中山大学概率统计第3习题解(3)

2?A ??2(r2?1)????A,

0故A?1/?.

EX???????????xp(x,y)dxdy???x/(x???????1????2?y2?1)2dxdy

?? ?类似地,EY?0.

1??????????????x/(x2?y2?1)2dxdy??1????0?dy?0,

2E|X2|???????x2p(x,y)dxdy?2?x???????1????/(x2?y2?1)2dxdy

x?rcos?,y?rsin??1??0cos?d??2??0r3dr???,

(r2?1)2故?XX?DX?E(X?EX)2?EX2不存在,类似地?YY亦不存在.

23. 设(X,Y)服从区域D?{(x,y):0?x?1,0?y?1?x}上的均匀分布,求相关系数?. 解 (X,Y)有密度p(x,y)?2ID(x,y). pX(x)?? EX??2??????p(x,y)dy??10????ID(x,y)dy?I[0,1](x)?1?x02dy?2(1?x)I[0,1](x),

??xpX(x)dx??2x(1?x)dx?(x2?2x3/3)123010?1/3,

410 EX????2xpX(x)dx????2x(1?x)dx?(2x/3?x/2)?1/6.

DX?EX2?(EX)2?1/6?(1/3)??1/18. 类似可得DY?1/18. EXY????????11?x01???2xyID(x,y)dxdy??2xdx?0ydy??x(1?x)2dx

0 ?(x2/2?2x3/3?x4/4)10?1/12,

cov(X,Y)?EXY?(EX)(EY)?1/12?(1/3)(1/3)??1/36 ??

24. 设(X,Y)为随机向量,?,?,a,b,c都是实数.证明:

cov(?X?a,?Y?b)???cov(X,Y), D(?X??Y?c)??2DX??2DY?2??cov(X,Y).

cov(X,Y)?1/36???1/2. 1/18DXDY

25. 已知DX?16,DY?25,???0.5.求cov(X,Y),D(X?Y)和D(3X?2Y?4). cov(X,Y)?DXDY?XY?16?25?(?0.5)??10,

D(X?Y)?DX?DY?2cos(X,Y)?16?25?2?(?10)?21,

习题3-11

D(3X?2Y?4)?32DX?22DY?2?3?(?2)cos(X,Y) ?9?16?4?25?(?12)?(?10)?364.

26. 设随机变量X有均值4和方差25.为了使得rX?s有均值0和方差1,应该怎样选择

r,s的值. 解 由题意得

0?E(rX?s)?rEX?s?4r?s, 1?D(rX?s)?r2DX?25r2,

解方程组

??4r?s?0 ?2??25r?1得r??1/5,s??4/5.

27. 设随机变量X1,X2,X3独立同分布,有有限的不等于零的方差.又设

Y?2X1?X2?2X3, Z?2X1?3X2?6X3,

求Y,Z的相关系数.

解 设DX1?DX2?DX3??2,则

DY?D(2x1)?DX2?D(2X3)?4?2??2?4?2?9?2, DZ?D(2x1)?D(3X2)?D(?6X3)?4?2?9?2?36?2?49?2, cov(Y,Z)?cov(2X1,2X1)?cov(2X1,3X2)?cov(2X1,?6X3) ?cov(X2,2X1)?cov(X2,3X2)?cov(X2,?6X3)? ?cov(2X3,2X1)?cov(2X3,3X2)?cov(2X3,?6X3)? ?cov(2X1,2X1)?cov(X2,3X2)?cov(2X3,?6X3) ?4?2?3?2?12?2??5?2.

?yz

cov(Y,Z)?5?2????5/21.

22DYDZ(9?)(49?)28. 设(X,Y)是二维正态随机向量,X和Y都有均值0和方差1,两者的相关系数为1/2.为了使得X和Y?kX相互独立,应该怎样选择常数k的值.

解 设Z?Y?kX,则(X,Z)服从正态分布,X,Z相互独立的充分必要条件是cov(X,Z)?0.由于

cov(X,Z)?cov(X,Y?kX)?cov(X,Y)?kcov(X,X) ?DXDY?XY?kDX?1?1?(1/2)?k?1?1/2?k. 故应选择k?1/2.

29. 分别求习题2.26中的随机变量X和Y的期望和方差,并求它们的协方差和相关系

习题3-12

数.

解 EX??k?0kP(X?k)?0?0.4?1?0.3?2?0.3?0.9, EX2??k?0k2P(X?k)?0?0.4?1?0.3?4?0.3?1.5, DX?EX2?(EX)2?1.5?0.81?0.69,

EY??k?0kP(Y?k)?0?0.1?1?0.2?2?0.3?3?0.4?2, EY2??k?0k2P(Y?k)?0?0.1?1?0.2?4?0.3?9?0.4?5, DY?EY2?(EY)2?5?4?1,

EXY??j?0?i?0ijP(X?i,Y?j)?1?0.1?2?0.1?3?0.1?4?0.1?6?0.2?2.2, cov(X,Y)?EXY?EXEY?2.2?0.9?2?0.4, ?XY?

30. 设(X,Y)服从二维正态分布,EX?EY?0,DX?a2,DY?b2,??0.求(X,Y)落在

322233cov(X,Y)0.4??0.4815.

