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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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  • $\begingroup$ What does the notation $F^{*2}(x)$ and $\overline{F}(x)$ mean? Also, if this happens to be homework, please add the homework tag. $\endgroup$
    – Macro
    May 22, 2012 at 0:49
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    $\begingroup$ Hint: $F(x) := \int_0^x \mathrm dF(t)$. $\endgroup$
    – cardinal
    May 22, 2012 at 1:32
  • $\begingroup$ 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? $\endgroup$
    – Dilip Sarwate
    May 22, 2012 at 2:25
  • $\begingroup$ You are right Dilip! By the way, I cannot answer my own question, not enough rep points. $\frac{F(x)-F^{*2}(x)}{\overline{F}(x)}=\frac{\int\limits_0^x dF(t) - \int\limits_0^x F(x-t)dF(t)}{\overline{F}(x)}=\frac{\int\limits_0^x 1- F(x-t)dF(t)}{\overline{F}(x)}=\frac{\int\limits_0^x \overline{F}(x-t)dF(t)}{\overline{F}(x)}=\int\limits_0^x \frac{\overline{F}(x-t)}{\overline{F}(x)}dF(t)$ $\endgroup$
    – Chris
    May 22, 2012 at 3:02
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    $\begingroup$ 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. $\endgroup$
    – cardinal
    May 22, 2012 at 9:17

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