Given two independent exponential RV $X,Y$, compute $E(X \mid \min(X,Y))$

Question

Please note that this question is related to How to find $\mathbb{E}[X\mid\min(X,Y)]$? in which the user Did explains the procedure and ontop of that gives the solution to this question.

Problem: Let $X,Y$ be two independent exponential RV with positive parameters $\lambda, \mu$ respectivly.
Compute $E(X \mid \min(X,Y))$

---

The solution given from the above link is: $$ E(X \mid \min (X,Y)) =\min(X,Y) + \frac{ \mu}{\lambda+\mu} \frac{1}{\lambda}$$

---

My approach to the problem: Following a similar routine as in the answer of the link I have already managed to compute $E ( \min(X,Y) \mid X)$ but on this task I fail.

I am aware that $\min(X,Y)$ has exponential distribution with parameter $\lambda+\mu$

If I fix an arbitrary $Z \in L^\infty ( \Omega, \sigma(\min(X,Y)), \mathbb{P})$ I can say that $Z=f(T=\min(X,Y))$ for a bounded borel function $f$ and I want to find $h(t)$ in $E(Xf(T))=E(h(t)f(T))$. Starting: \begin{align}E(Xf(T))&=E(Xf(\min(X,Y)))=E(Xf( \min(X,Y))1_{X<Y})+E(Xf(\min(X,Y))1_{X>Y}) \\ &= E(Xf(X)1_{X<Y})+E(Xf(Y)1_{X>Y}) \\ &\overset{1)}= \int xf(x)1_{x<y}P_X(dx)P_Y(dy) + \int x f(y) 1_{x>y} P_X(dx)P_Y(dy) \\ &\overset{2)}=\int_0^\infty \left( \int_x^\infty P_Y(dy) \right)xf(x)P_X(dx)+ \int_0^\infty \left(\int_y^\infty xP_X(dx) \right)f(y)P_Y(dy) \end{align} Where in 1) I used the fact that $X,Y$ are independent and in 2) Fubini-Lebesgue's Theorem. For the two integrals in braces above I obtain $$\int_x^\infty P_Y(dy) = \int_x^\infty \mu e^{-\mu y}dy= e^{-\mu x} $$ and $$ \int_y^\infty x P_X(dx) = \int_y^\infty x \lambda e^{-\lambda x} dx = \frac{1}{\lambda}e^{-\lambda y} (\lambda y+1) $$ If these calculations are right (I hope they are) then I would obtain that $$E(Xf(T))=E\left(Xf(X)e^{-\mu X} + \frac{1}{\lambda} e^{- \lambda Y}( \lambda Y +1)f(Y) \right) $$

Maybe everything I did was wrong so far, but even if not, I wouldn't know how to take it from here. My next step would be to introduce the indicator function again to obtain a $T= \min(X,Y)$ but then I just get stuck at this line.

Are my calculations wrong? if so, please show me where I went wrong and maybe provide a hint on how I can continue

Answer 1

Continuing your calculation, the right side of your last display can be written as \begin{equation*} \begin{split} \int_0^\infty &\left(tf(t)\lambda e^{-(\mu+\lambda)t}+\mu\lambda^{-1}(\lambda t+1)f(t)e^{-(\mu+\lambda)t}\right)dt=\\ &=\int_0^\infty f(t) [t(\lambda+\mu) +\mu\lambda^{-1}]e^{-(\mu+\lambda)t}dt=\\ &=\frac{1}{\mu+\lambda}\int_0^\infty f(t) [t(\lambda+\mu) +\mu\lambda^{-1}]P_T(dt)=\\ &=\frac{1}{\mu+\lambda}E(f(T)\cdot[T(\lambda+\mu)+\mu\lambda^{-1}]) \end{split} \end{equation*} This means that $E(X|T)=T+{\mu\over\lambda(\mu+\lambda)}$.

Answer 2

Note that the density function of $\min(X, Y)$ is given by $(\mu+\lambda)e^{-(\mu+\lambda)x}$. Moreover, \begin{align*} E\big(Xf(\min(X, Y))\big) &=E\big(\min(X, Y)f(\min(X, Y)) + [X-\min(X, Y)] f(\min(X, Y))\big)\\ &=E\big(\min(X, Y)f(\min(X, Y)) + [X-Y] f(Y)1_{X>Y}\big). \end{align*} Furthermore, \begin{align*} E\big((X-Y) f(Y)1_{X>Y}\big)&=E\Big(E\big((X-Y) f(Y)1_{X>Y}\mid Y\big)\Big)\\ &=E\Big(f(Y)E\big((X-Y) 1_{X>Y}\mid Y\big)\Big)\\ &=E\left(f(Y)\int_Y^{\infty}(x-Y) \lambda e^{-\lambda x}dx \right)\\ &=E\left(f(Y)\frac{1}{\lambda}e^{-\lambda Y}\right)\\ &=\frac{1}{\lambda}\int_0^{\infty}f(y)e^{-\lambda y}\mu e^{-\mu y} dy\\ &=\frac{\mu}{\lambda(\lambda +\mu)}\int_0^{\infty}f(y)(\lambda +\mu)e^{-(\lambda +\mu) y} dy\\ &=\frac{\mu}{\lambda(\lambda +\mu)}E\big(f(\min(X, Y)) \big). \end{align*} That is, \begin{align*} E\big(Xf(\min(X, Y))\big) &=E\left(\min(X, Y)f(\min(X, Y)) + \frac{\mu}{\lambda(\lambda +\mu)}f(\min(X, Y))\right). \end{align*} Therefore, \begin{align*} E\big(X \mid \min(X, Y) \big) = \min(X, Y)+\frac{\mu}{\lambda(\lambda +\mu)}. \end{align*}