Finding the conditional expectation of independent exponential random variables

Let $$X$$ and $$Y$$ be independent exponential random variables with respective rates $$\lambda$$ and $$\mu$$. Let $$M = \text{min}(X,Y)$$. Find

(a) $$E(MX|M=X)$$

(b) $$E(MX|M=Y)$$

(c) Cov$$(X,M)$$

(a) I first tried $$\displaystyle E(MX|M=X) = E(X^2) = \int_{0}^{\infty} x^2 f(x) dx = \int_{0}^{\infty} x^2 \lambda e^{-\lambda x} \, dx = \frac{2}{\lambda ^2}$$, which does not agree with the textbook answer.

I then tried $$\displaystyle E(MX|M=X) = E(M^2) = \int_0^\infty m^2 f(m) \,dm$$, where $$f(m)$$ is the pdf of $$M$$ which I know from here is equal to $$\displaystyle (\lambda + \mu) e^{-(\lambda + \mu)m}$$

$$\therefore E(MX|M=X) = \int_{0}^{\infty} m^2 (\lambda + \mu) e^{-(\lambda + \mu)m} dm = \frac{2}{(\lambda+\mu)^2},$$ which agrees with the textbook answer. Why is my first attempt not correct?

\begin{align} \\[15pt] \end{align}

(b) On this part I first tried $$E(MX|M=Y) = E(XY)$$ and using the fact that

$$E(g(X,Y)) := \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(x,y)f(x,y) \, dxdy \tag{*}$$

to write $$E(MX|M=Y) = E(XY) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xy f(x,y) dx dy$$

but I was not able to find the joint pdf $$f(x,y)$$.

Alternatively, I tried $$E(MX|M=Y) = E(MY|Y, but couldn't figure out where to go from here. My guess is to use the memoryless property of exponentials, but I'm not sure how to apply that.

\begin{align} \\[15pt] \end{align}

(c) $$\;\text{Cov}(X,M) = E(MX)-E(X)E(M)$$, where $$\displaystyle E(X) = \frac{1}{\lambda}$$ and $$E(M) = \int_{0}^{\infty} m f(m) dm = \frac{1}{\lambda + \mu}$$

I'm not sure how to calculate $$E(MX)$$. If I use equation (*), then I would again be stuck trying to find the joint pdf $$f(x,m)$$ like in part (b).

Using a different approach: $$M = \text{min}(X,Y) = \frac{X+Y-|X-Y|}{2}$$ so that \begin{align} E(MX) &= E\left(\frac{ X^2 + XY - X(|X-Y|) }{2}\right) \\ &= \frac{1}{2} \left( E(X^2) + E(XY) - E \left( X\sqrt{(X-Y)^2} \right)\right) \\ &= \frac{1}{2} \left( E(X^2) + E(X)E(Y) - \iint_{0}^{\infty} x\sqrt{(x-y)^2} f(x,y) dxdy \right), \end{align} and again I'm stuck.

• You cannot drop the conditions in the calculation. It should be $E(MX\mid M=X)=E(X^2\mid M=X)=E(X^2\mid X\le Y)$ and similarly for other expectations. – StubbornAtom Apr 23 at 7:17
• My book only provides ways to compute conditional expectations of the form $E(X|Y=y)$. Is true that $E(X^2 | X \leq Y) = \frac{E(X^2)}{Pr(X \leq Y)}$ ? – Michael Tagle Apr 23 at 22:32
• No, it is $\frac{E(X^2\mathbf1_{X\le Y})}{P(X\le Y)}$ where $\mathbf1_A$ is an indicator variable (equals $1$ when $A$ is true and equals zero otherwise). – StubbornAtom Apr 24 at 6:59
• Ah! Based on your comments and @KaviRamaMurthy 's comments I computed $$E(X^2 I_{X \leq Y }) = \int_{0}^{\infty} \int_{x}^{\infty} x^2 f(x,y) \;dy dx$$ and similarly for the other expectation, $$E(XY I_{X \geq Y} ) = \int_{0}^{\infty} \int_{y}^{\infty} xy f(x,y) \, dx dy$$ These allowed me to obtain the correct answers – Michael Tagle Apr 24 at 7:29

Hints: $$E(MX|M=X)=\frac {EMX I_{M=X}} {P(M=X)}$$ by definition. Note that $$\{M=X\}=\{X\leq Y\}$$. Now use the joint distribution of $$X,Y$$ to calculate $$E(MX|M=X)$$. $$E(MX|M=Y)$$ is similar.
For c) $$EMX=EX^{2}I_{\{ X \leq Y\}}+EXYI_{\{ X >Y\}}$$
• I calculated $P(X\leq Y) = \frac{\lambda}{\lambda + \mu}$, but I'm not sure if $EMX I_{M=X} = P(MX)$ or if $EMX I_{M=X} = E(MX)$. My book doesn't seem to use this notation. Moreover, it provides ways to compute conditional expectations only of the form $E(X|Y=y)$ – Michael Tagle Apr 23 at 22:50
• @michael $E(X|A)$ stands for $\frac {EXI_A} {P(A)}$. – Kavi Rama Murthy Apr 23 at 23:14