Inequality convolution distribution function

Question

$F$ 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

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

Hint: Use monotonicity twice. First, for an arbitrary distribution function $F$ and, second, for the integral.
Note also that the nonnegativity restriction allows for an alternate simple probabilistic proof requiring no calculation at all. What might that be?
@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).
It is also possible to derive the result from integration by parts.
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.

Chris · Accepted Answer · 2012-05-24 14:14:38Z

up vote 5 down vote accepted

Answer by Dilip Sarwate and cardinal:

$F^{*2}(x)=P(X+Y \leq x) \leq P(X \leq x, Y \leq x)=F^{2}(x).$

Thank you very much!

answered May 24 '12 at 14:14

Chris
26219

	(+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
	+1 I am glad you wrote up the hints succinctly. – Dilip Sarwate May 24 '12 at 14:35

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