信息论与编码理论第二章习题答案(王育民)
部分答案,仅供参考。
2.1信息速率是指平均每秒传输的信息量
点和划出现的信息量分别为3log ,2
3log ,
一秒钟点和划出现的次数平均为
4
15314.0322.01=
?+?
一秒钟点和划分别出现的次数平均为4
5.410
那么根据两者出现的次数,可以计算一秒钟其信息量平均为2
53log 4
153log 4
52
3log 4
10-=+
2.3 解:
(a)骰子A 和B ,掷出7点有以下6种可能:
A=1,B=6; A=2,B=5; A=3,B=4; A=4,B=3; A=5,B=2; A=6,B=1 概率为6/36=1/6,所以信息量
-log(1/6)=1+log3≈2.58 bit
(b) 骰子A 和B ,掷出12点只有1种可能: A=6,B=6
概率为1/36,所以信息量
-log(1/36)=2+log9≈5.17 bit
2.5解:
出现各点数的概率和信息量:
1点:1/21,log21≈4.39 bit ; 2点:2/21,log21-1≈3.39 bit ; 3点:1/7,log7≈2.81bit ; 4点:4/21,log21-2≈2.39bit ; 5点:5/21,log (21/5)≈2.07bit ; 6点:2/7,log(7/2)≈1.81bit 平均信息量:
(1/21)×4.39+(2/21)×3.39+(1/7)×2.81+(4/21)×2.39+(5/21)×2.07+(2/7)×1.81≈2.4bit
2.7解:
X=1:考生被录取; X=0:考生未被录取; Y=1:考生来自本市;Y=0:考生来自外地; Z=1: 考生学过英语;Z=0:考生未学过英语
P(X=1)=1/4, P(X=0)=3/4; P(Y=1/ X=1)=1/2; P(Y=1/ X=0)=1/10;
P(Z=1/ Y=1)=1, P(Z=1 / X=0, Y=0)=0.4, P(Z=1/ X=1, Y=0)=0.4, P(Z=1/Y=0)=0.4 (a) P(X=0,Y=1)=P(Y=1/X=0)P(X=0)=0.075, P(X=1,Y=1)= P(Y=1/X=1)P(X=1)=0.125
P(Y=1)= P(X=0,Y=1)+ P(X=1,Y=1)=0.2
P(X=0/Y=1)=P(X=0,Y=1)/P(Y=1)=0.375, P(X=1/Y=1)=P(X=1,Y=1)/P(Y=1)=0.625
I (X ;Y=1)=∑∑=====x
x )P()
1Y /(P log )1Y /(P )1Y (I )1Y /(P x x x x;x
=1)
P(X )
1Y /1X (P log
)1Y /1X (P 0)P(X )1Y /0X (P log
)1Y /0X (P =====+=====
=0.375log(0.375/0.75)+0.625log(0.625/0.25)=(5/8)log5-1≈0.45bit
(b) 由于P(Z=1/ Y=1)=1, 所以 P (Y=1,Z=1/X=1)= P (Y=1/X=1)=0.5 P (Y=1,Z=1/X=0)= P (Y=1/X=0)=0.1
那么P (Z=1/X=1)= P (Z=1,Y=1/X=1)+ P (Z=1,Y=0/X=1)=0.5+ P (Z=1/Y=0,X=1)P (Y=0/X=1)=0.5+0.5*0.4=0.7
P(Z=1/X=0)= P (Z=1,Y=1/X=0)+ P (Z=1,Y=0/X=0)=0.1+P(Z=1/Y=0,X=0)P(Y=0/X=0)=0.1+0.9*0.4=0.46
P (Z=1,X=1)= P (Z=1/X=1)*P(X=1)=0.7*0.25=0.175 P (Z=1,X=0)= P (Z=1/X=0)*P(X=0)= 0.46*0.75=0.345 P(Z=1) = P(Z=1,X=1)+ P(Z=1,X=0) = 0.52 P(X=0/Z=1)=0.345/0.52=69/104 P(X=1/Z=1)=35/104
I (X ;Z=1)=∑∑=====x
x )P()
1Z /(P log )1Z /(P )1Z (I )1Z /(P x x x x;x
=1)P(X )1Z /1X (P log )1Z /1X (P 0)P(X )1Z /0X (P log )1Z /0X (P =====+=====
=(69/104)log(23/26)+( 35/104)log(35/26) ≈0.027bit
