CDF of a sum of independent random variables

Let $X,Y$ be two independent (and identically distributed) random variables. Let $Z:=X+Y$.

It's easy to check that the moment generating function $\phi_Z(t):=\mathbb{E}[\,e^{itZ}\,]$ can be expressed as $\phi_Z=\phi_X\cdot\phi_Y$.

Is there a way to express the cumulative distribution function $F_Z(z):=\mathbb{P}(Z\leq z)$ using the cumulative distribution functions of $X$ and $Y$ ?

Edit: note I don't assume that $X$ and $Y$ have a density.

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\begin{eqnarray*} F_Z \left( z \right) & = & \int F_X \left( z - y \right) dF_Y \left( y \right)\\ & = & \int F_Y \left( z - x \right) dF_X \left( x \right) \end{eqnarray*}

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I don't assume $X$ and $Y$ have a density, I edit the question to make it clear. Does the second formula you wrote always hold? The integral is a Riemann-Stietjes integral? Is it possible to generalize the formula to a sum of $n$ i.i.d. random variables? –  qwertyuio Dec 23 '12 at 14:22
Yes, the integral is Lebesgue-Stieltjes, and it holds generally. I guess it could be generalized to $n$ variables. –  Learner Dec 23 '12 at 14:24

Yes

$$F_Z(z) = P(Z\leq z) = P(X+Y\leq z) = \int_{\mathbb{R}}\int_{-\infty}^{z-x}f_{X,Y}(x,y)dydx = \int_{\mathbb{R}}F_Y(z-x)f(x)dx.$$

BTW, if differentiate it with respect to $z$ you obtain $$f_Z(z) = (f_X*f_Y)(z)$$ where $*$ stands for convolution.

EDIT: Followed by Dilip comment, we also have the relation

$$F_Z(z) = (F_X*f_Y)(z) = (f_X*F_Y)(z)$$

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+1 In fact, $F_Z(z) = (F_X*f_Y)(z) = (f_X*F_Y)(z)$. –  Dilip Sarwate Dec 23 '12 at 14:17
I don't assume $X$ and $Y$ have a density, I edit the question to make this point clear. –  qwertyuio Dec 23 '12 at 14:19
In this case, Learner answer is what you seek for (after "This could have been written also as").. –  user39097 Dec 23 '12 at 14:20