第一章-第二章课后作业答案.docx
信息论基础（于秀兰陈前斌王永）课后作业答案为随机变量'概率P（X = X）是X的函数，所以P（X）仍为关于X的随机变量 文中如无特别说明，那么以此类推。J大丁入灯肛机义里,第一章1.6P(xy) =S"R 舞父=四6 0.04尸仙1。2)尸仙2。2)b.120.48JP(y)=仍应)P(Z?2) = 0,48 0.52P(x)y)l = (&%)P(a2161)1 _ r 0,750,25 P(Qi也)尸(Q21b2)J - ho77 0.923J第二章1(B) = TogP(B) = -log- = 3(bit) o注：此处P(B)表示事件B的概率。(2) 设信源为X,H(X) = E-logP(X) = _/og；_ 2 f 04_L = 1.75(bit/symbol)1 75一砌"2.2P(3和5同时出现)=1/181I = Tog记4.17(bit)P(两个2同时出现)=i/361="Iog36 x 5.17(bit)向上点数和为5时(14, 23, 41, 32)有4种,概率为1/9,1I = - log- x 3.17(bit)9(4)设两个点数和为X,那么X23456789101112P(X)1/362/363/364/365/366/365/364/363/362/361/36H(X) = E-logP(X) x 327(bit/symbol) (5)5 511P(两个点数至少有一个1) = 1- 7'7 = T7o o 3o11I = - log x 1.71 (bit)DO(6)相同点数有6种，概率分别为1/36；不同点数出现有15种，概率分别为1/18；H = 6 ' log36 + 15 ' logl8 x 4.34(bit/symbol)36182.9(1)33H(X,Y) = ElogP(X, Y) = W W p(卬x 23(bit/sequence) 1=1 j=l(2) H(Y) = ElogP(Y) x 1.59(bit/symbol)(3) H(X|Y) = H(X, Y) H(Y) = 0.71(bit/symbol) 2.12(1)22 11H(X) = E-logP(X) = -log-log- 0.92(bit/symbol)JJ JJY的分布律为：1/2,1/34/6；H(Y) = ElogP(Y) x 1.46(bit/symbol)(2) H(Y%) = E-logP(Y|X)|X = % = - P(4 1。1乂。"(瓦也)3311=-log 彳一彳 log 7 y 0.81 (b it / symbol)44 44H(Y|a2) = E-logP(Y|X)|X = a2 = - P| a2)logP (bt a2)1111=-ylog-log-= 1 (bit/symbol)0.87 (bit / symbol)(3)0.87 (bit / symbol)Z21P(af)H(Y|af)=-0.81+-lD32.13(1) H(X) = H(0.3,0.7) x Q.S3(bit/symbol)二次扩展信源的数学模型为随机矢量X2 = (X/2),其中X、X2和X同分布，且 相互独立，那么H(X2) = 2H(X) = 1.7 6(bit / sequence)平均符号焙H2(X2) = H(X) x 0.88(bit/symbol)(2) 二次扩展信源的数学模型为随机矢量X2 = (X/2),其中为、X2和X同分布，且 Xi、X2相关，H(X2I1) = E-10gP(X2li) = 一 W Wp(%i，2)logp(%2%)%1 X21 122 21371=lOg3lOg340lOg440l°94'084(bit/symbol)H(X2) = H(XnX2) = H(X2|XJ + H(XJ = 0.84 + 0.88 = 1.72(bit/sequence)H2 (X2) = H(X2)/2 = 0.86(bit / symbol) 2.14(1) 令无记忆信源为X,/I 313H(X) = H , ) = 7 x 2 + x 0.415 工 0.81(bit/symbol)4 4/44r/im/3100-m1(X1。)= -logP(X100 =%ioo) = Tog =2m + (2 20g3)(100 m) = 200 (100 m)log3 (bit)H(X100) = 100H(X) = 81(bit / sequence) 2.15因为信源序列符号间相互独立，且同分布，所以信源为一维离散平稳信源。