Y2Y3|X2) = I(2;X1Y3|X2) − I(2;X1Y3|X2) − I(2;Y2|X1X2Y3) =  − I(2;Y2|X1X2Y3)
where (a) follows since 2 → (X2Y2) → (X1Y3) form Markov chain. This completes the proof of the compress-forward lower bound.
There are several other coding schemes that achieve the compress-forward lower bound, most notably the noisy network coding scheme described in Section 18.3

3.7.2 Compress-Forward for the Gaussian RC

The conditional distribution F(x1)F(x2)F(2|y2, x2) that attains the compress-forward lower bound in Theorem 3.7↑ is not known for the Gaussian RC in general. Let X1 ~  N(0, P), X2 ~ N(0, P), and Z ~ N(0, N) be mutually independent and \mathchoice2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z (see 9↓). Substituting in the compress-foward lower bound in Theorem 3.7↑ and optimizing over N, we obtain the lower bound
(20) \mathchoiceC ≥ CS31 + (S21S32)/(S31 + S21 + S32 + 1)C ≥ CS31 + (S21S32)/(S31 + S21 + S32 + 1)C ≥ CS31 + (S21S32)/(S31 + S21 + S32 + 1)C ≥ CS31 + (S21S32)/(S31 + S21 + S32 + 1)
figure Figure 16.8.png
Figure 9 Compress-forward for the Gaussian RC.
C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
X1 ~  N(0, P), X2 ~ N(0, P), and Z ~ N(0, N) be mutually independent and \mathchoice2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z
C ≥ CS31 + (S21S31)/(S31 + S21 + S32 + 1) 2 → (X2Y2) → (X1Y3)
Y3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2
\mathchoiceI(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3) = H(Y3) − H(Y3|X1X2) ≤ (1)/(2)log(g31P + g32P + N) − (1)/(2)logN = (1)/(2)log(1 + (g31 + g32)(P)/(N)) = (1)/(2)log1 + (S31 + S32)/(N) = C(S31 + S32)/(N)
\mathchoiceI(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2) = H(2, Y3|X2) − H(2, Y3|X1X2) = 
 = H(g21X1 + Z2 + Z, g31X1 + g32X2 + Z3 + Z|X2) − H(g21X1 + Z2 + Z, g31X1 + g32X2 + Z3 + Z|X1X2) = 
 = H(g21X1 + Z2 + Z|X2) + H(g31X1 + Z3 + Z|X2) − (1)/(2)log2πe(2N) − (1)/(2)log2πe(2N) = 
 = (1)/(2)log2πe(g21X1 + 2N) + (1)/(2)log2πe(g31X1 + 2N) − (1)/(2)log2πe(2N) − (1)/(2)log2πe(2N) = 
 = log(g21P ⁄ 2N + 1) + (1)/(2)log(g31P ⁄ 2N + 1) = (1)/(2)log(S21 ⁄ 2N + 1) + (1)/(2)log(S31 ⁄ 2N + 1) = (1)/(2)log(S21 ⁄ 2N + 1)(S31 ⁄ 2N + 1) = (1)/(2)log((S21 + 2N)(S31 + 2N))/(4N2)
\mathchoiceI(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3) = H(2|X1X2Y3) − H(2|X1X2Y3Y2) = H(Y2 + Z|X1X2Y3) − H(Y2 + Z|X1X2Y3Y2) = 
 = H(g21X1 + Z2 + Z|X1X2Y3) − (1)/(2)log(2πe)(N) = (1)/(2)log(2πe)(2N) − (1)/(2)log(2πe)(N) = (1)/(2)log2 = (1)/(2)
maxminC(S31 + S32)/(N) − (1)/(2), (1)/(2)log((S21 + 2N)(S31 + 2N))/(4N2)
f(N) = (1)/(2)log((S21 + 2N)(S31 + 42N))/(4N2)
(d)/(dN)(f(N)) = 0 → N0 = ( − S21S31)/(S21 + S31)
f(N0) = 1 ⁄ 2 log2 − 1 ⁄ 4 ((S21 − S31)2)/(S21S31) = (1)/(2)log2 − (1)/(4)((S221 − 2S21S31 + S231))/(S21S31) = (1)/(2)log2((2S21S31 − S221 − S231))/(4⋅S21S31) = (1)/(2)log2(1)/(2) − (S221 + S231)/(4⋅S21S31)
C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
X1 ~  N(0, P), X2 ~ N(0, P), and Z ~ N(0, N) be mutually independent and \mathchoice2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z
C ≥ CS31 + (S21S31)/(S31 + S21 + S32 + 1) 2 → (X2Y2) → (X1Y3)
Y3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + Z
\mathchoiceI(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3) = H(Y3) − H(Y3|X1X2) ≤ (1)/(2)log(2πe)(g31P + g32P + 1) − (1)/(2)log(2πe) = (1)/(2)log(1 + S31 + S32) = C(S31 + S32)
\mathchoiceI(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2) = H(2, Y3|X2) − H(2, Y3|X1X2) = 
 = H(g21X1 + Z2 + Z, g31X1 + g32X2 + Z3 + Z|X2) − H(g21X1 + Z2 + Z, g31X1 + g32X2 + Z3 + Z|X1X2) = 
 = H(g21X1 + Z2 + Z|X2) + H(g31X1 + Z3 + Z|X2) − (1)/(2)log2πe(N + 1) − (1)/(2)log2πe(N + 1) = 
 = (1)/(2)log2πe(g21X1 + N + 1) + (1)/(2)log2πe(g31X1 + N + 1) − (1)/(2)log2πe(N + 1) − (1)/(2)log2πe(N + 1) = 
 = log(g21P ⁄ (N + 1) + 1) + (1)/(2)log(g31P ⁄ (N + 1) + 1) = (1)/(2)log(S21 ⁄ (N + 1) + 1) + (1)/(2)log(S31 ⁄ (N + 1) + 1) = (1)/(2)log(S21 ⁄ (N + 1) + 1)(S31 ⁄ (N + 1) + 1) = (1)/(2)log((S21 + N + 1)(S31 + N + 1))/((N + 1)2)
\mathchoiceI(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3) = H(2|X1X2Y3) − H(2|X1X2Y3Y2) = H(Y2 + Z|X1X2Y3) − H(Y2 + Z|X1X2Y3Y2) = 
 = H(g21X1 + Z2 + Z|X1X2Y3) − (1)/(2)log(2πe)(N) = (1)/(2)log(2πe)(N + 1) − (1)/(2)log(2πe)(N) = (1)/(2)log(N + 1)/(N)
maxmin(1)/(2)log(1 + S31 + S32) − (1)/(2)log(N + 1)/(N), (1)/(2)log((S21 + N + 1)(S31 + N + 1))/((N + 1)2)
(1)/(2)log(1 + S31 + S32)(N)/(N + 1) = (1)/(2)log((S21 + N + 1)(S31 + N + 1))/((N + 1)2)
(1 + S31 + S32)(N)/(N + 1) = ((S21 + N + 1)(S31 + N + 1))/((N + 1)2)

Cover Theorem 6 (Важи ако S21 ≤ S31)
\mathchoiceR*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2)
\mathchoiceI(X2;Y3) ≥ I(Y2;2|X2, Y3)I(X2;Y3) ≥ I(Y2;2|X2, Y3)I(X2;Y3) ≥ I(Y2;2|X2, Y3)I(X2;Y3) ≥ I(Y2;2|X2, Y3)
\mathchoiceY3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + ZY3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + ZY3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + ZY3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + Z \mathchoicep(x1)p(x2)p(2|x2y2)p(x1)p(x2)p(2|x2y2)p(x1)p(x2)p(2|x2y2)p(x1)p(x2)p(2|x2y2)
I(X2;Y3) = H(Y3) − H(Y3|X2) = h(g31X1 + g32X2 + Z3) − h(g31X1 + g32X2 + Z3|X2) = 
 = (1)/(2)log(2πe)(S31 + S32 + 1) − log(2πe)(S31 + 1) = (1)/(2)log((S31 + S32 + 1))/((S31 + 1))
I(Y2;2|X2, Y3) = H(2|X2Y3) − H(2|Y2X2Y3) = (1)/(2)log(2πe)(g21X1 + Z2 + Z|X2Y3) − (1)/(2)log(2πe)N = (1)/(2)log(S21 + 1)/(N) + 1
((S31 + S32 + 1))/((S31 + 1)) ≥ (S21 + 1)/(N) + 1 (S32)/((S31 + 1)) + 1 ≥ (S21 + 1)/(N) + 1 (S32)/((S31 + 1)) ≥ (S21 + 1)/(N) S32 ≥ (S21 + 1)/(N)(S31 + 1) \mathchoiceN ≥ (S21 + 1)/(S32)(S31 + 1)N ≥ (S21 + 1)/(S32)(S31 + 1)N ≥ (S21 + 1)/(S32)(S31 + 1)N ≥ (S21 + 1)/(S32)(S31 + 1)
I(X1;Y2̂, Y3, |X2) = (1)/(2)log((S21 + N + 1)(S31 + N + 1))/((N + 1)2) = (1)/(2)log(S21 + (S21 + 1)/(S32)(S31 + 1) + 1S31 + (S21 + 1)/(S32)(S31 + 1) + 1)/((S21 + 1)/(S32)(S31 + 1) + 12)
——————————————————————————–—————————
Y3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + Z
Горе имав грешка во пресметката на I(X1;2, Y3|X2). Вака треба:
\mathchoiceI(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2) = H(2, Y3|X2) − H(2, Y3|X1X2) = H(2|X2) + H(Y3|X22) − H(2|X1X2) − H(Y3|X1X22)
 = H(g21X1 + Z2 + Z|X2) + H(g31X1 + g32X2 + Z3|X22) − H(g21X1 + Z2 + N|X1X2) − H(g31X1 + g32X2 + Z3|X1X22) = 
 = H(g21X1 + Z2 + Z) + H(g31X1 + Z3) − H(Z2 + N) − (1)/(2)log2πe(1) = (1)/(2)log2πe(S21 + 1 + N) + (1)/(2)log2πe(S31 + 1) − (1)/(2)log2πe(1 + N) − (1)/(2)log2πe = 
 = (1)/(2)log((S21 + 1 + N))/(N + 1) + (1)/(2)log(S31 + 1) = (1)/(2)log(S21)/(\mathchoiceNNNN + 1) + 1(S31 + 1) = (1)/(2)log(S21)/((S21 + 1)/(S32)(S31 + 1) + 1) + 1(S31 + 1) = (1)/(2)log(S21S32)/((S21 + 1)(S31 + 1) + S32) + 1(S31 + 1) = (*)
Expand[(S21 + 1)(S31 + 1)] = S31S21 + S21 + S31 + 1
(*) = (1)/(2)log(S21S32)/(S31S21 + S21 + S31 + 1 + S32) + 1(S31 + 1)
(*) = \mathchoice(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1 = (1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + (S31 + 1) = \mathchoice(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1
Изразов погоре во Cyan е многу сличен на резултатот за RFD Gaussian Channel 26↓.
Ако ја занемарам 1-цата во именителот се добива:
(S21 + 1)/(S32)(S31 + 1) > 1 → S32 ≤ (S21 + 1)(S31 + 1)
(1)/(2)log((S21 + 1 + N))/(N + 1) + (1)/(2)log(S31 + 1) = (1)/(2)log(S21)/(\mathchoiceNNNN + \cancel1) + 1(S31 + 1) = (1)/(2)log(S21)/((S21 + 1)/(S32)(S31 + 1)) + 1(S31 + 1) = (1)/(2)log(S21S32)/((S21 + 1)(S31 + 1)) + 1(S31 + 1) = (1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1)) + (S31 + 1)
(1)/(2)log(S21S32)/((S21 + 1)) + S31 + 1

Под претпоставка дека N ≥ N2, N3
(1)/(2)log2πe(S21 + N) + (1)/(2)log2πe(S31 + 1) − (1)/(2)log2πe(N) − (1)/(2)log2πe = (1)/(2)log((S21 + N))/(N) + (1)/(2)log(S31 + 1) = (1)/(2)log(S21)/(N) + 1(S31 + 1) = (*)
N ≥ (S21 + 1)/(S32)(S31 + 1)
(*) = (1)/(2)log(S21)/((S21 + 1)/(S32)(S31 + 1)) + 1(S31 + 1) = (1)/(2)log(S21S32)/(S21 + 1) + S31 + 1
Да пробам откако ја отстранив грешката да одам без користење на Theorem 6 од Capacity Theorem.
C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
X1 ~  N(0, P), X2 ~ N(0, P), and Z ~ N(0, N) be mutually independent and \mathchoice2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z
C ≥ CS31 + (S21S31)/(S31 + S21 + S32 + 1) 2 → (X2Y2) → (X1Y3)
Y3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + Z
\mathchoiceI(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3) = H(Y3) − H(Y3|X1X2) ≤ (1)/(2)log(2πe)(g31P + g32P + 1) − (1)/(2)log(2πe) = (1)/(2)log(1 + S31 + S32) = C(S31 + S32)
\mathchoiceI(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2) = H(2, Y3|X2) − H(2, Y3|X1X2) = H(2|X2) + H(Y3|X22) − H(2|X1X2) − H(Y3|X1X22)
 = H(g21X1 + Z2 + Z|X2) + H(g31X1 + g32X2 + Z3|X22) − H(g21X1 + Z2 + N|X1X2) − H(g31X1 + g32X2 + Z3|X1X22) = 
 = H(g21X1 + Z2 + Z) + H(g31X1 + Z3) − H(Z2 + N) − (1)/(2)log2πe(1) = (1)/(2)log2πe(S21 + 1 + N) + (1)/(2)log2πe(S31 + 1) − (1)/(2)log2πe(1 + N) − (1)/(2)log2πe = 
 = (1)/(2)log((S21 + 1 + N))/(N + 1) + (1)/(2)log(S31 + 1) = (1)/(2)log(S21)/(\mathchoiceNNNN + 1) + 1(S31 + 1) = (1)/(2)log(S21)/((S21 + 1)/(S32)(S31 + 1) + 1) + 1(S31 + 1) = (1)/(2)log(S21S32)/((S21 + 1)(S31 + 1) + S32) + 1(S31 + 1) = 
 = \mathchoice(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1 = (1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + (S31 + 1) = \mathchoice(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1
\mathchoiceI(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3) = H(2|X1X2Y3) − H(2|X1X2Y3Y2) = H(Y2 + Z|X1X2Y3) − H(Y2 + Z|X1X2Y3Y2) = 
 = H(g21X1 + Z2 + Z|X1X2Y3) − (1)/(2)log(2πe)(N) = (1)/(2)log(2πe)(N + 1) − (1)/(2)log(2πe)(N) = (1)/(2)log(N + 1)/(N)
maxmin(1)/(2)log(1 + S31 + S32) − (1)/(2)log(N + 1)/(N), (1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1
(1)/(2)log(1 + S31 + S32)(N)/(N + 1) = (1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1
(1 + S31 + S32)(N)/(N + 1) = (S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1
Со користење на Maple се добива:
(S312 + S21S312 + 2 \mathchoiceS31S31S31S31 + 2 \mathchoiceS21S31S21S31S21S31S21S31 + 1 + \mathchoiceS21S21S21S21 + S31S32 + S32S21S31 + \mathchoiceS32S32S32S32 + S21S32)/(\mathchoiceS31S31S31S31 + \mathchoiceS21S31S21S31S21S31S21S31 + 1 + \mathchoiceS21S21S21S21 + \mathchoiceS32S32S32S32)

