概率论与数理统计(经管类)第四章课后习题答案

1、习题4.11. 设随机变帚x的概率密度为(2x, 0 < x< 1,(l)f(x) =0 其他(2)f(x) =-e"lxl, -oo<x < +oo求 E(X)解：(l)E(X)=仁二xf(x)dx = £ x 2xdx = 2 学(2) E(X) = J二xf(x)dx =Ixl =02. 设连续熨随机变量X的分布函数为(0, x < -lfF(x) = a + b arcsinx, 1 x < 1,(1, x> 1.arcsinx的导数为拮试确定常数a,b,并求E(X). 解：1Iarctanx的导数为 f(x) = F

2、9;(x) = Vi*b ,-l <X< 10,其他f1 b f(x)dx = Idx = b arcsinx乂因当一 1Wxv1时PX1 1.1. I x一， dx = 一 arcsinx = -in 71 -x2 nITJi = bn = l,L!卩 b =HF(X) = f(x)dx =J-l.111- arcsinx + H|J a =- n22SHEET 1 OF 10SHEET # OF 10 E(X) = J二xf(x)dx =7=? = 03. 设轮船横向摇摆的随机振幅X的概率密度为(1f(x) = &e 迅 x>0,(0, x < 0.求 E(

3、X).解:E(X)=匸二xf(x)dx =右x e zdx = 14. 设X】,X. Xn独立同分布,均值为人且设Y =扌5X1 Xi，求E(Y).解:E(Y) = EXiliXi) =E(XJL1Xi) = -ni = i5.设(X,Y)的概率密度为f(x,y) = f0 S x V l,y > 0,0, 其他.求 E(X+Y).解:E(X + Y)=亡 Ux + y)f(x,y)dxdy =(x + y)eydxdy =肿扌.萨y + y edy = |SHEET # OF 106.设随机变疑Xb X2相互独立，且X】,X2的概率密度分别为fi(x) = 2e"2x# x

4、> 0,0, x SO,x > 0,x < 0,该题服从指数分布,II:故 E(X)=|人求：(1)E(2X1+3X2); (2)E(2X1 - 3X22); (3)E(X1X2).解：(1) E(2Xi + 3X2) = 2E(XJ + 3E(X2) = 2*- + 3*-=223(2)E(2X1-3X22) = 2E(X1)-3E(X22)= 1-3*+ ©OI x2 3e3xdx0r +«*>I x2 d(e"3x) 丿o=1 3 * x2 e"3x"3x dx2=1 - 3 * 0 +=1-3*II2 1 = 1

5、_3*3*3I e"3x 2xdx 丿o+ xe"3x 3x dx(3)E(X1X2) = E(X1)E(X2) = i*i = iX01210.10.20.120.30.10.27.己知二维随机变m(XzY)的分布律为求 E(X).解:E(X) =Xj XiPij = 0 * 0.1 4- 0 * 0.3 + 1 * 0.2 4- 1 *+ 2 * 0.1 4- 2 * 0.2 = 0.98.设随机变量X的概率密度为0 < x < 1, 其他.SHEET 2 OF 10且E(X)-0.75,求常数c和a.解：E(X)=二 xf(x)dx = Jq1 X cxa

6、dx = 0.75习题421.设离散型随机变帚X的分布律为X-100.512p0.10.50.10.10.2求 E(X),E(X2),D(X)解：E(X) = (-1) *0.1 + 0* 0.5 + 0.5 *0.14-1*0.1+2* 0.2 = 0.45E(X2) = (-1)2 *0.1 + 0* 0.5 + (0.5)2 * 0.1 + l2 * 0.1 + 22 * 0.2 = 1.025D(X) = (-1 一 0.45)2 * 0.1 + (0- 0.45)2 * 0.5 + (0.5 一 0.45)2 * 0.1 + (1 - 0.45)2 * 0.1 + (2 - 0.45

7、)2 * 0.2 = 0.8225P(X = 0 = | = 0.1PX = 2 = | =0.32. 盒中有5个球,其中有3个白球,2个黑球，从中任取两个球，求白球数X的期望和方差. 解:X的可能取值为0,1,2注盘此处不可以用二项分布式：IPX = k =蹲 pkqti-k:E(X) = 0*0.1 + 1*0.6 + 2*03 = 1.2D(X) = (0- 1.2)2 * 0.1 + (1 - 1.2尸 * 0.6 + (2 - 1.2)2 * 0.3 = 0.144 + 0.024 + 0.192 = 0.363. 设随机变最X,Y相互独立，他们的概率密度分别为2e"2x,

8、 x > 0,0, x S 0,00,其他，SHEET 3 OF 10SHEET # OF 10求 D(X+Y).解:D(X 4-Y) = D(X) + D(Y) = $ + 玄竺= '7、' V 722121924. 设随机变鼠X的概率密度为(x(x) = -e-|xLoo < x < +8,2求 D(X)解:E(X) = f=0oo 2严00 x2=I e_|x| dx = 2 丿一8 £D(X)=E(X2)- E(X)2 = 2设随机变量X与Y相互独立，且D(X)=1,D(Y)=2,求D(X-Y). 解：D(X 一 Y) = D(X) + D(

