概率论第三章

1、(完好版)概率论第三章答案(完好版)概率论第三章答案21/21(完好版)概率论第三章答案习题3-1已知随机变量X1和X2的概率分布分别为X1-101P111424X201P1122并且PX1X201.求X1和X2的联合分布律.解由PX1X201知PX1X200.所以X1和X2的联合分布必形如X201piX1-1P110140P21P22121P31014pj11122于是依据边沿概率密度和联合概率分布的关系有X1和X2的联合分布律X201piX1-110144001122110144pj11122(2)注意到PX10,X200,而PX10PX2010,所以X1和X24不独立.一盒子中有3只黑球

2、、2只红球和2只白球,在此中任取4只球.以X表示取到黑球的只数,以Y表示取到红球的只数.求X和Y的联合分布律.解从7只球中取4球只有C7435种取法.在4只球中,黑球有i只,红球有j只(余下为白球4ij只)的取法为C3iC2jC24ij,i0,1,2,3,j0,1,2,ij4.于是有PX0,Y2C30C22C221PX1,Y1C31C21C21635,35,3535PX1,Y2C31C22C2162,Y0C32C20C22335,PX35,3535PX2,Y1C32C21C21122,Y2C32C22C20335,PX35,3535PX3,Y0C33C20C212PX3,Y1C33C21C20

3、235,35,3535PX0,Y0PX0,Y1PX1,Y0PX3,Y20.分布律的表格形式为X0123Y00032353510612235353521630353535设随机变量(X,Y)的概率密度为k(6xy),0 x2,2y4,f(x,y)其余.0,求:(1)常数k;(2)PX1,Y3;(3)PX1.5;(4)PXY4.解(1)由f(x,y)dxdy1,得424124dyxy)dxk(6y)xx2dyk(10yy2)8k,1k(6202202所以k1.811(2)PX1,Y3f(x,y)dxdy3dyxy)dx(6x1,y3208131113(113(6y)xx2dyy)dy.822082

4、281.5fX(x)dx1.5(3)PX1.5dxf(x,y)dy41.51xy)dxdy(62081411.5(6y)xx2dy822014(633y)dy828227.32(4)作直线xy4,并记此直线下方地域与f(x,y)0的矩形地域(0,2)(0,4)的交集为G.即G:0 x2,0y4x.见图3-8.所以PXY4P(X,Y)Gf(x,y)dxdy4dy4x1xy)dx20(68G1414x(6y)x2dy8x22014(6y)(4y)1(42dy822y)142(4y)1(4y)2dy8221142(4y)2(4y)3.8623图3-8第4题积分地域二维随机变量(X,Y)的概率密度为k

5、xy,x2f(x,y)y1,0 x1,0,其余.试确立k,并求P(X,Y)G,G:x2yx,0 x1.11k1x4)dxk,解由1f(x,y)dxdydx2kxydyx(10 x206解得k6.1x12x4)dx1因此P(X,Y)Gdxx26xydy3x(x.004设二维随机变量(X,Y)概率密度为4.8y(2x,x),0 x1,0yf(x,y)其余.0,求关于X和Y边沿概率密度.解(X,Y)的概率密度f(x,y)在地域G:0 x1,0yx外取零值.因此,有xx)dy,0 x1,fX(x)f(x,y)dy4.8y(200,其余.2.4(2x)x2,0 x1,0,其余.10y1,fY(y)f(x

6、,y)dx4.8y(2x)dx,y其余.0,2.4y(34yy2),0y1,0,其余.假设随机变量U在区间-2,2上遵从均匀分布,随机变量1,若U1,1,若UY1,X若U1,1,若U1.1,试求：(1)X和Y的联合概率分布；(2)PXY1.解(1)见本章第三节三(4).(2)PXY11PXY11PX1,Y1131.习题3-2441.设(X,Y)的分布律为Y1234X10.100.1020.300.10.2300.200求:(1)在条件X=2下Y的条件分布律;PX2Y2.解(1)因为PX20.300.10.20.6,所以在条件X=2下Y的条件分布律为PY1|XPX2,Y10.31220.6,PX

