夏冬虫
幼苗
共回答了19个问题采纳率:89.5% 举报
由性质得:
F(+∞,+∞)=1,
则
A(B+arctan x/2)(C+arctan Y/3) =A(B+π/2)(C+π/3)
F(-∞,+∞)=0
A(B+arctan x/2)(C+arctan Y/3) =A(B-π/2)(C+π/3)
F(+∞,-∞)=0
A(B+arctan x/2)(C+arctan Y/3) =A(B+π/2)(C-π/2)
解得:A=6/(11π),B=π/2,C=π/2
(X,Y)的联合概率密度:
6/(11π)(π/2+arctan x/2)(π/2+arctan Y/3)
边缘分布函数及边缘概率密度:
f(x)=∫f(x,y)dy
f(y)=∫f(x,y)dx
f(x,y)=d^2(F(x,y))/dxdy
所以f(x)=d(F(x,y))/dx=6/(11π)*2/(x^2+4)*(π/2+arctan Y/3)
f(y)=d(F(x,y))/dy=6/(11π)*(π/2+arctan x/2)*3/(x^2+9)
1年前
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