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- Thread starter suvadip
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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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- Jan 29, 2012

- 1,151

[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].

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- Feb 7, 2012

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