中山大学概率统计第3习题解(2)

上式中

r?1rk?r ?k?rkCk?1r/p, ?1pqr?1rk?r??k?r ?k?rk(k?1)Ck?1pq???(k?1)k(k?1)!rk?rpq

(k?r)!(r?1)! ?r(r?1)?(k?1)!r(r?1)?r(r?1)r?1k?rr?1r?2k?rpq?Cpq? ?k?r(k?r)!(r?1)!?k?rk?1p2p2p2由此得

r(r?1)rr2rqEY???2?2,

ppp2p2因而

r2rqr2rqDY?EY?(EY)?2?2?2?2.

pppp22

11. 设X服从?分布,即它的密度为

pX(x)??(???)??1x(1?x)??1I(0,1)(x),

?(?)?(?)10其中??0,??0.求EX和DX.(提示:称B(s,t)??us?1(1?u)t?1dx为?函数,由微积分的知识知B(s,t)??(s)?(t)/?(s?t)) 解 (见p.239,命题2.1)

12. 分别以下的几种情况,求Z?X2?Y2的均值.请用两种方法分别计算,即利用1.10式直接计算,以及先求Z的密度,再利用1.4式计算. 1) (X,Y)有联合密度p(x,y)? 2) (X,Y)有联合密度p(x,y)?1?e2?1x2?y2. ?e?(x2?y2).

3) (X,Y)有联合密度p(x,y)?4xye?(x解

1) 方法1

2?y2)ID(x,y),其中D?(x,y):x?0,y?0).

FZ(z)?P(Z?z)?P(X2?Y2?z)?I(0,??)(z)x?rco?sy?rsi?n??X2?Y2?z1?e2?x2?y2dxdy

?2?z?r1I(0??(z)d?,)?0?0redr?I2??r(???(0,z))redr?0z?(1e?z?ze?zI)??(z0,, ())?(z)?ze?zI(0,??)(z), pZ(z)?FZEZ??

????zpZ(z)dz????2?zzedz0??(3)?2.

习题3-6

方法2 EZ?E2X?2Y???????????22x?22y(,p)xy dxdyx?rcos?y?rsin? ?12???????????x2?y2e?x?ydxdy?12??02?d??1??2?rredr0??(3)?2.

2) 方法1

FZ(z)?P(Z?z)?P(X2?Y2?z)?I(0,??)(z)x?rco?sy?rsi?n??X2?Y2?z?2e?x2?y2dxdy

?1?I(0??,(z))?2?0d??re?rdr?2I0z2?rdr???(0,z(?))re02z?z?(e1I2)(z??0,, ())?(z)?2ze?zI(0,??)(z), pZ(z)?FZEZ?? 方法2 EZ?E????zpZ(z)dz?2???z?t??2?z2zedz?02???1/2?ttedz0??(3/2)??/2.

2X?2Y?????????2x?2y(,p)xy dxdyx?rcos?y?rsin? ?1???????????x2?y21?e?(x2?y2)dxdy?1??02?d????2?r2redr0

?2?r?t??2?r2redr?02???1/2?ttedz0??(3/2)??/2.

3) 方法1

FZ(z)?P(Z?z)?P(X2?Y2?z)?I(0,??)(z)x?rcos?y?rsin???X2?Y2?z4xye?(x2?y2)ID(x,y)dxdy

?I(0,??)(z)?z20?/204sin?cos?d??r3e?rdr?2I(0,??)(z)?r3e?rdr

0022z2z2 ?I(0,??)(z)?te?tdt?(1?e?z?z2e?z)I(0,??)(z),

2?(z)?2z3e?zI(0,??)(z), pZ(z)?FZEZ?? 方法2 EZ?E??????????zpZ(z)dz??????02ze4?z2dz?z?t???3/2?ttedt0??(5/2)?3?/4.

2X?2Y?????2????2x?2y(,p)xy dxdyx?rco?sy?rsi?n ?????4x2?y2xye?(x?y2)ID(x,y)dxdy??0?/24sin?cos?d????4?r2redr 0 ??

??02re4?r2dz?r?t???3/2?ttedt0??(5/2)?3?/4.

习题3-7

13. 设X~N(0,?2),求EXn. 解 EX0?? 由于xe????pX(x)dx?1. 是奇函数,???0?x2/2x?e?x2/(2?2)dx???x??t????t/2edt0?2???,故

EX1??当n?2时, EX??n??nxpX??????xpX(x)dx????22xe?x/(2?)dx?0. 2??(x)dx???x/(2?)22????n?122xn?x2/(2?2)2???xedx???de?x/(2?) ??2??2???x ?2?2n?1??de????2?????(n?1)xn?2?x2/(2?2)edx?(n?1)?2EXn?2. 2?由此得

??0EXn??n??(n?1)(n?3)?3?1??

n为奇数n为偶数.

14. 设球的直径服从[a,b]上的均匀分布,求球体积的期望.

11解 设球的直径为X,球的体积为Y.则Y??X3,X有密度pX(x)?I[a,b](x),而

6b?a??1b11?x4333EY?E(?X)???xpX(x)dx???xdx???6a6(b?a)624(b?a)b?a?(b4?a4)24(b?a).

15. 点随机地落在中心在原点、半径为R的圆周上,并对弧长是均匀分布的.求落点横坐标的期望和方差.

