Question

二维随机变量（x，y）服从平面区域D={0<=x<1,|y|<=x}上的均匀分布，求随机变量Z=X+Y的概率密度函数

Answer 1

http://zhidao.baidu.com/question/1605291166027624587

答: f(z) = 1-(z/2), 0<z<2; =0, 其它.

证明一(阶跃函数法): 先回忆一下阶跃函数的定义: u(x)=1, x>0; =0, x<0.

f(x,y)=u(x)u(1-x)u(x-y)u(y-(-x))= u(x)u(1-x)u(x-y)u(y+x)

这里: u(x-y)表明y要小于x. 

u(y-(-x))表明y要大于-x.

Z=X+Y. 有公式: f(z)=∫[-∞,∞]f(w,z-w)dw  

f(z)=∫[-∞,∞] u(w)u(1-w)u(w-z+w)u(z-w+w)dw

=∫[-∞,∞] u(w)u(1-w)u(2w-z)u(z)dw

= {∫[0,1] u(2w-z)dw}u(z)

= {∫[0,1] u(w-(z/2))dw}u(z)

= {∫[z/2,1] 1dw}u(z)u(1-(z/2))

= (1- z/2)u(z)u(2-z)

= 1 - (z/2), 0<z<2; =0, 其它. 

证毕!

证明二(初等几何法): 

F(z) = P(Z<z) = 图中梯形面积 = 1-(1-(z/2)^2) = z-(1/4)z^2

f(z) = dF(z)/dz = 1-(z/2), 0<z<2; =0, 其它.