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Can someone show me why for a distribution function $F$ on $(0,\infty)$, $0 < y \leq x$, the following holds:

$\frac{F(x)-F^{*2}(x)}{\overline{F}(x)}=\int\limits_{0}^{x}\frac{\overline{F}(x-t)}{\overline{F}(x)}dF(t)$

Edit: $F^{*2}$ stands for the convolution of $F$ with itself:

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

$\overline{F}=1-F$

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What does the notation $F^{*2}(x)$ and $\overline{F}(x)$ mean? Also, if this happens to be homework, please add the homework tag. – Macro May 22 '12 at 0:49
Thank you, I've edited it. This not homework. – Chris May 22 '12 at 0:53
2  
Hint: $F(x) := \int_0^x \mathrm dF(t)$. – cardinal May 22 '12 at 1:32
Chris, @cardinal's astute hint gives the solution immediately, so why don't you write up an answer yourself and maybe even accept your own answer if you like it? – Dilip Sarwate May 22 '12 at 2:25
1  
Chris, you can answer your own question; you may just have to wait a few hours. But, I think you've likely exceeded the necessary threshold already. Cheers. – cardinal May 22 '12 at 9:17
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