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