 = 1 + (S231 + S21S231 + S31 + S21S31 + S31S32 + S32S21S31 + S21S32)/(S31 + S21S31 + 1 + S21 + S32) =  = 1 + ((S31 + S21S31 + 1 + S21 + S32)S31 + S32S21S31 + S21S32)/(S31 + S21S31 + 1 + S21 + S32) = 

 = 1 + S31 + (S32S21(S31 + 1))/(S31 + S21S31 + 1 + S21 + S32) = 1 + S31 + (S32S21(S31 + 1))/(S31(1 + S21) + (1 + S21) + S32) = \mathchoice1 + S31 + (S32S21(S31 + 1))/((S31 + 1)(1 + S21) + S32)1 + S31 + (S32S21(S31 + 1))/((S31 + 1)(1 + S21) + S32)1 + S31 + (S32S21(S31 + 1))/((S31 + 1)(1 + S21) + S32)1 + S31 + (S32S21(S31 + 1))/((S31 + 1)(1 + S21) + S32)

1 + S31 + (S32S21)/(1 + S21 + (S32)/(S31 + 1))

Се добива истиот израз како во пристапот со користење на Теорема 6 од Capacity Theorem.
(S32S21(S31 + 1))/(S31 + S21S31 + 1 + S21 + S32)

S21 ≤ S31
S32 ≥ S21(1 + S21) − (1 + S31)
S31 + (S21S32)/(S31 + S21 + S32 + 1)
Видов една грешка погоре. Сум заменил N но нема потреба. Сепак пак се добива истиот резултат
C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
X1 ~  N(0, P), X2 ~ N(0, P), and Z ~ N(0, N) be mutually independent and \mathchoice2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z
C ≥ CS31 + (S21S31)/(S31 + S21 + S32 + 1) 2 → (X2Y2) → (X1Y3)
Y3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + Z
\mathchoiceI(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3) = H(Y3) − H(Y3|X1X2) ≤ (1)/(2)log(2πe)(g31P + g32P + 1) − (1)/(2)log(2πe) = (1)/(2)log(1 + S31 + S32) = C(S31 + S32)
\mathchoiceI(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2) = H(2, Y3|X2) − H(2, Y3|X1X2) = H(2|X2) + H(Y3|X22) − H(2|X1X2) − H(Y3|X1X22)
 = H(g21X1 + Z2 + Z|X2) + H(g31X1 + g32X2 + Z3|X22) − H(g21X1 + Z2 + N|X1X2) − H(g31X1 + g32X2 + Z3|X1X22) = 
 = H(g21X1 + Z2 + Z) + H(g31X1 + Z3) − H(Z2 + N) − (1)/(2)log2πe(1) = (1)/(2)log2πe(S21 + 1 + N) + (1)/(2)log2πe(S31 + 1) − (1)/(2)log2πe(1 + N) − (1)/(2)log2πe = 
 = (1)/(2)log((S21 + 1 + N))/(N + 1) + (1)/(2)log(S31 + 1) = \mathchoice(1)/(2)log(S21)/(\mathchoiceNNNN + 1) + 1(S31 + 1)(1)/(2)log(S21)/(\mathchoiceNNNN + 1) + 1(S31 + 1)(1)/(2)log(S21)/(\mathchoiceNNNN + 1) + 1(S31 + 1)(1)/(2)log(S21)/(\mathchoiceNNNN + 1) + 1(S31 + 1) = (1)/(2)log(S21)/((S21 + 1)/(S32)(S31 + 1) + 1) + 1(S31 + 1) = (1)/(2)log(S21S32)/((S21 + 1)(S31 + 1) + S32) + 1(S31 + 1) = 
 = \mathchoice(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/((S21 + 1)(S31 + 1) + S32) + S31 + 1 = (1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + (S31 + 1) = \mathchoice(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1(1)/(2)log(S21S32)/(((S21 + 1)(S31 + 1) + S32)/((S31 + 1))) + S31 + 1
\mathchoiceI(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3) = H(2|X1X2Y3) − H(2|X1X2Y3Y2) = H(Y2 + Z|X1X2Y3) − H(Y2 + Z|X1X2Y3Y2) = 
 = H(g21X1 + Z2 + Z|X1X2Y3) − (1)/(2)log(2πe)(N) = (1)/(2)log(2πe)(N + 1) − (1)/(2)log(2πe)(N) = (1)/(2)log(N + 1)/(N)
maxmin(1)/(2)log(1 + S31 + S32) − (1)/(2)log(N + 1)/(N), (1)/(2)log(S21)/(\mathchoiceNNNN + 1) + 1(S31 + 1)
(1)/(2)log(1 + S31 + S32)(N)/(N + 1) = (1)/(2)log(S21)/(\mathchoiceNNNN + 1) + 1(S31 + 1)
(1 + S31 + S32)(N)/(N + 1) = (S21)/(N + 1) + 1(S31 + 1)
N0 = (1 + S21 + S21S31 + S31)/(S32)
((1 + S31 + S32)(1 + S21 + S21S31 + S31))/(S32(1 + S21 + S21S31 + S31)/(S32) + 1) = ((1 + S31 + S32)(1 + S21 + S21S31 + S31))/((1 + S21 + S21S31 + S31 + S32))

Expand[(1 + S31 + S32)(1 + S21 + S21S31 + S31)] = S21S231 + S231 + 2S21S31 + S21S32S31 + S32S31 + 2S31 + S21 + S21S32 + S32 + 1
(S21S231 + S231 + 2S21S31 + S21S32S31 + S32S31 + 2S31 + S21 + S21S32 + S32)/((1 + S21 + S21S31 + S31 + S32)) = 1 + (S21S231 + S231 + S21S31 + S21S32S31 + S32S31 + S31 + S21S32)/((1 + S21 + S21S31 + S31 + S32)) = 

(S21S231 + S231 + S21S31 + S21S32S31 + S32S31 + S31 + S21S32 + 1)/((1 + S21 + S21S31 + S31 + S32))

 = (S31(S21S31 + S31 + S21 + S21S32 + S32 + 1) + S21S32)/((1 + S21 + S21S31 + S31 + S32)) = 1 + S31 + (S31S21S32 + S21S32)/((1 + S21 + S21S31 + S31 + S32)) = 1 + S31 + ((S31 + 1)S21S32))/(1 + S21 + S21S31 + S31 + S32)
C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
X1 ~  N(0, P), X2 ~ N(0, P), and Z ~ N(0, N) be mutually independent and \mathchoice2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z
C ≥ CS31 + (S21S31)/(S31 + S21 + S32 + 1) 2 → (X2Y2) → (X1Y3)
Y3 = g31X1 + g32X2 + Z3 Y2 = g21X1 + Z2 Y2̂ = Y2 + Z = g21X1 + Z2 + Z
\mathchoiceI(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3)I(X1, X2;Y3) = H(Y3) − H(Y3|X1X2) ≤ (1)/(2)log(2πe)(g31P + g32P + 1) − (1)/(2)log(2πe) = (1)/(2)log(1 + S31 + S32) = C(S31 + S32)
\mathchoiceI(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3) = H(2|X1X2Y3) − H(2|X1X2Y3Y2) = H(Y2 + Z|X1X2Y3) − H(Y2 + Z|X1X2Y3Y2) = 
 = H(g21X1 + Z2 + Z|X1X2Y3) − (1)/(2)log(2πe)(N) = (1)/(2)log(2πe)(N + 1) − (1)/(2)log(2πe)(N) = (1)/(2)log(N + 1)/(N)
\mathchoiceI(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2)I(X1;2, Y3|X2) = H(2, Y3|X2) − H(2, Y3|X1X2) = H(2|X2) + H(Y3|X22) − H(2|X1X2) − H(Y3|X1X22)
 = H(g21X1 + Z2 + Z|X2) + H(g31X1 + g32X2 + Z3|X22) − H(g21X1 + Z2 + N|X1X2) − H(g31X1 + g32X2 + Z3|X1X22) = 
 ≤ (1)/(2)log(E[Var(g21X1)] + 1 + N) + (1)/(2)log(E[Var(g31X1|2)] + 1) − (1)/(2)log2πe(1 + N) − (1)/(2)log2πe
 = (1)/(2)log2πe(S21 + 1 + N) + \mathchoice(1)/(2)log2πeS31 − (g231E2(X1))/(E[2]) + 1(1)/(2)log2πeS31 − (g231E2(X1))/(E[2]) + 1(1)/(2)log2πeS31 − (g231E2(X1))/(E[2]) + 1(1)/(2)log2πeS31 − (g231E2(X1))/(E[2]) + 1 − (1)/(2)log2πe(1 + N) − (1)/(2)log2πe = 
(1)/(2)log((S21 + 1 + N))/(N + 1) + (1)/(2)logS31 − (g231E2(X1(g21X1 + Z2 + Z)))/(E[2]) + 1 = (1)/(2)log((S21 + 1 + N))/(N + 1) + (1)/(2)logS31 − (g231g221E2(X21))/(S21 + 1 + N) + 1 = 
 = (1)/(2)log((S21 + 1 + N))/(N + 1) + (1)/(2)logS31 − (S31S21)/(S21 + 1 + N) + 1 = (1)/(2)log((S21 + 1 + N))/(N + 1)S31 − (S31S21)/(S21 + 1 + N) + 1
maxmin(1)/(2)log(1 + S31 + S32) − (1)/(2)log(N + 1)/(N), (1)/(2)log((S21 + 1 + N))/(N + 1)S31 − (S31S21)/(S21 + 1 + N) + 1
(1 + S31 + S32)(N)/(N + 1) − ((S21 + 1 + N))/(N + 1)S31 − (S31S21)/(S21 + 1 + N) + 1 = 0
\mathchoiceN0 = (1 + S31 + S21)/(S32)N0 = (1 + S31 + S21)/(S32)N0 = (1 + S31 + S21)/(S32)N0 = (1 + S31 + S21)/(S32)

((1 + S31 + S32)(1 + S31 + S21))/(S32(1 + S31 + S21)/(S32) + 1) = ((1 + S31 + S32)(1 + S31 + S21))/((1 + S31 + S21 + S32))

Expand[(1 + S31 + S32)(1 + S31 + S21)] = S231 + S21S31 + S32S31 + 2S31 + S21 + S21S32 + S32 + 1
((1 + S31 + S32)(1 + S31 + S21))/((1 + S31 + S21 + S32)) = (S231 + S21S31 + S32S31 + 2S31 + S21 + S21S32 + S32 + 1)/((1 + S31 + S21 + S32)) = 1 + (S231 + S21S31 + S32S31 + S31 + S21S32)/((1 + S31 + S21 + S32)) = 

 = 1 + (S31(S31 + S21 + S32 + 1) + S21S32)/((1 + S31 + S21 + S32)) = \mathchoice1 + S31 + (S21S32)/((1 + S31 + S21 + S32))1 + S31 + (S21S32)/((1 + S31 + S21 + S32))1 + S31 + (S21S32)/((1 + S31 + S21 + S32))1 + S31 + (S21S32)/((1 + S31 + S21 + S32))  Q.E.D!!!
Q.E.D!!!
This bound becomes tight as S32 tends to infinity. When S21 is small, the bound can be improved via time sharing on the sender side.
figure Figure 16.9a.png figure Figure 16.9b.png
Figure 10 Comparision of the cutset bound RCS, the decode-foward lower bound RDF and compress-forward lower bound RCF on the capacity of the Gaussian relay channel
Figure 16.9 compares the cutset (upper) bound, the decode forward lower bound and the compress-forward lower bound as a function of S31 for different values of S21 and S32. Note that in general compress-forward outperforms decode-forward when the channel form the sender to the relay is weaker than that to the receiver, i.e., S21 < S31, or when the channel from the relay to the receiver is sufficiently strong, specifically if S32 ≥ S21(1 + S21) ⁄ S31 − (1 + S31). Decode-forward outperforms compress-forward (when the latter is evaluated using Gaussian distributions) in other regimes. In general, it can be shown that both decode-forward and compress-forward achieve rates within half a bit of the cutset bound.

