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Say I have two independent variables $X$ and $Y$ that are exponentially distributed with respective rates $\lambda_X$ and $\lambda_Y$. How do I compute $\mathbb{E}[X\mid \min\{X,Y\}]$?

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  • $\begingroup$ Which approaches to compute conditional expectations do you know? $\endgroup$ – Did Apr 27 '14 at 21:45
  • $\begingroup$ I understand how to compute conditional expectations when the conditioned variable is independent of the other variables. For example $E[aX + bZ|Y]$. Also I can compute the answer if $Y$ takes a finite amount of values. $\endgroup$ – Laxy_economist Apr 27 '14 at 22:23
  • $\begingroup$ No general formula/definition? $\endgroup$ – Did Apr 27 '14 at 22:28
  • $\begingroup$ This is the only formula that come to mind:$$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$ integrated over the correct bounds. $\endgroup$ – Laxy_economist Apr 27 '14 at 22:36
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    $\begingroup$ After a while, some serious definitions ARE useful... For a good introduction, quite readable, you might try Probability with martingales by David Williams. $\endgroup$ – Did Apr 27 '14 at 22:41
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(Answering a query made in the comments.)

Recall that, for every integrable random variable $X$, $E(X\mid Z)$ is defined as the almost surely unique random variable $u(Z)$ such that $E(Xv(Z))=E(u(Z)v(Z))$ for every bounded measurable function $v$.

In words, $E(X\mid Z)$ is at the same time measurable with respect to $Z$ and similar to $X$ in that $E(X\mid Z)$ and $X$ give the same expectation when integrated against any function of $Z$.

In the present case, $Z=\min\{X,Y\}$ hence one needs the distribution of $Z$ to compute $E(u(Z)v(Z))$. Next, the task is to compute $E(Xv(Z))$. And finally one must equate these, that is, find the unique $u$ such that these coincide for every $v$...

The final result is $$ E(X\mid\min\{X,Y\})=\min\{X,Y\}+\frac{\mu}{\lambda+\mu}\,\frac1{\lambda}, $$ which has some nice interpretations in terms of marked Poisson processes or other similar notions.

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  • $\begingroup$ I guess you mean "Marked Poisson processes". $\endgroup$ – Arash Apr 28 '14 at 22:06
  • $\begingroup$ @Arash I do. Thanks. $\endgroup$ – Did Apr 29 '14 at 5:52

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