中山大学概率统计第3习题解(4)

EX??2lm2P{Xm?0?m}??mn?mCMCN?Mlm(m?1)?nm?0CN??mn?mCMCN?Mlm?nm?0CN.

上式中 ? ?mn?mCMCN?Mlm?nm?0CN?nM. N??M(M?1)m?2n?mCM?2CN?Mlm(m?1)m(m?1)?N(N?1)n?2m?2CN?2n(n?1)mn?mCMCN?Mlm(m?1)?nm?2CN

mn?2?mm?2n?mM(M?1)n(n?1)lCMM(M?1)n(n?1)l?2CM?2C(N?2)?(M?2)?2CN?M ? ??m?2Cn?2?m?0n?2N(N?1)N(N?1)CN?2N?2 ?M(M?1)n(n?1).

N(N?1)EX2?22故

M(M?1)n(n?1)nM?,

N(N?1)N2M(M?1)n(n?1)nM?nM?nM(N?n)(N?M). DX?EX?(EX)??????2N(N?1)N?N?N(N?1)

39*. X的密度函数p(x)满足p(c?x)?p(c?x),其中c为一常数.又设?收敛.求证:EX?c(提示:xp(c?x)是关于变元x的奇函数,故???????|x|p(x)dxxp(c?x)dx?0;令

t?c?x,代入积分得要证的结果).

解 EX??x?c?t????xp(x)dx??????????cp(x)dx????0????(x?c)p(x)dx?c??0??????(x?c)p(x)dx

?c??tp(c?t)dt?c??t??stp(c?t)dt??tp(c?t)dt,

上式的最后的一项

???故EX?c.

0tp(c?t)dt?????0sp(c?s)ds?????0tp(c?t)dt,

40*. 1) 求泊松分布的矩母函数.

2) 利用矩母函数求泊松分布期望和方差.

3) 利用矩母函数证明泊松分布的加法定理:设随机变量X,Y独立,X~Po(?),Y~Po(?),则X?Y~Po(???).

解 1） 设X~Po(?),则P(X?k)??ke??/k!,k?0,1,2,?.

MX(t)?EeXt??k?0ekP(X?k)??k?0ekt?ke??/k!?e???k?0(et?)k/k!

?e??ee??e(e

tt????1)?.

习题3-16

t(e 2） M?X(t)??eet?1)???e(et?1)??t?(t)??(?et?1)e(e, M?Xt?1)??t,故

2?(0)??(??1), DX?EX2?(EX)2??. EX?M?X(0)??, EX?M?X 3） MX(t)?e(et?1)?, MY(t)?e(et?1)?,

(et?1)?(et?1)?MX?Y(t)?MX(t)MY(t)?e故X?Y~Po(???).

e?e(et?1)(???),

41.* 设随机变量X1,X2,X3,X4,X5相互独立同分布且仅取正值.证明

??X1?X2?X3E???3/5. X?X?X?X?X2345??1解 设Yijk?Xi?Xj?XkX1?X2?X3?X4?X5,1?i?j?k?5,由于X1,X2,X3,X4,X5相互独立

同分布,故各个Yijk有相同的分布,因而有相同的期望.又

EY123?EY124?EY125?EY134?EY135?EY145?EY234?EY235?EY245?EY345 ?E(Y123?Y124?Y125?Y134?Y135?Y145?Y234?Y235?Y245?Y345) ?6(X1?X2?X3?X4?X5)? ?E???6,

X?X?X?X?X12345????X1?X2?X3故E???EY123?6/10?3/5.

X?X?X?X?X2345??1

2242*. 设(X,Y)服从二维正态分布,EX??1,EY??2,DX??1,DY??2,?XY??.求

(X,Y)落在区域

??(x??1)22?(x??1)(y??2)(y??2)?2?D??(x,y):???k? 222??????12?12?中的概率.

解1 (X,Y)有密度 p?x,y??12??1?2?1?exp??21??21??2????x???22??x????y????y???2???1122?????. 22??????1212?????? P{(X,Y)?D}???p(x,y)dxdy

D ???Dx??1?s,12??1?2y??2?1?exp??21??21??2??????x???22??x????y????y???2???1122?????dxdy 22???2??12??1????1??2?t??1??22??exp?s?2?st?t???2?1??2??dsdt, 2?21????D*??1?? 习题3-17

其中D*?(s,t):s2?2?st?t2?k2. 记a?1??,b?1??,则

a2?b2?2, a2?b2?2?, (a?b)2?(a?b)2?2(a2?b2)?4.

???u??s?1?a?ba?b??u?1?a?b?a?b??s?令???????,则???????,

v?a?ba?btta?ba?b2ab2???????????v?J??(s,t)?s?1?a?ba?b?1222???????[(a?b)?(a?b)]?ab?1??, ?(u,v)?t?2?a?ba?b?411 s2?t2?[(a?b)u?(a?b)v]2?[(a?b)u?(a?b)v]2

441 ?[(a?b)2(u2?v2)?(a?b)2(u2?v2)?4(a?b)(a?b)uv]

41 ?[4(u2?v2)?8?uv]?(u2?v2)?2?uv,

411 st?{(a?b)(a?b)(u2?v2)?[(a?b)2?(a?b)2]uv}??(u2?v2)?uv

42 s2?2?st?t2?(u2?v2)?2?uv??2(u2?v2)?2?uv?(u2?v2)(1??2). ??1?22?s?2?st?t?? P{(X,Y)?D}???exp????dsdt 2?221????D*2?1????1???s?1?a?ba?b??u?????????t?2?a?ba?b??v???u?v1?exp???2??2??D**22???dudv??k2u?rcos?v?rsin??12??02?d??k2/(1??2)0re?r2/2dr

??k2/(1??2)0re?r2/2?dr?1?e2(1??2).

其中D**?(u,v):u2?v2?k2/(1??2).

解2 把(X,Y)改记为(X1,X2),以X记(X1,X2)的列向量的形式,即

?X?X?(X1,X2)T??1?,

?X2???由题目知

22X~N(?1,?2,?1,?2,?),

记x?(x1,x2)T,??(?1,?2)T,则X的密度可记为

pX(x)?1?1?T?1exp?(x??)?(x??)??

2?|?|1/2?2??1其中?????

???是X的协方差矩阵. 1?习题3-18

由线性代数理论知,存在正定矩阵B,满足B2??.令U?(U1,U2)T?(X??)B?1, 则U服从正态分布,

EU?E(X??)B?1?(EX??)B?1?0, varU?D[(X??)B?1]?(B?1)T?B?1?I, 故U~N(0,I).设u?(u1,u2)T?(x??)B?1,则

(x1??1)2(x1??1)(x2??2)(x2?2?2?12????2)21?2?2?k 2?(x??)??1(x??)T?k2 2(1??2)

?(x??)B?1(B?1)T(x??)T?k22(1??2) ?uuT?k22(1??2)

X?D?UUT?k2 2(1??2)

P{X?D}?P(UUT?k22(1??2))???pU1U2(u1,u2)du1du2

u12?u22?k22(1??2) ???12?exp????12(u22?1?u2)??du1du2

2u12?u22?k2(1??2)u?rcosv?rsin??2

?12?)r2/22??0d??k2/(1??0re?dr

222?k ??k/(1??)0re?r2/2dr?1?e2(1??2).

习题3-19

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