Conditional variance of bivariate normal

Let's assume that $$X_1$$ and $$X_2$$ are bivariate normal, then the standard result for the conditional distribution gives me $$X_{1}\mid X_{2}=a\ \sim \ {\mathcal {N}}\left(\mu _{1}+{\frac {\sigma _{1}}{\sigma _{2}}}\rho (a-\mu _{2}),\,(1-\rho ^{2})\sigma _{1}^{2}\right)$$ using standard notation for the variance and correlation.

I am trying to make sense of the following scenario: Before $$X_1$$ and $$X_2$$ are drawn, somebody asks you about the variance of $$X_1$$ and you confidently answer $$\sigma_1^2$$. Then $$X_2$$ is drawn but not revealed to you and you are asked again about the variance of $$X_1$$. The formula above tells you that, no matter which value $$X_2$$ took, the new variance is $$(1-\rho ^{2})\sigma _{1}^{2}$$.

So even if you don't know the value of $$X_2$$, does the variance change? How is this possible? So how can the variance change from $$\sigma_1^2$$ to $$(1-\rho²)\sigma_1^2$$ if the only difference between the scenarios is that I know $$X_2$$ has realized, but don't know its value?

Thanks!

• What value would you use for "$a$" in this formula in the second scenario?? – whuber May 17 '20 at 11:04
• My question is that the new variance doesn't depend on $a$, so it seems like it does not matter and you could choose any value. – Alex May 17 '20 at 11:33
• Because $a$ is undefined, the distribution is undefined and therefore it makes no sense even to refer to its variance. – whuber May 17 '20 at 11:50

Let's take your argument to the extreme, $$\rho = 1$$. Then the conditional variance is zero, i.e. $$X_1$$ is known, which is absurd.
• $$X_2$$ is not drawn yet
• $$X_2$$ is drawn, but not revealed to you