3.7.3 RFD - Relay Channel with Orthogonal Receiver Components

As a dual model to the DMRC with orthogonal sender components (see 3.6.2↑), consider the DMRC with orthogonal receiver components depicted in 11↓
figure Figure16.10.png
Figure 11 Relay channel with orthogonal receiver components.
Here Y3 = (Y3’, Y3’’) and \mathchoicep(y2, y3|x1x2) = p(y3’, y2|x1)p(y3’’|x2)p(y2, y3|x1x2) = p(y3’, y2|x1)p(y3’’|x2)p(y2, y3|x1x2) = p(y3’, y2|x1)p(y3’’|x2)p(y2, y3|x1x2) = p(y3’, y2|x1)p(y3’’|x2), decoupling the broadcast channel form the sender to the relay and the receiver form the direct channel from the relay to the receiver.
The capacity of the DMRC with orthogonal receiver components is not known in general. The cutset bound in Theorem 3.2↑ simplifies to
(21) \mathchoiceC ≤ maxp(x1)p(x2)min{I(X1;Y3) + I(X2;Y3’’), I(X1;Y2, Y3)}C ≤ maxp(x1)p(x2)min{I(X1;Y3) + I(X2;Y3’’), I(X1;Y2, Y3)}C ≤ maxp(x1)p(x2)min{I(X1;Y3) + I(X2;Y3’’), I(X1;Y2, Y3)}C ≤ maxp(x1)p(x2)min{I(X1;Y3) + I(X2;Y3’’), I(X1;Y2, Y3)}
p(y2, y3|x1x2) = p(y3’, y2|x1)p(y3’’|x2)
p(x, y, z) = p(x)p(y|x)p(z|x, y)
p(y2, y3’, y3’’|x1x2) = p(y3’’|x1x2)p(y3’, y2|x1x2y3’’) = p(y2, y3’|x1x2)p(y3’’|x1x2y2y3’’) = p(y3’’|x2)p(y3’, y2|x1)
Следи дека y3’’ зависи само од x2 т.е. не зависи од x1, y2, y3’’ (y2 не содржи повеќе информација за y3’’ од онаа што е содржана во x2), и y3’, y2 зависат само од x1 т.е. не зависи од x2 и y3’’.
I(X1X2;Y3) = I(X1X2;Y3’, Y3’’) = I(X1X2;Y3) + I(X1X2;Y3’’|Y3) = H(Y3) − H(Y3’|X1X2) + H(Y3’’) − H(Y3’’|X1X2Y3)
 = H(Y3) − H(Y3’|X1) + H(Y3’’) − H(Y3’’|X2) = I(X1;Y3) + I(X2;Y3’’)
I(X1;Y2Y3Y3’’|X2) = I(X1;Y2Y3’|X2) + I(X1;Y3’’|Y2Y3’|X2) = I(X1;Y2Y3’|X2) + H(Y3’’|Y2Y3X2) − H(Y3’’|Y2Y3X1X2) = 
 = H(Y2, Y3’|X2) − H(Y2, Y3’|X2X1) + H(Y3’’|Y2Y3X2) − H(Y3’’|Y2Y3X1X2) = H(Y2, Y3) − H(Y2, Y3’|X1) + H(Y3’’|X2) − H(Y3’’|X2) = I(X1;Y2Y3)
Let C0 = maxp(x2)I(X2, Y3’’) denote the capacity of the channel form the relay to the receiver. Then the cutset bound can be expressed as
(22) C ≤ maxp(x1)min{I(X1;Y3) + C0, I(X1;Y2, Y3)}
By comparison, the compress-foward lower bound in Theorem 3.7↑ simplifies to:
(23) \mathchoiceC ≥ maxp(x1)p(2|y2)min{I(X1;Y3) − I(Y2;2|X1Y3) + C0, I(X1;2, Y3)}C ≥ maxp(x1)p(2|y2)min{I(X1;Y3) − I(Y2;2|X1Y3) + C0, I(X1;2, Y3)}C ≥ maxp(x1)p(2|y2)min{I(X1;Y3) − I(Y2;2|X1Y3) + C0, I(X1;2, Y3)}C ≥ maxp(x1)p(2|y2)min{I(X1;Y3) − I(Y2;2|X1Y3) + C0, I(X1;2, Y3)}
C ≥ maxmin{I(X1, X2;Y3) − I(Y2;2|X1X2Y3), I(X1;2, Y3|X2)}
I(X1;2, Y3|X2) = I(X1;2, Y3Y3’’|X2) = I(X1;2, Y3’|X2) + I(X1;Y3’’|X22, Y3) = I(X1;2, Y3’|X2) + H(Y3’’|X22, Y3) − H(Y3’’|X1X22, Y3) = 
 = I(X1;2, Y3’|X2) + H(Y3’’|X22, Y3) − H(Y3’’|X22, Y3) = I(X1;2, Y3’|X2) = H(2Y3’|X2) − H(2, Y3’|X2X1) = 
 = H(2Y3) − H(2, Y3’|X1) = I(X1;2Y3)
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I(X1X2;Y3) = I(X1;Y3) + I(X2;Y3’’) = I(X1;Y3) + C0
I(Y2;2|X1X2Y3) = H(2|X1X2Y3Y3’’) − H(2|X1X2Y2Y3Y3’’)
\mathchoiceI(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3)I(Y2;2|X1X2Y3) = I(Y2Y3;2|X1X2) − I(Y3;2|X1X2) = H(Y2Y3|X1X2) − H(Y2Y3|2X1X2) − I(Y3;2|X1X2) = 
 = H(Y2Y3Y3’’|X1X2) − H(Y2Y3Y3’’|2X1X2) − I(Y3Y3’’;2|X1X2) = 
 = H(Y2Y3’|X1X2) + H(Y3’’|X1X2Y2Y3) − H(Y2Y3’|2X1X2) − H(Y3’’|2X1X2Y2Y3) − I(Y3’;2|X1X2) − I(Y3’’;2|X1X2Y3) = 
 = H(Y2Y3’|X1) + H(Y3’’|X2Y2Y3) − H(Y2Y3’|2X1) − H(Y3’’|2X2Y2Y3) − H(Y3’|X1X2) + H(Y3’|2X1X2) − H(Y3’’|X1X2Y3) + H(Y3’’|2X1X2Y3) = 
 = H(Y2Y3’|X1) + H(Y3’’|X2Y2Y3) − H(Y2Y3’|2X1) − H(Y3’’|2X2Y2Y3) − H(Y3’|X1) + H(Y3’|2X1) − H(Y3’’|X2Y3) + H(Y3’’|2X2Y3) = 
 = \cancelH(Y3’|X1) + H(Y2|X1Y3) + H(Y3’’|X2Y2Y3) − H(Y2Y3’|2X1) − H(Y3’’|2X2Y2Y3) − \cancelH(Y3’|X1) + H(Y3’|2X1) − H(Y3’’|X2Y3) + H(Y3’’|2X2Y3) = 
 = H(Y2|X1Y3) + H(Y3’’|X2Y2Y3) − \cancelH(Y3’|2X1) − H(Y2|2X1Y3) − H(Y3’’|2X2Y2Y3) + \cancelH(Y3’|2X1) − H(Y3’’|X2Y3) + H(Y3’’|2X2Y3) = 
 = H(Y2|X1Y3) + H(Y3’’|X2Y2Y3) − H(Y2|2X1Y3) − H(Y3’’|2X2Y2Y3) − H(Y3’’|X2Y3) + H(Y3’’|X2Y3) = 
 = H(Y2|X1Y3) + H(Y3’’|X2Y2Y3) − H(Y2|2X1Y3) − \cancelH(Y3’’|X2Y3) − H(Y3’’|X2Y3) + \cancelH(Y3’’|X2Y3) = 
 = H(Y3’’|X2\cancelY2Y3) + H(Y2|X1Y3) − H(Y2|2X1Y3) − H(Y3’’|X2Y3) = 
 = \cancelH(Y3’’|X2Y3) + I(Y2;2|X1Y3) − \cancelH(Y3’’|X2Y3) = \mathchoiceI(Y2;2|X1Y3)I(Y2;2|X1Y3)I(Y2;2|X1Y3)I(Y2;2|X1Y3) Q.E.D!
За да изглезе доказот поркрај p(y2, y3|x1x2) = p(y3’, y2|x1)p(y3’’|x2) треба да се претпостави дека Y3’’ не зависи од Y2 и 2. Да тоа правилна претпоставка!!! Види ја дискусијата во претходниот box.
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Скратена верзија на доказот:
I(Y2;2|X1X2Y3) = I(Y2Y3;2|X1X2) − I(Y3;2|X1X2) = H(Y2Y3|X1X2) − H(Y2Y3|2X1X2) − I(Y3Y3’’;2|X1X2) = 
 = H(Y2Y3Y3’’|X1X2) − H(Y2Y3Y3’’|2X1X2) − I(Y3’;2|X1X2) − I(Y3’’;2|X1X2Y3) = 
 = H(Y2Y3’|X1X2) + H(Y3’’|X1X2Y2Y3) − H(Y2Y3’|2X1X2) − H(Y3’’|2X1X2Y2Y3) − H(Y3’|X1X2) + H(Y3’|2X1X2) − H(Y3’’|X1X2Y3) + H(Y3’’|2X1X2Y3) = 

 = H(Y2Y3’|X1) + \cancelH(Y3’’|X2) − H(Y2Y3’|2X1) − \cancelH(Y3’’|2X2) − H(Y3’|X1) + H(Y3’|2X1) − \cancelH(Y3’’|X2) + \cancelH(Y3’’|2X2) = 

 = I(2;Y2Y3’|X1) − I(2;Y3’|X1) = \cancelI(2;Y3’|X1) + I(2;Y2|X1Y3) − \cancelI(2;Y3’|X1) = I(2;Y2|X1Y3)
Q.E.D!!!
This two bounds coincide for the deterministic relay channel with orthogonal receiver components where Y2 is a function of (X1, Y3). The proof follows by setting 2 = Y2 in the compress-forward lower bound 23↑ and using the fact that H(Y2|X1Y3) = 0. Note that in general, the capacity itself depends on p(y3’’|x2) only through C0.
The following example shows that the cutset bound is not tight in general.
(Modulo-2 sum relay channel).
Consider the DMRC with orthogonal receiver components depicted in Figure 12↓, where
figure Figure 16.11.png
Figure 12 Modulo-2 sum relay channel
Y3’ = X1Z3, 
Y2 = Z2Z3, 
and \mathchoiceZ2 ~ Bern(p)Z2 ~ Bern(p)Z2 ~ Bern(p)Z2 ~ Bern(p) and \mathchoiceZ3 ~ Bern(1 ⁄ 2)Z3 ~ Bern(1 ⁄ 2)Z3 ~ Bern(1 ⁄ 2)Z3 ~ Bern(1 ⁄ 2) are independent of each other and of X1.
For C0 ∈ [0, 1], the capacity of this relay channel is
C = 1 − H(p*H − 1(1 − C0))
C ≤ maxp(x1)min{I(X1;Y3) + C0, I(X1;Y2, Y3)}
X1, Z3|X1Z3 0 1 00 1 0 01 0 1 10 0 1 11 1 0 X1, Z3|Y3’ = X1Z3 0 1 00 p1 ⁄ 2 0 01 0 p1 ⁄ 2 10 0 (1 − p1) ⁄ 2 11 (1 − p1) ⁄ 2 0 p(X1Z3) 1 ⁄ 2 1 ⁄ 2 X1, Z3, X1Z3 000 011 101 110 ()
\mathchoiceI(X1;Y3)I(X1;Y3)I(X1;Y3)I(X1;Y3) = H(Y3) − H(Y3’|X1) = 1 − H(X1Z3|X1) = 1 − H(Z3) = 0
Z2, Z3|Y2 = Z2Z3 0 1 00 1 0 01 0 1 10 0 1 11 1 0 Z2, Z3|Y2 = Z2Z3 0 1 00 p ⁄ 2 0 01 0 p ⁄ 2 10 0 (1 − p) ⁄ 2 11 (1 − p) ⁄ 2 0 p(Z2Z3) 1 ⁄ 2 1 ⁄ 2
C0 = maxp(x2)I(X2, Y3’’)
\mathchoiceI(X2, Y3’’)I(X2, Y3’’)I(X2, Y3’’)I(X2, Y3’’) = H(Y3’’) − H(Y3’’|X2) = 1
\mathchoiceI(X1;Y2, Y3)I(X1;Y2, Y3)I(X1;Y2, Y3)I(X1;Y2, Y3) = H(Y2, Y3) − H(Y2, Y3’|X1) = \cancelto1H(Y2) + \cancelto1H(Y3) − H(Y3’|X1) − H(Y2|X1, Y3) = 1 − \overset(m)H(Z2Z3|X1, X1Z3) = 1 − H(Z2) = 1 − H(p)
C ≤ maxp(x1)min{I(X1;Y3) + C0, I(X1;Y2, Y3)} = maxp(x1)min{0 + C0, 1 − H(p)} = \mathchoicemin{C0, 1 − H(p)}min{C0, 1 − H(p)}min{C0, 1 − H(p)}min{C0, 1 − H(p)}
Oва е за 22↑ a треба за 23↑!!! (Првиот е без компресија, а вториов со компресија)
(m)-Овде вака е резонот. Штом го знаеш X1 тогаш можеш да го елиминираш во X1Z3 и ќе добиеш чисто Z3.

Е сега ќе рачунам за 23↑.
C ≥ maxp(x1)p(2|y2)min{I(X1;Y3) − I(Y2;2|X1Y3) + C0, I(X1;2, Y3)}
\mathchoiceI(X2, Y3’’)I(X2, Y3’’)I(X2, Y3’’)I(X2, Y3’’) = H(Y3’’) − H(Y3’’|X2) = 1
C0 = maxp(x2){I(X2;Y3’’)} = 1
V ~ Bern(α)
—————————————
\mathchoiceI(Y2;2|X1Y3)I(Y2;2|X1Y3)I(Y2;2|X1Y3)I(Y2;2|X1Y3) = I(Y2;Y2V|X1Y3) = H(Y2|X1Y3) − H(Y2V|Y2, X1Y3) = H(Z2Z3|X1Y3) − H(Y2V|Y2, X1Y3) = 
 = H(Z2) − H(V) = H(p) − H(α)
———————————————
И ова горе сум го згрешил. Треба
I(Y2;2|X1Y3) = H(2|X1Y3) − H(2|X1, Y3’, Y2) = H(Y2V|X1Y3) − H(Y2V|X1Y3Y2)
 = H(Z2Z3V|X1Y3) − H(V|X1Y3) =  = H(1 ⁄ 2) − H(V) = 1 − H(α)
—————————————————
I(X1;Y3) − I(Y2;2|X1Y3) + C0 = 1 − H(p) + H(α) + 1 = 2 − H(p) + H(α)

I(X1;2, Y3) = I(X1;Y2V, Y3) = H(Y2V, Y3) − H(Y2V, Y3|X1) = H(Z2Z3V, X1Z3) − H(Z2Z3V, X1Z3|X1) = 
 = H(Z2Z3V, X1Z3) − H(Z2Z3V, Z3) = H(Z2Z3V, X1Z3) − H(Z3) − H(Z2Z3V|Z3) = 
 = H(Z2Z3V, X1Z3) − 1 − H(Z2V) ????

!! V, Z2|VZ2 0 1 00 1 0 01 0 1 10 0 1 11 1 0 V, Z1|VZ1 0 1 00 αp 0 01 0 α(1 − p) 10 0 (1 − α)p 11 (1 − α)(1 − p1) 0 p(VZ2) 1 − α(1 − p) + (1 − α)p α(1 − p) + (1 − α)p 1 − αp \mathchoice\alphap\alphap\alphap\alphap
H(Z2V) = H(αp)
————————————
(*)
I(X1;2, Y3) = I(X1;2) + I(X1;2|Y3) = H(Y2V) − H(Y2V|X1) + H(2|Y3) − H(2|Y3X1) = H(Y2V) − H(Y2V|X1) + H(Y2V|X1Z3) − H(Y2V|X1Z3, X1)
 = H(Z2Z3V) − H(Z2Z3V|X1) + H(Z2Z3V|X1Z3) − H(Z2Z3V|X1Z3, X1) = 
 = H(Z2Z3V) − \cancelH(Z2Z3V) + \cancelH(Z2Z3V) − H(Z2V) = H(Z2Z3V) − H(Z2V)
——————————————–
Ова горе не е добро, грешно сум ја развил формулата за здружена ентропија. Треба како подoлолу.
I(X1;2, Y3) = H(2, Y3) − H(Y2, Y3’|X1) = H(2) + H(2|Y3) − H(2|X1) − H(Y2|Y3X1) = H(Y2V) + H(Y2V|Y3) − H(Y2V|X1) − H(Y2V|Y3X1) = 
H(Z2Z3V) + H(Z2Z3V|Y3) − H(Z2Z3V|X1) − H(Z2Z3V|X1Z3, X1) = H(1 ⁄ 2) + \cancelH(1 ⁄ 2) − \cancelH(1 ⁄ 2) − H(Z2V) = 1 − H(Z2V)
Веројатно може да се докаже и со горниот пристап (*) но кога се навратив на овој проблем заборавив да земам во предвид дека ја знам дистрибуцијата на X1Z3 што ја имам најдено на почеток на овој box.
——————————————–
V, Z2|VZ2Z3 0 1 000 1 0 001 0 1 010 0 1 011 1 0 100 0 1 101 1 0 110 1 0 111 0 1 V, Z2, Z3|VZ2Z3 0 1 000 αp ⁄ 2 0 001 0 αp ⁄ 2 010 0 α(1 − p) ⁄ 2 011 α(1 − p) ⁄ 2 0 100 0 (1 − α)p ⁄ 2 101 (1 − α)p ⁄ 2 0 110 (1 − α)(1 − p) ⁄ 2 0 111 0 (1 − α)(1 − p) ⁄ 2 p(VZ2Z3) 1 ⁄ 2 1 ⁄ 2
Горниов резултат е апсолутно логичен зошто Z2Z3 можеш да ги сметаш како една променлива со веројатност  ~ Bern(1 ⁄ 2) па горниот кејс ќе се сведе на кејсот Z2Z3.
Expand[αp ⁄ 2 + α(1 − p) ⁄ 2 + (1 − α)p ⁄ 2 + (1 − α)(1 − p) ⁄ 2] = (1)/(2)
H(Z2Z3V) = H(1)/(2) = 1
\mathchoiceI(X1;2, Y3)I(X1;2, Y3)I(X1;2, Y3)I(X1;2, Y3) = 1 − H(Z2V) = 1 − H(αp)
C ≥ maxp(x1)p(2|y2)min{2 − H(p) + H(α), 1 − H(αp)} = 1 − H(αp)
\mathchoiceC = 1 − H(p*H − 1(1 − C0))C = 1 − H(p*H − 1(1 − C0))C = 1 − H(p*H − 1(1 − C0))C = 1 − H(p*H − 1(1 − C0))  since α = H − 1(1 − C0) Q.E.D!!!
where H − 1(v) ∈ [0, 1 ⁄ 2] is inverse of the binary entropy function. Te proof of achievability follows by setting \mathchoice2 = Y2V2 = Y2V2 = Y2V2 = Y2V where V ~ Bern(α) is independent of (X1, Z2, Z3) and α = H − 1(1 − C0), in the compress-forward lower bound in 23↑. For the proof of the converse, consider
nR ≤ I(Xn1;Y3n, Y3’’n) + nϵn\overset(a) = I(Xn1;Y3n|Y3’’n) + nϵn ≤ n − H(Y3n|Xn1, Y3’’n) + nϵn = n − H(Zn3|Y3’’n) + nϵn\overset(b) ≤ 
 ≤ n − nH(p*H − 1(H(Yn2|Y3’’n) ⁄ n)) + nϵn\overset(c) ≤ n − nH(p*H − 1(1 − C0)) + nϵn, 
I(Xn1;Y3n, Y3’’n) = \overset0I(Xn1;Y3’’n) + I(Xn1;Y3’|Y3’’n) = I(Xn1;Y3’|Y3’’n)
I(Xn1;Y3n|Y3’’n) = H(Y3n|Y3’’n) − H(Y3n|Y3’’n, Xn1) = ni = 1H(Y3i|Y3’’n, Yi − 131) − H(Y3n|Y3’’n, Xn1) ≤ 
 ≤ ni = 1H(Y3i|Yi − 131) − H(Y3n|Y3’’n, Xn1) = n − H(Y3n|Y3’’n, Xn1) = n − H(Xn1 + Zn3|Y3’’n, Xn1) = n − H(Zn1|Y3’’n)

