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Let $E_0\sim$ Exp($\lambda$) and $E_1\sim$ Exp($1$) be independetn r.v. and let $\lambda>1$ and $x>0$.

I am not sure if the bounds of the following integral should be $(0,1)$ or $(0,\infty)$ $$\Bbb P(E_0+E_1>x)=\int\limits_0^1 \Bbb P(E_0>x-s)f_{E_1}(s)ds$$ where $f_{E_1}$ is the density function of $E_1$

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  • $\begingroup$ Why do you think the bounds should be $0$ and $1$? $\endgroup$ – angryavian Jun 12 at 7:47
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    $\begingroup$ @Henry But $s$ is the dummy variable $\endgroup$ – angryavian Jun 12 at 7:48
  • $\begingroup$ Oh the right bound should be $x$ right? I realize now that $1$ doesn't make sense $\endgroup$ – John Cataldo Jun 12 at 7:51
  • $\begingroup$ $P(E_0 >x-s)$ is not $0$ when $x-s$ is negative. It is $1$. So don't ignore $s >x$. $\endgroup$ – Kabo Murphy Jun 12 at 7:54
  • $\begingroup$ related: math.stackexchange.com/questions/474775/… $\endgroup$ – Math-fun Jun 12 at 8:01
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It is $\int_x^{\infty} e^{-s}ds+\int_0^{x}\int_{x-s}^{\infty} \lambda e^{-\lambda x}dx e^{-s} ds$. [$s>x$ cannot be ignored in the computation].

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The integral in theory should be over $(-\infty,\infty)$

In practice this reduces to an integral over $[0,x]$ plus another over $(x, \infty)$ since

  • $f_{E_1}(s)=0$ for $s \lt 0$ and
  • $\Bbb P(E_0>x-s) = 1$ for $s \gt x$
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  • $\begingroup$ Wrong answer. The probability in the last line is $1$, not $0$. $\endgroup$ – Kabo Murphy Jun 12 at 7:58
  • $\begingroup$ @KaviRamaMurthy - Corrected - thank you $\endgroup$ – Henry Jun 12 at 8:01

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