DXDY0.69?1??x2y2??区域D??(x,y):2?2?k2?中的概率.

ab????解 (X,Y)有密度p(x,y)?1?(x2/a2?y2/b2)/2, e2ab?21e?(x2ab?,?)D?}??px(ydxdy,)??? P{(XYDD/a2?y2/b2)/2dxdy

x?arsin?y?brcos??12??02?d??re?r0k2/2dr?1?e?k2/2.

31. 设随机变量X和Y独立.证明D(XY)?(DX)(DY)?(EX)2DY?(EY)2DX.

32. 设随机变量X1,?,Xn相互独立有相同的分布,且有有限的不等于零的方差.记

X?(X1???Xn)/n.

1) 证明对i?1,?,n,Xi?X与X不相关. 2) 对i?j,求Xi?X与Xj?X的相关系数. 解 设各个随机变量有共同的方差?2. 1) cov(Xi?X,X)?cov(Xi,X)?cov(X,X)

11 ?cov(Xi,X1???Xn)?2cov(X1???Xn,X1???Xn)

nn11 ?cov(Xi,Xi)?2[cov(X1,X1)???cov(Xn,Xn)]

nn

习题3-13

?1111DXi?2[DX1???DXn)]??2?2?n?2?0, nnnn故Xi?X与X不相关.

2) cov(Xi?X,Xj?X)?cov(Xi?X,Xj)?cov(Xi?X,X)?cov(Xi?X,Xj)

?cov(X??cov(X,X11i,Xj)?cov(X,Xj)j)??ncov(XjXj)??n?2,

D(Xi?X)?cov(Xi?X,Xi?X)?cov(Xi?X,Xi)?cov(Xi?X,X)

?cov(XX?DX1i?X,i)?cov(Xi,Xi)?cov(X,Xi)i?ncov(XiXi)

??2?1n?12n?2?n?,

类似地,D(Xi?X)?n?1n?2.故Xi?X与Xj?X的相关系数为 ??cov(Xi?X,Xj?X)??2D(XX)D(X?/n2??1. i?j?X)(n?1)?/nn?1

33. 设某一概率分布有密度12e?x/2I(0,??)(x).求这一分布的3/4分位数.

解 设这一分布的3/4分位数为?,根据定义,

3/4???1e?x/2I?1?(0,??)(x)dx??e?x/2dx??e?x/2?/2??202.

0?1?e?因而??2ln4?2.7726.

34. 连续型随机变量X有密度p(x)?x8I(0,4)(x).求X的中位数. 解 设X的中位数为?,根据定义,

1/2??????2??p(x)dx??x?xx2??8I(0,4)(x)dx??08dx?16?.

016因而??22.

35. 设X有密度函数p(x)??e??xI(0,??)(x).求X的k阶矩EXk. 解 EXk????????xkXp()x?d?x?kx??ex??(0??,I()?x?d??)0x?k??xx e x??t/?1??1)!?k?0xke?tdt??(k??k?k?k.

36. 设X~N(0,1),Y?Xn,n是整数.求?XY.

解 EX?0,DX?1.由习题13可得,EY2?EX2n?(2n?1)(2n?3)???3?1.

习题3-14

dx 当n为偶数时,EY?EXn?(n?1)(n?3)???3?1,

DY?EY2?(EY)2?(2n?1)(2n?3)???3?1?[(n?1)(n?3)???3?1]2. EXY?EXn?1?0,cov(X,Y)?EXY?(EX)(EY)?0 ?XY?cov(X,Y)?0.

DXDY 当n为基数时,EY?0,DY?EY2?(EY)2?EY2?(2n?1)(2n?3)???3?1. EXY?EXn?1?n(n?2)???3?1,

cov(X,Y)?EXY?(EX)(EY)?EXY?n(n?2)???3?1, ?XY?

?137. 直接验证:若DX?0,b?0,Y?a?bX,则?????1b?0. b?0cov(X,Y)n(n?2)???3?1n!!??. DXDY1?(2n?1)(2n?3)???3?1(2n?1)!!证 DY?b2DX,cov(X,Y)?bcov(X,X)?bDX,故

??

?1cov(X,Y)bDX???DXDYb2(DX)2??1b?0. b?0mn?mn38.* 设X服从参数为N,M,n(n?N?M)的超几何分布,即P{X?m}?CM, CN?M/CNm?0,1,2,?,l.这里l?min(M,n).试证明:

1) ?m?0P{X?m}?1; 2) E(X)?nM/N; 3) DX?lnM(N?n)(N?M)N2(N?1).

提示:展开恒等式(1?x)N?(1?x)M(1?x)N?M两边的二项式,比较便得1);利用1)

r?1及等式Crs?Crs?1可得2)和3).

s证 1） 展开恒等式(1?x)N?(1?x)M(1?x)N?M两边的二项式得

?NCkxkk?0Nmmii??m?0CMx?i?0CN?Mx,

MN?M上式两边xn的系数相等,故

?nCmCn?m??m?0MN?MnCN??Mmn?m???m?0CMCN?M因而?lP{Xm?0n?Mn?M,

?m}??mn?mCMCN?Mlnm?0CNmn?mn?(?m?0CMCN?M)/CN?1.

l 2） EX??lmP{Xm?0?m}??mn?mCMCN?Mlm?nm?0CN??Mlmm?m?1m?1n?mCM?1CN?MnCN

nM ?N

?m?1(n?1)?(m?1)CMl?1C(N?1)?(M?1)n?1m?1CN?1nM?N?m(n?1)?ml?1CM?1C(N?1)?(M?1)n?1m?0CN?1?nM, N习题3-15

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