Vector Mrs Gerbers Lemma (MGL).
If Zn is a vector of independent and identically distributed Bern(p) random variables independent of (Xn, U) and Yn = XnZn, then
(H(Yn|U))/(n) ≥ HH − 1(H(Xn|U))/(n)*p

Zn3 = Yn2Zn2 важи модуло 2 и во обратната насока. Види ги табелите во боксот погоре (на пример ())
H(Zn3|Y3’’n) ≥ nH(p*H − 1((Yn2|Y’’n3) ⁄ n))
I(Xn1;Y3n|Y3’’n) = n − H(Zn1|Y3’’n) = n − nH(p*H − 1((Yn2|Y’’n3) ⁄ n))
nC0 ≥ I(Xn2;Y’’n3)\overset(d) ≥ I(Yn2, Y3’’n) = H(Yn2) − H(Yn2|Y3’’n) = n − H(Yn2|Y3’’n)
due to data processing inequality
Y2 → X2 → Y’’3
where (a) follows by the independence of Xn1 and (Zn2, Zn3, Xn2, Y3’’n), (b) follows by the vector MGL (Mrs Gerber’s Lemma) with Zn3 = Yn2Zn2, which yields H(Zn3|Y’’n3) ≥ nH(p*H − 1(H(Yn2|Y3’’n) ⁄ n)), and (c) follows since nC0 ≥ I(Xn2;Y’’n3) ≥ I(Yn2;Y’’n3) = n − H(Yn2|Y’’n3). Note that the cutset bound in 22↑ simplifies to min{1 − H(p), C0} Докажано е во box-от погоре , which is strictly larger Во Maple покажав дека за било кое α ∈ [0, 1] H(αp) ≥ H(p). Еднаквоста се достигнува за α=0 и α = 1. than the capacity if p ≠ 1 ⁄ 2 and 1 − H(p) ≤ C0. Hence, the cutset bound is not tight in general.

3.8 RFD Gaussian Relay channel

The receiver frequency-division (RFD) depicted in 13↓ has orthogonal receiver components.
figure Figure 16.12.png
Figure 13 Receiver frequency-division Gaussian Relay channel
In this half-duplex model, the channel form the relay to the receiver uses a different frequency band form the broadcast channel from the sender to the relay and the receiver. More specifically, in this model Y3 = (Y3’, Y3’’) and
Y2 = g21X1 + Z2, 
Y3’ = g31X1 + Z3’, 
Y3’’ = g32X2 + Z3’’, 
where g21, g31,   and g32 are channel gains, and \mathchoiceZ2 ~ N(0, 1)Z2 ~ N(0, 1)Z2 ~ N(0, 1)Z2 ~ N(0, 1) and \mathchoiceZ3 ~ N(0, 1)Z3 ~ N(0, 1)Z3 ~ N(0, 1)Z3 ~ N(0, 1) are independent noise components. Assume average power constraint P on each of X1 and X2.
The capacity of this channel is not known in general. The cutset upper bound in theorem 3.2↑ (under the power constraints) simplifies to
(24) \mathchoiceC ≤  C(S31) + C(S32) if S21 ≥ S32(S31 + 1)    C(S21 + S31) otherwise C ≤  C(S31) + C(S32) if S21 ≥ S32(S31 + 1)    C(S21 + S31) otherwise C ≤  C(S31) + C(S32) if S21 ≥ S32(S31 + 1)    C(S21 + S31) otherwise C ≤  C(S31) + C(S32) if S21 ≥ S32(S31 + 1)    C(S21 + S31) otherwise
C ≤ maxp(x1)min{I(X1;Y3) + C0, I(X1;Y2, Y3)}; C0 = maxp(x2)I(X2;Y3’’); p(y2, y3|x1x2) = p(y3’, y2|x1)p(y3’’|x2)
\mathchoiceY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’
I(X1, X2;Y3) = H(Y3) − H(Y3|X1X2) = H(Y3Y3’’) − H(Y3Y3’’|X1X2)
I(X2;Y3’’) = H(Y3’’) − H(Y3’’|X2) = (1)/(2)log(S32 + 1) = C(S32)
I(X1;Y3) = H(Y3) − H(Y3’|X1) = (1)/(2)log2πe(g231P + 1) − (1)/(2)log(2πe) = (1)/(2)log(1 + S31) = C(S31)
\mathchoiceI(X1;Y3) + C0I(X1;Y3) + C0I(X1;Y3) + C0I(X1;Y3) + C0 = \mathchoiceC(S31) + C(S32)C(S31) + C(S32)C(S31) + C(S32)C(S31) + C(S32) = (1)/(2)log(1 + S31)(S32 + 1) = (1)/(2)log(1 + S31)(S32 + 1) = (1)/(2)log(S32S31 + S31 + S32 + 1) = C(S32S31 + S31 + S32)
Expand[(1 + S31)(S32 + 1)] = S32S31 + S31 + S32 + 1

\mathchoiceI(X1;Y2, Y3)I(X1;Y2, Y3)I(X1;Y2, Y3)I(X1;Y2, Y3) = H(Y2Y3) − H(Y2Y3’|X1) = H(Y2) + H(Y3’|Y2) − H(Y2|X1) − H(Y3’|Y2X1) = 
 ≤ (1)/(2)log(2πe)(S21 + 1) + (1)/(2)log(2πe)E[Y’23] − (E2[Y3Y2])/(E[Y22]) − (1)/(2)⋅log(2πe) − (1)/(2)⋅log(2πe) = 
 = (1)/(2)log(S21 + 1) + (1)/(2)logS31 + 1 − (E2[(a31X1 + Z3)(a21X1 + Z3)])/(E[Y22]) = (1)/(2)log(S21 + 1) + (1)/(2)logS31 + 1 − (a231a221E2[X21])/(S21 + 1) = 
 = (1)/(2)log(S21 + 1) + (1)/(2)logS31 + 1 − (S21S31)/(S21 + 1) = (1)/(2)log(S21 + 1)S31 + 1 − (S21S31)/(S21 + 1) = (*)
(S21 + 1)S31 + 1 − (S21S31)/(S21 + 1) = ((S31 + 1)(S21 + 1) − S21S31) = S31S21 + S21 + S31 + 1 − S21S31 = 1 + S31 + S21
(*) = \mathchoice(1)/(2)log(1 + S31 + S21)(1)/(2)log(1 + S31 + S21)(1)/(2)log(1 + S31 + S21)(1)/(2)log(1 + S31 + S21)

C ≤ maxp(x1)min{I(X1;Y3) + C0, I(X1;Y2, Y3)} = maxp(x1)min(1)/(2)log(1 + S31) + (1)/(2)log(1 + S32), (1)/(2)log(1 + S31 + S21)
(1 + S31)(1 + S32)\overset? = (1 + S31 + S21) S32S31 + \cancelS31 + S32 + 1\overset? = 1 + \cancelS31 + S21 S32S31 + S32\overset? = S21 \mathchoiceS32(S31 + 1)\overset? = S21S32(S31 + 1)\overset? = S21S32(S31 + 1)\overset? = S21S32(S31 + 1)\overset? = S21
Expand[(1 + S31)(1 + S32)] = S32S31 + S31 + S32 + 1
C ≤  C(S31) + C(S32)  if S21 ≥ S32(S31 + 1)       C(S21 + S31)  otherwise Q.E.D!!!

Q.E.D!!!
C ≤ maxp(x1x2)min{I(X1, X2;Y3), I(X1;Y2, Y3|X2)} (*)
Да проверам со општиот израз (*):
I(X1X2;Y3) = I(X1X2;Y3’, Y3’’) = I(X1X2;Y3) + I(X1X2;Y3’’|Y3) = H(Y3) − H(Y3’|X1X2) + H(Y3’’) − H(Y3’’|X1X2Y3)
 = H(Y3) − H(Y3’|X1) + H(Y3’’) − H(Y3’’|X2) = I(X1;Y3) + I(X2;Y3’’)
I(X1;Y2Y3Y3’’|X2) = I(X1;Y2Y3) + I(X1;Y3’’|Y2Y3) = I(X1;Y2Y3) + H(Y3’’|Y2Y3) − H(Y3’’|Y2Y3X1) = 
 = I(X1;Y2Y3) + H(Y3’’|Y2Y3) − H(Y3’’|Y2Y3) = I(X1;Y2Y3)
I(X1X2;Y3) = I(X1X2;Y3’, Y3’’) = I(X1X2;Y3) + I(X1X2;Y3’’|Y3) = H(Y3) − H(Y3’|X1X2) + H(Y3’’) − H(Y3’’|X1X2Y3)
 = H(Y3) − H(Y3’|X1) + H(Y3’’) − H(Y3’’|X2) = |I(X1;Y3) + I(X2;Y3’’)| = 
 = (1)/(2)log2πe(S21 + 1) − (1)/(2)log2πe + (1)/(2)log2πe(S32 + 1) − (1)/(2)log2πe = C(S21) + C(S32)

I(X1;Y2Y3Y3’’|X2) = I(X1;Y2Y3) + I(X1;Y3’’|Y2Y3) = C(S31S21 + S21 + S31)
Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’
I(X1;Y3’’|Y2Y3) = H(Y3’’|Y2Y3) − H(Y3’’|Y2Y3’, X1) = 0
The decode-forward lower bound in 3.4.3↑ simplifies to
(25) C ≤  C(S31) + C(S32)  if S21 ≥ S32(S31 + 1)       C(S21)  otherwise
Доказ на изразот 12↑
Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’
I(X1X2;Y3) = I(X1X2;Y3’, Y3’’) = I(X1X2;Y3) + I(X1X2;Y3’’|Y3) = H(Y3) − H(Y3’|X1X2) + H(Y3’’) − H(Y3’’|X1X2Y3)
 = H(Y3) − H(Y3’|X1) + H(Y3’’) − H(Y3’’|X2) = I(X1;Y3) + I(X2;Y3’’) = 
 = (1)/(2)log2πe(S31 + 1) − (1)/(2)log2πe + (1)/(2)log2πe(S32 + 1) − (1)/(2)log2πe = C(S31) + C(S32)

I(X1;Y2|X2) = H(Y2|X2) − H(Y2|X2X1) ≤ (1)/(2)log2πe(S21 + 1) − (1)/(2)log2πe = C(S21)
C ≤ maxp(x1x2)min{I(X1, X2;Y3), I(X1;Y2|X2)} = maxp(x1x2)min{C(S31) + C(S32), C(S21)}
log(1 + S31) + (1)/(2)log(1 + S32)(1)/(2)⋅log(1 + S21)
(1 + S31)(1 + S32)(1 + S21)
1 + S31 + S32(1 + S31)≶1 + S21
S31 + S32(1 + S31)S21
Q.E.D!!!
When S21 ≥ S31 + S32(S31 + 1) the bounds in 24↑ and 25↑ coincide and the capacity C = C(S31) + C(S32) is achieved by decode-forward. If S21 ≤ S31 , the decode-forward lower bound is worse than the direct-transmission lower bound. As in the full-duplex case, the partial decode-forward lower bound reduces to the maximum of the direct transmission and decode-forward lower bounds, which is in sharp contrast to the sender frequency-division Gaussian RC; see 3.6.3↑ and 3.6.3↑. The compress-forward lower bound in 3.7↑ with X1 ~ N(0, P),  X2 ~ N(0, P) and Z ~ N(0, N), independent of each other, and \mathchoice2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z2 = Y2 + Z, simplifies (after optimizing over N) to
(26) \mathchoiceC ≥ CS31 + (S21S32(S31 + 1))/(S21 + (S31 + 1)(S32 + 1)).C ≥ CS31 + (S21S32(S31 + 1))/(S21 + (S31 + 1)(S32 + 1)).C ≥ CS31 + (S21S32(S31 + 1))/(S21 + (S31 + 1)(S32 + 1)).C ≥ CS31 + (S21S32(S31 + 1))/(S21 + (S31 + 1)(S32 + 1)).
C ≥ maxp(x1)p(2|y2)min{I(X1;Y3) − I(Y2;2|X1Y3) + C0, I(X1;2, Y3)}
2 = Y2 + Z X1 ~ N(0, P),  X2 ~ N(0, P) and Z ~ N(0, N)
Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’
I(X1;2, Y3) = I(X1;2) + I(X1;Y3’|2) = H(2) − H(2|X1) + I(X1;Y3’|2) = H(Y2 + Z) − H(Y2 + Z|X1) + I(X1;Y3’|2) = 
 = (1)/(2)log(2πe)(S21 + 1 + N) − (1)/(2)log(2πe)(1 + N) + I(X1;Y3’|2) = (*)
I(X1;Y3’|2) = H(Y3’|Y2 + Z) − H(Y3’|X1Y2 + Z) = H(Y3) − H(Y3’|X1) =  = (1)/(2)log(2πe)(S31 + 1) − (1)/(2)log(2πe) = C(S31)
(*) = (1)/(2)log(2πe)(S21 + 1 + N) − (1)/(2)log(2πe)(1 + N) + (1)/(2)⋅log(S31 + 1) = (1)/(2)log(S21)/(1 + N) + 1 + (1)/(2)⋅log(S31 + 1) = (1)/(2)log(S21)/(1 + N) + 1(S31 + 1)
I(X1;Y3) − I(Y2;2|X1Y3) + C0 = ???
I(X1;Y3) = (1)/(2)log(S31 + 1)
C0 = I(X2;Y3’’) = (1)/(2)log(S32 + 1)
I(Y2;2|X1Y3) = H(2|X1X2) − H(2|X1X2Y2) = H(Y2 + Z|X1X2) − H(Y2 + Z|X1X2Y2) = (1)/(2)log2πe(1 + N) − (1)/(2)log2πeN = (1)/(2)log(1)/(N) + 1
I(X1;Y3) − I(Y2;2|X1Y3) + C0 = (1)/(2)log(S31 + 1) + (1)/(2)log(S32 + 1) + (1)/(2)log(1)/(N) + 1 = (1)/(2)log(S31 + 1)(S32 + 1)(1)/(N) + 1
(S31 + 1)(S32 + 1)(1)/(N) + 1 = (S21)/(1 + N) + 1(S31 + 1) (S32 + 1)(1)/(N) + 1 = (S21)/(1 + N) + 1 (S32 + 1)(1 + N)/(N) = (S21 + 1 + N)/(1 + N)
(S32 + 1)(1 + N)/(N)(1 + N)/(S21 + 1 + N) = 1
Simplify(S32 + 1)(1 + N)/(N)(1 + N)/(S21 + 1 + N) = ((N + 1)2(S32 + 1))/(N(N + S21 + 1))
Solve((N + 1)2(S32 + 1))/(N(N + S21 + 1)) − 1 =  = 0, N = N → (S21 − 2S32 − (S221 − 4S32S21 − 2S21 + 1) − 1)/(2S32), N → (S21 − 2S32 + (S221 − 4S32S21 − 2S21 + 1) − 1)/(2S32)
Ги заменив овие вредности во мапле и добив:
I(X1;2, Y3) = (1)/(2)log(4 S32S21 + 2 S32 − 1 + S21 + ( − 4 S32S21 + 1 − 2 S21 + S212))/(2 S32 − 1 + S21 + ( − 4 S32S21 + 1 − 2 S21 + S212))(S31 + 1)
C ≥ maxp(x1)p(2|y2)min{I(X1;Y3) − I(Y2;2|X1Y3) + C0, I(X1;2, Y3)}
2 = Y2 + Z X1 ~ N(0, P),  X2 ~ N(0, P) and Z ~ N(0, N)
\mathchoiceY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + Z
I(X1;2, Y3) = I(X1;2) + I(X1;Y3’|2) = H(2) − H(2|X1) + I(X1;Y3’|2) = H(Y2 + Z) − H(Y2 + Z|X1) + I(X1;Y3’|2) = 
 = (1)/(2)log(2πe)(S21 + 1 + N) − (1)/(2)log(2πe)(1 + N) + I(X1;Y3’|2) = (*)

