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Let $X$ be an exponential random variable with rate $\lambda$. Let $Y$ be a generic continuous random variable, which is independet of $X$. Define random variable $Z = \min\{X,Y\}$. I am interested in $E[Z|Z = X]$ and $E[Z|Z = Y]$.

I think, if $Y$ is an exponential distribution with rate $\mu$, then the two expected values are equal to $1/(\lambda + \mu)$---I could not show it though. However, I do not think the equality holds if $Y$ is not exponential. I would appreciate if someone could help me derive the two expected values for the generic continuous $Y$.

Salivan
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1 Answers1

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Partial answer:

Suppose that $Y$ is an exponential distribution with rate $\mu$. Note that for $z>0$, we have \begin{align*} P(Z>z|Z=X)&=P(Z>z|X\leq Y)=\frac{P(Z>z,X\leq Y)}{P(X\leq Y)}\\&=\frac{P(z<X\leq Y)}{P(X\leq Y)}\\&=\frac{\int_z^\infty\int_x^\infty\lambda\mu e^{-\lambda x}e^{-\mu y}\,dy\,dx}{\int_0^\infty\int_x^\infty\lambda\mu e^{-\lambda x}e^{-\mu y}\,dy\,dx}\\&=e^{-(\lambda+\mu)z}=P(X>z,Y>z)\\&=P(Z>z). \end{align*} So $Z$ and $\{Z=X\}$ are independent, hence $$E[Z|Z=X]=E[Z]=\frac1{\lambda+\mu}.$$ For the same reason, $$E[Z|Z=Y]=E[Z]=\frac1{\lambda+\mu}.$$

Feng
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  • Thanks, @Feng Shao It might help me find my answer for the second part of my question as well. Just one note: I think, you need to change the order of integration in your integrals. Imean, dydx, instead of dxdy. – Salivan Aug 30 '19 at 13:35
  • @Amir Fixed now. Some ideas:$E[Z|Z=X]=E[Z|Z=Y] \iff EZ=E[Z|Z=X]P(Z=X)+E[Z|Z=Y]P(Z=Y)=EZ|Z=X=E[Z|Z=X]$ so you just need to find $Y$ such that $EZ=E[Z|Z=X]$. – Feng Aug 30 '19 at 14:38
  • please note that my question is not about finding $Y$, such that $E[Z|Z=X]$ and $E[Z|Z=Y]$ are equal. My question is about finding a closed-form solution for these two expected values when $Y$ is a generic continuous random variable. Re. your comment: you can factor out $E[Z|Z=X]$, only if $E[Z|Z=X] = E[Z|Z=Y]$, which we do not know if it holds for a generic $Y$---in your proof above, you showed it only when $Y$ is exponential. – Salivan Aug 30 '19 at 15:50