《信息论与编码理论》(王育民 李晖 梁传甲)课后习题答案 高等教育出版社

H(X/Y)H(Y/Z)H(X/Y)H(Y/Z)???H(XY)H(YZ)H(Y)?H(X/Y)H(Y)?H(Z/Y)H(X/Y)H(Y/Z)??H(Y)?H(X/Y)?H(Z/Y)H(Y)?H(Z/Y)?H(X/Y)H(X/Y)?H(Y/Z)?H(Y)?H(X/Y)?H(Z/Y)H(X/Y)?H(Y/Z)?H(YZ)?H(X/Y)H(X/Y)?H(Y/Z)?H(X/Y)?H(Y/Z)?H(Z)?H(X/Y)?H(Y/Z)?H(X/Z)?0,H(Z)?0H(X/Y)H(Y/Z)???H(XY)H(YZ)H(X/Y)?H(Y/Z)H(X/Y)?H(Y/Z)?H(Z)H(X/Z)?H(X/Z)?H(Z) H(X/Z)?H(XZ)注：

?a1?a2?0,b?0?a1b?a2b?a1b?a1a2?a2b?a1a2?a1a2 ?a1?ba2?b2.15 解: (a)

d(X,X)?H(X/X)?H(X/X)?0

d(X,Y)?H(X/Y)?H(Y/X)?0(b)

d(X,Y)?H(X/Y)?H(Y/X)?H(Y/X)?H(X/Y)?d(Y,X)

(c)

d(X,Y)?d(Y,Z)?H(X/Y)?H(Y/X)?H(Y/Z)?H(Z/Y)H(X/Y)?H(Y/Z)?H(X/YZ)?H(Y/Z)?H(XY/Z)?H(X/Z)同理H(Z/Y)?H(Y/X)?H(Z/X)?d(X,Y)?d(Y,Z)?H(X/Z)?H(Z/X)?d(X,Z)

2.16 解: (a)

I(X,Y)?H(X)?H(Y)?H(XY)?H(X)?H(Y)?H(XY)?H(X/Y)?H(Y/X)?H(XY)?S(X,Y)?I(X,Y)?1H(XY)

又由互信息的非负性，即I(X;Y)?0 有S(X;Y)?0，所以

0?S(X;Y)?1

(b) S(X,X)?I(X,X)H(X)?H(X/X)H(X)???1 H(XX)H(XX)H(X)(c) 当且仅当X和Y独立时，I（X；Y）=0，所以 当且仅当X和Y独立时，S(X,Y)? 2.23 解: (a)

?pX(x)???1I(X,Y)?0。 H(XY)12,?1?x?10,其它

HC(X)???1log1dx?1比特 22?1(b) 令y?x2,dx1? dy2y?1,y?1? pY(y)??2y??0,其它HC(X)???pY(y)logpY(y)dy??12?????102ylog12ydy

?1?log2e??0.443比特(c) 令z?x3,dx1?2?z3 dz3dxdzpZ(z)?pX(x) ?1?23?z,z?1??6??0,其它HC(X3)???pZ(z)logpZ(z)dz??221?21?2333??zlog(6z)dz??zlog(6z3)dz 66?10??01?log26?2log2e??0.3比特2.28 解: (a) 由已知，

,?3?y?1?14 p(yx??1)??其它?0,,?1?y?3?1 p(yx?1)??4其它?0,w(y)??pxy(xy)??px(x)pyx(yx)xx?px(x??1)pyx(yx??1)?px(x?1)pyx(yx?1) ,?3?y??1?18?1?,?1?y?1??14?8,1?y?3?其它?0,（b）

?11183H(Y)??3?log28dy??1log24dy??1log28dy48?11

?2.5bits1314H(YX)??2bits12?3?log24dy?12?1?14log24dy

?I(X;Y)?H(Y)?H(Y/X)?0.5bit

(c)由

?1,y?1?v??0,?1?y?1 ??1,y??1?可求得V的分布为

??101?V???111??

?424??1,y?1?再由p(y/x)及v??0,?1?y?1可求得V的条件分布为

??1,y??1?,(v,x)??(?1,?1),(0,?1),(0,?1),(?1,?1)??1 p(v/x)??20,(v,x)??(?1,?1),(1,?1)???H(V)?2?1log24?1log22?1.5bit42H(V/X)???p(x??1)p(v/x??1)log2p(v/x??1)??p(x?1)p(v/x?1)log2p(v/x?1)vv?1bitI(V;X)?H(V)?H(V/X)?0.5bit可见I(X;Y)?I(X;V),Y?V变换没有信息损失.