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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?

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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
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3 Answers 3

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.

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+1 Nice and clean. –  Sasha Feb 28 '13 at 1:07
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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 $E[X\min(X,Y)]$.

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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:

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