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 $Y$.
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, $$E[Z|Z=Y]=E[Z]=\frac1{\lambda+\mu}.$$