I(X1;Y3’|2) = H(Y3’|Y2 + Z) − H(Y3’|X1Y2 + Z) = H(Y3) − H(Y3’|X1) =  = (1)/(2)log(2πe)(S31 + 1) − (1)/(2)log(2πe) = C(S31)
(*) = (1)/(2)log(2πe)(S21 + 1 + N) − (1)/(2)log(2πe)(1 + N) + (1)/(2)⋅log(S31 + 1) = (1)/(2)log(S21)/(1 + N) + 1 + (1)/(2)⋅log(S31 + 1) = (1)/(2)log(S21)/(1 + N) + 1(S31 + 1)
I(X1;Y3’|2) мислам дека не е добро пресметано и дека треба да биде како во 3.8↓!!!

I(X1;Y3) − I(Y2;2|X1Y3) + C0 = ???
I(X1;Y3) = (1)/(2)log(S31 + 1)
C0 = I(X2;Y3’’) = (1)/(2)log(S32 + 1)
I(Y2;2|X1Y3) = H(2|X1Y3) − H(2|X1Y3Y2) = H(g21X1 + Z2 + Z|X1Y3) − H(Y2 + Z|X1Y3Y2) = (1)/(2)log2πe(1 + N) − (1)/(2)log2πeN = (1)/(2)log(1)/(N) + 1
\mathchoiceI(X1;Y3) − I(Y2;2|X1Y3) + C0I(X1;Y3) − I(Y2;2|X1Y3) + C0I(X1;Y3) − I(Y2;2|X1Y3) + C0I(X1;Y3) − I(Y2;2|X1Y3) + C0 = (1)/(2)log(S31 + 1) + (1)/(2)log(S32 + 1) − (1)/(2)log(1)/(N) + 1 = \mathchoice(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1)(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1)(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1)(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1)
Ако се оди со решението за кое мислам дека не е добро се добива:
N = ((1 + S21))/(S32)
C = ((S32 + 1)(S31 + 1)(1 + S21))/(1 + S21 + S32)
C = (S31S32 + S31S32S21 + S31 + S31S21 + \mathchoiceS32S32S32S32 + S32S21 + \mathchoice1111 + \mathchoiceS21S21S21S21)/(1 + S21 + S32)
C = 1 + (S31S32 + S31S32S21 + S31 + S31S21 + S32S21)/(1 + S21 + S32) = 1 + S31 + (S31S32S21 + S32S21)/(1 + S21 + S32) = \mathchoice1 + S31 + ((S31 + 1)S32S21)/(1 + S21 + S32)1 + S31 + ((S31 + 1)S32S21)/(1 + S21 + S32)1 + S31 + ((S31 + 1)S32S21)/(1 + S21 + S32)1 + S31 + ((S31 + 1)S32S21)/(1 + S21 + S32)

\mathchoiceY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + Z
I(X1;Y3’|2) = H(Y3’|2) − H(Y3’|X12) = (*)
H(Y3’|2) = E[H(Y3’|2)] ≤ E(1)/(2)log2πe(Var(Y3|2)) ≤ (1)/(2)log2πe(E[Var(Y3|2)]) ≤ (1)/(2)log2πeE[Y’23] − (E2[Y3’, 2])/(E[22])
\mathchoiceE[Y2’3] − (E2[Y32])/(E[22])E[Y2’3] − (E2[Y32])/(E[22])E[Y2’3] − (E2[Y32])/(E[22])E[Y2’3] − (E2[Y32])/(E[22]) = S31 + 1 − (E2[(g31X1 + Z3)(g21X1 + Z2 + Z)])/(S21 + 1 + N) = S31 + 1 − (E[(g31X1 + Z3)(g21X1 + Z2 + Z)])/(S21 + 1 + N) = S31 + 1 − (g231g221P2)/(S21 + 1 + N) = 
 = \mathchoiceS31 + 1 − (S31S21)/(S21 + 1 + N)S31 + 1 − (S31S21)/(S21 + 1 + N)S31 + 1 − (S31S21)/(S21 + 1 + N)S31 + 1 − (S31S21)/(S21 + 1 + N)
(*) = (1)/(2)log(2πe)S31 + 1 − (S31S21)/(S21 + 1 + N) − (1)/(2)log2πe = (1)/(2)logS31 + 1 − (S31S21)/(S21 + 1 + N) = 
C = min(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1), (1)/(2)logS31 + 1 − (S31S21)/(S21 + 1 + N)
Solve(S31 + 1)(S32 + 1)(N)/(N + 1) − S31 − 1 + (S31S21)/(S21 + 1 + N) =  = 0, N = 
N → ( − S21S31 − S21S32S31 − S32S31 + S31 − S21S32 − S32 − ((S21S31 + S21S32S31 + S32S31 − S31 + S21S32 + S32 − 1)2 − 4( − S21 − S31 − 1)(S31S32 + S32)) + 1)/(2(S31S32 + S32)), 
N → ( − S21S31 − S21S32S31 − S32S31 + S31 − S21S32 − S32 + ((S21S31 + S21S32S31 + S32S31 − S31 + S21S32 + S32 − 1)2 − 4( − S21 − S31 − 1)(S31S32 + S32)) + 1)/(2(S31S32 + S32))
C = (S31 + 1)(S32 + 1) − 
 − 2⋅((S31 + 1)2(S32 + 1)S32)/(S21 − (S31 + 1)(S32 + 1)(S21 + 1) + 2 (S31 + 1)(S32 + 1) + (((S31 + 1)(S32 + 1) − 1)2(S21 + 1)2 + ( − 2 + 6 (S31 + 1)(S32 + 1) − 4 (S31 + 1)2(S32 + 1))(S21 + 1) + 1 + 4 (S31 + 1)(S32(S31) + 1)))
- Да пробам со користење на Theorem 6 од Capacity Theorem
2 = Y2 + Z X1 ~ N(0, P),  X2 ~ N(0, P) and Z ~ N(0, N)
\mathchoiceY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + Z
\mathchoiceR*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2)R*1 = supI(X1;Y2̂, Y3, |X2)
\mathchoiceI(X2;Y3) ≥ I(Y2;2|X2, Y3)I(X2;Y3) ≥ I(Y2;2|X2, Y3)I(X2;Y3) ≥ I(Y2;2|X2, Y3)I(X2;Y3) ≥ I(Y2;2|X2, Y3)
I(Y2;2|X1Y3) = H(2|X1Y3) − H(2|X1Y3Y2) = H(g21X1 + Z2 + Z|X1Y3) − H(Y2 + Z|X1Y3Y2) = (1)/(2)log2πe(1 + N) − (1)/(2)log2πeN = (1)/(2)log(1)/(N) + 1
I(X2;Y3) = I(X2;Y3Y3’’) = I(X2;Y3’’) + I(X2;Y3’|Y3’’) = H(Y3’’) − H(Y3’’|X2) + H(Y3’|Y3’’) − H(Y3’|X2Y3’’) = \mathchoice(1)/(2)⋅log(S32 + 1)(1)/(2)⋅log(S32 + 1)(1)/(2)⋅log(S32 + 1)(1)/(2)⋅log(S32 + 1)
(1)/(2)log(1)/(N) + 1 ≤ (1)/(2)⋅log(S32 + 1) (1)/(N) + 1 ≤ (S32 + 1) S32 ≤ (1)/(N) N ≤ (1)/(S32)
I(X1;Y2̂, Y3, |X2) = \mathchoiceI(X1;2|X2)I(X1;2|X2)I(X1;2|X2)I(X1;2|X2) + \mathchoiceI(X1;Y3|X22)I(X1;Y3|X22)I(X1;Y3|X22)I(X1;Y3|X22) = I(X1;Y3|X2) + I(X1;2|X2Y3)
I(X1;2|X2) = H(2|X2) − H(2|X1X2) = (1)/(2)log(S21 + 1 + N) − (1)/(2)log(1 + N) = \mathchoice(1)/(2)log(S21)/(1 + N) + 1(1)/(2)log(S21)/(1 + N) + 1(1)/(2)log(S21)/(1 + N) + 1(1)/(2)log(S21)/(1 + N) + 1
I(X1;Y3|X22) = I(X1;Y3’|X22) + I(X1;Y3’’|X22Y3) = (1)/(2)log(S31 + 1) + H(Y3’’|X22Y3) − H(Y3’’|X22Y3X1) = \mathchoice(1)/(2)log(S31 + 1)(1)/(2)log(S31 + 1)(1)/(2)log(S31 + 1)(1)/(2)log(S31 + 1)
\mathchoiceI(X1;Y2̂, Y3, |X2)I(X1;Y2̂, Y3, |X2)I(X1;Y2̂, Y3, |X2)I(X1;Y2̂, Y3, |X2) = (1)/(2)log(S21)/(1 + N) + 1 + (1)/(2)log(S31 + 1) = (1)/(2)log(S21)/(1 + N) + 1(S31 + 1) ≥ (1)/(2)log(S21)/(1 + (1)/(S32)) + 1(S31 + 1) = (1)/(2)log(S21S32)/(1 + S32) + 1(S31 + 1) = 
 = \mathchoice(1)/(2)log(S21S32(S31 + 1))/(1 + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/(1 + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/(1 + S32) + S31 + 1(1)/(2)log(S21S32(S31 + 1))/(1 + S32) + S31 + 1

Алтернативна пресметка на I(X2;Y3):
I(X2;Y3) = I(X2;Y3Y3’’) = I(X2;Y3) + I(X2;Y3’’|Y3) = H(Y3) − H(Y3’|X2) + H(Y3’’|Y3) − H(Y3’’|X2Y3) = H(Y3’’|Y3) − H(Y3’’|X2Y3)
 ≤ (1)/(2)⋅log(2πe)E[Y3’’2] − (E2[Y3Y3’’])/(E[Y3]) − (1)/(2)log(2πe) = (1)/(2)⋅logE[Y3’’2] − (E2[Y3Y3’’])/(E[Y3]) = (1)/(2)⋅logS32 + 1 − (\cancelto0E2[(g31X1 + Z3)(g32X2 + Z3’’)])/(E[Y3]) = \mathchoice(1)/(2)⋅log(S32 + 1)(1)/(2)⋅log(S32 + 1)(1)/(2)⋅log(S32 + 1)(1)/(2)⋅log(S32 + 1)
Ова е многу важна констатација. И да згрешиш дека постои зависност меѓу две посложени променливи пристапов со Шварцово неравенство ќе ти го покаже тоа дека тие се независни и ќе стигнеш до резултатот кој би го добил доколку уште веднаш си претпоставел дека тие се независни!!!
Во условните ентропии, онаму каде што имаш јасна зависност таа е доминантна, т.е. другите условни случајни променливи не носат ништо повеќе информација од случајната променлива со јасна зависност кон усовената слујана променлива.

Алтернативна пресметка на I(X1;Y3|X22)
\mathchoiceY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + Z
\mathchoiceI(X1;Y3|X22)I(X1;Y3|X22)I(X1;Y3|X22)I(X1;Y3|X22) = I(X1;Y3’|X22) + I(X1;Y3’’|X22Y3) = (1)/(2)log(S31 + 1) + H(Y3’’|X22Y3) − H(Y3’’|X22Y3X1) = (1)/(2)log(S31 + 1) ???
I(X1;Y3’|X22) = H(Y3’|X22) − H(Y3’|X2Y2X1) ≤ (1)/(2)logE(Var(Y3’|2) = (1)/(2)logE[Y32] − (E2[Y32])/(E[22]) = (1)/(2)logS31 + 1 − (E2[(g31X1 + Z3)(g21X1 + Z2 + Z)])/(S21 + 1 + N) = 

 = (1)/(2)logS31 + 1 − (g231g221P2)/(S21 + 1 + N) = (1)/(2)logS31 + 1 − (S21S31)/(S21 + 1 + N)

I(X1;Y3|X22) = (1)/(2)logS31 + 1 − (S21S31)/(S21 + 1 + N)
\mathchoiceI(X1;Y2̂, Y3, |X2)I(X1;Y2̂, Y3, |X2)I(X1;Y2̂, Y3, |X2)I(X1;Y2̂, Y3, |X2) = (1)/(2)log(S21)/(1 + N) + 1 + (1)/(2)logS31 + 1 − (S21S31)/(S21 + 1 + N) = (1)/(2)log(S21)/(1 + N) + 1S31 + 1 − (S21S31)/(S21 + 1 + N) = (1)/(2)log(S21S32)/(1 + S32) + 1S31 + 1 − (S21S31)/(S21 + 1 + (1)/(S32))
(S21S32 + 1 + S32)/(1 + S32)S31 + 1 − (S21S31S32)/(S21S32 + S32 + 1) = (S21S32 + 1 + S32)/(1 + S32)((S31 + 1)(S21S32 + S32 + 1) − S21S31S32)/(S21S32 + S32 + 1) = ((S21S32 + 1 + S32)[(S31 + 1)(S21S32 + S32 + 1) − S21S31S32])/((1 + S32)(S21S32 + S32 + 1))
(S21S32 + 1 + S32)/(1 + S32)(S31 + 1) − (\cancelS21S32 + 1 + S32)/(1 + S32)(S21S31S32)/(\cancelS21S32 + S32 + 1) = (S21S32 + 1 + S32)/(1 + S32)(S31 + 1) − (S21S31S32)/(1 + S32)

