概率论基础(第三版)-李贤平-试题+答案-期末复习 下载本文

EX??k?k?0??kk!?e??

??e???(k?1)!

k?12?k?1??e???e???EX??k?2k?0??kk!e????e

???k?(k?1)!k?1???k?1??e???[(k?1)?1]?(k?1)!k?1??k?1

??e[?????k?2k?2(k?2)!???](k?1)!k?1?k?1??e??[?e??e?]??2?? DX?EX?(EX)???????? 4.解:由题设可得

2222EX??k?pq

k?1?k?1?p?kqk?1k?1??p??111?p??(1?q)2p2p

EX??k2?pqk?12k?1???[k(k?1)?k]pqk?1k?1

?pq?k(k?1)qk?2?k?2?p?kqk?1

k?1??pq21?p(1?q)3(1?q)22q?p?p22q?p12q?()?2 2pppbDX?EX2?(EX)2?5.解:由题设可得 EX??????x?f(x)dx??x?a1a?b dx?b?a2 EX?2?????x?f(x)dx??22ba1a2?ab?b2x?dx?

b?a32a2?ab?b2a?b2(b?a)2?()? DX?EX?(EX)?

321226.解:由题设可得

EX????0????x?f(x)dx????0x??e??xdx??0??(?x)de??x?(?x)e??xEX??2??????e??xd(?x)?02??x??1?x?f(x)dx??22??0x??edx?2

?2DX?EX2?(EX)2?7.解:由题设可得

?2??11?()2?2??EX??令

????x?f(x)dx????x?1e?2??(x??)22?2dx

x????t 则有

2t?1EX??(?t??)?e2?dt???2???????t2?tedtEX?e?dt ??????2?2?????0??2???2?2?????t222DX?E(X?EX)2??(x??)2?f(x)dx?????1??(x??)2?e???2???(x??)22?2

dx 令

x????t 则有

t?122DX???t?e2?dt???2???2????22??2????t2edt?t2?2?t22?22??????(?t)de?t222?[(?t)e?????edt]?????t2?2

?2?(0?2?)??22?8.解:由题设可得

EX????x??12????x?f(x)dx121?1?x??2??dx?0

EX??1x?f(x)dx??x2??111?1?x2dx

1?2?x2?dx?02?1?x21DX?EX2?(EX)2?9.解:由题设可得

??11?0? 22EX??????????x?f(x)dx

1?xx?edx?02????EX2??????0x2?f(x)dx????0????1?xx2?edx2x2?e?xdx??(?x2)de?x??0?(?x2)e?x??0??e?xd(?x2)0??0??

?0??2xe?xdx?2?(?x)de?x?2?e?xdx?20??DX?EX2?(EX)2?2?0?2

10.解:由题设可得

E(X?e?2X)??(x?e?2x)f(x)dx??????(x?e?2x)e?xdx0??????0xedx??e0?x???3xdx

?1?11.解:由题设可得

14?33EX??k?k?02???kk!2e??e?????(k?1)!??e??e???

?k?1??k?1EX??k?k?0?kk!e????2??

E[(X?1)(X?2)]?EX2?3EX?2?(???)?3??2?2 ??2??0

22

故 ??2 (??0舍去) 12.解:(1)记以(0,1),(1,0),(1,1)为顶点的三角形区域为D,则区域D的面积

为SD?1, 从而(X,Y)的联合概率密度为 2

?1??2,(x,y)?D f(x,y)??SD?0,(x,y)?D? (2)

E(X?Y)?? ?D?????????(x?y)f(x,y)dxdy1101?x??2?(x?y)dxdy?2?dx?(x?y)dy

10?2?(x?124x)?2313.解:(1)根据数学期望的性质,有

E(X?Y)?EX?EY?0?0?0

(2)根据方差与协方差及相关系数的性质,有

cov(X,Y)cov(X,Y)??0.5DXDY22cov(X,Y)?2?0.5?1R(X,Y)?D(X?Y)?DX?DY?2cov(X,Y)?2?2?2?1?6