Two dimensional normal distribution

[Suvadip]

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]

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]

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]

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 $$x^2 - 2rxy + y^2 = \tfrac12(1-r)u^2 + \tfrac12(1+r)v^2.$$

 

[blue_raver22]

The expectation of the greater of X and Y can be calculated by integrating the joint probability density function over the appropriate region. In this case, we are interested in the region where X > Y. Using the properties of the normal distribution, we can rewrite this expectation as:

E[max(X,Y)] = ∫∫ max(x,y) * f(x,y) dx dy

Where f(x,y) is the joint probability density function of (X,Y). Since X and Y have zero means and unit variances, we can write the joint probability density function as:

f(x,y) = (1/2π√(1-r^2)) * e^(-(x^2+y^2-2rx^y)/2(1-r^2))

Substituting this into the expectation equation and using the fact that max(x,y) = x when x > y, we get:

E[max(X,Y)] = ∫∫ x * (1/2π√(1-r^2)) * e^(-(x^2+y^2-2rx^y)/2(1-r^2)) dx dy

Integrating this expression over the region where X > Y, we get:

E[max(X,Y)] = ∫∫∫ x * (1/2π√(1-r^2)) * e^(-(x^2+y^2-2rx^y)/2(1-r^2)) dx dy dz

= ∫∫ (1/2π√(1-r^2)) * e^(-(x^2+y^2-2rx^y)/2(1-r^2)) dx dy

= (1/2π√(1-r^2)) * ∫∫ x * e^(-(x^2+y^2-2rx^y)/2(1-r^2)) dx dy

Using the substitution u = x^2+y^2-2rx^y, we can rewrite this integral as:

E[max(X,Y)] = (1/2π√(1-r^2)) * ∫∫ √(u+2ry) * e^(-u/2(1-r^2)) du dy

= (1/2π√(1-r^2)) * ∫∫ √(u+2ry) * e^(-u/2