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 $E(MX|M=Y) = E(MY|Y<X)$, 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.

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    $\begingroup$ 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. $\endgroup$ – StubbornAtom Apr 23 at 7:17
  • $\begingroup$ My book only provides ways to compute conditional expectations of the form $E(X|Y=y)$. Is true that $E(X^2 | X \leq Y) = \frac{E(X^2)}{Pr(X \leq Y)}$ ? $\endgroup$ – Michael Tagle Apr 23 at 22:32
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    $\begingroup$ 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). $\endgroup$ – StubbornAtom Apr 24 at 6:59
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    $\begingroup$ 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 $\endgroup$ – 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) $EMX=EX^{2}I_{\{ X \leq Y\}}+EXYI_{\{ X >Y\}}$

  • $\begingroup$ 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 $E(X|Y=y)$ $\endgroup$ – Michael Tagle Apr 23 at 22:50
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    $\begingroup$ @michael $E(X|A)$ stands for $\frac {EXI_A} {P(A)}$. $\endgroup$ – Kavi Rama Murthy Apr 23 at 23:14
  • $\begingroup$ Ok I understand now, thank you very much. $\endgroup$ – Michael Tagle Apr 24 at 7:34

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