(c)H (X )=0.25*log(1/0.25)+0.75*log(1/0.75)=2-(3/4)log3=0.811bit H(Y/X)=-P(X=1,Y=1)logP(Y=1/X=1) -P(X=1,Y=0)logP(Y=0/X=1)
-P(X=0,Y=1)logP(Y=1/X=0) -P(X=0,Y=0)logP(Y=0/X=0)
=-0.125*log0.5-0.125*log0.5-0.075*log0.1-0.675*log0.9
=1/4+(3/40)log10-(27/40)log(9/10)≈0.603bit
H(XY)=H(X)+H(Y/X)=9/4+(3/4)log10-(21/10)log3=1.414bit
P(X=0,Y=0,Z=0)= P(Z=0 / X=0, Y=0)* P( X=0, Y=0)=(1-0.4)*(0.75-0.075)=0.405 P(X=0,Y=0,Z=1)= P(Z=1 / X=0, Y=0)* P( X=0, Y=0)=0.4*0.675=0.27
P(X=1,Y=0,Z=1)= P(Z=1/ X=1,Y=0)* P(X=1,Y=0)=0.4*(0.25-0.125)=0.05 P(X=1,Y=0,Z=0)= P(Z=0/ X=1,Y=0)* P(X=1,Y=0)=0.6*0.125=0.075 P(X=1,Y=1,Z=1)=P(X=1,Z=1)- P(X=1,Y=0,Z=1)=0.175-0.05=0.125 P(X=1,Y=1,Z=0)=0 P(X=0,Y=1,Z=0)=0
P(X=0,Y=1,Z=1)= P(X=0,Z=1)- P(X=0,Y=0,Z=1)= 0.345-0.27=0.075
H(XYZ)=-0.405*log0.405-0.27*log0.27-0.05*log0.05-0.075*log0.075-0.125*log0.125-0.075*log 0.075=(113/100)+(31/20)log10-(129/50)log3 =0.528+0.51+0.216+0.28+0.375+0.28=2.189 bit
H(Z/XY)=H(XYZ)-H(XY)= -28/25+(4/5)log10-12/25log3 =0.775bit
2.9 解:
A ,
B ,
C 分别表示三个筛子掷的点数。 X=A, Y=A+B, Z=A+B+C
由于P(A+B+C/ A+B)=P(C/A+B)=P(C)
所以H(Z/Y)=H(A+B+C/ A+B)=H (C )=log6 =2.58bit
一共36种情况,每种情况的概率为1/36,即P(A=a,Y=y)=1/36
H(X/Y)=H(A/Y)=(1/36)[(-1*log1-2*log(1/2)-3*log(1/3)-4*log(1/4)-5*log(1/5) )*2-6*log(1/6)]=1 .89bit
由于P(A+B+C/ A+B,A)=P(C/A+B,A)=P(C)
H(Z/XY)=H(C) =log6 =2.58bit
由于P(A=x,A+B+C=z/A+B=y)=P(A=x,C=z-y/ A+B=y)=P(A=x/A+B=y)P(C=z-y/A+B=y)=
P(A= x / A+B=y)P(C=z-y)=P(A/Y)P(C)
一共216种情况,每种情况的概率为1/216,即P(XYZ)=1/216
H(XZ/Y)=
(1/216)[(-6*log(1/6)-12*log(1/12)-18*log(1/18)-24*log(1/24)-30*log(1/30))*2-36*log(1/36)]= (1/36)*[(log6+2log12+3log18+4log24+5log30)*2+6log36]=4.48 bit
由于P(Z/X)=P(B+C/A)=P(B+C)
(/)()log (/)
()(/)log (/)
()()log ()()
xyz
abc
a
bc
H Z X p xz p z x p a p a b c a p a b c a p a p b c p b c H B C =-=-++++=-++=+∑∑∑∑
= (1/36)*{[log36+2log(36/2)+ 3log(36/3)+ 4log(36/4)+ 5log(36/5)]*2+6log(36/6)}bit
2.11解:P(0/0)=P(1/1)=1- p , P(1/0)=P(0/1)= p (a) P (u l)=1/8
