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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.

2 Answers 2

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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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    Yes, the integral is Lebesgue-Stieltjes, and it holds generally. I guess it could be generalized to $n$ variables.2012-12-23
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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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    In this case, Learner answer is what $y$ou seek for (after "This could have been written also as")..2012-12-23