金庸全集.doc

文档介绍：解释为什么 H(X) >log6 不满足信源熵的极值性。解:585 .26 log )( /657 .2 )17 .0 log 17 .016 .0 log 16 .017 .0 log 17 .018 .0 log 18 .019 .0 log 19 . log ( )( log )()( 2 222222 62????????????? XH symbol bit xpxpXH i ii 不满足极值性的原因是 107 .1)( 6??? i ixp 。 证明: H(X 3/X 1X 2)≤H(X 3/X 1),并说明当 X 1,X 2,X 3是马氏链时等式成立。证明: 0 log 1)/()( log )()/()( log 1)/( )/()( )/( )/( log )( )/( log )()/( log )( )/( log )()/( log )( )/()/(2 123 1321 2 123 321 123 1321 123 2213 13321 123213 13321 123 13321 123 213321 13 1331 123 213321 13213????????????????????????????????????????????????????????????????????? exxpxxp exxxpxxpxxp exxxp xxpxxxp xxxp xxpxxxp xxpxxxpxxxpxxxp xxpxxpxxxpxxxp XXHXXXH iii iiii iii iii iii iiii iiiiii iiiii iiiiii iiiii iii iiiii iii iiiiii ii iiii iii iiiiii氏链是马等式成立的条件是时等式成立当_,, )/()/()/( )()/()/()( )()/()/()( )/()/( 01)/( )/( )/()/( 321 1321312 32113121 212131321 21313 213 13 13213XXX xxxpxxpxxp xxxpxxpxxpxp xxpxxxpxxpxxp xxxpxxp xxxp xxp XXHXXXH iiiiiii iiiiiiii iiiiiiiii iiiii iii ii????????????? 证明: H(X 1X 2。。。X n)≤H(X 1)+H(X 2)+…+H(X n)。证明:)(...)()()()...( ).../()(0)...;( ... )/()(0);( )/()(0);( ).../(...)/()/()()...( 32121 121 121 2133213 12212 **********N N NNN NN NN NXHXHXHXHXXXH XXXXHXHXXXXI XXXHXHXXXI XXHXHXXI XXXXHXXXHXXHXHXXXH??????????????????????? 设有一个信源,它产生 0,1 序列的信息。它在任意时间而且不论以前发生过什么符号,均按 P(0) = ,P(1) = 的概率发出符号。(1) 试问这个信源是否是平稳的? (2) 试计算 H(X 2),H(X 3/X 1X 2)及H ∞; (3) 试计算 H(X 4)并写出 X 4信源中可能有的所有符号。解: (1) 这个信源是平稳无记忆信源。因为有这些词语:“它在任意时间而且不论以前发生过什么符号……”(2) symbol bit XHH symbol bit xpxpXHXXXH symbol bit XHXH i ii/ 971 .0)( / 971 .0) log log ()( log )()()/( / 942 .1) log log (2)(2)( 2223213 22 2?????????????????(3) 1011 1111 1110 1101 1100 1010 1001 1000 0111 0110 0101 0100 0011 0010 0001 0000 的所有符号: /884 .3) log log (4)(4)( 4 22 4X symbol bit XHXH?????? 一阶马尔可夫信源的状态图如下图所示。信源 X的符号集为{0, 1,2}。(1) 求平稳后信源的概率分布; (2) 求信源的熵 H ∞。解: (1)??????????????????????????????????????????????3/1)( 3/1)( 3/1)( 1)()()( )()()( )()()( )()(