设二维随机变量（X,Y）的联合密度函数为。。。求概率等。

Question

就是这道题，求这三问的详细解答。谢谢~~

Answer 1

1)

P(xy<1)很简单，就是对下图阴影的面积求二重积分

∫(1/2~2)∫(1/2~1/y) 1/(4x²y³) dxdy

= ∫(1/2~2) 1/(4(1/2)y³)-1/(4(1/y)y³) dy

=  ∫(1/2~2) 1/(2y³)-1/(4y²) dy

=(-1/(4y²)+1/(4y))|(1/2~2)

=(-1/16+1/8)-(-1+1/2)

=1/16+1/2

=9/16

3)

 ∫(1~3)∫(1/2~) 1/(4x²y³) dydx

= ∫(1~3) {1/(8x²(1/4))}dx

=-1/(2x) |(1~3)

=1/2-1/6

=1/3

追问

亲，还有第二问呢

追答

你为什麽自己划了。还以为你自己做了
x>1/2,y>1/2时
F(x,y)
= ∫(1/2~y)∫(1/2~x)1/4x²y³ dxdy
=∫(1/2~y) 1/(2y³)-1/(4xy³) dy
=-1/(4y²)+1/(8xy²)|(1/2~y)
=1-1/(4y²)+1/(8xy²)-1/(2x)
=(1-(1/4y²))(1-1/(2x))

x<1/2或y<1/2时
F(x,y)=0

边缘分布函数
fx(x)=dF(x,无穷)/dx=(1-1/(2x))'=1/2x²
fy(y)=dF(无穷，y)/dy=(1-(1/4y²))'=1/2y³

所以
f(x,y)=1/(4x²y³)=fx(x)fy(y)

x,y相互独立