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I have two independent exponential random variable $v$ and $u$ with parameters $\lambda$ and $\mu$. What is the mean of $v$, knowing that $v$ happened before $u$ i.e. $E[v|v<u]$?

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  • $\begingroup$ It depends on the full joint distribution of $(v,u).$ If you intend them to be independent, please specify that. $\endgroup$
    – whuber
    Commented Feb 18, 2022 at 18:23
  • $\begingroup$ Yes, they are independent $\endgroup$
    – simone123
    Commented Feb 18, 2022 at 18:34
  • $\begingroup$ We have so many closely related posts that I'm sure you can find several solution methods already explained here: try this site search. $\endgroup$
    – whuber
    Commented Feb 18, 2022 at 18:50

1 Answer 1

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By the law of iterated expectations,

$$\begin{align}E[v{\bf1}_{v<u}]&=E[E[v{\bf1}_{v<u}|{\bf1}_{v<u}]]\\ &=E[v{\bf1}_{v<u}|{\bf1}_{v<u}=1]P({\bf1}_{v<u}=1)+E[v{\bf1}_{v<u}|{\bf1}_{v<u}=0]P({\bf1}_{v<u}=0)\\ &=E[v|v<u]P(v<u)\\ \implies E[v|v<u]&=\frac{E[v{\bf1}_{v<u}]}{P(v<u)}, \end{align}$$ where $$\begin{align}E[v{\bf1}_{v<u}]&=\int_\Omega v f(u,v)dudv,\\ P(v<u)&=\int_\Omega f(u,v)dudv,\\ \Omega&:={\{(u,v)\in \mathbb{R^2}:u>v\}}. \end{align}.$$

Can you take it from here?

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  • $\begingroup$ I think it should not be a problem compute the two integrals. Alternatively, can i say that $E[v|v<u]$ is equalt to $E[\min(v,u)]$, thus since the $\min$ is distributed as an exponential r.v. with parameter $\lambda+\mu$ the result is $\frac{1}{\lambda+\mu}$? $\endgroup$
    – simone123
    Commented Feb 18, 2022 at 19:19
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    $\begingroup$ Considerations of units of measurement tell us the answer must remain the same when $\lambda$ and $\mu$ are scaled by the same positive amount. Since that's not the case for $1/(\mu+\lambda),$ that cannot be a correct answer. $\endgroup$
    – whuber
    Commented Feb 18, 2022 at 19:49
  • $\begingroup$ @simone123 $E[\min(u,v)]=E[v|v<u]P(v<u)+E[u|u<v]P(u>v)$ so the mean of the min is different than what you want. $\endgroup$ Commented Feb 18, 2022 at 20:33
  • $\begingroup$ Weird because I have computed the first integral which is $\lambda/(\lambda+\mu)^2$ (It should be right since I have checked by computing the integral online for fixed values). Regarding the second one it is a known result and it is $\lambda/(\lambda+\mu)$, thus the result is $1/(\lambda+\mu)$ $\endgroup$
    – simone123
    Commented Feb 18, 2022 at 21:16
  • $\begingroup$ Also if I consider the equation suggested by @Golden_Ratio, with the result I found, the equation is satisfied. I think my answer is correct. $\endgroup$
    – simone123
    Commented Feb 18, 2022 at 21:30

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