How can I find the m(x) (that is, EY|X=x), in a non-parametric regression, when Y and X have a joint bivariant normal pdf with both mean 0 and variance 1 (unknown correlation)? $f(x,y)=\frac{\exp \left\{ -\frac 1{2(1-\rho ^2)}\left[ \left( \frac{x-\mu _x% }{\sigma _x}\right) ^2-2\rho \left( \frac{x-\mu _x}{\sigma _x}\right) \left( \frac{y-\mu _y}{\sigma _y}\right) +\left( \frac{y-\mu _y}{\sigma _y}\right) ^2\right] \right\} }{2\pi \sigma _x\sigma _y\sqrt{1-\rho ^2}} $
I have tried to integrate respect to dy the f(x,y), that is, a normal pdf of X divided by f(x,y) to find the conditional probability, and then make the integral of y times the conditional probability here, but I believe something is wrong.
UPDATED:
If you find the conditional probability, that is $\frac{f(x, y)}{f(x)}$, and simplify, you would found a pdf of a Normal with mean $\rho x$ and variance $1-\rho^2$, so m(x) = E(Y|X)= $\rho x$