(S32S31 + S31 + S21S32 + S32 + 1)/(1 + S32) = ((S32 + 1)S31 + S21S32 + (S32 + 1))/(1 + S32) = \mathchoice1 + S31 + (S21S32)/(1 + S32)1 + S31 + (S21S32)/(1 + S32)1 + S31 + (S21S32)/(1 + S32)1 + S31 + (S21S32)/(1 + S32)

Expand[(S21S32 + 1 + S32)(S31 + 1) − S21S31S32] = S32S31 + S31 + S21S32 + S32 + 1
C ≥ maxp(x1)p(2|y2)min{I(X1;Y3) − I(Y2;2|X1Y3) + C0, I(X1;2, Y3)}
2 = Y2 + Z X1 ~ N(0, P),  X2 ~ N(0, P) and Z ~ N(0, N)
\mathchoiceY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + ZY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’ 2 = Y2 + Z = g21X1 + Z2 + Z
\mathchoiceI(X1;2, Y3)I(X1;2, Y3)I(X1;2, Y3)I(X1;2, Y3) = I(X1;2) + \mathchoiceI(X1;Y3’|2)I(X1;Y3’|2)I(X1;Y3’|2)I(X1;Y3’|2) = H(2) − H(2|X1) + I(X1;Y3’|2) = H(g21X1 + Z2 + Z) − H(g21X1 + Z2 + Z|X1) + I(X1;Y3’|2) = 
 = (1)/(2)log(2πe)(S21 + 1 + N) − (1)/(2)log(2πe)(1 + N) + I(X1;Y3’|2) = (*)

\mathchoiceI(X1;Y3’|2)I(X1;Y3’|2)I(X1;Y3’|2)I(X1;Y3’|2) = H(Y3’|2) − H(Y3’|X12) = ()
The received signal at the relay and at the receiver y3 are correlated since they are copies of the same signal x1 obtained form two independent paths with noise and path loss.
H(Y3’|2) = E[H(Y3’|2)] ≤ E(1)/(2)log2πe(Var(Y3|2)) ≤ (1)/(2)log2πe(E[Var(Y3|2)]) ≤ (1)/(2)log2πeE[Y’23] − (E2[Y32])/(E[22])
\mathchoiceE[Y’23] − (E2[Y32])/(E[22])E[Y’23] − (E2[Y32])/(E[22])E[Y’23] − (E2[Y32])/(E[22])E[Y’23] − (E2[Y32])/(E[22]) = S31 + 1 − (E2[(g31X1 + Z3)(g21X1 + Z2 + Z)])/(S21 + 1 + N) = S31 + 1 − (E[(g31X1 + Z3)(g21X1 + Z2 + Z)])/(S21 + 1 + N) = S31 + 1 − (g231g221P2)/(S21 + 1 + N) =  = \mathchoiceS31 + 1 − (S31S21)/(S21 + 1 + N)S31 + 1 − (S31S21)/(S21 + 1 + N)S31 + 1 − (S31S21)/(S21 + 1 + N)S31 + 1 − (S31S21)/(S21 + 1 + N)
() = (1)/(2)log2πeS31 + 1 − (S31S21)/(S21 + 1 + N) − (1)/(2)log2πe = \mathchoice(1)/(2)logS31 + 1 − (S31S21)/(S21 + 1 + N)(1)/(2)logS31 + 1 − (S31S21)/(S21 + 1 + N)(1)/(2)logS31 + 1 − (S31S21)/(S21 + 1 + N)(1)/(2)logS31 + 1 − (S31S21)/(S21 + 1 + N)

(*) = (1)/(2)log(2πe)(S21 + 1 + N) − (1)/(2)log(2πe)(1 + N) + (1)/(2)⋅logS31 + 1 − (S31S21)/(S21 + 1 + N) = (1)/(2)log(S21)/(1 + N) + 1 + (1)/(2)⋅logS31 + 1 − (S31S21)/(S21 + 1 + N) = 

 = \mathchoice(1)/(2)log(S21)/(1 + N) + 1S31 + 1 − (S31S21)/(S21 + 1 + N)(1)/(2)log(S21)/(1 + N) + 1S31 + 1 − (S31S21)/(S21 + 1 + N)(1)/(2)log(S21)/(1 + N) + 1S31 + 1 − (S31S21)/(S21 + 1 + N)(1)/(2)log(S21)/(1 + N) + 1S31 + 1 − (S31S21)/(S21 + 1 + N)

I(X1;Y3) − I(Y2;2|X1Y3) + C0 = ???
I(X1;Y3) = (1)/(2)log(S31 + 1)
C0 = I(X2;Y3’’) = (1)/(2)log(S32 + 1)
I(Y2;2|X1Y3) = H(2|X1Y3) − H(2|X1Y3Y2) = H(g21X1 + Z2 + Z|X1Y3) − H(Y2 + Z|X1Y3Y2) = (1)/(2)log2πe(1 + N) − (1)/(2)log2πeN = (1)/(2)log(1)/(N) + 1
\mathchoiceI(X1;Y3) − I(Y2;2|X1Y3) + C0I(X1;Y3) − I(Y2;2|X1Y3) + C0I(X1;Y3) − I(Y2;2|X1Y3) + C0I(X1;Y3) − I(Y2;2|X1Y3) + C0 = (1)/(2)log(S31 + 1) + (1)/(2)log(S32 + 1) − (1)/(2)log(1)/(N) + 1 = \mathchoice(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1)(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1)(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1)(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1)

C = min(1)/(2)log(S31 + 1)(S32 + 1)(N)/(N + 1), (1)/(2)log(S21)/(1 + N) + 1S31 + 1 − (S31S21)/(S21 + 1 + N)
(S31 + 1)(S32 + 1)(N)/(N + 1) − (S21)/(1 + N) + 1S31 + 1 − (S31S21)/(S21 + 1 + N) = 0
((S32 + 1)(1 + S31 + S21))/(S32(1 + S31 + S21)/(S32(S31 + 1)) + 1) = ((S32 + 1)(1 + S31 + S21))/(S32(1 + S31 + S21)/(S32(S31 + 1)) + (S32(S31 + 1))/(S32(S31 + 1))) = ((S32 + 1)(1 + S31 + S21)(S31 + 1))/((1 + S31 + S21 + S32(S31 + 1)))

((S32 + 1)(1 + S31 + S21)(S31 + 1))/((1 + S31 + S21 + S32(S31 + 1))) = (S32S231 + S231 + S21S31 + S21S32S31 + 2S32S31 + 2S31 + S21 + S21S32 + S32 + 1)/(1 + S31 + S21 + S32S31 + S32) = 

 = 1 + (S32S231 + S231 + S21S31 + S21S32S31 + S32S31 + S31 + S21S32)/(1 + S31 + S21 + S32S31 + S32) = 1 + (S31(S32S31 + S31 + S21 + S32 + 1) + S21S32S31 + S21S32)/(1 + S31 + S21 + S32S31 + S32) = 

 = 1 + S31 + (S21S32S31 + S21S32)/(1 + S31 + S21 + S32S31 + S32) = 1 + S31 + (S21S32(S31 + 1))/(1 + S31 + S21 + S32S31 + S32) = \mathchoice1 + S31 + (S21S32(S31 + 1))/(S21 + (S31 + 1)(S32 + 1))1 + S31 + (S21S32(S31 + 1))/(S21 + (S31 + 1)(S32 + 1))1 + S31 + (S21S32(S31 + 1))/(S21 + (S31 + 1)(S32 + 1))1 + S31 + (S21S32(S31 + 1))/(S21 + (S31 + 1)(S32 + 1))  Q.E.D!!!

Expand[(S32 + 1)(1 + S31 + S21)(S31 + 1)] = S32S231 + S231 + S21S31 + S21S32S31 + 2S32S31 + 2S31 + S21 + S21S32 + S32 + 1
(1 + S31 + S21 + S32(S31 + 1)) = S21 + (1 + S32)(S31 + 1) = S21 + (S32 + 1)(S31 + 1)
Q.E.D!!!
This bound becomes asymptotically tight as either S31 or S32 approaches infinity. At low S21, that is, low SNR for the channel form the sender to the relay, compres-forward outperforms both direct transmission and decode-forward. Furthermore, the compress-forward rate can be improved via time sharing at the sender, that is, by having the sender transmit at power P ⁄ α for fraction α ∈ [0, 1] of the time and at zero power for the rest of the time.

3.8.1 Linear Relaying for RFD Gaussian RC

Consider the RFD Gaussian RC with the relaying functions restricted to being linear combinations of past received symbols. Note that under the orthogonal receiver components assumption, we can eliminate the delay in relay encoding simply by relabeling the transmission time for the channel from x2 to Y3’’ . Hence, we equivalently consider relaying functions of the form x2i = ij = 1aijy2ji ∈ [1:n] or in vector notation of the form Xn2 = AYn2 where A is and nxn lower triangular matrix.
x21 x22 ... x2n  =  a11 0 0 0 a21 a22 0 0 ... ... ... ... an1 an2 ... ann y21 y22 ... y23
This scheme reduces the relay channel to a point-to-point Gaussian channel with input Xn1 and output Yn3 = (Y3n, Y3’’n). Note that linear relaying is conisderably simpler to implement in practice than decode-forward and compress-forward. It turns out that its performance also compares well with these more complex schemes under certain high SNR conditions.
The capacity with linear relaying, CL, is characterized by the multiletter expression
CL = limk → ∞C(k)L
(27) C(k)L = supF(xk1), A(1)/(k)I(Xk1;Yk3)
The suppremum is over all cdfs F(xk1) and lower triangular matrices A that satisfy the sender and the relay power constraints P; see Problems 16.10 and 16.12 for multiletter characterizations of the capacity of the DM and Gaussian relay channels. It can be easily shown that C(k)Lis attained by a Gaussian input Xk1 that satisfies the power constraint.

3.8.2 Amplify-Forward

Consider C(1)L, which is the maximum rate achieved via a simple amplify-forward relaying scheme. It can be shown that C(1)L is attained by X1 ~ N(0, P) and X2 = Y2(P ⁄ (S21 + 1)).Therefore
C(1)L = CS31 + (S21S32)/(S21 + S32 + 1)
C ≤ maxp(x1)p(x2)min{I(X1;Y3) + I(X2;Y3’’), I(X1;Y2, Y3)}

y3 = g31x1 + g32x2 + z3 y2 = g21x1 + z2 z3 ~ N(0, 1) z2 ~ N(0, 1)

x2 = ((P)/(S21 + 1))y2 = g21((P2)/(S21 + 1))x1 + ((P2)/(g221P1 + 1))z2 y3 = g31x1 + g32x2 + z3 = g31x1 + g32g21((P2)/(g221P1 + 1))x1 + g32((P2)/(g221P1 + 1))z2 + z3 = 

I(X1;Y3) = h(Y3) − h(Y3|X1) = h(Y) − h(g31x1 + g32g21((P2)/(g221P1 + 1))x1 + g32((P2)/(g221P1 + 1))z2 + z3|X1) = 

 = h(g31x1 + g32g21((P2)/(g221P1 + 1))x1 + g32((P2)/(g221P1 + 1))z2 + z3) − h(g32((P2)/(g221P1 + 1))z2 + z3) = 

 = (1)/(2)log2πeg231P + g232g221(P2)/(g221P1 + 1)P1 + g232(P2)/(g221P1 + 1) + 1 − (1)/(2)log2πeg232(P2)/(g221P1 + 1) + 1 = 

 = (1)/(2)log(g231P + g232g221(P2)/(g221P1 + 1)P1 + \mathchoiceg232(P2)/(g221P1 + 1) + 1g232(P2)/(g221P1 + 1) + 1g232(P2)/(g221P1 + 1) + 1g232(P2)/(g221P1 + 1) + 1)/(g232(P2)/(g221P1 + 1) + 1) = (1)/(2)log1 + (g231P + g232g221(P2)/(g221P1 + 1)P1)/(g232(P2)/(g221P1 + 1) + 1)

(1)/(2)log1 + (S31 + (S32S21)/(S21 + 1))/((S32)/(S21 + 1) + 1) = \mathchoice(1)/(2)log1 + (S31(S21 + 1) + S32S21)/(S21 + S31 + 1)(1)/(2)log1 + (S31(S21 + 1) + S32S21)/(S21 + S31 + 1)(1)/(2)log1 + (S31(S21 + 1) + S32S21)/(S21 + S31 + 1)(1)/(2)log1 + (S31(S21 + 1) + S32S21)/(S21 + S31 + 1) = \mathchoice(1)/(2)log1 + ((S31 + S32)S21 + S31)/(S21 + S31 + 1)(1)/(2)log1 + ((S31 + S32)S21 + S31)/(S21 + S31 + 1)(1)/(2)log1 + ((S31 + S32)S21 + S31)/(S21 + S31 + 1)(1)/(2)log1 + ((S31 + S32)S21 + S31)/(S21 + S31 + 1)

