I have $(X,Y_1)$ with joint bivariate normal distribution. Also, $Y_1$ is conditional on $X$. Therefore $\rho$ is non-zero. Suppose I have $Y_2$ that is also normally distributed and $\mu_{Y_2} = \mu_{Y_1}$ and $\sigma_{Y_2}=\sigma_{Y_1}$. What is the relationship between $\text{Var}(Y_1|X)$ and $\text{Var}(Y_2)$ ?

I've been generating samples from $Y_1|X$ and $Y_2$ and it seems that for large sample sizes, the difference $|\text{Var}(Y_1|X)$ - $\text{Var}(Y_2)$| converges to a particular (small) number. This disagrees with theoretical result that I got by calculating $\sigma_{Y_1|X} = \sigma_{Y_1}\sqrt{1-\rho^2}$.

What is the interpretation of that?

  • 1
    $\begingroup$ I must not being understanding something. You specified in the problem that $\sigma_{Y_2}=\sigma_{Y_1}$. So the relationship is just that they're the same? If this isn't the case then please provide some more detail as to why not. $\endgroup$
    – Dason
    Nov 7, 2011 at 20:06
  • $\begingroup$ Sorry, I wasnt clear, I'm interested in comparing samples from Y1|X and Y2. $\endgroup$
    – Saideira
    Nov 7, 2011 at 20:18
  • $\begingroup$ Could you describe your simulation experiment which disagrees with the theoretical result? Thanks. $\endgroup$
    – Xi'an
    Nov 9, 2011 at 6:31

1 Answer 1


First, "$Y_1$ is conditional on $X$" has no specific probabilistic meaning. Do you mean $X$ and $Y_1$ are not independent?

Second, if $Y_1$ and $Y_2$ have the same marginal normal distribution, then $$ \text{var}(Y_1|X) \le \text{var}(Y_1)=\text{var}(Y_2)\,. $$ More exactly, as a generic property, $$ \text{var}(Y_1)=\text{var}(Y_2) = \text{var}(\mathbb{E}[Y_1|X])+\mathbb{E}[\text{var}(Y_1|X)]\,, $$ which means here that $$ \text{var}(Y_1)= \sigma_{Y_1}^2 = \rho^2\dfrac{\sigma_{Y_1}^2}{\sigma_X^2} \sigma_X^2+(1-\rho^2)\sigma_{Y_1}^2\,. $$ Third, I do not understand your simulation experiment: if you generate from the distribution of $Y_1$ conditional on $X$, this means you generate from a $$ \mathcal{N}\left(\rho \sigma_{Y_1} X / \sigma_X, \sqrt{1-\rho^2}\sigma_{Y_1}^2\right) $$ distribution (assuming $\mu_X=\mu_Y=0$). What am I missing?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.