Capacity of AWGN Channels

Chapter 3 Capacity of AWGN channels In this chapter we prove that the capacity of an AWGN channel with bandwidth W and signal-tonoise ratio SNR is W log2 (1+SNR) bits per second (b/s). The proof that reliable transmission is possible at any rate less than capacity is based on Shannon’s random code ensemble, typical-set decoding, the Chernoff-bound law of large numbers, and a fundamental result of large-deviation theory. We also sketch a geometric proof of the converse. Readers who are prepared to accept the channel capacity formula without proof may skip this chapter. 3.1 Outline of proof of the capacity theorem The first step in proving the channel capacity theorem or its converse is to use the results of Chapter 2 to replace a continuous-time AWGN channel model Y (t) = X(t) + N (t) with bandwidth W and signal-to-noise ratio SNR by an equivalent discrete-time channel model Y = X + N with a symbol rate of 2W real symbol/s and the same SNR, without loss of generality or optimality. We then wish to prove that arbitrarily reliable transmission can be achieved on the discretetime channel at any rate (nominal spectral efficiency) ρ < C[b/2D] = log2 (1 + SNR) b/2D. This will prove that reliable transmission can be achieved on the continuous-time channel at any data rate R < C[b/s] = W C[b/2D] = W log2 (1 + SNR) b/s. We will prove this result by use of Shannon’s random code ensemble and a suboptimal decoding technique called typical-set decoding. Shannon’s random code ensemble may be defined as follows. Let Sx = P/2W be the allowable average signal energy per symbol (dimension), let ρ be the data rate in b/2D, and let N be the code block length in symbols. A block code C of length N , rate ρ, and average energy Sx per dimension is then a set of M = 2ρN/2 real sequences (codewords) c of length N such that the expected value of ||c||2 under an equiprobable distribution over C is N Sx . For example, the three 16-QAM signal sets shown in Figure 3 of Chapter 1 may be regarded as three block codes of length 2 and rate 4 b/2D with average energies per dimension of Sx = 5, 6.75 and 4.375, respectively. In Shannon’s random code ensemble, every symbol ck of every codeword c ∈ C is chosen independently at random from a Gaussian ensemble with mean 0 and variance Sx . Thus the average energy per dimension over the ensemble of codes is Sx , and by the law of large numbers the average energy per dimension of any particular code in the ensemble is highly likely to be close to Sx . We consider the probability of error under the following scenario. A code C is selected randomly from the ensemble as above, and then a particular codeword c0 is selected for transmission. The channel adds a noise sequence n from a Gaussian ensemble with mean 0 and variance Sn = N0 /2 per symbol. At the receiver, given y = c0 + n and the code C, a typical-set decoder implements the following decision rule (where ε is some small positive number): • If there is one and only one codeword c ∈ C within squared distance N (Sn ± ε) of the received sequence y, then decide on c; • Otherwise, give up. A decision error can occur only if one of the following two events occurs: • The squared distance ||y − c0 ||2 between y and the transmitted codeword c0 is not in the range N (Sn ± ε); • The squared distance ||y − ci ||2 between y and some other codeword ci = c0 is in the range N (Sn ± ε). Since y − c0 = n, the probability of the first of these events is the probability that ||n||2 is not in the range N (Sn − ε) ≤ ||n||2 ≤ N (Sn + ε). Since N k } is an iid zero-mean Gaussian � = {N 2 2 sequence with variance Sn per symbol and ||N|| = k Nk , this probability goes to zero as N → ∞ for