P (u l ,0)=P (u l)×P (0/u l)=(1/8)×(1-p ) 接收的第一个数字为0的概率:
P (0)=P (u l)×P (0/u l)+ P (u 2)×P (0/u 2)+……. P (u 8)×P (0/u 8)
=4×(1/8)×(1-p )+ 4×(1/8)×p =1/2
I(u l; 0)=log[ P (u l ,0)/P(0)P(u l)]=1+log(1-p ) (b) P (u l ,00)=P (u l)×P (00/u l)=(1/8)×(1-p )2
P (00)=P (u l)×P (00/u l)+ P (u 2)×P (00/u 2)+……. P (u 8)×P (00/u 8) =2×(1/8)×(1-p )2 +4×(1/8)×p (1-p )+ 2×(1/8)×p 2
=1/4
I(u l; 00)=log[ P (u l ,00)/P(00)P(u l)]= 2+2log(1-p ) (c) P (u l ,000)=P (u l)×P (000/u l)=(1/8)×(1-p )3
P (000)=P (u l)×P (000/u l)+ P (u 2)×P (000/u 2)+……. P (u 8)×P (000/u 8) = (1/8)×(1-p )3 +3×(1/8)×p (1-p ) 2+3×(1/8)×p 2 (1-p ) +(1/8)×p 3
=1/8
I(u l; 000)=log[ P (u l ,000)/P(000)P(u l)]= 3+3log(1-p ) (d) P (u l ,0000)=P (u l)×P (0000/u l)=(1/8)×(1-p )4
P (0000)=P (u l)×P (0000/u l)+ P (u 2)×P (0000/u 2)+……. P (u 8)×P (0000/u 8) = (1/8)×(1-p )4 +6×(1/8)×p 2 (1-p ) 2+ (1/8)×p 4
I(u l; 0000)=log[ P (u l ,0000)/P(0000)P(u l)]=
4224
(1)
3log{22}log{
}(1)6(1)p p p p p p
-=-+-+-+
2.12解:
I(X;Z)= H(Z)-H(Z/X) I(XY ;Z)=H(Z)-H(Z/XY)
I(Y;Z/X)=I(XY;Z)-I(X;Z)
I(X;Z/Y)= I(XZ;Y)-I(Y;Z)= H(XZ)-H(XZ/Y) -I(Y;Z)=H(X)+H(Z/X) -H(XZ/Y) -I(Y;Z)
以上可以根据2.9的结果求出
2.27解:考虑到约束条件
()1,
()q x xq x m ∞
∞
==?
?
采用拉格朗日乘子法
1212
120
12120
0(())()log ()[()][()1]
()()log ()log ()[1]()()
c x x H q x q x q x dx xq x dx m q x dx e a m q x dx m e q x dx
q x q x λλλλλλλλλλ∞
∞
∞
----∞
∞=-----=++≤++?-????
?
当且仅当12
()x q x a λλ--=时,等式成立。
将12
()x q x a
λλ--=带入
()1,
()q x dx xq x dx m ∞
∞
==?
?
: 211
,ln a m m a
λλ-=
= 实现最大微分熵的分布ln ()ln ln 111()x x x
a m a m a
m q x a
e e m m m
---===,相应的熵值log (me )
2.29证明: (a)
1
2
()()(1)()(1)1Q x Q x Q x λλλλ=+-=+-=∑∑
所以Q(x)为概率分布。 (b) 即证明熵的凸性。
1112121212121212()(1)()()
111
()log (1)()log (()(1)())log ()()()()()
()log
(1)()log ()()
()()
log ()[
1]log (1)()[1]0()()
H U H U H U Q x Q x Q x Q x Q x Q x Q x Q x Q x Q x Q x Q x Q x Q x Q x e Q x e Q x Q x Q x λλλλλλλλλλ+--=+--+-=+-≤?-+?--=∑∑∑∑∑∑∑