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 $E_1$

  • $\begingroup$ Why do you think the bounds should be $0$ and $1$? $\endgroup$ – angryavian Jun 12 at 7:47
  • 1
    $\begingroup$ @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

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].


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 $s \gt x$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.