概率统计，二维连续型随机变量，已知二维随机变量（X，Y）的联合概率密度为 概率统计，二维连续型随机变量，已知二维随机变量（X，Y）的联...
\u6982\u7387\u7edf\u8ba1\u95ee\u9898\uff0c\u4e8c\u7ef4\u8fde\u7eed\u578b\u968f\u673a\u53d8\u91cf\uff0c\u5df2\u77e5\u4e8c\u7ef4\u968f\u673a\u53d8\u91cf\uff08X\uff0cY\uff09\u7684\u8054\u5408\u6982\u7387\u5bc6\u5ea6\u4e3a5\u9898\uff1a
f(x,y)=ke^(-y), 0<x<y.
(1) k \u222b [0, \u221e]{ \u222b [x, \u221e] e^(-y)}dx = k \u222b [0, \u221e] e^(-x) = 1. --> k=1.
(2) P(X+Y\u22641) = \u222b [0, 1/2]{ \u222b [x,1-x] e^(-y)}dx . \u4ee5\u4e0b\u81ea\u5df1\u7b97\u4e0b\u3002
(3) f(x) = \u222b [x, \u221e] e^(-y)dy = e^(-x), x>0.
f(y) = \u222b [0,y] e^(-y)dx = ye^(-y), y>0.
(4) f(x|y) = f(x,y)/f(y) = 1/y, 0<x<y.
6\u9898\uff1af(x,y) = 1/[(Ax^2)y], x>1, (1/x) <y<x.
(1) \u222b [1, \u221e]{ \u222b [1/x, x] (1/[(Ax^2)y]) }dx
\u6ce8\u610f\uff1a\u5916\u79ef\u5206\u4e3ax\u4ece1\u5230\u65e0\u7a77\u3002\u5185\u79ef\u5206\u4e3ay\u4ece(1/x)\u5230x.
I do not know why. I cannot post pictures and cannot write more in Chinese.
题不难，计算太复杂，给你个思路
(1) 1=∫(0,+∞)dx∫(x,+∞)ke^(-y)dy=k
(2)P(X+Y<=1)=∫(0,1/2)dx∫(x,1-x)e^(-y)dy
(3)fX(x)=∫(x,+∞)e^(-y)dy=e^(-x) (x>0)
(4)fY(y)=∫(0,y)e^(-y)dx=ye^(-y) (y>0)
f(x|y)=f(x,y)/fY(y)=1/y (0<x<y)
(5)因f(x,y)≠fX(x)*fY(y)，所以不独立
(6)P(Y>=1/2|X>=1/2)=P(X>=1/2,Y>=1/2)/P(X>=1/2)
=∫(1/2,+∞)dx∫(x,+∞)e^(-y)dy/∫(1/2,+∞)e^(-x)dx
(7)f(y|x)=f(x,y)/fX(x)=e^(-y)/e^(-x)=e^[-(y-x)] (0<x<y)
P(Y>=1/2|X=1/2)=∫(1/2,+∞)e^[-(y-1/2)]dy
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