概率论与数理统计(茆诗松)第二版课后第三章习题(15)

习题3.3

1． 设二维随机变量(X, Y ) 的联合分布列为

YX12

123

0.070.110.220.040

.070.09

试分布求U = max{X, Y } 和V = min{X, Y } 的分布列．

解：因P{U = 1} = P{X = 0, Y = 1} + P{X = 1, Y = 1} = 0.05 + 0.07 = 0.12；

P{U = 2} = P{X = 0, Y = 2} + P{X = 1, Y = 2} + P{X = 2, Y = 2} + P{X = 2, Y = 1}

= 0.15 + 0.11 + 0.07 + 0.04 = 0.37；

P{U = 3} = P{X = 0, Y = 3} + P{X = 1, Y = 3} + P{X = 2, Y = 3} = 0.20 + 0.22 + 0.09 = 0.51； 故U的分布列为

UP123

0.120.370.51

因P{V = 0} = P{X = 0, Y = 1} + P{X = 0, Y = 2} + P{X = 0, Y = 3} = 0.05 + 0.15 + 0.20 = 0.40； P{V = 1} = P{X = 1, Y = 1} + P{X = 1, Y = 2} + P{X = 1, Y = 3} + P{X = 2, Y = 1}

= 0.07 + 0.11 + 0.22 + 0.04 = 0.44；

P{V = 2} = P{X = 2, Y = 2} + P{X = 2, Y = 3} = 0.07 + 0.09 = 0.16； 故V的分布列为

V012

P0.400.440.16

2． 设X和Y是相互独立的随机变量，且X ~ Exp(λ )，Y ~ Exp(µ )．如果定义随机变量Z如下

1,当X≤Y,

Z=

0,当X>Y.

求Z的分布列．

解：因(X, Y ) 的联合密度函数为

λµe (λx+µy),x>0,y>0,

p(x,y)=pX(x)pY(y)=

0,其他.

则P{Z=1}=P{X≤Y}=∫

+∞

+∞

+∞

dx∫λµe (λx+µy)dy=∫dx ( λ)e (λx+µy)

x

+∞+∞

x

=∫λe (λ+µ)xdx=

λλ+µ

，

e (λ+µ)x

+∞

=

λλ+µ

，

P{Z=0}=1 P{Z=1}=

故Z的分布列为

µλ+µ

P

λ+µ

λ+µ