Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

having trouble with this one. The exact questions is the $\operatorname{cov}(X, \max(X,Y))$ 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?

share|cite|improve this question
Welcome to MSE. In probability theory, there is a somewhat important distinction between $X$ and $x$ (using the usual notation). Please try to keep these straight and consistent in submitting a question. – gnometorule Feb 19 '13 at 1:37
Are $X,Y$ independent? – Nate Eldredge Feb 19 '13 at 1:55

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.

share|cite|improve this answer
+1 Nice and clean. – Sasha Feb 28 '13 at 1:07

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 $E[X\min(X,Y)]$.

share|cite|improve this answer

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:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.