Tutorial 4. b a. h(f) = a b a ln 1. b a dx = ln(b a) nats = log(b a) bits. = ln λ + 1 nats. = log e λ bits. = ln 1 2 ln λ + 1. nats. = ln 2e. bits.

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1 Tutoril 4 Exercises on Differentil Entropy. Evlute the differentil entropy h(x) f ln f for the following: () The uniform distribution, f(x) b. (b) The exponentil density, f(x) λe λx, x 0. (c) The Lplce density, f(x) λe λ x. (d) The sum of X nd X, where X nd X re independent norml rndom vribles with men µ i nd vrince σi, i,. () Uniform Distribution (b) Exponentil distribution. h(f) b ln b dx h(f) ln(b ) nts log(b ) bits 0 0 λe λx ln λe λx dx ln λ + nts λe λx [ln λ λxdx (c) Lplce density. h(f) log e λ bits λe λ x ln λe λ x dx λe λ x [ln + ln λ λ x ]dx ln ln λ + ln e λ log e λ (d) The sum of two norml distributions. nts bits The sum of two norml rndom vribles is lso norml, so pplying the result derived the clss for the norml distribution, since X + X (µ + µ, σ + σ ), h(f) log πe(σ + σ ) bits. Remrk: If X (µ, σ ), then h(x) log(πeσ ).

2 . Consider X is continuous rndom vrible defined over intervl [, b]. () Wht is the mximum vlue of h(x)? (b) Wht is the corresponding distribution of X? Let u(x) b be the uniform probbility density function over [, b], nd nd let p(x) be the probbility mss function for X. Then Since D(p u) 0, D(p u) h(x) p(x) log p(x) u(x) dx p(x) log p(x)dx p(x) log log(b ) h(x) h(x) log(b ) p(x) log u(x)dx b dx where the equlity holds when p(x) is the uniform probbility density function over [, b]. 3. Consider dditive chnnel whose input lphbet X {0, ±, ±}, nd whose output Y X + Z, where Z is uniformly distributed over the intervl [, ]. Thus the input of the chnnel is discrete rndom vrible, while the output is continuous. Clculte the cpcity C mx p(x) I(X; Y ) of the chnnel. We cn expnd the mutul informtion nd h(z) log, since Z U(, ). I(X; Y ) h(y ) h(y X) h(y ) h(z) The output Y is sum of discrete nd continuous rndom vrible, nd if the probbility of X re p, p,, p, then the output distribution of Y hs uniform distribution with weight p for 3 Y when X, uniform with weight p for Y 0 when X, uniform with weight p 0 for Y when X 0, uniform with weight p for 0 Y when X, nd uniform with weight p for Y 3 when X. Thus we hve the density function of Y s follows p Y (y) p y [ 3, ) p +p y [, ) p +p 0 y [, 0) p 0 +p y [0, ) p +p y [, ) p y [, 3)

3 Given tht Y rnges from [ 3, 3], the mximum entropy tht it cn hve is n uniform over this rnge. This cn be chieved if the distribution of X is (/3, 0, /3, 0, /3). Then h(y ) log 6 nd the cpcity of this chnnel is C log 6 log log 3 bits. 4. Suppose tht (X; Y ; Z) re jointly Gussin nd tht X Y Z forms Mrkov chin. Let X nd Y hve correltion coefficient ρ xy nd let Y nd Z hve correltion coefficient ρ yz. Find I(X;Z). ote tht for constnt, h( + X) h(x). Thus, without loss of generlity, we ssume tht the mens of X, Y nd Z re zero. Let Λ ( σ x ρ xz ρ xz be the covrince mtrix of X nd Z where ρ xz is the correltion coefficient between X nd Z. Then we hve σ z ) I(X; Z) h(x) + h(z) h(x, Z) Since (X, Y, Z) re jointly Gussin, X nd Z re individully mrginlly Gussin, nd (X, Z) is jointly Gussin. Thus, we hve ow, We cn conclude tht I(X; Z) h(x) + h(z) h(x, Z) log(πeσ x) + log(πeσ z) log(πe Λ ) log( ρ xz) ρ xz E[XZ] E[E[XZ Y ]] E[E[X Y ]E[Z Y ]] σxρxy E[ σ y Y ]E[ σzρyz σ y Y ] E[ σxσzρxyρyz Y ] σy σ xσ zρ xyρ yz σ y σ xσ zρ xyρ yz σ y ρ xy ρ yz E[Y ] Vr(Y ) I(X; Y ) log( ρ xyρ yz) 3

