概率论与数理统计第一章习题参考答案(3)

2019-01-19 14:58

(2) F?x???x???0?x1?dt?0?8f?t?dt??2?1dt???08?1?916?xt82dt?0?10?x?2?x?8??2?x2?x?4?16?x?4?1x?0716x?00?x?2

2?x?4x?4P?1?X?3??F?3??F?1??P?X?3??F?3??916?18?

7?P?X?1X?3??P?1?X?3?P?X?3?1n?716? 991613、解:P?X?i,Y?j???1n?1 i?j,i,j?1,2,??,n

P?X?i,Y?i??0 当n=3时,(X,Y)联合分布律为

14、(1) P{X?1,Y?1}?0.2,

P{X?1,Y?1}?P{X?0,Y?0}?P{X?0,Y?1}?P{X?1,Y?0}?P{X?1,Y?1} Y X 1 2 3 1 0 2 3 1/6 1/6 1/6 0 1/6 1/6 1/6 0 ?0.10?0.08?0.04?0.20?0.42

(2) 1?P{X?0,Y?0}?1?0.10?0.90

(3) P{X?Y}?P{X?0,Y?0}?P{X?1,Y?1}?P{X?2,Y?2}

?0.10?0.20?0.30?0.60

P{X?Y?2}?P{X?0,Y?2}?P{X?1,Y?1}?P{X?2,Y?0} ?0.06?0.20?0.02?0.28 15、?1??c?8

?????f?x?dx???0????0ce??2x?4y?dxdy??c8e?2x??0??e??4y???0?c8

P?X?2????x?2f?x,y?dxdy????2dx???08e??2x?4y?dy??e?2x??0??e??4y???0?e?4

11

D:

0?x??0?y?x

P?X?Y????x?yf?x,y?dxdy????0dx?8e0x??2x?4y?dy????0e?2x??2e??4y?x0dx

????2e0???6x?2e?2x1?dx???e?3?6x?e?2x???0???23

D:

0?x?10?y?1?x

1?x??2x?4y?P?X?Y?1????10dx?2e08edy dx??1?2x0??e??4y2x?41?x0??2e01?2x?2e2x?4?dx

??e1??2x?ex2?10?1?e1??22?

,(x,y)?G,其他16、(1)s??0(x?22)dx??6, f(x,y)??6?0

(2)fX(x)??xx222?3x2,0?x?1 6dy??,其他?0?2y??y6dx?6(2y???1 fY(x,y)???6dx?6(1?y),y??0,其他??y),120?y??y?112

17、(1)

12

Y X 0 1 2 (2)

0 1 2 P{X=xi} 0.24 0.38 0.38 1 ???e?ydy?f?x,y?dy???x??00.10 0.08 0.06 0.04 0.20 0.14 0.02 0.06 0.30 P{Y=yi} 0.16 0.34 0.50 ?fX?x?????x?0x?0??

D:

0?x???x?y???0?y???0?x?y?e?x???0fY?y??x?0x?0

??????ye?ydx?f?x,y?dx???0??0y?0y?0

或:

?ye?y???0y?0y?0

22、(1) Y1 Y2 -1 -1 0 1 ??2?2??24 ?2?1??? ??2?2??24 0 ?2?1??? ??1???2 ?2?2?1??? ?1 ??2?2?24?1??? ??2?2?24 且P?Y1?Y2??P?Y1??1,Y2??1??P?Y1?0,Y2?0??P?Y1?1,Y2?1?

??24?1???2???24?32?2?2??1

(2)P?X?0,Y?0??0.10 又?P?X?0??P?Y?0??0.0384

P?X?0,Y?0??P?X?0??P?Y?0?

∴X与Y不相互独立

??8y23、?X~U?0,1? fY?y????0?

0?y?其它12 且X与Y相互独立

13

??8y 则f?x,y??fX?x??fY?y????0?0?x?1,0?y?其它1212

D:

0?y?y?x?1 14

1P{X?Y}???x?y8ydxdy??20(8y?y)dy?(4y?228312y)|0?323

24

X pk -2 15-1 160 151 1153 1130 Y?X2?1 5 2 2 1 5 2 10 10

Y pk 1 15 16?115 15 1130 即

Y pk 1 152 7305 1510 1130 25、U=|X|,当y?0时,FU(y)?P(Y?y)?P(|X|?y)?0

当y?0时,FU(y)?P(Y?y)?P(|X|?y)?P(?y?X?y)?FX(y)?FX(?y)?2?(y)?1

U?|X|的概率概率密度函数为y?2?故 ?e?fU(y)????0,22,y?0y?0

26、(1)Y?X,当y?0时,FY(y)?P(Y?y)?P(X?y)?0

X?y)?P(X?y)2

当y?0时,FY(y)?P(Y?y)?P(?FX(y)2

Y?X的概率概率密度函数为2故

?y?,y?0?2yefY(y)???y?0?0,

15


概率论与数理统计第一章习题参考答案(3).doc 将本文的Word文档下载到电脑 下载失败或者文档不完整,请联系客服人员解决!

下一篇:室外管线综合

相关阅读
本类排行
× 注册会员免费下载(下载后可以自由复制和排版)

马上注册会员

注:下载文档有可能“只有目录或者内容不全”等情况,请下载之前注意辨别,如果您已付费且无法下载或内容有问题,请联系我们协助你处理。
微信: QQ: