设随机变量(X,Y)的概率密度为f(x,y)=x+y,0
1个回答
展开全部
F(z)=P(Z<z)=p(xy<z)
(1) z<=0时：F(z)=0,f(z)=0
(2) 0<z<=1时：f(z)=p(xy =z)
=1-∫(z,1)dx∫(z/x,1)(x+y)dy
=1-∫(z,1)(x+1/2-z-1/2*z^2/x^2)dx
=1-(1-2z+z^2)=2z-z^2
f(z)=2-2z
(3) z>1时：F(z)=1,f(z)=0</z<=1时：f(z)=p(xy </z)=p(xy<z)
(1) z<=0时：F(z)=0,f(z)=0
(2) 0<z<=1时：f(z)=p(xy =z)
=1-∫(z,1)dx∫(z/x,1)(x+y)dy
=1-∫(z,1)(x+1/2-z-1/2*z^2/x^2)dx
=1-(1-2z+z^2)=2z-z^2
f(z)=2-2z
(3) z>1时：F(z)=1,f(z)=0</z<=1时：f(z)=p(xy </z)=p(xy<z)
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