4 Remrk: If (X, Y ) is jointly Gussin, the conditionl distribution of X given Y y is s follows. ( X Y y µ x + σ ) x ρ xy (y µ y ), ( ρ σ xy)σx y Exercises on Gussin Chnnel. Let Y nd Y be conditionlly independent nd conditionlly identiclly distributed given X. () Show I(X; Y, Y ) I(X; Y ) I(Y ; Y ). (b) Conclude tht the cpcity of the chnnel X (Y, Y ) is less thn twice the cpcity of the () chnnel X Y. I(X; Y, Y ) H(Y, Y ) H(Y, Y X) H(Y ) + H(Y ) I(Y ; Y ) (H(Y X) + H(Y X, Y )) H(Y ) + H(Y ) I(Y ; Y ) H(Y X) H(Y X) H(Y ) H(Y X) + H(Y ) H(Y X) I(Y ; Y ) I(X; Y ) + I(X; Y ) I(Y ; Y ) I(X; Y ) I(Y ; Y ) (b) The cpcity of the single look chnnel X Y is The cpcity of the chnnel X (Y, Y ) is C mx p(x) I(X; Y ) C mx p(x) I(X; Y, Y ) mx p(x) I(X; Y ) I(Y ; Y ) mx p(x) I(X; Y ) C Hence, the two independent looks cnnot be more thn twice s good s one look.. Consider the ordinry Gussin chnnel with two correlted looks t X, i.e., Y (Y, Y ), where Y X + Z Y X + Z 4

5 with power constrint P on X, nd (Z, Z ) (0, K), where [ ] ρ K. ρ Find the cpcity C for () ρ (b) ρ 0 (c) ρ It is cler tht the input distribution tht mximizes the cpcity is X (0, P ). Evluting the mutul informtion for the distribution, C mx I(X; Y, Y ) h(y, Y ) h(y, Y X) h(y, Y ) h(z, Z X) h(y, Y ) h(z, Z ) ow since ( [ (Z, Z ) 0, ρ ρ ]), we hve h(z, Z ) log(πe) K log(πe) ( ρ ). Since Y X + Z nd Y X + Z, we hve ( [ (Y, Y ) 0, P + P + ρ P + ρ P + ]), nd Hence the cpcity is h(y, Y ) log(πe) K log(πe) ( ( ρ ) + P ( ρ)). C h(y, Y ) h(z, Z ) ) ( log P +. ( + ρ) () ρ. In this cse, C log( + P ), which is the cpcity of single look chnnel. This is not surprising, since in this cse Y Y. (b) ρ 0. In this cse, C ( log + P ), which corresponds to using twice the power in single look. cpcity of the chnnel X (Y + Y ). The cpcity is the sme s the 5

6 (c) ρ 0. In this cse, C, which is not surprising since if we dd Y nd Y, we cn recover X Remrk: exctly. X (Y + Y ). The cpcity of the bove chnnel in ll cses is the sme s the cpcity f the chnnel 3. Output power constrint. Consider n dditive white Gussin noise chnnel with n expected output power constrint P. Thus Y X + Z, Z (0, ), Z is independent of X, nd E[Y ] P. Find the chnnel cpcity. C mx I(X; Y ) p(x):e[(x+z) ] P mx h(y ) h(y X) p(x):e[(x+z) ] P mx h(y ) h(z X) p(x):e[(x+z) ] P mx h(y ) h(z) p(x):e[(x+z) ] P Given constrint on the output power of Y, the mximum differentil entropy is chieved by norml distribution, nd we cn chieve this by hve X (0, P ), nd in this cse, C log πep log πe log P. 4. Fding Chnnel. Consider n dditive fding chnnel Y XV + Z, where Z is dditive noise, V is rndom vrible representing fding, nd Z nd V re independent of ech other nd of X. Argue tht knowledge of the fding fctor V improves cpcity by showing I(X; Y V ) I(X; Y ). Expnding I(X; Y, V ) in two wys, we get I(X; Y, V ) I(X; V ) + I(X; Y V ) I(X; Y ) + I(X; V Y ) i.e. I(X; V ) + I(X; Y V ) I(X; Y ) + I(X; V Y ) I(X; Y V ) I(X; Y ) + I(X; V Y ) I(X; Y V ) I(X; Y ) 6

7 5. Consider the dditive whiter Gussin chnnel Y i X i + Z i where Z i (0, ), nd the input signl hs verge power constrint P. () Suppose we use ll power t time, i.e. E[X ] np nd E[X i ] 0 for i, 3,, n. Find I(X n ; Y n ) mx p(x n ) n where the mximiztion is over ll distributions p(x n ) subject to the constrint E[X ] np nd E[Xi ] 0 for i, 3,, n. (b) Find () nd compre to prt (). E[ n I(X n ; Y n ) mx n i X i ] P n (b) I(X n ; Y n ) I(X ; Y ) mx mx p(x n ) n p(x n ) n log ( + np ) n where the first equlity comes from the constrint tht ll our power, np, be used t time, nd the second equlity comes from tht fct tht given Gussin noise nd power constrint np, I(X; Y ) log( + np ). I(X n ; Y n ) ni(x; Y ) mx mx p(x n ) n p(x n ) n mx I(X; Y ) p(x n ) ( log + P ) where the first equlity comes from the fct tht the chnnel is memoryless. otice tht the quntity in prt () goes to 0 s n while the quntity in prt (b) stys constnt. Remrk: The impulse scheme is suboptiml. 7

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