习题3-1
1. 已知随机变量X1和X2的概率分布分别为 X1 -1 0 11 P 42 X2 0 1 P 2而且P{X1X2?0}?1. 求X1和X2的联合分布律.
解 由P{X1X2?0}?1知P{X1X2?0}?0. 因此X1和X2的联合分布必形如 2 1 141 1 2 XX1 -1 0 1 0 1 0 pi· 141214 1 P11 P21 P31 12 P22 0 p·j 12 于是根据边缘概率密度和联合概率分布的关系有X1和X2的联合分布律
X2 0 1 pi· X1 11 -1 0 44110 0 2211 1 0 44p·j 12 12 141 ?0, 所以
(2) 注意到P{X1?0,X2?0}?0, 而P{X1?0}?P{X2?0}?X1和X2不独立.
2. 设随机变量(X,Y)的概率密度为
?k(6?x?y),0?x?2,2?y?4, f(x,y)??0,其它.?求: (1) 常数k; (2) P{X?1,Y?3}; (3) P{X?1.5}; (4) P{X?Y≤4}.
解 (1) 由?4220?????????f(x,y)dxdy?1, 得
421??dy?k(6?x?y)dx?k?所以 k?12??2(6?y)x?xdy?k(10y?y)?8k, ??2??02241. 8x?1,y?3(2) P{X?1,Y?3}???3f(x,y)dxdy??dy?2131018(6?x?y)dx
131131?? ??(6?y)x?x2dy??(?y)dy?. 822882?2???01(3) P{X?1.5}? ? ??11.5??fX(x)dx??dx???1.5????f(x,y)dy
?42dy?41.5018(6?x?y)dx
1.58?214633??(?y)dy 828227?. 32(4) 作直线x?y?4, 并记此直线下方区域与f(x,y)?0的矩形区域(0,2)?(0,4)的交集为G. 即G:0?x?2,0?y≤4?x.见图3-8. 因此
P{X?Y≤4}?P{(X,Y)?G}
12??(6?y)x?x?dy ?2?0? ???Gf(x,y)dxdy??dy?244?x018(6?x?y)dx
??11[(6?y)(4?y)?(4?y)2]dy ?822141??[2(4?y)?(4?y)2]dy 8228?214412??(6?y)x?x?dy ?2?0?4?x?2???(4?y)?(4?y)??. 8?6?231?2143
图3-8 第4题积分区域
3. 二维随机变量(X,Y)的概率密度为
?kxy,x2≤y≤1,0≤x≤1, f(x,y)??其它.?0,试确定k, 并求P{(X,Y)?G},G:x≤y≤x,0≤x≤1. 解 由1?解得k?6.
因而 P{(X,Y)?G}?2??????????f(x,y)dxdy??dx?2kxydy?0x11k1k4,x(1?x)dx??026?10dx?26xydy?3?x(x2?x4)dx?x0x11. 44. 设二维随机变量(X, Y)概率密度为
?4.8y(2?x),0≤x≤1,0≤y≤x, f(x,y)??其它.?0,求关于X和Y边缘概率密度.
解 (X,Y)的概率密度f(x,y)在区域G:0≤x≤1,0≤y≤x外取零值.因而, 有
概率论与数理统计03-第三章作业及答案
