What's the integral of int(-inf->X)int(-inf->Y)x*exp(-z/2*(1-rho^2))dxdy
where z = (x^2 -2rho*x*y + y^2)
Expressed in terms of Normal Distribution and Bivariate Normal distribution.
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It looks to me like the "hint" is pointing toward doing a change of
variable related to completing the square, i.e.

exp( -(x^2(1-rho^2) + (y - rho*x)^2)/2 * (1-rho^2) )

and substituting y = u + rho*x.  Now with an inner integral over y, x
is considered a constant, so dy = du.  But you have a change in the
limits of integration.

I'll assume for the sake of notational consistency that the limits of
integration on y are from -oo to Y.  Then the inner integral on u
would range from -oo to Y - rho*x, and the factors x * exp( -x^2/2 )
can be moved out of this inner integral.

regards, mathtalk-ga