MHB Two dimensional normal distribution

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The discussion focuses on proving that the expectation of the greater of two correlated normally distributed variables, \(X\) and \(Y\), with zero means and unit variances, is given by \(\sqrt{(1-r)\pi}\). The conditional mean for \(X\) when \(X > Y\) is derived through integration over the appropriate half-plane. The correlation coefficient \(r\) is emphasized as a crucial factor in the density function of the joint distribution, which is derived from the correlation matrix. A change of variables is suggested to facilitate the integration needed to compute the expectation. The conversation highlights the importance of considering correlation when analyzing the relationship between \(X\) and \(Y\).
Suvadip
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If $$( X,Y )$$ has the normal distributions in two dimensions with zero means and unit variances and the correlation coefficient $$ r$$, then how to prove that the expectation of the greater of $$X$$ and $$Y$$ is $$\sqrt{(1-r)\pi}$$?
 
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suvadip said:
If $$( X,Y )$$ has the normal distributions in two dimensions with zero means and unit variances and the correlation coefficient $$ r$$, then how to prove that the expectation of the greater of $$X$$ and $$Y$$ is $$\sqrt{(1-r)\pi}$$?

Without loss of generality suppose $X=x>y$ then the conditional mean for $x$ is:

$\displaystyle \overline{X}_y=\int_{x=y}^{\infty} x p(x|y)\, dx$

So:

$\displaystyle \overline{X}=\int_{y=-\infty}^{\infty} \overline{X}_y p(y)\, dy=\int_{y=-\infty}^{\infty}\int_{x=y}^{\infty} x p(x|y)p(y)\, dxdy =\int_{y=-\infty}^{\infty}\int_{x=y}^{\infty} x p(x,y)\, dx dy$

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Geometrically, y= x is a line through the origin. x> y is the half-plane to the right. Given that x> y we want to integrate over that half plane. We can, like zzephod, say that taking y from -\infty to \infty and, for each y, x from y to \infty:
\overline{X}= \frac{1}{2\pi}\int_{-\infty}^\infty\int_y^\infty xe^{-(x^2+ y^2}dxdy.

Or you can take x from -\infty to \infty and y from -\infty to x: \overline{Y}= \frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^x ye^{-x^2+ y^2}dy dx.
 
suvadip said:
If $$( X,Y )$$ has the normal distributions in two dimensions with zero means and unit variances and the correlation coefficient $$ r$$, then how to prove that the expectation of the greater of $$X$$ and $$Y$$ is $$\sqrt{(1-r)\pi}$$?
I am not a statistician, so I may be misunderstanding something here. But it seems to me that the previous two comments overlook the fact that $X$ and $Y$ are supposed to be correlated, with correlation coefficient $r$. According to Multivariate normal distribution - Wikipedia, the free encyclopedia, the density function in that case is given by $f(x,y) = \frac1{2\pi\sqrt{1-r^2}}\exp\bigl(-\frac12[x\:\:y]\Sigma^{-1}\bigl[{x\atop y}\bigr]\bigr)$, where $\Sigma$ is the correlation matrix $\Sigma = \begin{bmatrix}1&r \\r&1 \end{bmatrix}.$ Then $\Sigma^{-1} = \frac1{1-r^2}\begin{bmatrix}1&-r \\-r&1 \end{bmatrix},$ giving $$f(x,y) = \frac1{2\pi\sqrt{1-r^2}}\exp\biggl(\frac{-(x^2 - 2rxy + y^2)}{2(1-r^2)}\biggr).$$ To integrate that over the region $x>y$ I would make the change of variables $u = x+y$, $v = x-y$, using the fact that $$x^2 - 2rxy + y^2 = \tfrac12(1-r)u^2 + \tfrac12(1+r)v^2.$$
 
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