概率论与数理统计教程答案(徐建豪版).

即 Fma(xz)? FX(z)?FY(z)

0,x?0??x??1?e?y,y?0?1?e,0?x?1,F(其中FX(x)?? Yy)???11?e?0,y?0?1,x?1??0,u?0??u2??(1?e),0?u?1 因此FU(u)???1?1?e1,u?1??4.设随机变量X、Y相互独立,且服从同一分布,试证明:

P{a?min{X,Y}?b}?[P{X?a}]2?[p{X?b}]2.

证明:由N?min(X,Y)>z 等价于X?z 且Y?z 可得

Fmin(z)?P{N?z}?1?P{N?z}?1?P{X?z,Y?z}?1?P{X?z}?P{Y?z}?1?[1?P{X?z}]?[1?P{Y?z}]?FXz()?]F[Y1z ( 即 Fmi(nz)?1?[1

=1?[P{X?z}]2 代入得

P{a?min{X,Y}?b}?1?[P{X?b}]2?1?[P{X?a}]2 ?[P{X?a}]2?[P{X?b}]2??复习题三(A)

1. 在一箱子中装有12只开关,其中2只是次品,在其中取两次,每次任取一只,考虑两种试验:(1)有放回抽样;(2)不放回抽样.我们定义随机变量X,Y如下:

?0,若第一次取出的是正品?0,若第二次取出的是正品X??; Y??.

?1,若第一次取出的是次品?1,若第二次取出的是次品试分别就(1),(2)两种情况,写出X和Y的联合分布律.

解:(1)有放回抽样,X和Y的联合分布律为

10?1025P{X?0,Y?0}??

12?1236 66

P{X?1,Y?0}?2?1012?12?536 P{X?0,Y?1}?10?2512?12?36 P{X?1,Y?1}?2?2112?12?36 X 0 1

Y 0 25/36 5/36 1 5/36 1/36 (2)不放回抽样,X和Y的联合分布律为 P{X?0,Y?0}?10?94512?11?66 P{X?1,Y?0}?2?101012?11?66 P{X?0,Y?1}?10?21012?11?66 P{X?1,Y?1}?2?112?11?166 X 0 1

Y 0 45/66 10/66 1 10/66 1/66

2. 假设随机变量Y服从参数为1的指数分布,随机变量:

X?0,Y?kk???1,Y?k(k?1,2),

求(X1,X2)的联合分布律与边缘分布律. 解:因为Y~E(1)

X?0,若Y?11???1,若Y?1

所以有

P{X1?1}?P{Y?1}????e?y1dy?e?1

P{X1?0}?P{Y?1}??1e?ydy?1?e?10

同理P{X??2?1}?P{Y?2}??2e?ydy?e?2

P{X2?0}?P{Y?2}?1?e?2

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所以P{X1?1,X2?1}?P{Y?2}?e?2

P{X1?1,X2?0}?P{X1?1}?P{X1?1,X2?1}?e?1?e?2 P{X1?0,X2?0}?P{Y?1}?1?e?1

P{X1?0,X2?1}?P{X1?0}?P{X1?0,X2?0}?0

故(X1,X2)的联合分布律与边缘分布律

X1 X2 0 1 P{ X1=i}

0 0 1?e?1 1?e?1

1 e?1?e?2 e?2 e?1

P{ X2=j} 1?e?2 e?2

3.元旦茶会上,每人发给一袋水果,内装3只句子,2只苹果,3只香蕉.今从袋中随机抽出四只,以X记桔子数,Y记苹果数,求(X,Y)的分布律 解:X可取的值为0,1,2,3.Y可取的值为0,1,2

P{X?0,Y?0}?P{?}?0

013P{X?0,Y?1}?C3C2C3/C84?2/70 022P{X?0,Y?2}?C3C2C3/C84?3/70 103P{X?1,Y?0}?C3C2C3/C84?3/70 112P{X?1,Y?1}?C3C2C3/C84?18/70 121P{X?1,Y?2}?C3C2C3/C84?9/70 02P{X?2,Y?0}?C32C2C3/C84?9/70 11P{X?2,Y?1}?C32C2C3/C84?18/70 20P{X?2,Y?2}?C32C2C3/C84?3/70 301P{X?3,Y?0}?C3C2C3/C84?3/70 310P{X?3,Y?1}?C3C2C3/C84?2/70

P{X?3,Y?2}?P{?}?0

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所以,(X,Y)的分布律为

Y X 0 1

0 0 3/70

1 2/70 18/70

2 3/70 9/70

4.随机变量(X,Y)的分布律如右表,求: Y (1) a值;

(2) (X,Y)的分布函数F(x,y); (3) (X,Y)关于X,Y的边缘分布函数 FX(x)与FY(y).

解:(1)由分布律的性质?pij?1得a?i,j2 9/70 18/70 3/70 3 3/70 2/70 0 X 1 2 —1 1/4 1/6 0 1/4 a 1 3(2)因F(x,y)?P{X?x,Y?y} ①x??1或y??1时,F(x,y)?0

②1?x?2,?1?y?0时,F(x,y)?P{X?1,Y??1}?1 45 12③x?2,?1?y?0时,F(x,y)?P{X?1,Y??1}?P{X?2,Y??1}?④1?x?2,y?0时,F(x,y)?P{X?1,Y??1}?P{X?1,Y?0}?⑤

1 2x?2,y?0时,F(x,y)?P{X?1,Y??1}?P{X?2,Y??1}?P{X?1,Y?0}?P{X?2,Y?0}?1

综上所述,(X,Y)的分布函数F(x,y)为

0,x?1或y??1??1/4,1?x?2,?1?y?0?1?(1)a?(2)F(x,y)??5/12,x?2,?1?y?0;

3?1/2,1?x?2,y?0?1,x?2,y?0??(X,Y)关于X,Y的边缘分布律为 X -1 0 P 5/12 7/12

Y 1 2 P 1/2 1/2

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故(X,Y)关于X,Y的边缘分布函数FX(x)与FY(y)为

?0,x?1?0,y??1(3)F?X(x)??1/2,1?x?2,Fy)???Y(?5/12,?1?y?0.

?1,x?2??1,y?05.设:

f(x,y)???1,0?x?2,max(0,x?1)?y?min(1,x)?0,其它, 求FX(x)与FY(y).

解;max(0,x?1)???0,x?1?x,x?1?x?1,x?1, min(1,x)???1,x?1,

所以,f(x,y)有意义的区域可分为

{0?x?1,0?y?x},{1?x?2,1?x?y?1}

Y=x Y=x-1 ?1,0?x?1,0?y?xf(x,y)???1,1?x?2,x?1?y?1

??0,其他所以

?x??0dy,0?x?1?x,0?x?1f(x)???1?X??x?1dy,1?x?2??2?x,1?x?2

?0,其他?0,其他??

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