Question

设二维随机变量(X,Y)的概率密度为f(x,y)=2-x-y ，0<x<1,0<y<1, 求Z=X+Y的概率密度

Answer 1

设T=X-Y
则X=(Z+T)/2
Y=(Z-T)/2
f(z,t)=f(x(z,t),y(z,t))*|det(jacobian)|
jacobian=[(dx/dz,dx/dt),(dy/dz,dy/dt)]=[(1/2,1/2)(1/2,-1/2)]
|det(jacobian)|=|-1/4-1/4|=1/2

f(x(z,t),y(z,t))=2-(x+y)=2-z
f(z,t)=(2-z)/2
T=X-Y~(0,1)
fZ(z)=∫(t~(0,1))f(z,t)dt=(2-z)t/2](t~(0,1))=(2-z)/2
0<z<2

Answer 2

设T=X-Y
则X=(Z+T)/2
Y=(Z-T)/2
f(z,t)=f(x(z,t),y(z,t))*|det(jacobian)|
jacobian=[(dx/dz,dx/dt),(dy/dz,dy/dt)]=[(1/2,1/2)(1/2,-1/2)]
|det(jacobian)|=|-1/4-1/4|=1/2

f(x(z,t),y(z,t))=2-(x+y)=2-z
f(z,t)=(2-z)/2
T=X-Y~(0,1)
fZ(z)=∫(t~(0,1))f(z,t)dt=(2-z)t/2](t~(0,1))=(2-z)/2
0<z<2