any ε > 0 by the weak law of large numbers. In fact, by the Chernoff bound of the next section, this probability goes to zero exponentially with N . For any particular other codeword ci ∈ C, the probability of the second event is the probability that a code sequence drawn according to an iid Gaussian pdf pX (x) with symbol variance Sx and a received sequence drawn independently according to an iid Gaussian pdf pY (y) with symbol variance Sy = Sx + Sn are “typical” of the joint pdf pXY (x, y) = pX (x)pN (y − x), where here we define “typical” by the distance ||x − y||2 being in the range N (Sn ± ε). According to a fundamental result of large-deviation theory, this probability goes to zero as e−N E , where, up to terms of the order of ε, the exponent E is given by the relative entropy (Kullback-Leibler divergence) � pXY (x, y) . D(pXY ||pX pY ) = dx dy pXY (x, y) log pX (x)pY (y) If the logarithm is binary, then this is the mutual information I(X; Y ) between the random variables X and Y in bits per dimension (b/D). In the Gaussian case considered here, the mutual information is easily evaluated as � � Sy 1 1 (y − x)2 log2 e 1 y 2 log2 e = log2 + log2 2πSy + I(X; Y ) = EXY − log2 2πSn − 2 2Sy Sn 2Sn 2 2 b/D. Since Sy = Sx + Sn and SNR = Sx /Sn , this expression is equal to the claimed capacity in b/D. Thus we can say that the probability that any incorrect codeword ci ∈ C is “typical” with respect to y goes to zero as 2−N (I(X;Y )−δ(ε)) , where δ(ε) goes to zero as ε → 0. By the union bound, the probability that any of the M − 1 < 2ρN/2 incorrect codewords is “typical” with respect to y is upperbounded by Pr{any incorrect codeword “typical”} < 2ρN/2 2−N (I(X;Y )−δ(ε)) , which goes to zero exponentially with N provided that ρ < 2I(X; Y ) b/2D and ε is small enough. In summary, the probabilities of both types of error go to zero exponentially with N provided that ρ < 2I(X; Y ) = log2 (1 + SNR) = C[b/2D] b/2D and ε is small enough. This proves that an arbitrarily small probability of error can be achieved using Shannon’s random code ensemble and typical-set decoding. To show that there is a particular code of rate ρ < C[b/2D] that achieves an arbitrarily small error probability, we need merely observe that the probability of error over the random code ensemble is the average probability of error over all codes in the ensemble, so there must be at least one code in the ensemble that achieves this performance. More pointedly, if the average error probability is Pr(E), then no more than a fraction of 1/K of the codes can achieve error probability worse than K Pr(E) for any constant K > 0; e.g., at least 99% of the codes achieve performance no worse than 100 Pr(E). So we can conclude that almost all codes in the random code ensemble achieve very small error probabilities. Briefly, “almost all codes are good” (when decoded by typical-set or maximum-likelihood decoding). 3.2 Laws of large numbers The channel capacity theorem is essentially an application of various laws of large numbers. 3.2.1 The Chernoff bound The weak law of large numbers states that the probability that the sample average of a sequence of N iid random variables differs from the mean by more than ε > 0 goes to zero as N → ∞, no matter how small ε is. The Chernoff bound shows that this probability goes to zero exponentially with N , for arbitrarily small ε. Theorem 3.1 (Chernoff bound) Let SN be the sum of N iid real random variables Xk , each with the same probability distribution pX (x) and mean X = EX [X]. For τ > X, the probability that SN ≥ N τ is upperbounded by Pr{SN ≥ N τ } ≤ e−N Ec (τ ) , where the Chernoff exponent Ec (τ ) is given by Ec (τ ) = max sτ − µ(s), s≥0 where µ(s) denotes the semi-invariant moment-generating function, µ(s) = log EX [esX ]. Proof. The indicator function Φ(SN ≥ N τ ) of the event {SN ≥ N τ } is bounded by Φ(SN ≥ N τ ) ≤ es(SN −N τ ) for any s ≥ 0. Therefore Pr{SN ≥ N τ } = Φ(SN ≥ N τ ) ≤ es(SN −N τ ) , s ≥ 0, � = k Xk and that the Xk are where the overbar denotes expectation. Using the facts that SN independent, we have � es(Xk −τ ) = e−N (sτ −µ(s)) , es(SN −N τ ) = k where µ(s) = log esX . Optimizing the exponent over s ≥ 0, we obtain the Chernoff exponent Ec (τ ) = max sτ − µ(s). s≥0 We next show that the Chernoff exponent is positive: Theorem 3.2 (Positivity of Chernoff exponent) The Chernoff exponent Ec (τ ) is positive when τ > X, provided that the random variable X is nondeterministic. Proof. Define X(s) as a random variable with the same alphabet as X, but with the tilted probability density function q(x, s) = p(x)esx−µ(s) . This is a valid pdf because q(x, s) ≥ 0 and � � q(x, s) dx = e−µ(s) esx p(x) dx = e−µ(s) eµ(s) = 1. Evidently µ(0) = log EX [1] = 0, so q(x, 0) = p(x) and X(0) = X. Define the moment-generating (partition) function µ(s) Z(s) = e sX = EX [e � ]= esx p(x) dx. Now it is easy to see that Z  (s) = � xesx p(x) dx = eµ(s) Similarly, Z  (s) = � � xesx q(x, s) dx = Z(s)X(s). x2 esx p(x) dx = Z(s)X 2 (s). Consequently, from µ(s) = log Z(s), we have µ (s) = µ (s) = Z  (s) = X(s); Z(s) �  �2 Z (s) Z  (s) 2 − = X 2 (s) − X(s) . Z(s) Z(s) Thus the second derivative µ (s) is the variance of X(s), which must be strictly positive unless X(s) and thus X is deterministic. We conclude that if X is a nondeterministic random variable with mean X, then µ(s) is a strictly convex function of s that equals 0 at s = 0 and whose derivative at s = 0 is X. It follows that the function sτ − µ(s) is a strictly concave function of s that equals 0 at s = 0 and whose derivative at s = 0 is τ − X. Thus if τ > X, then the function sτ − µ(s) has a unique maximum which is strictly positive. Exercise 1. Show that if X is a deterministic random variable— i.e., the probability that X equals its mean X is 1— and τ > X, then Pr{SN ≥ N τ } = 0. The proof of this theorem shows that the general form of the function f (s) = sτ − µ(s) when X is nondeterministic is as shown in Figure 1. The second derivative f  (s) is negative everywhere, so the function f (s) is strictly concave and has a unique maximum Ec (τ ). The slope f  (s) = τ − X(s) therefore decreases continually from its value f  (0) = τ − X > 0 at s = 0. The slope becomes equal to 0 at the value of s for which τ = X(s); in other words, to find the maximum of f (s), keep increasing the “tilt” until the tilted mean X(s) is equal to τ . If we denote this value of s by s∗ (τ ), then we obtain the following parametric equations for the Chernoff exponent: Ec (τ ) = s∗ (τ )τ − µ(s∗ (τ )); τ = X(s∗ (τ )). 6 f (s) Ec (τ ) 0  0   slope    s∗ (τ ) τ −X slope 0 s - Figure 1. General form of function f (s) = sτ − µ(s) when τ > X. We will show below that the Chernoff exponent Ec (τ ) is the correct exponent, in the sense that log Pr{SN ≥ N τ } = Ec (τ ). lim N →∞ N The proof will be based on a fundamental theorem of large-deviation theory We see that finding the Chernoff exponent is an exercise in convex optimization. In convex optimization theory, Ec (τ ) and µ(s) are called conjugate functions. It is easy to show from the properties of µ(s) that Ec (τ ) is a continuous, strictly convex function of τ that equals 0 at τ = X and whose derivative at τ = X is 0. 