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?
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(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.
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(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.
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) $EMX=EX^{2}I_{\{ X \leq Y\}}+EXYI_{\{ X >Y\}}$