10
$\begingroup$

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.

$\endgroup$
13
$\begingroup$

\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*}

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – qwertyuio
    Dec 23 '12 at 14:22
  • 1
    $\begingroup$ Yes, the integral is Lebesgue-Stieltjes, and it holds generally. I guess it could be generalized to $n$ variables. $\endgroup$
    – Learner
    Dec 23 '12 at 14:24
  • 2
    $\begingroup$ Can you explain why this is true? $\endgroup$
    – Jake
    Apr 20 '20 at 17:31
6
$\begingroup$

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

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

$\endgroup$
2
  • 1
    $\begingroup$ I don't assume $X$ and $Y$ have a density, I edit the question to make this point clear. $\endgroup$
    – qwertyuio
    Dec 23 '12 at 14:19
  • $\begingroup$ In this case, Learner answer is what you seek for (after "This could have been written also as").. $\endgroup$
    – user39097
    Dec 23 '12 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.