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Chapter 5 Differential Entropy and Gaussian Channels Chapter 5 Differential Entropy and Gaussian Channels Po-Ning Chen, Professor Institute of Communications Engineering National Chiao Tung University Hsin Chu, Taiwan 30010, R.O.C. Continuous sources I: 5-1 • Model {Xt ∈X,t∈ I} – Discrete sources ∗ Both X and I are discrete. – Continuous sources ∗ Discrete-time continuous sources ·Xis continuous; I is discrete. ∗ Waveform sources · Both X and I are continuous. • We have so far examined information measures and their operational charac- terization for discrete-time discrete-alphabet systems. In this chapter, we turn our focus to discrete-time continuous-alphabet (real-valued) sources. Information content of continuous sources I: 5-2 • If the random variable takes on values in a continuum, the minimum number of bits per symbol needed to losslessly describe it must be infinite. • This is illustrated in the following example and validated in Lemma 5.2. Example 5.1 – Consider a real-valued random variable X that is uniformly distributed on the unit interval, i.e., with pdf given by 1ifx ∈ [0, 1); fX(x)= 0otherwise. – Given a positive integer m, we can discretize X by uniformly quantizing it into m levels by partitioning the support of X into equal-length segments 1 of size ∆ = m (∆ is called the quantization step-size) such that: i i − 1 i q (X)= , if ≤ X< , m m m m for 1 ≤ i ≤ m. – Then the entropy of the quantized random variable qm(X)isgivenby m 1 1 H(q (X)) = − log2 =log2 m (in bits). m m m i=1 Information content of continuous sources I: 5-3 – Since the entropy H(qm(X)) of the quantized version of X is a lower bound to the entropy of X (as qm(X) is a function of X) and satisfies in the limit lim H(qm(X)) = lim log2 m = ∞, m→∞ m→∞ we obtain that the entropy of X is infinite. 2 • The above example indicates that to compress a continuous source without incurring any loss or distortion requires an infinite number of bits. • Thus when studying continuous sources, the entropy measure is limited in its effectiveness and the introduction of a new measure is necessary. • Such a new measure is obtained upon close examination of the entropy of a uniformly quantized real-valued random-variable minus the quantization accuracy as the accuracy increases without bound. Information content of continuous sources I: 5-4 Lemma 5.2 Consider a real-valued random variable X with support [a, b) and pdf fX such that (i) −f log f is (Riemann-)integrable,and X 2 X − b (ii) a fX(x)log2 fX (x)dx is finite. Then a uniform quantization of X with an n-bit accuracy (i.e., with a quanti- zation step-size of ∆ = 2−n) yields an entropy approximately equal to b − fX(x)log2 fX(x)dx + n bits a for n sufficiently large. In other words, b lim [H(q (X)) − n]=− f (x)log2 f (x)dx →∞ n X X n a where qn(X) is the uniformly quantized version of X with quantization step- size ∆ = 2−n. Information content of continuous sources I: 5-5 Proof: Step 1: Mean-value theorem. Let ∆ = 2−n be the quantization step-size, and let a + i∆,i=0, 1, ··· ,j− 1 ti:= b, i = j where b − a j = . ∆ From the mean-value theorem, we can choose xi ∈ [ti−1,ti]for1≤ i ≤ j such that ti pi := fX(x)dx = fX(xi)(ti − ti−1)=∆· fX(xi). ti−1 Information content of continuous sources I: 5-6 Step 2: Definition of h(n)(X). Let j j (n) −n h (X):= − [fX(xi)log2 fX(xi)]∆ = − [fX(xi)log2 fX(xi)]2 . i=1 i=1 Since −fX(x)log2 fX (x) is (Riemann-)integrable, b (n) h (X) →− fX(x)log2 fX(x)dx as n →∞. a Therefore, given any ε>0, there exists N such that for all n>N, b (n) − fX(x)log2 fX(x)dx − h (X) <ε. a Information content of continuous sources I: 5-7 Step 3: Computation of H(qn(X)). The entropy of the (uniformly) quan- tized version of X, qn(X), is given by j H(qn(X)) = − pi log2 pi i=1 j = − (fX(xi)∆) log2(fX(xi)∆) i=1 j −n −n = − (fX(xi)2 )log2(fX(xi)2 ) i=1 where the pi’s are the probabilities of the different values of qn(X). Information content of continuous sources I: 5-8 (n) Step 4: H(qn(X)) − h (X) . From Steps 2 and 3, j (n) −n −n H(qn(X)) − h (X)=− [fX(xi)2 ]log2(2 ) i=1 j ti = n fX(x)dx t −1 