9、Y) =1 + 2 = 36.若连续型随机变iftX的概率密度为E(X2)+ 00x2e_x = 200广扌eT刃dxJ o© 乙此为奇函数，故=0U汕正负无穷带入结果都一样，故5.z、 fax2 + bx + c 0 < x < 1, f(x) = I 0,其他且 E(X)=0.5rD(X)=0.15.求常数 a,b,c. 解：E(X)c十厂0.5f19a=I x(ax + bx + c) dx =云 +Jo1丄a bcE(X2) = I x2(ax2 4-bx-l-c)dx = -4- + - = 0.15 十(0.5)2 = 0.4 JqS °d严 8a

10、bJ f(x)dx = J (ax2 + bx十 c)dx = 3 + ? + °= 1解得 a=12,b=-12,c=3.SHEET 4 OF 10SHEET # OF 10习题431. 设两个随机变帚:X,Y相互独立，方差分别为4A. 8B. 16C. 28 D. 442. 设二维随机变帚(X,Y)的概率密度为“、(；(x + y) 0*xs2,0SyS2,f(x, y) = f 8(0,其他和2,则随机变帚3X-2Y的方差是D(a + b + cx)dx求 Cov(X,Y).解：x2:=(ax + bx + c) 2+ 8OOSHEET # OF 10270 = 622 %r2

11、 x2xE(X) = J J -(x+y)dydx = J (-y+ -E(Y) = J J 才(x + y)dxdy = £e(xy)= J 牛& + y)dyx = ?4 7 71Cov(KY) = E(XY) 一 E(X)E(Y) = 了一三康三=一土3 0 0303.设二维随机变量(X,Y)的概率密度为f(& y)=ye(x+y), x > 0,y > 0,0,其他求X与Y的相关系数pxy. 解：SHEET 5 OF 10SHEET # OF 10E(X) =+°°xye 一(x+y)dy)dx = 1SHEET # OF 10

12、SHEET # OF 10r+E(Y) = I (I y2e(x+y)dx)dy 丿0丿or+<* r+°°=I ( y2e_xeydx)dy 丿0丿0r+°°=I y2eydy 丿0r + °°=-y2d(ey) 丿0SHEET 6 OF 10SHEET # OF 10+(T十旷皿3)r + °°= 04-1 e_y 2y dy丿0:运用分部积分法.:fe-y-ydy服从的折数分布SHEET # OF 10SHEET # OF 10=2 | e_y y dy = 2 丿0E(XY)=xy2e_(x+y)dy

13、)dx = 2SHEET # OF 10Cov(X,Y) = E(XY) 一 E(X)E(Y) =2-2*l = 0 所以Cov(X, Y)pxy =厂 一= 0x/DQOx/DW4. 设二维随机变量(X,Y)服从二维正态分布，且E(X)=O, E(Y)=O? D(X)=16# D(Y)=25, Cov(X,Y)“乙求(X,Y)的联合概 率密度函数f(x,y).解：f(s)=融冶1 r(X-Pl)2 “(X-Pl)(y-U2)(y-U2)2、 e 2(l-p2)<_- z p-H o22 j E(X) = 0, E(Y) = 0 b = 0，n2 = 0, D(X) = 16,D(Y)

14、= 25 <i = 4, a2 = 5 Cov(X,Y) = 12Cov(X, Y)123:.p=#(x)>/n(n4*551_25rx_3xy, Af(x,y) =e辽(矿药十刃5. 证明 D(X-Y)=D(X)+D(Y)-2Cov(X/Y).证：D(X- Y) = EX-Y - E(X 一 Y)2= E(X-E(X)-(Y-E(Y)2=E (X - E(X)2 一 2EX 一 E(X) EY 一 E(Y) + E (Y 一 E(Y)2=D(X) + D(Y) 一 2Cov(X, Y)6. 设(X,Y)的协方差矩阵为C =(二：),求X与Y的相关系数pxy.解:弋=(爲3) Co

15、v(X, Y) = -3, D(X) = 4, D(Y) = 9Cov(X, Y) -31 pxy = =-=-yn(X)>/D(Y)2*32自测题4一、选择题1.设随机变帚:X服从参数为0.5的指数分布，则下列各项中正确的是A. E(X)=0.5, D(X)=0.25B. E(X)=2, D(X)=4C. E(X)=0.5, D(X)=4D. E(X)=2, D(X)=0.25解：指数分布的E(X) = i D(X)=令SHEET 7 OF 10SHEET # OF 102.34.设随机变帚X,Y相互独立，且XB(16,0.5),Y服从参数为9的泊松分布，则D(X-2Y+1)=_CA.

