Conditional mean of an exponential random variable

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?

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

By the law of iterated expectations,

\begin{align}E[v{\bf1}_{v where \begin{align}E[v{\bf1}_{vv\}}. \end{align}.

Can you take it from here?

• 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 $\frac{1}{\lambda+\mu}$? Commented Feb 18, 2022 at 19:19
• 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.
– whuber
Commented Feb 18, 2022 at 19:49
• @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. Commented Feb 18, 2022 at 20:33
• 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 $1/(\lambda+\mu)$ Commented Feb 18, 2022 at 21:16
• 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. Commented Feb 18, 2022 at 21:30