i=1 i b = n fX(x)dx a = n. Hence, we have that for n>N, b (n) − fX(x)log2 fX(x)dx + n − ε<H(qn(X)) = h (X)+n a b < − fX (x)log2 fX(x)dx + n + ε, a yielding that − − b 2 limn→∞ [H(qn(X)) n]= a fX(x)log2 fX(x)dx. Information content of continuous sources I: 5-9 Lemma 5.2 actually holds not limited for support [a, b) but for any support SX . Theorem 5.3 [340, Theorem 1](R´enyi 1959) For any real-valued random vari- able with pdf fX,if j − pi log2 pi i=1 is finite, where the pi’s are the probabilities of the different values of uniformly quantized qn(X)oversupportSX,then lim [H(qn(X)) − n]=− fX(x)log2 fX(x)dx n→∞ SX provided the integral on the right-hand side exists. 1 This suggests that fX (x)log2 dx could be a good information measure for SX fX(x) continuous-alphabet sources. 5.1 Differential entropy I: 5-10 Definition 5.4 (Differential entropy) The differential entropy (in bits) of a continuous random variable X with pdf f and support S is defined as X X h(X):= − fX(x) · log2 fX (x)dx = E[− log2 fX(X)], SX when the integral exists. Example 5.5 A continuous random variable X with support SX =[0, 1) and pdf fX (x)=2x for x ∈ SX has differential entropy equal to 1 2 1 x (log2 e − 2log2(2x)) −2x · log2(2x)dx = 0 2 0 1 = − log2(2) = −0.278652 bits. 2ln2 5.1 Differential entropy I: 5-11 Next, we have that qn(X)isgivenby i i − 1 i q (X)= , if ≤ X< , n 2n 2n 2n for 1 ≤ i ≤ 2n. Hence, i (2i − 1) Pr q (X)= = , n 2n 22n which yields 2n 2i − 1 2i − 1 H(q (X)) = − log2 n 22n 22n i=1 2n 1 n = − (2i − 1) log2(2i − 1) + 2 log2(2 ) . 22n i=1 5.1 Differential entropy I: 5-12 n H(qn(X)) H(qn(X)) − n 1 0.811278 bits −0.188722 bits 2 1.748999 bits −0.251000 bits 3 2.729560 bits −0.270440 bits 4 3.723726 bits −0.276275 bits 5 4.722023 bits −0.277977 bits 6 5.721537 bits −0.278463 bits 7 6.721399 bits −0.278600 bits 8 7.721361 bits −0.278638 bits 9 8.721351 bits −0.278648 bits 5.1 Differential entropy I: 5-13 Example 5.7 (Differential entropy of a uniformly distributed ran- dom variable) Let X be a continuous random variable that is uniformly dis- tributed over the interval (a, b), where b>a; i.e., its pdf is given by 1 ∈ b−a if x (a, b); fX(x)= 0otherwise. So its differential entropy is given by b − 1 1 − h(X)= − log2 − =log2(b a)bits. a b a b a • Note that if (b − a) < 1 in the above example, then h(X)isnegative. 5.1 Differential entropy I: 5-14 Example 5.8 (Differential entropy of a Gaussian random variable) Let X ∼N(µ, σ2); i.e., X is a Gaussian (or normal) random variable with finite mean µ, variance Var(X)=σ2 > 0andpdf 2 (x−µ) 1 − 2 fX(x)=√ e 2σ 2πσ2 for x ∈ R. Then its differential entropy is given by 2 1 2 (x − µ) h(X)= fX (x) log2(2πσ )+ 2 log2 e dx R 2 2σ 1 2 log2 e 2 = log2(2πσ )+ E[(X − µ) ] 2 2σ2 1 2 1 = log2(2πσ )+ log2 e 2 2 1 2 = log2(2πeσ )bits. (5.1.1) 2 • Note that for a Gaussian random variable, its differential entropy is only a function of its variance σ2 (it is functionally independent from its mean µ). • This is similar to the differential entropy of a uniform random variable, which only depends on difference (b − a) but not the mean (a + b)/2. 5.2 Joint & cond. diff. entrop., diverg. & mutual info I: 5-15 n Definition 5.9 (Joint differential entropy) If X =(X1,X2, ··· ,Xn)isa continuous random vector of size n (i.e., a vector of n continuous random variables) n with joint pdf fXn and support SXn ⊆ R , then its joint differential entropy is defined as n h(X ):=− fXn(x1,x2, ··· ,xn)log2 fXn(x1,x2, ··· ,xn) dx1 dx2 ··· dxn SXn n = E[− log2 fXn(X )] when the n-dimensional integral exists. 5.2 Joint & cond. diff. entrop., diverg. & mutual info I: 5-16 Definition 5.10 (Conditional differential entropy) Let X and Y be two jointly distributed continuous random variables with joint pdf fX,Y and support 2 SX,Y ⊆ R such that the conditional pdf of Y given X,givenby fX,Y (x, y) fY |X(y|x)= , fX (x) is well defined for all (x, y) ∈ SX,Y ,wherefX is the marginal pdf of X.Thenthe conditional entropy of Y given X is defined as h(Y |X):= − fX,Y (x, y)log2 fY |X(y|x) dx dy = E[− log2 fY |X(Y |X)], SX,Y when the integral exists.
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