(1) H(X) = H(0.2,0.8) h QJ2(bit/symbol)H(X2) = 2H (X) = lA4bit / sequence) H(X3|X1X2) = 7/(X3) = H(X) = 0.72(bit/symbol) “8 = "(X) = 0,72(bit / symbol)2.16 (1)H(X2I1) = EHogPlXj = -P(%i，x2)logP(x2 |xx)%269212288T7：loq loq loq loq x 0.55(bit/svmbol) 10 10 3010 30 10 30101 /)EKX3IX2X1) = H(X3|X2) = H(Xz|Xi) h 0.55(bit/symbol)H(X41X3X2) = H(X4|X3) = H(X2%) x 0.55(bit/symbol)Hqo = H(X2Ii) x 0.55(bit/symbol)0.55log20.55log2=45%H(X) = H (L) 0.92 bit/symbol)H8 4H(X),二维离散平稳信源的极限：!：商不大于其单符号信源的燧，说明离散单 符号信源扩展后的单符号平均焰是非增的。2.18(1) _A是状态集；P(%i+i|Si =七)表示i时刻状态为, i + 1时刻输出修+1。 该马尔科夫链的状态转移矩阵为/、1rV2 1/4 i/4iP = P(Ea EaA = P8+1M) = P(%i+il* = %) = 2/301/3，12/3 1/30 _7/12 5/24 5/24-P2 = 5/9 5/18 1/6 ,.5/91/65/18.所以该链为齐次遍历马尔科夫链。令P® = k) = pi(k),那么pi pi Pi =口/2 1/4 1/4,历2 P222(3)=历1 Pi PiP1 - 4 o 1 - 31-22-32-3_ 一1 - 41-41-3 O5424J2因为P1(D P1(2) P1(3) W历2(1) P2(2) P2(3),所以该信源不是离散平稳 信源。(2) 当信源的输出序列足够长，马尔科夫链到达平稳分布时，该信源可以看作离散平 稳信源。(3) H(%+i|Si =%)=一乏尸(勺+1 =。应=Ei)/ogP(%i+i = ast = %) a1 111=-log - - 2 - - log - = l.S(bit/symbol) 2244同理得：H(Xi+i|Sj = E2) x 0.92(比/symbol)H(Xj+i|Sj = E3) x 092(bit/symbol)设极限分布为P(Ei) P(F2) P(E3),那么122Pi)=?P(i)+P(F2)+-P(F3) 乙DD/、11P(E2)=-P(E1)+-P(E3) 44D/、11P(E3)=-P(E1)+-P(F2)P(E1) + P(E2) + P(E3) = 1解得P(Ei) = 4/7, P(E2) = 3/14, P(E3) = 3/14 43“8 = "(Xi+i|s。= -xl.5 + 2x x 0.92 x l2S(bit / symbol) 714(4) Ho = log3 x 1.59(bit/symbol) /I 1 1% = 1.5 (bit/symbol)1.5曷 = 1 -7左右 5.66% L3 V524524247H2 = H(121.42 = 1795%-1. 0 O JzP2(3)P31144.历3(1) P3(2) P3(3) = p2(l) P2(2)4 41° 314 41° 311 117一-21IL4131.72 144413131H2 H( ) x 1A2(bit/symbol)72 144 1441 421 = 1一手-1069% L3 V0.80.200000.50.50.50.500000.20.82.20 (1) 状态转移矩阵p = p(即用)=P(E4)，那么(2)P(E4)，那么由P知此马尔科夫链存在极限分布， 设极限分布为P(Ei) P(F2) P(a) P(Ei) = 0.8P(Ei) + 0.5P(E3) P(F2) = 0,2P(Ei) + 0.5P(E3) P(E3) = 0.5P(E2) + 0.2P(E4) P(F4) = 0.5P(E2) + 0.8P(E4) P(瓦)+ P(%) + P(殳)+ P(K) = 1 解得P(Ei) = 5/14, P(E2) = 1/7, P(E3) = 1/7, P(F4) = 5/14(3) H(Xj+i|Sj = E4) = H(Xj+i|Sj = Ei) = -0.8/og0.8 0.2/og0.2x 0.72(bit/symbol)H(Xj+i|Sj = E3) = H(Xj+i|Sj = F2) = -2 * H520go.5 = l(bit/symbol) 51“8 = H(Xi+i|s。= 2 x x 0.72 + 2 x x 1 = O.Qbit/symbol) 147P(0) = 2 P(°底)P(&) = 08 x 五 + o.5 x ； + 0.5 x J + 0.2 x=0.5 i/714P(l) = 0.5(4) 初始时刻的P(0),P(l)和(4)中不一样，所以初始时刻的信源不是平稳信源。
