随机变量X与Y相互独立且服从区间（0，a）上的均匀分布，求随机变量函数Z=XY的概率密度

Question

如题所述

Answer 1

f(x,y)=1/a^2 (0<x<a,0<y<a)

Fz(z)
=P(Z<=z)
=P(XY<=z)
=1-P(XY>z)
=1-∫(z/a~a)∫(z/y~a)(1/a^2) dxdy
=1- ∫(z/a~a) (a-z/y)/a^2 dy
=1- (ay-zln(y))/a^2 (z/a~a)
=1- (a^2-zln(a)-z+zln(z/a))/a^2
=1-(a^2-zln(a)-z+zln(z)-zln(a))/a^2
=1-1-(-z)(2ln(a)+1-ln(z))/a^2
=z(2ln(a)+1-ln(z))/a^2

fz(z)=f'z(z)
=(2ln(a)+1-ln(z)+z(-1/z))/a^2
=(2ln(a)-ln(z))/a^2

可以化成ln(a^2/z)/a^2
(0<z<a^2)

fz(z)=0(其他 z)本回答被网友采纳