16、-14 B. 13 C. 40 D. 41解:D(X) = npq = 16 * 0.5 * 0.5 = 4, D(Y)=入=9D(X 一 2Y 十 1) = D(X) + 4D(Y)十 D(l) = 4 + 4 邓 9 十 0 = 40己知 D(X)=25/D(Y)=1/ pxy=0.4,则 D(X-Y)= BA.6 B. 22 C. 30 D. 46设(X,Y)为二维连续随机变鼠则X与Y不相关的充分必耍条件是_.A. X 与 Y 相互独立B. E(X+Y)=E(X)+E(Y)C. E(XY)= E(X)E(Y)D. (XM-NZbei勺 °22>0)SHEET 8 OF 1

17、0SHEET # OF 10解：tX与Y不相关 pxy = 0, Cov(X, Y) = 0E(XY)= E(X)E(Y)5.设二维随机变駅(X,Y)N(1,1,4,9,机则 Cov(X/Y)=_BSHEET # OF 10SHEET # OF 10A-|B. 3C. 18D. 36SHEET # OF 10解："繆=討島綸=等，Cov(X,Y) = 3SHEET # OF 106.7.二、1.2.3.4.5.6.7.&三、已知随机变最X与Y相互独立，且它们分别在区间卜1,3和2,4上服从均匀分布，则E(XY)= AA. 3 B. 6C. 10D. 12解:XU(-1,3),

18、YU(2,4)E(X)=Z=1,E(Y) =SHEET 9 OF 10SHEET # OF 10E(XY) = E(X)E(Y) =1*3 = 3设二维随机变量(X,Y)N(O,O,l,l,O),0(x)为标准任态分布两数，则下列结论中错误的是CA. X与Y都服从N(0,l)正态分布B. X与Y相互独立C. Cov(XzY)=lD. (X,Y)的分布函数是(D(x)(y)填空题若二维随机变帚(X,Y)N(b，畑6勺O22, 0),且X与Y相互独立，则0=0解：vCov(X/Y)=0设随机变量X的分布律为3X1012p0.10.20.30.4令 Y=2X+1,则 E(Y)=3解：E(2X+l)=

19、(2*-l+l)*0.1+(2*0+l)*0.2+(2*l+l)*0.3+(2*2+l)*0.4=3 己知随机变量X服从泊松分布，且D(X)=1,则PX=l=_e1_.解:D(X) = A = 1 PX = 1 = e设随机变帚X与Y相互独立，且D(X)= D(Y)=1,则D(X-Y) =2己知随机变最X服从参数为2的泊松分布,E(X2)=6解：E(X)=入=2, D(X)=入=2, E(X2) = E2(X) + D(X) = 4 + 2 = 6设 X 为随机变鼠,IL E(X)=2, D(X)=4,则E(X2)=8.已知随机变量X的分布函数为0, x V 0X0 < x < 4

20、1, x>4则 E(X) = 2.0 < x < 4解：f(x) = F”(x) = 40,其他严xE(X) = J。群乂 = 0设随机变暈X与Y和互独立，且D(X)=2, D(Y)=1Z则D(X-2Y+3)=_6_ 设随机变量X的概率密度函数为SHEET # OF 10-1 < x < 1,0,其他SHEET 10 OF 10试求:(l)E(XL D(X); (2)P|X 一 E(X)| < 2D(X).解：(1) E(X) =|x3 dx = 0ri3彳 3D(X) = E(X2)-E2(X) = J 認x4=z丁 匕祚 P|X- E(x)| <

21、2D(X) = P |X| <| = J%f(x)dx = J；|x2 dx = 1四、设随机变量X的概率密度为'x 0 x < 1f(x) = 2 x, 1 < x < 20,其他试求:(l)E(XL D(X); (2)E(Xn),其中n为正整数.解： E(X) = J： x2 dx + J；x(2 - x) dx =扌 + 扌=1D(X) = E(X2) 一 E2(X)=x2(2 - x) - 1 = - 4-SHEET 11 OF 10SHEET # OF 10(2) E(Xn) = xn+1 dx + f2 xn(2 一 x)=严“7' 7 v

22、7 JoJl v 7(n+l)(n+2)五、设随机变帚 XA 与 X2 相互独立，且 XrN(p, a2), X2N(H，*)令 x= X1+X2/ Y= XrX2.求:D(X), D(Y);(2)X与Y的相关系数pxy.解：(1) D(X) = D(X十 X2) = D(Xx) + D(X2) = a2 + a2 = 2o7D(Y) = D(Xx 一 X2) = D(Xj + D(X2) = 2a2(2) Cov(X, Y) = E(XY) 一 E(X)E(Y) = 0Cov(X, Y) npxy = 厂= 0x/D(X)/D(Y)六、设随机变帚x的概率密度为x > 0, x兰0SHEET # OF 10(I)求 E(X),D(X);令Y =求Y的概率密度fY(y)."D(X)解: E(X) =Q°°2x严dx = #D(X) =E(X2) 一 E2(X)r+OO=12x2e"2xdx11114 _ 24 4由Y=2X-1得X =号巳Xz=|：