概率论与数理统计第三章课后习题与参考 答案.pdf

文档介绍：(X,Y)只能取下列数组中的值:(0,0),(1,1),(1,)及(2,0),31115且取这几组值的概率依次为,,和,求二维随机变量(X,Y):由二维离散型随机变量分布律的定义知,(X,Y)的联合分布律为1XY01311,其中来自理科的2名,,Y分别为主席来自理科、工科的人数,求:(1)(X,Y)的联合分布律;(2):(1)由题意,X的可能取值为0,1,2,Y的可能取值为0,1,2,3,则C31C2C193,33,P(X0,Y0)3P(X0,Y1)3C856C856C1C29C3133,3,P(X0,Y2)3P(X0,Y3)3C856C856C1C23C1C1C1923,233,P(X1,Y0)3P(X1,Y1)3C828C828C1C2323,,P(X1,Y2)3P(X1,Y3)0C828C2C13C2C1323,23,P(X2,Y0)3P(X2,Y1)3C856C856P(X2,Y2)0,P(X2,Y3)0.(X,Y)的联合分布律为:1YX0123pi1991505656565614393151028282828333200565628515151pj128282856(2)(X,Y)的概率密度为k(6xy),0x2,2y4,f(x,y)0,:(1)常数k;(2)P(X1,Y3);(3)P(Y);(4)P(XY4).解:方法1:42(1)1f(x,y)dxdyk(6xy)dxdy2044122k(6xxyx)|0dyk(102y)dy8k,2221k.831311(2)P(X1,Y3)f(x,y)dxdy(6xx2yx)dxdy2023**********(6xxyx)|0dy(yy)|2.8228228(3)P(X)P(X,Y)(6xy)dxdy(6xy)dxdy(6xx2yx)dy2088222163327(yy2)|4.884232124x(4)P(XY4)f(x,y)dxdydx(6xy)dy02xy482112121322(128xx)dx(12x4xx)|0.8201633方法2:(1)同方法1.(2)0x2,2y4时,yxyx1F(x,y)f(u,v)dudv(6uv)dudv208yy112x112(6uuuv)|0dv(6xxxv)dv822822111111(6xvx2vxv2)|y(6xyx2yxy210x2),8222822其他,F(x,y,)0,111(6xyx2yxy210xx2),0x2,2y4,F(x,y)8220,(X1,Y3)F(1,3).8(3)P(X)P(X,Y)P(X,2Y4)P(X,Y4)P(X,Y2)27F(,4)F(,2).32(4)(X,Y)的概率密度为Aex2y,x0,y0,f(x,y)0,:(1)常数A;(2)(X,Y)的联合分布函数.解:(1)1f(x,y)dxdyAex2ydxdy00x2yx12yAAedxedyA(e)|0(e)|0,0022A(2)x0,y0时,yxyxF(x,y)f(u,v)dudv2eu2vdudv0012(eu)|x(e2v)|y(1ex)(1e2y),020其他,F(x,y)0,(1ex)(1e2y),x0,y0,F(x,y).0,(X,Y)的概率密度为Axy,0x1,0y1,f(x,y)0,:(1)常数A;(2)(X,Y)的联合分布函数.1111解:(1)1f