I would like to find the conditional bivariate normal distribution. There are two dependent normal variables with the same distribution and the correlation coefficient $\rho$: $X,Y \sim N(\mu, \sigma^2)$. I would like to get $P(X|Y>M)$.

I found the conditional expectation of $X$ given that $Y$ is bigger than $M$: $E(X|Y>M)= \mu + \rho \sigma \frac{\phi(\frac{M-\mu}{\sigma})}{1-\Phi(\frac{M-\mu}{\sigma})}$.

But what is the conditional variance of $var(X|Y>M)$? Is it $(1-\rho^2)\sigma^2 $, as it would be in the case of $var(X|Y=M)$, where variance does not depends on $M$?

And is the conditional distribution $N(E(X|Y>M),var(X|Y>M))$?

  • $\begingroup$ Please search our site for more about these questions. $\endgroup$
    – whuber
    Commented Jan 9, 2021 at 22:36
  • $\begingroup$ Please don't repost a closed question: that caused useful information about duplicate questions to be removed, which is unfair to all subsequent readers. $\endgroup$
    – whuber
    Commented Jan 10, 2021 at 15:52
  • $\begingroup$ The entire distribution is obtained explicitly, with explanations, in at least one of the duplicates. Given that, do you still see any use in having an expression for the conditional variance? If so, then I suppose your question has not been answered--but as John L's answer here indicates, it might not be useful. $\endgroup$
    – whuber
    Commented Jan 10, 2021 at 21:53

1 Answer 1


The conditional variance depends on $M$.

I am not able to find a closed form for the conditional variance, but I can find a closed form for the density. I found it by starting with the conditional cumulative distribution function using the definition of conditional probability, then differentiated to find the conditional density.

The density using Mathematica input form is:

(((mu*(-1 + rho) - rho*t)*Erf[Sqrt[-((mu*(-1 + rho) - rho*t)^2/((-1 + rho^2)*s^2))]/Sqrt[2]])/Sqrt[-((mu*(-1 + rho) - rho*t)^2/((-1 + rho^2)*s^2))] - 
  ((M + mu*(-1 + rho) - rho*t)*Erf[Sqrt[-((M + mu*(-1 + rho) - rho*t)^2/((-1 + rho^2)*s^2))]/Sqrt[2]])/Sqrt[-((M + mu*(-1 + rho) - rho*t)^2/((-1 + rho^2)*s^2))] + 
  (1 + Erf[Sqrt[(2*s^2 - 2*rho^2*s^2)^(-1)]*(mu - mu*rho + rho*t)])/Sqrt[(s^2 - rho^2*s^2)^(-1)])/(2*E^((mu - t)^2/(2*s^2))*Sqrt[2*Pi]*Sqrt[(1 - rho^2)*s^4]*(1 - Erfc[(-M + mu)/(Sqrt[2]*s)]/2))

Your formula for the conditional mean is correct.

I know the conditional variance depends on $M$ because I calculated it by numerical integration.

  • $\begingroup$ Is the conditional density still normal? $\endgroup$
    – Stasya7
    Commented Jan 10, 2021 at 18:05
  • $\begingroup$ it seems like it is not. The older questions @whuber listed above are helpful for your question, particularly this one I think: stats.stackexchange.com/questions/444925/… $\endgroup$
    – John L
    Commented Jan 10, 2021 at 20:23

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