3.2.2 Chernoff bounds for functions of rvs If g : X → R is any real-valued function defined on the alphabet X of a random variable X, then g(X) is a real random variable. If {Xk } is a sequence of iid random variables Xk with the same distribution as X, then {g(Xk )} is a sequence of iid random variables g(Xk ) with the same thus applies to the sequence {g(Xk )}, and shows that distribution as g(X). The Chernoff bound 1 � the probability that the sample mean N k g(Xk ) exceeds τ goes to zero exponentially with N as N → ∞ whenever τ > g(X). Let us consider any finite set {gj } of such functions gj : X → R. Because the Chernoff bound decreases � exponentially with N , we can conclude that the probability that any of the sample means N1 k gj (Xk ) exceeds its corresponding expectation gj (X) by a given fixed ε > 0 goes to zero exponentially with N as N → ∞. We may define a sequence {Xk } to be ε-typical with respect to a function gj : X → R if � k gj (Xk ) < gj (X) + ε. We can thus conclude that the probability that {Xk } is not ε-typical with respect to any finite set {gj } of functions gj goes to zero exponentially with N as N → ∞. � A simple application of this result is that the probability that the sample mean N1 k gj (Xk ) is not in the range gj (X) ± ε goes to zero exponentially with N as N → ∞ for any ε > 0, � because this probability is the sum of the two probabilities Pr{ k gj (Xk ) ≥ N (gj (X) + ε)} and � Pr{ k −gj (Xk ) ≥ N (−gj (X) + ε)}. 1 N More generally, if the alphabet X is finite, then by considering the indicator functions of each possible value of X we can conclude that the probability that all observed relative frequencies in a sequence are not within ε of the corresponding probabilities goes to zero exponentially with N as N → ∞. Similarly, for any �alphabet X , we can conclude that the probability of any finite number of sample moments N1 k Xkm are not within ε of the corresponding expected moments X m goes to zero exponentially with N as N → ∞. In summary, the Chernoff bound law of large numbers allows us to say that as N → ∞ we will almost surely observe a sample sequence x which is typical in every (finite) way that we might specify. 3.2.3 Asymptotic equipartition principle One consequence of any law of large numbers is the asymptotic equipartition principle (AEP): as N → ∞, the observed sample sequence x of an iid sequence whose elements are chosen according to a random variable X will almost surely be such that pX (x) ≈ 2−N H(X) , where H(X) = EX [− log2 p(x)]. If X is discrete, then pX (x) is its probability mass function (pmf ) and H(X) is its entropy; if X is continuous, then pX (x) is its probability density function (pdf) and H(X) is its differential entropy. The AEP is proved by observing that − log2 pX (x) is a sum of iid random variables − log2 pX (xk ), so the probability that − log2 pX (x) differs from its mean N H(X) by more than ε > 0 goes to zero as N → ∞. The Chernoff bound shows that this probability in fact goes to zero exponentially with N . A consequence of the AEP is that the set Tε of all sequences x that are ε-typical with respect to the function − log2 pX (x) has a total probability that approaches 1 as N → ∞. Since for all sequences x ∈ Tε we have pX (x) ≈ 2−N H(X ) — i.e., the probability distribution pX (x) is approximately uniform over Tε — this implies that the “size” |Tε | of Tε is approximately 2N H(X) . In the discrete case, the “size” |Tε | is the number of sequences in Tε , whereas in the continuous case |Tε | is the volume of Tε . In summary, the AEP implies that as N → ∞ the observed sample sequence x will almost surely lie in an ε-typical set Tε of size ≈ 2N H(X) , and within that set the probability distribution pX (x) will be approximately uniform. 