解 从点(1,0)沿反时针方向到落点的弧长为S,落点横坐标为X,则X?Rcos密度pS(s)?S,S有2?1I[0,2?](s).因而 2?2???2?1Sss1sEX?E[Rcos]??RcospS(s)ds??Rcosds?R?2?sin??02?2?2?2?2?2??0,

0S EX2?E[Rcos2?R2 ?4?2?]?s????s[Rcos2Sp]2?s?(d?)s02?2?12?2R2s dscos2??02?R2s(1?cos)ds?(s?sincos)?4??0R2?. 2DX?EX2?(EX)2?R2/2.

16. 设X~N(?,?2),Y?aX,其中a?0,a?1.求Y的密度,期望和方差.

习题3-8

解 FY(y)?P(Y?y)?P(aX?y).

当y?0时,FY(y)?0,pY(y)?FY?(y)?0.

当y?0,a?0时,FY(y)?P(X?lny/lna)?1?FX(lny/lna),

pY(y)?FY?(y)?((lny)?/lna)pX(lny/lna)??lna?(lny/lna)2/(2?2)e.

y2?? 当y?0,a?0时,FY(y)?P(X?lny/lna)?FX(lny/lna),

pY(y)?FY?(y)?((lny)?/lna)pX(lny/lna)?22lnae?(lny/lna??)/(2?).

y2??由上知Y有密度

pY(y)?FY?(y)?((lny)?/lna)pX?|lna|?(lny/lna??)2/(2?2)eI(0,??)(y).

y2??2

17. 设轮船横向摇摆的振幅X是随机变量,有密度p(x)?Axe?x的期望和方差,并求振幅大于其期望的概率. 解 1??????/2?2I(0,??)(x).求A和Xp(x)dx??xe?x2????2Axe?x2/2?2I(0,??)(x)dx A?2????tedt0 ?A?故A?1/?2. EX?? ?2??0/(2?)dxt?x2/(2?2)??A?2.

????xp(x)dx??????1?2x2e?x2?2?2/2?2I(0,??)(x)dx

221e?x/2?dx???/2. 2??1?2???2?x2/2?2xedx0????2x??EX????2xp(x)dx????????1?2xe3?x2/2?2I(0,??)(x)dxx2/(2?2)?t?2?2???0te?tdt?2?2.

DX?EX2?(EX)2?(2??/2)?2.

18. 设等腰直角三角形的直角边长X为随机变量,服从[0,1]上的均匀分布.求这个三角形的面积的期望.

解 X有密度pX(x)?I[0,1](x),这个三角形的面积S?X2/2. ES?E(X2/2)??

19. 设园的面积服从指数分布,有密度p(x)??e??xI(0,??)(x).求这个园的半径的期望. 解 设园的面积为X,则这个园的半径R?X/?. ER?E(X/?)??

????????(x2/2)p(x)dx??????(x2/2)I[0,1](x)dx??(x2/2)dx?1/6.

01x/?pX(x)dx??????x/??e??xI(0,??)(x)dx

习题3-9

??

??0x/??e??xdx?t??x1???0??te?tdt?1???(3/2)?11??.

??22??120. 设X,Y独立,分别有密度pX(x)?求Z的期望和方差. 解 EX??????1xI[1,3](x)和pY(x)?2e?2yI[0,??)(y),又设Z?XY.4311xpX(x)dx????1??4xI[1,3](x)dx??4x2dx?13/6,

313xdx14??0 EX2?? EY????2xpX(x)dx??????????13xI[1,3](x)dx??4???5,

????ypY(y)dy??y?2e?2yI[0,??)(y)dy????2ye?2y?1/2,

EY2?? ??????y2pY(y)dy????y2?2e?2yI[0,??)(y)dy

??02y2e?2ydy?1??2?t1tedt??(3)?1/2. ?044 EZ?E(XY)?EXEY?(13/6)(1/2)?13/12, EZ2?E(X2Y2)?EX2EY2?5?(1/2)?5/2, DZ?EZ2?(EZ)2?5/2?(13/12)2?191/144.

21. 设某人在3天中共收到5份电子邮件,每份电子邮件在这3天中的那一天被收到都是等可能的.设这3天中有X天当天都至少收到一份电子邮件,求X的期望.(提示:设

?1Yi???0第i天至少收到一份电子邮件, 则X?Y1?Y2?Y3).

第i天没有收到电子邮件解 对于i?1,2,3,设

?1Yi???0第i天至少收到一份电子邮件,

第i天没有收到电子邮件则X?Y1?Y2?Y3,

P(Yi?0)?(2/3)5?32/243, P(Yi?1)?1?P(Yi?0)?1?32/243?211/243, 故Yi~B(1,65/81).因而

EX?EY1?EY2?EY3?211/243?211/243?211/243?211/81?2

49. 8122222. 设(X,Y)有联合密度p(x,y)?A/(x,其中A是常数.求出A的值,并问?y?1)?XX与?YY是否存在?

解 1??

?????????p(x,y)dxdy??2?0A/(x?????????2?y2?1)2dxdy

x?rcos?,y?rsin??A?d????0??r1dr?2?Adr2 ?222202(r?1)(r?1)习题3-10

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