# Two dimensional normal distribution

##### Member
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
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
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: $$\overline{Y}= \frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^x ye^{-x^2+ y^2}dy dx$$.

#### Opalg

##### MHB Oldtimer
Staff member
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 $$x^2 - 2rxy + y^2 = \tfrac12(1-r)u^2 + \tfrac12(1+r)v^2.$$