7、2PY2|XPX2,Y200,220.6PXPY3|XPX2,Y30.11220.6,PX6PY4|XPX2,Y40.21220.6,PX3或写成Yk1234PYk|X21011263注意到PY2PY1PY20.10.3000.20.6.而PX2,Y2PX2,Y1PX2,Y2PX3,Y1PX3,Y2所以0.3000.20.5.PX2Y2PX2,Y20.55PY20.6.62.设平面地域D由曲线y1及直线y0,x1,xe2所围成,二维随机变量(X,Y)x在地域D上遵从均匀分布,求(X,Y)关于X的边沿概率密度在x=2处的值.21ee2解由题设知D的面积为SD1xdxlnx12.1(x,y)D,所

8、以,(X,Y)的密度为f(x,y),20，其余.由此可得关于X的边沿概率密度fX(x)f(x,y)dy.明显,当x1或xe2时,fX(x)0;当1xe2时,fX(x)x11dy1022x.故1fX(2).4设二维随机变量(X,Y)的概率密度为1,0 x1,0f(x,y)其余.0,求：(1)(X,Y)的边沿概率密度fX(x),fY(y)；(2)PY解(1)当0 x1时,fX(x)f(x,y)dy当x0时或x1时,()0fXx.故fX(x)2x,0 x1,0,其余.1当0y2时,fY(y)f(x,y)dxydx122x,1X.22xdy2x;0y;2当y0时或y2时,fY(y)0.1y,0y2,故

9、fY(y)20,其余.(2)当z0时,FZ(z)0;当z2时,FZ(z)1;当0z0),试求随机变量和Z=X+Y的概率密度.解已知X和Y的概率密度分别为2N(,),Y遵从均匀分布1(x)21,y(a,a),.fX(x)2x(,);fY(y)2ae2,20,y(a,a).因为X和Y互相独立,所以1a1(zy)222fZ(z)fX(zy)fY(y)dyedy2aa21zaza=()().2a设随机变量X和Y的联合分布是正方形G=(x,y)|1x3,1y3上的均匀分布,试求随机变量U=|X-Y|的概率密度f(u).解由题设知,X和Y的联合概率密度为1113,f(x,y),x3,y40,其余.记F(u

10、)为U的分布函数,拜见图3-7,则有当u0时,F(u)P|XY|u=0;当u2时,F(u)1;当0u2Y;(2)求Z=X+Y的概率密度fZ(z).117解(1)PX2Yf(x,y)dxdy2dy(2xy)dx.x2y02y24(2)方法一:先求Z的分布函数:FZ(z)P(XYZ)f(x,y)dxdy.xyz当z0时,FZ(z)0;FZ(z)zzyxy)dx当0z1时,f(x,y)dxdydy(2D100=z2-1z3;311当1z2时,FZ(z)1f(x,y)dxdy1dy(2xy)dxD2z1zy=1-当z2时,FZ(z)=1.故Z=X+Y的概率密度为13(2-z)3;2zz2,0z1,fZ

11、(z)FZ(z)(2z)2,1z2,0,其余.方法二:利用公式fZ(z)f(x,zx)dx:f(x,zx)2x(zx),0 x1,0zx1,0,其余2z,0 x1,xz1x,0,其余.当z0或z2时,fZ(z)=0;fZ(z)zz)dxz(2z);当0z1时,(201(2z)dx(2z)2.当1z1,11PYX及PY|X.22解(1)当x0或y0时,(x,y)=0,当0 x1,0y2时,(x,y)=x2+1xy,3xy所以F(x,y)(u,v)dudv所以F(x,y)=0.xy1uv)dvdu(u20031x3y1x2y2.12当02时,xy(u,v)dudvxyx2F(x,y)(u,v)dv

12、du(u,v)dvdu0000 x221uv)dvdu1(2x1)x2.(u0033当x1,01,y2时,1.122F(x,y)(uuv)dvdu1003综上所述,分布函数为F(x,y)当0 x1时,X(x)故X(x)当0y2时,Y(y)故Y(y)0,x0或y0,1x2y(xy),0 x1,0y2,341x2(2x1),0 x1,y2,31y(4y),x1,0y2,121,x1,y2.(x,y)dy2(x2xy)dy2x22x,03322x2x,0 x1,30,其余.(x,y)dx(x2xy)dx11y,10336113y,0y2,60,其余.当0y2时,X关于Y=y的条件概率密度为(x,y)6x22xy(x|y)2.Y(y)y当0 x1时,Y关于X=x的条件概率密度为(y|x)(x,y)3xy.