Suppose that $X$ and $Y$ are independent continuous RVs with PDFs $f_X$ and $f_Y$ respectively.
I want to find the PDF of $Z = X + Y$.
The CDF of $X + Y$ is
$$F_{X+Y}(z) = P(X + Y \leq z)$$ $$=\int\int_{\{(x,y):x+y\leq z\}}f_X(x)f_Y(y)dx dy$$ $$=\int_{-\infty}^\infty f_Y(y) \bigg\{{\int_{-\infty}^{z-y}f_X(x)dx\bigg\}dy}$$
$$=\int_{-\infty}^\infty F_X(z-y)f_Y(y)dy$$
Now this is all jolly good, but my problem is with deriving a PDF from this. I know that we have:
$$f_{X+Y}(z) = \frac{d}{dz} F_{X+Y}(z)$$
But after this point, I am confused. My professor says the derivation continues like so:
$$f_{X+Y}(z)=\int_{-\infty}^\infty \frac{d}{dz} F_X(z-y)f_Y(y)dy$$ $$=\int_{-\infty}^\infty f_X(z-y)f_Y(y)dy$$
So my two questions are:
1) Why does the $\frac{d}{dz}$ jump inside the integral? Why are we not writing: $$f_{X+Y}(z)= \frac{d}{dz}\int_{-\infty}^\infty F_X(z-y)f_Y(y)dy$$
2) Why does $\frac{d}{dz} F_X(z-y) = f_X(z-y) $?
Under some conditions (which most "decent" functions follow) you can interchange derivation and integration operations. If the extremes of integration depend on the variable $z$ (not the case here) then it's gets a littler more complex. For the general form, and for the precise conditions, see Leibniz integral rule
You are supposed to know that $F_X$ a primitive of $f_X$, then $F_X'=f_X$
Because here the funcion is applied, not to the variable $z$ but to a function of it, we apply the chain rule.
$$\frac{d}{dz} F_X(g(z)) = f_X(g(z)) \frac{d g(z)}{dz} $$
Here $g(z)=z-y$, and it's derivative is $1$.