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?


  • 1
    $\begingroup$ What value would you use for "$a$" in this formula in the second scenario?? $\endgroup$
    – whuber
    May 17, 2020 at 11:04
  • $\begingroup$ 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. $\endgroup$
    – Alex
    May 17, 2020 at 11:33
  • 1
    $\begingroup$ Because $a$ is undefined, the distribution is undefined and therefore it makes no sense even to refer to its variance. $\endgroup$
    – whuber
    May 17, 2020 at 11:50

1 Answer 1


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.

For the purposes of uncertainty there is no difference between

  • $X_2$ is not drawn yet
  • $X_2$ is drawn, but not revealed to you

In both cases, its value is completely unknown to you. If your friend has seen the outcome, his variance will drop (due to conditioning), but yours won't.


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