(S31S21 + S32 + S32S21)/(S21 + S31 + 1) = (S31S21 + \cancelto0(S32)/(S21) + S32S21)/(S21 + S31 + 1) = (S31S21 + S32S21)/(S21 + S31 + 1) = ((S31 + S32)S21)/(S21 + S31 + 1)
figure /home/jovan/Dropbox/Books Read/Notebooks/LyxNotebooks/Miscellaneous/Fig. Linear Relaying for FD-AWGN.png
I(X1;Y’, Ys) = H(Y’, Ys) − H(Y’, Ys|X1)
X1, X2, ... are i.i.d  ~ N(0, P), and xi’ = dyi where d is chossen to satisfy the relay power constraint
\mathchoiceX1 ~  N(0, P) Y’ = aX1 + Z1 X’ = dYYR = bX’ + ZR = bd(aX1 + Z1) + ZR = bdaX1 + bdZ1 + ZRX1 ~  N(0, P) Y’ = aX1 + Z1 X’ = dYYR = bX’ + ZR = bd(aX1 + Z1) + ZR = bdaX1 + bdZ1 + ZRX1 ~  N(0, P) Y’ = aX1 + Z1 X’ = dYYR = bX’ + ZR = bd(aX1 + Z1) + ZR = bdaX1 + bdZ1 + ZRX1 ~  N(0, P) Y’ = aX1 + Z1 X’ = dYYR = bX’ + ZR = bd(aX1 + Z1) + ZR = bdaX1 + bdZ1 + ZR
\mathchoiceYs = YD1 = X1 + ZsYs = YD1 = X1 + ZsYs = YD1 = X1 + ZsYs = YD1 = X1 + Zs
\mathchoiced = ((2αP)/(a22αP + N)) d2 = (P)/(a2P + N) = ((P)/(N))/(a2(P)/(N) + 1)d = ((2αP)/(a22αP + N)) d2 = (P)/(a2P + N) = ((P)/(N))/(a2(P)/(N) + 1)d = ((2αP)/(a22αP + N)) d2 = (P)/(a2P + N) = ((P)/(N))/(a2(P)/(N) + 1)d = ((2αP)/(a22αP + N)) d2 = (P)/(a2P + N) = ((P)/(N))/(a2(P)/(N) + 1)
I(X1;YD1YR1) = \mathchoiceH(YD1YR1)H(YD1YR1)H(YD1YR1)H(YD1YR1) − H(YD1YR1|X1) = H(YR1) + H(YD1|YR1) − H(YD1|X1) − H(YR1|X1YD1)
Не дели со два 27↑, затоа што за во доволно голем број на блокови (еднозначни блокови) b/b-1 → 1 т.е. може да се смета како да не испраќаш информацијата во два тајм слота туку во еден. Вториот тајм слот секако се користи за испраќање на наредниот знак до релето.
H(YR1) = H(bdaX1 + bdZ1 + ZR) = (1)/(2)log(2πe)(b2d2a2P + b2d2N + N)
H(YD1|YR1) = H(X1 + Z|YR1) = (1)/(2)log(2πe)E(Var(X1 + Z|YR1)) = (1)/(2)log(2πe)P − (b2d2a2P2)/(b2d2a2P + b2d2N + N) + N
E(Var(X1 + Z|YR1)) = E(Var(X1|YR1)) + N = P − (E2(X1YR1))/(E(Y2R1)) + N = P − (E2(X1(bdaX1 + bdZ1 + ZR)))/(E(Y2R1)) + N = P − (b2d2a2P2)/(b2d2a2P + b2d2N + N) + N
\mathchoiceH(YD1YR1)H(YD1YR1)H(YD1YR1)H(YD1YR1) = (1)/(2)log(2πe)(b2d2a2P + b2d2N + N) + (1)/(2)log(2πe)P − (b2d2a2P2)/(b2d2a2P + b2d2N + N) + N = 
 = (1)/(2)log(2πe)2(b2d2a2P + b2d2N + N)P − (b2d2a2P2)/(b2d2a2P + b2d2N + N) + N = (*)
(b2d2a2P + b2d2N + N)P − (b2d2a2P2)/(b2d2a2P + b2d2N + N) + N = (P + N)(b2d2a2P + b2d2N + N) − b2d2a2P2
Expand[(P + N)(b2d2a2P + b2d2N + N) − b2d2a2P2] = a2b2d2NP + b2d2N2 + b2d2NP + N2 + NP
(*) = \mathchoice(1)/(2)log(2πe)2(a2b2d2NP + b2d2N2 + b2d2NP + N2 + NP)(1)/(2)log(2πe)2(a2b2d2NP + b2d2N2 + b2d2NP + N2 + NP)(1)/(2)log(2πe)2(a2b2d2NP + b2d2N2 + b2d2NP + N2 + NP)(1)/(2)log(2πe)2(a2b2d2NP + b2d2N2 + b2d2NP + N2 + NP)
H(YD1|X1) = (1)/(2)log(2πe)(N)
H(YR1|X1YD1) = H(bdaX1 + bdZ1 + ZR|X1YD1) = (1)/(2)log(bdZ1 + ZR|YD1) =  = (1)/(2)log(b2d2N + N)
\mathchoiceI(X1;YD1YR1)I(X1;YD1YR1)I(X1;YD1YR1)I(X1;YD1YR1) = (1)/(2)log(2πe)2(a2b2d2NP + b2d2N2 + b2d2NP + N2 + NP) − (1)/(2)log(2πe)2N(b2d2N + N) = 
(1)/(2)log(2πe)2(a2b2d2NP + b2d2N2 + b2d2NP + N2 + NP) − (1)/(2)log(2πe)2N(b2d2N + N)
 = (1)/(2)log((a2b2d2NP + b2d2N2 + b2d2NP + N2 + NP))/(N(b2d2N + N)) = (1)/(2)log\undersetA((a2b2d2NP + \cancelb2d2N2 + b2d2NP + \cancelN2 + NP))/(b2d2N2 + N2) = (*)
A = 1 + ((a2b2d2NP + b2d2NP + NP))/(b2(P)/(a2P + N)N2 + N2) = 1 + (((a2 + 1)b2d2NP + NP))/(N2b2(P)/(a2P + N) + 1) = 1 + (NP((a2 + 1)b2d2 + 1))/(N2(b2P + a2P + N)/(a2P + N)) = 1 + (P(a2 + 1)b2(P)/(a2P + N) + 1)/(N(b2P + a2P + N)/(a2P + N)) = 

 = 1 + (P((a2 + 1)b2P + a2P + N)/(a2P + N))/(N(b2P + a2P + N)/(a2P + N)) = 1 + (P((a2 + 1)b2P + a2P + N))/(N(b2P + a2P + N)) = 1 + (P)/(N)((a2b2P + \cancelb2P + a2P + N))/((b2P + a2P + N)) = 1 + (P)/(N)1 + (a2b2P)/((a2 + b2)P + N)

(*) = (1)/(2)⋅log1 + (P)/(N)1 + (a2b2P)/((a2 + b2)P + N) = C(P)/(N)1 + (a2b2P)/((a2 + b2)P + N) Q.E.D!!!
C(P)/(N)1 + (a2b2P)/((a2 + b2)P + N) = C(P)/(N) + (a2b2)/((a2 + b2)(P)/(N) + 1)(P2)/(N2) = CS31 + (S21S32)/(S21 + S32 + 1) Ова е во согласност со изразот од El Gamal NIT.
Q.E.D!!!
The amplify-forward rate C(1)L is not convex in P. In fact, it is concave for small P and convex for large P. Hence the rate can be improved by time sharing between direct transmission and amplify-forward. Assuming amplify-forward is performed α of the time and the relay transmits at power P ⁄ α during this time, we can achieve the improved rate
max0 < α, β ≤ 1αC(βS)/(α) + (αC(β ⁄ α)(S31 + S21S32))/(βS21 + S32 + α)

C ≤ maxp(x1x2)min{I(X1, X2;Y3), I(X1;Y2, Y3|X2)}
p(y2, y3|x1x2) = p(y3’, y2|x1)p(y3’’|x2)
\mathchoiceY2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’Y2 = g21X1 + Z2,  Y3’ = g31X1 + Z3’,  Y3’’ = g32X2 + Z3’’
\mathchoiceX2 = Y2(P ⁄ (S21 + 1))X2 = Y2(P ⁄ (S21 + 1))X2 = Y2(P ⁄ (S21 + 1))X2 = Y2(P ⁄ (S21 + 1))
I(X1, X2;Y3) = H(Y3) − H(Y3|X1X2) = H(Y3Y3’’) − H(Y3Y3’’|X1X2) = I(X1;Y3) + I(X2;Y3’’)
I(X1;Y3) + C0 = C(S31) + C(S32)

Y2 = (X2)/((P ⁄ (S21 + 1)))
I(X1;Y2Y3Y3’’|X2) = I(X1;Y2Y3’|X2) + I(X1;Y3’’|X2Y2Y3) = (*)
I(X1;Y2Y3’|X2) = H(Y2Y3’|X2) − H(Y2Y3’|X2X1) = H(Y2|X2) + H(Y3) − H(Y2|X2X1) −  − H(Y3’|X2X1Y2)
 = \cancelto0H(Y2|X2) + H(Y3) − \cancelto0H(Y2|X2X1) − H(Y3’|X2X1Y2) = (1)/(2)⋅log2πe(S31 + 1) − (1)/(2)⋅log(2πe) = (1)/(2)⋅log(S31 + 1)
I(X1;Y3’’|X2Y2Y3) = H(Y3’’|X2Y2Y3) − H(Y3’’|X2Y2Y3X1) = 0
(*) = (1)/(2)⋅log(S31 + 1) = C(S31)

C ≤ maxp(x1x2)min{C(S31) + C(S32), C(S31)} = maxp(x1x2){C(S31)}
figure Figure 16.13a.png figure Figure 16.13.png
Figure 14 Comparison of the custet bound RCS, the decode and forward lower bound RDF, the compress-forward lower bound RCF, and amplify-foward lower bound RAF on the capacity of RFD Gaussian Channel
Figure 16.13 compares the cutset bound to the decode-forward, compress-forward and amplify-forward lower bounds for different SNRs. Note that compress-forward outperforms amplify-forward in general, but is significantly more complex to implement.

3.8.3 Linear Relaying Capacity of the RFD Gaussian RC

Капацитет на линеарен гаусов релеен канал со фреквентна распределба во приемникот
We can establish the following single letter characterization of the capacity with linear relaying.
The linear relaying capacity of the RFD Gaussian RC is
CL = maxα0C(β0P)/(α0) + 4j = 1αjC(βjP)/(αj)1 + (g221g232ηj)/(1 + g232ηj)
where the maximum is over αj, βj ≥ 0 and ηj ≥ 0 such that 4j = 0αj = 4j = 0βj = 1 and 4j = 1ηj(g221βj + αj) = P
Proof outline.
Assume without loss of generality that g31 = 1, Then C(k)L is the solution to the optimization problem
maximize:
(1)/(2k)log(||| I + KX1 g21g32KX1AT g21g32AKX1 I + g221g232AKX1AT + g32AAT |||)/(||| I 0 0 I + g232AAT |||)
subject to: KX1≽0,  tr(KX1) ≤ kP,  tr(g221KX1ATA + ATA) ≤ kP,  A lower trieangular,
where KX1 = E(Xk1(Xk1)T) and A are optimization variables. This is a nonconvex problem in (KX1, A) wiht k2 + k variables. For a fixed A, the problem is convex in KX1 and has a water-filling solution. However, finding A for a fixed KX1 is nonconvex problem.
Now it can be shown that it suffices to consider diagonal KX1 and A. Thus, the optimization problem simplifies to:
maximize:
(1)/(2k)logkj = 11 + σj1 + (g221g232a2j)/(1 + g232a2j)
subject to: σj ≥ 0,  j ∈ [1:k],  kj = 1σj ≤ kP,  kj = 1a2j(1 + g221σj) ≤ kP.
While this is still a nonconvex optimization problem, the problem now involves only 2k variables. Furthermore, it can be shown that at the optimum point, if σj = 0, then aj = 0, and conversely if aj = aj = 0, then σj = σj. Thus, the optimization problem can be further simplified to:
maximize
(1)/(2k)log1 + (kβ0P)/(k0)kj = 11 + σj1 + (g221g232a2j)/(1 + g232a2j)
subject to
aj > 0,  j ∈ [k0 + 1:k] kj = 1σj ≤ k⋅(1 − β0)⋅P,  kj = 1a2j(1 + g221σj) ≤ kP
By the KKT condition, it can be shown that at the optimum, there are no more than four distinct nonzero (σj, aj) pairs. Hence, C(k)L is the solution to the optimization problem
maximize
(1)/(2k)log1 + (kβ0P)/(k0)kj = 11 + σj1 + (g221g232a2j)/(1 + g232a2j)kj
subject to
aj > 0,  j ∈ [1:4] 4j = 1σj ≤ k⋅(1 − β0)⋅P,  4j = 1a2j(1 + g221σj) ≤ kP 4j = 0kj = k
where kj is a new optimization variable that denotes the number of times the pair (σj, aj) is used during transmission. Taking k → ∞ completes the proof.
This is a rare example for which there is no known single-letter mutual information characterization of the capacity, yet we are able to reduce the multiletter characterization directly to a computable characterization.

3.9 Look-ahead Relay Channels

Релејни канали со предвидување
The relay channel can be viewed as point-to-point communication system with side information about the input X1 available at the receiver through the relay. As we have seen in Chapters 7 and 11, the degree to which side information can help communication depends on it temporal availability (causal versus noncausal). In our discussion of the relay channel so far, we have assumed that the relay functions depend only on the past received relay sequence, and hence the side information at the receiver is available strictly causally. In this section, we study the relay channel with causal or lookahead relaying functions. The lookahead at the relay may be, for example, the result of a difference between the arrival times of the signal form the sender to the receiver and to the relay. If the signal arrives at both the relay and the receiver at the same time, then we obtain the strictly causal relaying functions assumed so far. If the signal arrives at the receiver later than at the relay, then we effectively have lookahead relaying functions.
To study the effect of lookahead on the capacity of the relay channel, consider the lookahead DMRC (X1 xX2, p(y2|x1)p(y3|x1, x2, y2), Y2 xY3, l) depicted in 15↓. The integer parameter l specifies the amount of relaying lookahead. Note that here we define the lookahead relay channel as p(y2|x1)p(y3|x1, x2, y2), since the conditionаl pmf p(y2, y3|x1x2) depends on the code due to the instantaneous or lookahead dependency of X2 on Y2.
figure Figure 16.14.png
Figure 15 Lookahead relay channel
The channel is memoryless in the sense that p(y2i|xi1, yi − 12, m) = pY2|X1(y2i|x1i) and p(y3i|xi1, xi2, yi2, yi − 13, m) = p(y3i|x1i, x2i, y2i).
A (2nR, n) code for the lookahead relay channel consists of
We assume that the message M is uniformly distributed over [1:2nR]. The definitions of probability of error, achievability, and capacity Cl are as before. Clearly, Cl is monotonically nondecreasing in l. The capacity Cl of the lookahead DMRC is not known in general for any finite or unbounded l.
The DMRC we studied earlier corresponds to lookahead parameter l =  − 1, or equivalently, a delay of 1. We denote its capacty C − 1 by C. In the following, we discus two special cases.

