So I have continuous random variables $X$ and $Y$ which have $\mu_x$ and $\mu_y$ as their means and variances $\sigma_x^2$ and $\sigma_y^2$ and correlation $\rho$.

Find $E(Y\mid X)$.

I know that $E(Y\mid X)=\int_{R} y \left[\frac{f(x,y)}{f_X(x)}\right]dy$.

I don't understand how to find the PDFs based on the given information. I realize that this is the regression function, but I don't understand how to approach this.


  • $\begingroup$ Do you know the distribution of $X$ and $Y$? Such as,are they normally distributed? $\endgroup$
    – Deep North
    Commented Oct 5, 2015 at 0:55
  • 2
    $\begingroup$ @DeepNorth Knowing that $X$ and $Y$ are normally distributed does not help in this problem. Knowing that $X$ and $Y$ are jointly normally distributed does; cf. the answer by Vimal. $\endgroup$ Commented Oct 5, 2015 at 1:36

1 Answer 1


Is this a homework question? If so please tag it accordingly.

You haven't specified the probability densities for the two random variables, but if you assume a multivariate normal distribution, you can easily compute the entire conditional distribution $p(Y|X=x)$. Its expectation is simply:

$E[Y|X=x] = \mu_y + \frac{\sigma_y}{\sigma_x} \rho (x - \mu_x)$.

See https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Bivariate_case.

  • $\begingroup$ That makes sense. Can you help me find a derivation of this. The wikipedia page just says "Proof: the result is obtained by taking the expectation of the conditional distribution X_1\mid X_2 above." $\endgroup$ Commented Oct 5, 2015 at 2:46
  • $\begingroup$ Does this have to do with $E(Y|X)=\int y(f(x,y)/f_x(x))dy$ $\endgroup$ Commented Oct 5, 2015 at 2:49
  • $\begingroup$ Yes, that is correct. A brute-force way is to substitute the value of $f(x,y)$ with the formula for a bivariate normal and simplify it. There's another clever way, which is easy to argue once you do the brute-force way. Both approaches are described in this post: stats.stackexchange.com/questions/30588/…. $\endgroup$
    – Vimal
    Commented Oct 5, 2015 at 4:34

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