Compute $\operatorname{cov}(X, \max(X,Y))$ and $\operatorname{cov}(X, \min(X,Y))$ when $X$ and $Y$ are standard normal 
Compute $\operatorname{cov}(X, \max(X,Y))$ and $\operatorname{cov}(X, \min(X,Y))$ where $X,Y \sim N(0,1)$. 

i think the way to calculate it is to get 
$$\begin{align}
\operatorname{cov}(X, \max(X, Y) + \min(X,Y)) & = \operatorname{cov}(X, X+Y) \\
& = \operatorname{cov}(X, \max(X,Y)) + \operatorname{cov}(x, \min(X,Y)) \\
\end{align}$$
and 
$$\begin{align}
\operatorname{cov}(X, \max(X,Y) - \min(X,Y)) & = \operatorname{cov}(X, \operatorname{abs}(X-Y)) \\
& = \operatorname{cov}(X, \max(X,Y)) - \operatorname{cov}(X, \min(X,Y)) \\
\end{align}$$
although this is pretty much as difficult to solve as $\operatorname{cov}(X, \max(X,Y))$  unless there is some particular trick. Anyone can help with this?
 A: Assume that the random variables $X$ and $Y$ are i.i.d. square integrable with a symmetric distribution (not necessarily gaussian).
Let $Z=\max(X,Y)$, then the covariance of $X$ and $Z$ is $\mathbb E(XZ)-\mathbb E(X)\mathbb E(Z)=\mathbb E(XZ)$. Using $Z=X\mathbf 1_{Y\lt X}+Y\mathbf 1_{X\lt Y}$, one sees that
$$
\mathbb E(XZ)=\mathbb E(X^2;Y\lt X)+\mathbb E(XY;X\lt Y).
$$
What is the value of the last term on the RHS? By symmetry, $\mathbb E(XY;X\lt Y)=\mathbb E(XY;Y\lt X)$ and the sum of these is $\mathbb E(XY)=\mathbb E(X)\mathbb E(Y)=0$ hence $\mathbb E(XY;X\lt Y)=0$. Thus,
$$
\mathbb E(XZ)=\mathbb E(X^2F(X)),
$$
where $F$ denotes the common CDF of $X$ and $Y$. Since $X$ is distributed as $-X$, $F(-X)=1-F(X)$ and
$$
\mathbb E(X^2F(X))=\mathbb E((-X)^2F(-X))=\mathbb E(X^2(1-F(X))=\mathbb E(X^2)-\mathbb E(X^2F(X)).
$$
This yields
$$
\mathrm{cov}(X,\max(X,Y))=\tfrac12\mathrm{var}(X).
$$
On the other hand, $\min(-X,-Y)=-\max(X,Y)$ hence, once again by symmetry,
$$
\mathrm{cov}(X,\min(X,Y))=\tfrac12\mathrm{var}(X).
$$
Edit: A much simpler proof is to note from the onset that, since $\max(X,Y)+\min(X,Y)=X+Y$, $\mathrm{cov}(X,\max(X,Y))+\mathrm{cov}(X,\min(X,Y))=\mathrm{cov}(X,X+Y)=\mathrm{var}(X)$, and that, by the symmetry of the common distribution of $X$ and $Y$ and the identity $\min(-X,-Y)=-\max(X,Y)$, $\mathrm{cov}(X,\max(X,Y))=\mathrm{cov}(X,\min(X,Y))$. These two elementary remarks yield the result and allow to skip nearly every computation.
A: Assuming that $X$ and $Y$ are independent random variables,
the direct way to solve this problem would be to compute $E[X\max(X,Y)]$ and
$E[X\min(X,Y)]$ via integration, breaking the double integral into two 
double integrals over the regions where the maximum is $y$ and where
the maximum is $x$, and then using a change to polar coordinates.
We have
$$\begin{align*}
E[X\max(X,Y)] &= \int_{y=-\infty}^\infty\int_{x=-\infty}^y
\frac{xy}{2\pi}e^{-(x^2+y^2)/2}\,\mathrm dx\,\mathrm dy
+ \int_{y=-\infty}^\infty\int_{x=y}^\infty
\frac{x^2}{2\pi}e^{-(x^2+y^2)/2}\,\mathrm dx\,\mathrm dy\\
&= \int_0^\infty\int_{\pi/4}^{5\pi/4}
\frac{r^2\cos\theta\sin\theta}{2\pi}e^{-r^2/2}\,r\,\mathrm d\theta\,\mathrm dr
+ \int_0^\infty\int_{-3\pi/4}^{\pi/4}
\frac{r^2\cos^2\theta}{2\pi}e^{-r^2/2}\,r\,\mathrm d\theta\,\mathrm dr
\end{align*}$$
and similarly for $E[X\min(X,Y)]$.
A: The OP states $X,Y$ ~ $N(0,1)$, but doesn't specify whether $X$ and $Y$ are independent or dependent. 
Whereas the other posters assume independence, instead  ... consider here the more general problem that nests same, namely $(X,Y)$ ~ standardBivariateNormal, with joint pdf $f(x,y)$:

The general solution to Cov[$X$, max$(X,Y)$] ... obtained here using the mathStatica / Mathematica combo ... is simply: 

The 'min' case is symmetrical, but here it is anyway:

