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Given the mean, variance, and covariance of two random variables $X$ and $Y$, how would I find $\operatorname{Var}(Y\mid X)$?

I know that I can find $E(Y\mid X)$ using the definition of covariance and finding $\mathbb{E}(XY)$ then dividing by $\mathbb{E}(X)$.

Would I try to find $\mathbb{E}[Y^2\mid X]?$ Or would I try to use the total conditional variance? I'm very confused.

Richard Hardy
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Rhubarb
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    It's not possible in general. Mean, variance and covariance do not entirely determine the joint law of $X$ and $Y$. This is the case if you assume that $(X,Y)$ follows a bivariate normal distribution. – Stéphane Laurent Sep 07 '17 at 08:09
  • How would this change if they are a joint normal? I'm not quite seeing it. – Rhubarb Sep 07 '17 at 08:16
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    A bivariate normal distribution is entirely determined by the means, the variances and the covariance. – Stéphane Laurent Sep 07 '17 at 08:17
  • Oh! That's right. So given the multivariate PDF of $f(x,y)$, to find the conditional variance, would I divide by the PDF of $Y$ to get $f_(Y|X)$ then derive the first and second moments to find the conditional variance? – Rhubarb Sep 07 '17 at 08:25
  • https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Bivariate_case does it help ? – Stéphane Laurent Sep 07 '17 at 08:35
  • You can readily estimate the conditional variance in terms of the statistics you have, because they are all you need in order to perform ordinary least squares regression. If the OLS assumptions are applicable (particularly that the model is correct and the errors are homoscedastic), this would solve your problem. It does not require the stronger assumption of bivariate normality. See https://stats.stackexchange.com/questions/107597 for further explanation and working code. – whuber Sep 07 '17 at 14:05
  • Does the second paragraph mean you think $E(Y \mid X) = E(XY)/E(X)$? If not, can you clarify how you intend to find $E(Y\mid X)$ from the information given? – Juho Kokkala Sep 08 '17 at 18:45

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