3.2.4 Fundamental theorem of large-deviation theory As another application of the law of large numbers, we prove a fundamental theorem of largedeviation theory. A rough statement of this result is as follows: if an iid sequence X is chosen according to a probability distribution q(x), then the probability that the sequence will be typical of a second probability distribution p(x) is approximately Pr{x typical for p | q} ≈ e−N D(p||q) , where the exponent D(p||q) denotes the relative entropy (Kullback-Leibler divergence) � � � p(x) p(x) dx p(x) log D(p||q) = Ep log . = q(x) q(x) X Again, p(x) and q(x) denote pmfs in the discrete case and pdfs in the continuous case; we use notation that is appropriate for the continuous case. Exercise 2 (Gibbs’ inequality). (a) Prove that for x > 0, log x ≤ x − 1, with equality if and only if x = 1. (b) Prove that for any pdfs p(x) and q(x) over X , D(p||q) ≥ 0, with equality if and only if p(x) = q(x). Given p(x) and q(x), we will now define a sequence x to be ε-typical with regard to log p(x)/q(x) if the log likelihood ratio λ(x) = log p(x)/q(x) is in the range N (D(p||q) ± ε), where D(p||q) = Ep [λ(x)] is the mean of λ(x) = log p(x)/q(x) under p(x). Thus an iid sequence X chosen according to p(x) will almost surely be ε-typical by this definition. The desired result can then be stated as follows: Theorem 3.3 (Fundamental theorem of large-deviation theory) Given two probability distributions p(x) and q(x) on a common alphabet X , for any ε > 0, the probability that an iid random sequence X drawn according to q(x) is ε-typical for p(x), in the sense that log p(x)/q(x) is in the range N (D(p||q) ± ε), is bounded by (1 − δ(N ))e−N (D(p||q)+ε) ≤ Pr{x ε−typical for p | q} ≤ e−N (D(p||q)−ε) , where δ(N ) → 0 as N → ∞. Proof. Define the ε-typical region Tε = {x | N (D(p||q) − ε) ≤ log p(x) ≤ N (D(p||q) + ε)}. q(x) By any law of large numbers, the probability that X will fall in Tε goes to 1 as N → ∞; i.e., � dx p(x) ≤ 1, 1 − δ(N ) ≤ Tε where δ(N ) → 0 as N → ∞. It follows that � � dx p(x)e−N (D(p||q)−ε) ≤ e−N (D(p||q)−ε) ; dx q(x) ≤ �Tε �Tε dx q(x) ≥ dx p(x)e−N (D(p||q)+ε) ≥ (1 − δ(N ))e−N (D(p||q)+ε) . Tε Tε Since we can choose an arbitrarily small ε > 0 and δ(N ) > 0, it follows the exponent D(p||q) is the correct exponent for this probability, in the sense that log Pr{x ε−typical for p | q} = D(p||q). N →∞ N lim Exercise 3 (Generalization of Theorem 3.3). (a) Generalize Theorem 3.3 to the case in which q(x) is a general function over X . State any necessary restrictions on q(x). (b) Using q(x) = 1 in (a), state and prove a form of the Asymptotic Equipartition Principle. As an application of Theorem 3.3, we can now prove: Theorem 3.4 (Correctness of Chernoff exponent) The Chernoff exponent Ec (τ ) is the correct exponent for Pr{SN ≥ N τ }, in the sense that lim N →∞ log Pr{SN ≥ N τ } = Ec (τ ), N � where SN = k xk is the sum of N iid nondeterministic random variables drawn according to some distribution p(x) with mean X < τ , and Ec (τ ) = maxs≥0 sτ − µ(s) where µ(s) = log esX . Proof. Let s∗ be the s that maximizes sτ − µ(s) over s ≥ 0. As we have seen above, for s = s∗ ∗ ∗ the tilted random variable X(s∗ ) with tilted distribution q(x, s∗ ) = p(x)es x−µ(s ) has mean X(s∗ ) = τ , whereas for s = 0 the untilted random variable X(0) with untilted distribution q(x, 0) = p(x) has mean X(0) = X. Let q(0) denote the untilted distribution q(x, 0) = p(x) with mean X(0) = X, and let q(s∗ ) ∗ ∗ denote the optimally tilted distribution q(x, s∗ ) = p(x)es x−µ(s ) with mean X(s∗ ) = τ . Then log q(x, s∗ )/q(x, 0) = s∗ x − µ(s∗ ), so D(q(s∗ )||q(0)) = s∗ τ − µ(s∗ ) = Ec (τ ). Moreover, the event that X is ε-typical with respect to the variable log q(x, s∗ )/q(x, 0) = s∗ x − µ(s∗ ) under q(x, 0) = p(x) is the event that s∗ SN − N µ(s∗ ) is in the range N (s∗ τ − µ(s∗ ) ± ε), since τ is the mean of X under