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Two dimensional normal distribution

suvadip

Member
Feb 21, 2013
69
If \(\displaystyle ( X,Y )\) has the normal distributions in two dimensions with zero means and unit variances and the correlation coefficient \(\displaystyle r\), then how to prove that the expectation of the greater of \(\displaystyle X\) and \(\displaystyle Y\) is \(\displaystyle \sqrt{(1-r)\pi}\)?
 

zzephod

Well-known member
Feb 3, 2013
134
If \(\displaystyle ( X,Y )\) has the normal distributions in two dimensions with zero means and unit variances and the correlation coefficient \(\displaystyle r\), then how to prove that the expectation of the greater of \(\displaystyle X\) and \(\displaystyle Y\) is \(\displaystyle \sqrt{(1-r)\pi}\)?
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:

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

.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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 [tex]-\infty[/tex] to [tex]\infty[/tex] and, for each y, x from y to [tex]\infty[/tex]:
[tex]\overline{X}= \frac{1}{2\pi}\int_{-\infty}^\infty\int_y^\infty xe^{-(x^2+ y^2}dxdy[/tex].

Or you can take x from [tex]-\infty[/tex] to [tex]\infty[/tex] and y from [tex]-\infty[/tex] to x: [tex]\overline{Y}= \frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^x ye^{-x^2+ y^2}dy dx[/tex].
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
If \(\displaystyle ( X,Y )\) has the normal distributions in two dimensions with zero means and unit variances and the correlation coefficient \(\displaystyle r\), then how to prove that the expectation of the greater of \(\displaystyle X\) and \(\displaystyle Y\) is \(\displaystyle \sqrt{(1-r)\pi}\)?
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 $$x^2 - 2rxy + y^2 = \tfrac12(1-r)u^2 + \tfrac12(1+r)v^2.$$