# Cumulative distribution function and convolution.

Hello, Given a probability distribution of a discrete variable p1(x) and a probability distribution of a discrete variable p2(y) defined by p2(y) = Sum_{x,x'} p1(x) p1(x') * KroneckerDelta((x+x')/2 = y). (1) Let F1(x) be the cumulative distribution function (CDF) of p1(x): F1(x) == Sum_{x'<=x} p1(x') and let F2(x) be the CDF p2(x).

Is there a way of expressing F2(x) ONLY in terms of F1(x)? If there is none, is there any known (tight) upper and lower bound for F2(x) that is a function ONLY of F1(x)?

Thank you! Best Michele

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Let $X$ be your given discrete variable, with distribution $p1$, and $Y$ the second one with distribution $p2$. Your definition of $Y$ means that $2Y$ is the sum of two independent variables distributed like $X$. Thus the distribution of $2Y$ is the convolution of the distribution of $X$ with itself.

That is: $$G(y)=\int F(2y-x)dF(x)=\sum F(2y-x_j)\lambda_j,$$ where $F,G$ are the cumulative distributions of $X$, $Y$, $x_j$ are the values of $X$ and $\lambda_j=p1(x_j)$.

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Dear Alexander, Thanks! But this doesn't answer my question... Best Michele –  Michele Sep 27 '12 at 2:04
Why? Did not I tell you how to express the distribution of Y in terms of the distribution of X? –  Alexandre Eremenko Sep 27 '12 at 11:40
Well the point is that in my question I specified that I need F2(y) as a function ONLY of F1(y). So maybe I was not clear enough but what I meant is that I want F2 as a function only of F1 and not of its finite differences (derivatives in the continuous case). The expression that you gave involves p1, which is the 'derivative' of F1. –  Michele Oct 8 '12 at 0:11

Let $X$, $Y$ be independent random variables and let $F_X$ and $F_Y$ denote their cumulative distribution functions. Let and $Z=X+Y$, then $F_Z$ can be expressed in terms of $F_X$ and $F_Y$: $$F_Z(x) = \frac{d}{dx}\int_{-\infty}^\infty F_X(t)F_Y(x-t)~dt.$$ Using the convolution '$*$', this can be written as $F_Z=(F_X*F_Y)'$.

With this we can express the $F_2$ in your problem in terms of $F_1$: $$F_2(x) = \frac{d}{dx}\int_{-\infty}^\infty F_1(t)F_1(2x-t)~dt.$$

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