q(x, s∗ ). This event is equivalent to SN being in the range N (τ ± ε/s∗ ). Since ε may be arbitrarily small, it is clear that the correct exponent of the event Pr{SN ≈ N τ } is Ec (τ ). This event evidently dominates the probability Pr{SN ≥ N τ }, which we have already shown to be upperbounded by e−N Ec (τ ) . Exercise 4 (Chernoff bound ⇒ divergence upper bound.) Using the Chernoff bound, prove that for any two distributions p(x) and q(x) over X , Pr{log p(x) ≥ N D(p||q) | q} ≤ e−N (D(p||q)) . q(x) [Hint: show that the s that maximizes sτ − µ(s) is s = 1.] 3.2.5 Proof of the forward part of the capacity theorem We now prove that with Shannon’s random Gaussian code ensemble and with a slightly different definition of typical-set decoding, we can achieve reliable communication at any rate ρ < C[b/2D] = log2 (1 + SNR) b/2D. We recall that under this scenario the joint pdf of the channel input X and output Y is 1 1 2 2 pXY (x, y) = pX (x)pN (y − x) = √ e−x /2Sx √ e−(y−x) /2Sn . 2πSn 2πSx Since Y = X + N , the marginal probability of Y is pY (y) = � 1 2 e−y /2Sy , 2πSy where Sy = Sx + Sn . On the other hand, since incorrect codewords are independent of the correct codeword and of the output, the joint pdf of an incorrect codeword symbol X  and of Y is 1 1  2 2 qXY (x , y) = pX (x )pY (y) = √ e−(x ) /2Sx � e−y /2Sy . 2πSx 2πSy We now redefine typical-set decoding as follows. An output sequence y will be said to be ε-typical for a code sequence x if λ(x, y) = log pXY (x, y) ≥ N (D(pXY ||pX pY ) − ε). pX (x)pY (y) Substituting for the pdfs and recalling that D(pXY ||pX pY ) = equivalent to ||y − x||2 ||y||2 ≤ + 2N ε. Sy Sn 1 2 log Sy /Sn , we find that this is Since ||y||2 /N is almost surely very close to its mean Sy , this amounts to asking that ||y −x||2 /N be very close to its mean Sn under the hypothesis that x and y are drawn according to the joint pdf pXY (x, y). The correct codeword will therefore almost surely meet this test. According to Exercise 4, the probability that any particular incorrect codeword meets the test λ(x, y) = log pXY (x, y) ≥ N D(pXY ||pX pY ) pX (x)pY (y) is upperbounded by e−N D(pXY ||pX pY ) = 2−N I(X;Y ) . If we relax this test by an arbitrarily small number ε > 0, then by the continuity of the Chernoff exponent, the exponent will decrease by an amount δ(ε) which can be made arbitrarily small. Therefore we can assert that the probability that a random output sequence Y will be ε-typical for a random incorrect sequence X is upperbounded by Pr{Y ε-typical for X} ≤ 2−N (I(X;Y )−δ(ε)) , where δ(ε) → 0 as ε → 0. Now if the random codes have rate ρ < 2I(X; Y ) b/2D, then there are M = 2ρN/2 codewords, so by the union bound the total probability of any incorrect codeword being ε-typical is upperbounded by Pr{Y ε-typical for any incorrect X} ≤ (M − 1)2−N (I(X;Y )−δ(ε)) < 2−N (I(X;Y )−ρ/2−δ(ε)) . If ρ < 2I(X; Y ) and ε is small enough, then the exponent will be positive and this probability will go to zero as N → ∞. Thus we have proved the forward part of the capacity theorem: the probability of any kind of error with Shannon’s random code ensemble and this variant of typical-set decoding goes to zero as N → ∞, in fact exponentially with N . 