# Inequality convolution distribution function

$F$ is the distribution function of a positive random variable. Why does the following hold for every $x \in (0,\infty)$:

$F^{*2}(x) \leq F^2(x)$, where

$F^{*2}=F*F=\int\limits_0^x F(x-y)dF(y)$

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Hint: Use monotonicity twice. First, for an arbitrary distribution function $F$ and, second, for the integral. –  cardinal May 23 '12 at 17:04
Note also that the nonnegativity restriction allows for an alternate simple probabilistic proof requiring no calculation at all. What might that be? –  cardinal May 23 '12 at 17:18
@cardinal +1 for a great comment! Once again, you have suggested the answer to Chris's question with a one-liner. (I like this one better than your earlier suggestion). –  Dilip Sarwate May 23 '12 at 18:26
It is also possible to derive the result from integration by parts. –  Stéphane Laurent May 23 '12 at 18:51
Since this discussion does not seem to be converging, let me add to @cardinal's hint. One of the two expressions is $P\{X+Y \leq x\}$ and the other is $P\{X \leq x, Y \leq x\}$ where $X$ and $Y$ are independent identically distributed positive random variables with common distribution function $F(\cdot)$. One of the two sets $\{X+Y \leq x\}, \{X \leq x, Y \leq x\}$ is a subset of the other. So no integral need be evaluated. –  Dilip Sarwate May 23 '12 at 21:09
$F^{*2}(x)=P(X+Y \leq x) \leq P(X \leq x, Y \leq x)=F^{2}(x).$
(+1) The key here is to recognize that $\{X+Y \leq x\} \subset \{X \leq x, Y \leq x\}$. A good follow-up is to come up with a (very!) simple counterexample to show that the result fails when $X$ can take on negative values. –  cardinal May 24 '12 at 14:25