3.9.1 Noncausal Relay Channels

The noncausal relay allows for arbitrarily large lookahead relaying functions. Its capacity C is not known in general. We establish upper and lower bounds on C that are tight in some, cases, and show that C can be strictly larger than the cutset bound on the capacity C of the (strictly causal) relay channel in Theorem 3.2↑. Hence C can be larger than the capacity C itself.
Cutset bound for the noncausal DMRC
The capacity of the non-causal DMRC p(y2|x1)p(y3|x1x2y2) is upper bounded as
\mathchoiceC ≤ maxp(x1)p(u|x1, x2), x2(u, y2)min{I(X1;Y2) + I(X1;Y3|X2Y2), I(U, X1;Y3)}C ≤ maxp(x1)p(u|x1, x2), x2(u, y2)min{I(X1;Y2) + I(X1;Y3|X2Y2), I(U, X1;Y3)}C ≤ maxp(x1)p(u|x1, x2), x2(u, y2)min{I(X1;Y2) + I(X1;Y3|X2Y2), I(U, X1;Y3)}C ≤ maxp(x1)p(u|x1, x2), x2(u, y2)min{I(X1;Y2) + I(X1;Y3|X2Y2), I(U, X1;Y3)}
The proof follows by noting that a (2nR, n) code for the noncausal relay channel induces a joint pmf of the form
p(m, xn1, xn2, yn2, yn3) = 2 − nRp(xn1|m)(ni = 1pY2|X1(y2i|x1i))p(xn2|yn2)(ni = 1pY3|X1X2Y2(y3i|x1ix2iy2i))
and using standard converse proof arguments.
p(y2|x1)p(y3|x1x2y2)
nR ≤ H(M) = H(Xn1) = I(Xn1;Yn3) + H(Xn1|Yn3) ≤ I(Xn1;Yn3) + nϵ = H(Yn3) − H(Yn3|Xn1) = ni = 1H(Y3i|Yi − 13) − ni = 1H(Y3i|Yi − 13Xn1) = 
 = ni = 1H(Y3i|Yi − 13) − ni = 1H(Y3i|Yi − 13Xn1) = ni = 1H(Y3i) − ni = 1H(Y3i|X1i) ≤ ni = 1H(Y3i) − ni = 1H(Y3i|X1i, X2i) = ni = 1I(X1iX2i;Y3i) = 
 = nR ≤ H(M) = H(Xn1) = I(Xn1;Yn2Yn3) + H(Xn1|Yn2Yn3) ≤ I(Xn1;Yn2Yn3) + nϵ = H(Yn3Yn2) − H(Yn3|Xn1) = 
 = ni = 1H(Y2iY3i|Yi − 13Yi − 12) − ni = 1H(Y2iY3i|Yi − 13Yi − 12Xn1) = 
 ≤ ni = 1H(Y2i, Y3i|Yi − 13Yi − 12) − ni = 1H(Y2i, Y3i|Yi − 13Yi − 12Xn1Xn2) = ni = 1H(Y3i, Y2i) − ni = 1H(Y2iY3i|X1iY2i) = ni = 1I(X1iX2i;Y2iY3i)
I(X1iX2i;Y2iY3i) = I(X1i;Y2iY3i) + \mathchoiceI(X2i;Y2iY3i|X1i)I(X2i;Y2iY3i|X1i)I(X2i;Y2iY3i|X1i)I(X2i;Y2iY3i|X1i)
I(X1i;Y2iY3i) = I(X1i;Y2i) + \mathchoiceI(X1i;Y3i|Y2i)I(X1i;Y3i|Y2i)I(X1i;Y3i|Y2i)I(X1i;Y3i|Y2i)
\mathchoiceI(X2i;Y2iY3i|X1i)I(X2i;Y2iY3i|X1i)I(X2i;Y2iY3i|X1i)I(X2i;Y2iY3i|X1i) = H(Y2iY3i|X1i) − H(Y2iY3i|X1iX2i) = H(Y2|X1) + H(Y3|Y2X1) − H(Y2|X1X2) − H(Y3|X1X2Y2) = 
 = H(Y2|X1) + H(Y3|Y2X1) − H(Y2|X1) − H(Y3|X1X2Y2) = H(Y3|Y2X1) − H(Y3|X1X2Y2) = I(Y2X1;Y3|X2)
I(X1iX2i;Y2iY3i) = I(X1i;Y2i) + \mathchoiceI(X1i;Y3i|Y2i)I(X1i;Y3i|Y2i)I(X1i;Y3i|Y2i)I(X1i;Y3i|Y2i) + \mathchoiceI(Y2X1;Y3|X2)I(Y2X1;Y3|X2)I(Y2X1;Y3|X2)I(Y2X1;Y3|X2)
\mathchoiceI(X1;Y3|Y2)I(X1;Y3|Y2)I(X1;Y3|Y2)I(X1;Y3|Y2) + \mathchoiceI(Y2X1;Y3|X2)I(Y2X1;Y3|X2)I(Y2X1;Y3|X2)I(Y2X1;Y3|X2) = H(Y3|Y2) − H(Y3|X1Y2) + H(Y3|X2) − H(Y3|X1Y2X2)
I(X1;Y3|Y2) + I(Y2X1;Y3|X2) = H(Y3|Y2) − H(Y3|X1Y2) + H(Y2X1|X2) − H(Y2X1|Y3X2) = 
 = H(Y3|Y2) − H(Y3|X1Y2) + H(X1|X2) + H(Y2|X1X2) − H(Y2X1|Y3X2) = 
 = H(Y3|Y2) − H(Y3|X1Y2) + H(X1|X2) + \cancelH(Y2|X1) − H(X1|Y3X2) − \cancelH(Y2|Y3X2X1) = \mathchoiceI(X1;Y3|Y2) + I(X1;Y3|X2)I(X1;Y3|Y2) + I(X1;Y3|X2)I(X1;Y3|Y2) + I(X1;Y3|X2)I(X1;Y3|Y2) + I(X1;Y3|X2)
I(X1;Y3|X2Y2) = H(Y3|X2Y2) − H(Y3|X1X2Y2)
H(Y3|Y2) − H(Y3|X1Y2) + H(X1|X2) − H(X1|Y3X2)
By extending decode-forward to the noncausal case, we can establish following lower bound.
(noncausal Decode-Forward Lower Bound).
The capacity of the noncausal DMRC p(y2|x1)p(y3|x1x2y2) is lower bounded as:
\mathchoiceC ≥ maxp(x1x2)min{I(X1;Y2), I(X1, X2;Y3)}C ≥ maxp(x1x2)min{I(X1;Y2), I(X1, X2;Y3)}C ≥ maxp(x1x2)min{I(X1;Y2), I(X1, X2;Y3)}C ≥ maxp(x1x2)min{I(X1;Y2), I(X1, X2;Y3)}
To prove achievability, fix the pmf p(x1, x2) that attains the lower bound and randomly and independently generate 2nR sequence pairs (xn1xn2)(m),  m ∈ [1:2nR], each according ni = 1pX1X2(x1i, x2i). Since the relay knows yn2 in advance, it decodes for the message m before transmission commences. The probability of error at the relay tends to zero as n → ∞ if R < I(X1;Y2) − δ(ϵ). The sender and the relay then cooperatively transmit (xn1xn2)(m) and the receiver decodes for m. The probability of error at the receiver tends to zero as n → ∞ if R ≤ I(X1, X2;Y3) − δ(ϵ) . Combining the two conditions completes the proof.
The lower bound is tight in some special cases.
(Noncausal Sato relay channel).
figure Ternary Relay Channel.png
Figure 16 Sato channel
Consider the noncausal version of the Sato relay channel in 3.4.3↑. We showed that the capacity for the strictly causal case is C = 1.1619 and coincides wiht the cutset bound in Theorem 16.1. Now for the noncausal case, note that by the degradedness of the cahnnel, I(X1;Y3|X2Y2) = 0 and p(x2|x1y2) = p(x2|x1) since Y2 = X1.
I(X1;Y3|X2Y2) = H(Y3|X2Y2) − H(Y3|X1X2Y2) = | degradedness | = H(Y3|X2Y2) − H(Y3|X2Y2) = 0
Hence, the noncausal decode-forward lower bound in Theorem 3.9.1↑ coincides with the cut-set bound in Theorem 16.6 (maximized by setting U = (X2, Y2) = (X1X2) and characterizes the capacity C. Optimizing the capacity expression C = maxp(x1x2)min{I(X1;Y2), I(X1, X2;Y3)} by setting p(x1, x2) as p(0, 1) = p(1, 0) = p(1, 1) = p(2, 0) = 1 ⁄ 18 and p(0, 0) = p(2, 1) = 7 ⁄ 18 we obtain C = log(9 ⁄ 4) = 1.1699 . Therefore, for this example C is strictly larger than the cutset bound on the capacity C of the relay channel in 3.2↑.
As another example, consider the noncausal version of the Gaussian relay channel in 3.5↑. By evaluating the cutset bound in Theorem 3.9.1↑ and the noncausal decode-forward lower bound in Theorem 3.9.1↑ with average power constraint P on each of X1 and X2 we can show that if S21 ≥ S31 + S32, the capacity is:
(28) C = C(S31 + S32 + 2(S31S32))
As in the Sato relay channel this capacity is strictly larger than the cutset bound for the Gaussian RC.

3.9.2 Causal Relay Channels

We now consider the causal DMRC, wehre for each i ∈ [1:n], the relay encoder assigns a symbol x2i(yi2) to every yi2 ∈ Yi2, that is, the relay lookajead is l = 0. Again the cpaacity C0 is not known in general. We provide upper and lower bounds on C0 and show that rates higher than the cutset bound on the capacity C of the corresponding DMRC can still be achieved.
(Cutset Bound for the Causal DMRC).
The capacity of the causal DMRC p(y2|x1)p(y3|x1, x2, y2) is upper bounded as:
\mathchoiceC0 ≤ maxp(u, x1)x2(u, y2)min{I(X1;Y2, Y3|U), I(U, X1;Y3)}C0 ≤ maxp(u, x1)x2(u, y2)min{I(X1;Y2, Y3|U), I(U, X1;Y3)}C0 ≤ maxp(u, x1)x2(u, y2)min{I(X1;Y2, Y3|U), I(U, X1;Y3)}C0 ≤ maxp(u, x1)x2(u, y2)min{I(X1;Y2, Y3|U), I(U, X1;Y3)}
where |U| ≤ |X1||X2| + 1.
Note that restricting x2 to be a function only of u reduces this bound to the cutset bound for the DMRC in Theorem 3.2↑.
I(X1;Y2, Y3|U), I(U, X1;Y3)
I(U, X1;Y3) = I(U;Y3) − H(X1;Y3|U) = I(X1, X2;Y3)
I(U;Y3) = H(Y3) − H(Y3|U) = H(Y3) − H(Y3|X2) = I(X2;Y3)
I(X1;Y2, Y3|U) = I(X1;Y2, Y3|X2)
Conversely, this upper bound can be expressed as a cutset bound C0 ≤ maxp(x1x2){I(X1;Y2, Y3|X2), I(X1, X2’;Y3)} for a DMRC with relay sender alphabet X2 that consist of all mappings x2’:Y2 → X2. This is analogus to the capacity expression for the DMC with DM state available caussaly at the encoder in Section 7.5 which is achieved via Shannon strategy.
Now we represent lower bound on the capacity of the causal DMRC. Note that any lower bound on the capacity of DMRC, for example using partial decode-forward or compress-forward, is a lower bound on the capacity of the causal relay channel. We expect, however, that higher rates can be achieved by using the present relay received symbol in addition to its past received symbols.
Instantaneous relaying lower bound. In this simple scheme, the relay transmitted symbol x2i = x2(y2i) for i ∈ [1:n], that is, x2i is function only of y2i. This simple scheme yields the lower bound
(29) C0 ≥ maxp(x1), x2(y2)I(X1, Y3)
(Causal Sato Relay channel).
Consider the causal version of the Sato relay channel in 3.4.3↑. We have shown that C = 1.1619 and C = log(9 ⁄ 4) = 1.1699. Consider the instantaneous relaying lower bound in 29↑ with pmf (3 ⁄ 9, 2 ⁄ 9, 4 ⁄ 9) on X1 and the function x2(y2) = 0 if y2 = 0 and x2(y2) = 1 if y2 = 1 or 2.
It can be shown that this choice yelds I(X1, Y3) = 1.1699. Hence the capacity of the causal Sato relay channel is C0 = C = 1.1699.
This result is not too surprising. Since the channel form the sender to the relay is noiseless, complete cooperation, which requires knowledge of the entire received sequence in advance, can be achieved simply via instantenous relaying. Since for this example C0 > C and C coincides with the cutset bound, instantaneous relaying alone can achieve a higher rate than the cutset bound on C!
Causal decode-forward lower bound.
The decode-forward lower bound for DMRC can be easily extended to incorporate the present received symbol at the relay. This yields the lower bound
C0 ≥ maxp(u, x1)x2(u, y2)min{I(X1;Y2|U), I(U, X1;Y3)}.
As for the cutset bound in 3.9.2↑, this bound can be viewed as a decode-forward lower bond for the DM-RC p(y2, y3|x1, x2) where the relay sender alphabet X2 consist of all mappings x2’:Y2 → X2. Note that the causal decode-forward lower bound coincides with the cut-set bound when the relay Chanel is degraded, i.e. p(y3|x1, x2, y2) = p(y3|x2, y2).
Now we investigate the causal version of the Gaussian relay channel in the previous subsections and in Section 3.5↑. Consider the amplify-forward relaying scheme, which is a special case of instantaneous relaying, where the relay in time i ∈ [1:n]. transmits a scaled version of its received signal, i.e., x2i = ay2i. To satisfy the relay power constraint, we must have a2 ≤ P ⁄ (g221P + 1). The capacity of the resulting equivalent point to point Gaussian channel with average received power (ag21g32 + g31)2P and noise power (a2g232 + 1) yields the lower bound on the capacity of the Gaussian relay channel
C0 ≥ C((ag21g32 + g31)2P)/(a2g232 + 1)
Now it can be shown that if S21(S21 + 1) ≤ S31S32, hten this boun is opimized by a* = g21 ⁄ g31g32 and simplifies to
(30) C0 ≥ C(S21 + S31)
Evaluating the cutset boudn on C in 3.9.1↑ for the Gaussian case with average power constraints yields
(31) C ≤ C(S21 + S31)
if S21 ≤ S32. Thus if S31 ≤ S32min{1, S31 ⁄ (S21 + 1)} , the bounds in 30↑ and 31↑ coincide and
(32) C0 = C = C(S21 + S31)
This shows that amplify-forward alone can be optimal for both causal and the noncausal Gaussian relay channels, which is surprising given the extreme simplicity of this relaying scheme. Note that the capacity in the above range of SNR values coincides with the cutset bound on C. This is not the case in general, however, and it can be shown using causal decode-forward that the capacity of the causal Gaussian relay channel can exceed this cutset bound.
Combinig 32↑ and 28↑, we have shown that the capacity of the noncausal Gaussian RC is known if S21 ≥ S31 + S32 or if S21 ≤ S32min{1, S31 ⁄ (S21 + 1)}

3.9.3 Coherent Cooperation

In studying the relay channel with and without lookahead we have encountered three different forms of coherent cooperation:
Although instantaneous relaying alone is sometimes optimal (e.g., for the causal Sato relay channel and for a class of causal Gaussian relay channels), utilizing past received symbol at the relay can achieve a higher rate in general.
1. Анализата на грешка поминија чекор по чекор со ставање на 2nR како што прават во Cover EIT и во Capacity Theorem.
2. Докажи го10↑ т.е. 12↑. Изведувањето во 3.6.3↑ мислам дека треба да помогне.
3. Докажи го 20↑. Иницијално не успеав да го докажам.
4. Главна работа што мора да ја сконтам е да го решам Sato каналот за кејсот со Markov Block Coding.
5. Докажи ја дискусијата за deterministic relay channel во 3.7.3↑
6. Не можев да го добијам 26↑
7. Во A. E. Gamal, Relay Networks with Delays го решава Sato каналот. Треба да се помине.
8. Обиди се да ја докажеш Converse на 3.9.1↑. Имам еден обид ама добив дека првиот член е I(X1;Y2) + I(X1;Y3|Y2) + I(X1;Y3|X2)
9. Докажи го 28↑.
10. Реши го 3.9.2↑
11. Докажи ги изразите за Amplify and forward 30↑-32↑
12. Помини ја главата за Kolmogorov Complexity во EIT.
13. Види ја пресметката на капацитет и брзина на дисторзија во Rate Distortion од EIT. Тоа мислам дека е клучот за пресметка на капацитетот во NIT што не можам да го откријам.
14. Прочитај ги решенијата на задачите во глава 15 EIT што не ги имам решено
15. Доказот за изразот за Amplify-And-Forward го има во докторската од Laneman.
16. Симулирај го капацитетот на гаусовиот канал
17. Дочитај го чланакот од Кафеџиски објавен на Телфор 2010
18. Докажи го уште еднаш вториот член од cutset upper bound-от за Gaussian RC 10↑. Откако го поминав делот од докторската на Khojastepour се чуствувам посигурен во естимацијат на горните граници на условните ентропии.
19. Ми текна сабајле дека за compress and forward ако земам N = 0 дека треба да го добијам Gaussian RC cutset bound т.е. decode and forward
20. Прочитај го чланакот: A. D. Wyner, On SourceCoding with Side Information at the Decoder [3]
21. Дочитај го чланакот: A. H. Madsen, Capacity bounds and power allocation of wireless relay channels

References

[1] A. E. Gamal, Y-H. Kim, Network Information Theory, Cambridge University Press, 2011.

[2] M. A. Khojastepour, Lower bounds on the capacity of gaussian relay channel, Conference on Information Sciences and Systems, 2004

[3] A. D. Wyner, “On source coding with side information at the decoder,” IEEE Trans. Inform. Theory, vol. IT-21, pp. 294-300, May 1975.