why is that $\frac{d}{dz}\int_{-\infty}^\infty\int_{-\infty}^{z-y}f_{X,Y}(x,y)\,dx\,dy=\int_{-\infty}^\infty f_{X,Y}(z-y,y)\,dy$

Question

This is part of proving the $X+Y=Z$,then,$f_z(z)=f_x(x)*f_y(y)$

$F_z(z)=\int_{-\infty}^\infty \int_{-\infty}^{z-y}f_{X,Y}(x,y)\,dx\,dy$

Then $f_Z(z)=\frac{dF_z(Z)}{dz}=\int_{-\infty}^\infty f_{X,Y}(z-y,y)\,dy$, i don't quite understand why is that $\frac{d}{dz}\int_{-\infty}^\infty \int_{-\infty}^{z-y} f_{X,Y}(x,y)\,dx\,dy=\int_{-\infty}^\infty f_{X,Y}(z-y,y)\,dy$,can anyone explain more about it to me? thankyou.

Answer 1

Let $h(z)=\int_{-\infty}^{\infty} f_{X,Y} (z-y,y)\, dy$. Then $\int_{-\infty}^{z} h(u) \, du =\int_{-\infty}^{z}\int_{-\infty}^{\infty} f_{X,Y} (u-y,y)\, dy du$. If you interchange the two integrals and put $x=u-y$ you get $\int_{-\infty}^{\infty} \int_{-\infty}^{z-y} f_{X,Y} (x,y)\, \, dxdy $. Hence $h$ is the derivative of this last expression.

Answer 2

The fundamental theorem of calculus states: $$\frac{d}{dx}\int_a^x f(t)\,dt = f(a).$$ Let's consider how this relates to your problem.

First, we can bring the derivative under the first integral sign (since the outer integral is only a function of $y$:

$$\int_{-\infty}^{\infty} \frac{d}{dz} \int_{-\infty}^{z-y}f_{X,Y}(x,y)\,dx\,dy.$$

Let $\tilde{f}_{X,Y}(x,y) = f_{X,Y}(x-y,y)$. Then the above expression equals $$\int_{-\infty}^{\infty} \frac{d}{dz} \int_{-\infty}^z \tilde{f}_{X,Y}(x,y)\,dx\,dy.$$

By the fundamental theorem of calculus, this equals, $$\int_{-\infty}^\infty \tilde{f}_{X,Y}(z,y)\,dx\,dy.$$

Replacing our definition of $\tilde{f}$, we get $$\int_{-\infty}^{\infty} f_{X,Y}(z-y,y)\,dy.$$