2

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.

Regards

Dilip Sarwate
  • 41,202
  • 4
  • 94
  • 200
jimduquettesucked
  • 51
  • 5
  • Do you know the distribution of $X$ and $Y$? Such as,are they normally distributed? – Deep North Oct 05 '15 at 00:55
  • 2
    @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. – Dilip Sarwate Oct 05 '15 at 01:36

1 Answers1

2

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.

michaelsnowden
  • 113
  • 7
Vimal
  • 1,047
  • 8
  • 16
  • 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." – jimduquettesucked Oct 05 '15 at 02:46
  • Does this have to do with $E(Y|X)=\int y(f(x,y)/f_x(x))dy$ – jimduquettesucked Oct 05 '15 at 02:49
  • 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: http://stats.stackexchange.com/questions/30588/deriving-the-conditional-distributions-of-a-multivariate-normal-distribution. – Vimal Oct 05 '15 at 04:34