Calculate the covariance between max(X,Y) and min(X,Y) Let $X$ and $Y$ be independent discrete random variables, each taking values $1$ and $2$ with probability $1/2$ each. How do I calculate the covariance between $max(X,Y)$ and $min(X,Y)$?
 A: First,
$$
\begin{array}{}
\mathrm{P}(\max(X,Y)=2)=\frac34\qquad\text{and}\qquad\mathrm{P}(\max(X,Y)=1)=\frac14\\
\mathrm{P}(\min(X,Y)=2)=\frac14\qquad\text{and}\qquad\mathrm{P}(\min(X,Y)=1)=\frac34\\
\end{array}
$$
Therefore,
$$
\mathrm{E}(\max(X,Y))=\frac74\qquad\text{and}\qquad\mathrm{E}(\min(X,Y))=\frac54
$$
Furthermore,
$$
\begin{align}
\mathrm{P}(\max(X,Y)\min(X,Y)=4)=\mathrm{P}(\min(X,Y)=2)&=\frac14\\
\mathrm{P}(\max(X,Y)\min(X,Y)=2)\hphantom{\ =\mathrm{P}(\min(X,Y)=2)}&=\frac12\\
\mathrm{P}(\max(X,Y)\min(X,Y)=1)=\mathrm{P}(\max(X,Y)=1)&=\frac14\\
\end{align}
$$
Therefore,
$$
\mathrm{E}(\max(X,Y)\min(X,Y))=\frac94
$$
Thus,
$$
\begin{align}
&\mathrm{Cov}(\max(X,Y),\min(X,Y))\\
&=\mathrm{E}(\max(X,Y)\min(X,Y))-\mathrm{E}(\max(X,Y))\mathrm{E}(\min(X,Y))\\
&=\frac1{16}
\end{align}
$$
A: Let $W=\min\{X,Y\}$ and $Z=\max\{X,Y\}$; then the desired covariance is $\mathrm{E}[WZ]-\mathrm{E}[W]\mathrm{E}[Z]$. $W=1$ unless $X=Y=2$, so $\mathrm{Pr}(W=1)=\frac34$ and $\mathrm{Pr}(W=2)=\frac14$, and therefore
$$\mathrm{E}[W]=\frac34\cdot1+\frac14\cdot2=\frac54\;.$$ 
The calculation of $\mathrm{E}[Z]$ is similar. So is that of $\mathrm{E}[WZ]$: just average the four possibilities according to their respective probabilities.