3.3 Geometric interpretation and converse For AWGN channels, the channel capacity theorem has a nice geometric interpretation in terms of the geometry of spheres in real Euclidean N -space RN . By any law of large numbers, the probability that the squared Euclidean norm ||X||2 of a random sequence X of iid Gaussian variables of mean zero and variance Sx per symbol falls in the range N (Sx ± ε) goes to 1 as N → ∞, for any ε > 0. Geometrically, the typical region Tε = {x ∈ RN | N (Sx − ε) ≤ ||x||2 ≤ N (Sx + ε)} is a spherical shell with outer squared radius N (Sx + ε) and inner squared radius N (Sx − ε). Thus the random N -vector X will almost surely lie in the spherical shell Tε as N → ∞. This phenomenon is known as “sphere hardening.” Moreover, the pdf pX (x) within the spherical shell Tε is approximately uniform, as we expect from the asymptotic equipartition principle (AEP). Since pX (x) = (2πSx )−N/2 exp −||x||2 /2Sx , within Tε we have (2πeSx )−N/2 e−(N/2)(ε/Sx ) ≤ pX (x) ≤ (2πeSx )−N/2 e(N/2)(ε/Sx ) . Moreover, the fact that pX (x) ≈ (2πeSx )−N/2 implies that the volume of Tε is approximately |Tε | ≈ (2πeSx )N/2 . More precisely, we have � pX (x) dx ≤ 1, 1 − δ(N ) ≤ where δ(N ) → 0 as N → ∞. Since |Tε | = Tε � Tε dx, we have 1 ≥ (2πeSx )−N/2 e−(N/2)(ε/Sx ) |Tε | ⇒ |Tε | ≤ (2πeSx )N/2 e(N/2)(ε/Sx ) ; 1 − δ(N ) ≤ (2πeSx )−N/2 e(N/2)(ε/Sx ) |Tε | ⇒ |Tε | ≥ (1 − δ(N ))(2πeSx )N/2 e−(N/2)(ε/Sx ) . Since these bounds hold for any ε > 0, this implies that lim N →∞ 1 log |Tε | = log 2πeSx = H(X), 2 N where H(X) = 12 log 2πeSx denotes the differential entropy of a Gaussian random variable with mean zero and variance Sx . We should note at this point that practically all of the volume of an N -sphere of squared radius N (Sx + ε) lies within the spherical shell |Tε | as N → ∞, for any ε > 0. By dimensional analysis, the volume of an N -sphere of radius r must be given by AN rN for some constant AN that does not depend on r. Thus the ratio of the volume of an N -sphere of squared radius N (Sx − ε) to that of an N -sphere of squared radius N (Sx + ε) must satisfy � � Sx − ε N/2 AN (N (Sx − ε))N/2 = → 0 as N → ∞, for any ε > 0. Sx + ε AN (N (Sx + ε))N/2 It follows that the volume of an N -sphere of squared radius N Sx is also approximated by eN H(X) = (2πeSx )N/2 as N → ∞. Exercise 5. In Exercise 4 of Chapter 1, the volume of an N -sphere of radius r was given as V⊗ (N, r) = (πr2 )N/2 , (N/2)! for N even. In other words, AN = π N/2 /((N/2)!). Using Stirling’s approximation, m! → (m/e)m as m → ∞, show that this exact expression leads to the same asymptotic approximation for V⊗ (N, r) as was obtained above by use of the asymptotic equipartition principle. The sphere-hardening phenomenon may seem somewhat bizarre, but even more unexpected phenomena occur when we code for the AWGN channel using Shannon’s random code ensemble. In this case, each randomly chosen transmitted N -vector X will almost surely lie in a spherical shell TX of squared radius ≈ N Sx , and the random received N -vector Y will almost surely lie in a spherical shell TY of squared radius ≈ N Sy , where Sy = Sx + Sn . Moreover, given the correct transmitted codeword c0 , the random received vector Y will almost surely lie in a spherical shell Tε (c0 ) of squared radius ≈ N Sn centered on c0 . A further consequence of the AEP is that almost all of the volume of this nonzero-mean shell, whose center c0 has squared Euclidean norm ||c0 ||2 ≈ N Sx , lies in the zero-mean shell TY whose squared radius is ≈ N Sy , since the expected squared Euclidean norm of Y = c0 + N is EN [||Y||2 ] = ||c0 ||2 + N Sn ≈ N Sy . “Curiouser and curiouser,” said Alice. We thus obtain the following geometrical picture. We choose M = 2ρN/2 code vectors at random according to a zero-mean Gaussian distribution with variance Sx , which almost surely puts them within the shell TX of squared radius ≈ N Sx . Considering the probable effects of a random noise sequence N distributed according to a zero-mean Gaussian distribution with variance Sn , we can define for each code vector ci a typical region Tε (ci ) of volume |Tε (ci )| ≈ (2πeSn )N/2 , which falls almost entirely within the shell TY of volume |TY | ≈ (2πeSy )N/2 . Now if a particular code vector c0 is sent, then the probability that the received vector y will fall in the typical region Tε (c0 ) is nearly 1. On the other hand, the probability that y will fall in the typical region Tε (ci ) of some other independently-chosen code vector ci is approximately equal to the ratio |Tε (ci )|/|TY | of the volume of Tε (ci ) to that of the entire shell, since if y is generated according to py (y) independently of ci , then it will be approximately uniformly distributed over TY . Thus this probability is approximately � �N/2 (2πeSn )N/2 Sn |Tε (ci )| = ≈ . Pr{Y typical for ci } ≈ N/2 Sy |TY | (2πeSy ) As we have seen in earlier sections, this argument may be made precise. It follows then that if ρ < log2 (1 + Sx /Sn ) b/2D, or equivalently M = 2ρN/2 < (Sy /Sn )N/2 , then the probability that Y is typical with respect to any of the M − 1 incorrect codewords is very small, which proves the forward part of the channel capacity theorem. On the other hand, it is clear from this geometric argument that if ρ > log2 (1 + Sx /Sn ) b/2D, or equivalently M = 2ρN/2 > (Sy /Sn )N/2 , then the probability of decoding error must be large. For the error probability to be small, the decision region for each code vector ci must include almost all of its typical region Tε (ci ). If the volume of the M = 2ρN/2 typical regions exceeds the volume of TY , then this is impossible. Thus in order to have small error probability we must have Sy Sx = log2 (1 + ) b/2D. 2ρN/2 (2πeSn )N/2 ≤ (2πeSy )N/2 ⇒ ρ ≤ log2 Sn Sn This argument may also be made precise, and is the converse to the channel capacity theorem. In conclusion, we obtain the following picture of a capacity-achieving code. Let TY be the N -shell of squared radius ≈ N Sy , which is almost the same thing as the N -sphere of squared radius N Sy . A capacity-achieving code consists of the centers ci of M typical regions Tε (ci ), where ||ci ||2 ≈ N Sx and each region Tε (ci ) consists of an N -shell of squared radius ≈ N Sn centered on ci , which is almost the same thing as an N -sphere of squared radius N Sx . As ρ → C[b/2D] = log2 (1 + SSnx ) b/2D, these regions Tε (ci ) form an almost disjoint partition of TY . This picture is illustrated in Figure 2. '$ nn nn &% Figure 2. Packing ≈ (Sy /Sn )N/2 typical regions Tε (ci ) of squared radius ≈ N Sn into a large typical region TY of squared radius ≈ N Sy . 3.3.1 Discussion It is natural in view of the above picture to frame the problem of coding for the AWGN channel as a sphere-packing problem. In other words, we might expect that a capacity-achieving code basically induces a disjoint partition of an N -sphere of squared radius N Sy into about (Sy /Sn )N/2 disjoint decision regions, such that each decision region includes the sphere of squared radius N Sn about its center. However, it can be shown by geometric arguments that such a disjoint partition is impossible as the code rate approaches capacity. What then is wrong with the sphere-packing approach? The subtle distinction that makes all the difference is that Shannon’s probabilistic approach does not require decision regions to be disjoint, but merely probabilistically almost disjoint. So the solution to Shannon’s coding problem involves what might be called “soft sphere-packing.” We will see that hard sphere-packing— i.e., maximizing the minimum distance between code vectors subject to a constraint on average energy— is a reasonable approach for moderate-size codes at rates not too near to capacity. However, to obtain reliable transmission at rates near capacity, we will need to consider probabilistic codes and decoding algorithms that follow more closely the spirit of Shannon’s original work.