这个问题可能很古老,但是我想提供另一个观点。它基于联合概率密度变量变化的公式。可以在2017年KTH的演讲笔记:概率和随机过程中找到。(Koski,T.,2017,pp 67),它本身是Analysens Grunder,del 2(Neymark,M.,1970,pp 148-168)中的详细证明:
X=(X1,X2,...,Xm)fX(x1,x2,...,xm)Y=(Y1,Y2,...,Ym)
Yi=gi(X1,X2,...,Xm),i=1,2,...,m
gi(g1,g2,...,gm)
Xi=hi(Y1,Y2,...,Ym),i=1,2,...,m
Y
fY(y1,y2,...,ym)=fX(h1(x1,x2,...,xm),h2(x1,x2,...,xm),...,hm(x1,x2,...,xm))|J|
J
J=∣∣∣∣∣∣∣∣∣∂x1∂y1∂x2∂y1⋮∂xm∂y1∂x1∂y2∂x2∂y2⋮∂xm∂y2......⋱...∂x1∂ym∂x2∂ym⋮∂xm∂ym∣∣∣∣∣∣∣∣∣
X1+X2
X=(X1,X2)fX(x1,x2)Y=(Y1,Y2)
Y1=g1(X1,X2)=X1+X2Y2=g2(X1,X2)=X2.
逆映射是
X1=h1(Y1,Y2)=Y1−Y2X2=h2(Y1,Y2)=Y2.
X1X2Y
fY(y1,y2)=fX(h1(y1,y2),h2(y1,y2))|J|=fX(y1−y2,y2)|J|=fX1(y1−y2)⋅fX2(y2)⋅|J|
J
J=∣∣∣∣∂x1∂y1∂x2∂y1∂x1∂y2∂x2∂y2∣∣∣∣=∣∣∣10−11∣∣∣=1
Y1=X1+X2
fY1=∫∞−∞fY(y1,y2)dy2=∫∞−∞fX(h1(y1,y2),h2(y1,y2))|J|dy2=∫∞−∞fX1(y1−y2)⋅fX2(y2)dy2
这就是我们找到您的卷积的地方:D