Suppose $X$ and $Y$ are bivariate normal with mean $\mu=(\mu_1,\mu_2)$ and covariance $\Sigma = \begin{bmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{12} & \sigma_{22} \\ \end{bmatrix}$. What is the probability $\Pr\left(X<Y|\min\left(X,Y\right)\right)$?

  • $\begingroup$ @whuber right thanks, deleted my thoughts as they aren't adding anything here. $\endgroup$
    – AdamO
    Jul 6, 2018 at 21:03
  • 1
    $\begingroup$ $$ \frac{Pr(m<Y|X=m)}{Pr(m<Y|X=m)+Pr(m<X|Y=m)}$$ $\endgroup$ Jul 6, 2018 at 21:10
  • $\begingroup$ useful link stats.stackexchange.com/questions/30588/… Is this a self-study question? $\endgroup$ Jul 6, 2018 at 22:26
  • $\begingroup$ You should share your thoughts on the problem, irrespective of the fact that this looks like a self-study question. $\endgroup$ Jul 7, 2018 at 4:59

2 Answers 2


Using the slightly more explicit notation $P(X<Y|\min(X, Y)=m)$, where $m$ is a real number, not a random variable. The set on which $\min(X,Y) = m$ is an L shaped path with two half-open segments: one going straight up from the point $(m,m)$ and another going straight to the right from this same point. It's clear that on the vertical leg, $x<y$ and on the horizontal leg $x>y$.

mu1=0, mu2=2, sigma11=0.5, sigma22=1, sigma12=0.2, m=1

Given this geometric intuition its easy to rewrite the problem in an equivalent form, where in the numerator we have only the vertical leg where $x<y$ and in the denominator we have the sum of the two legs.

$P(X<Y|\min(X, Y)) = \frac{ \displaystyle P(m<Y|X=m) }{ \displaystyle P(m<Y|X=m) + P(m<X|Y=m) } \tag{1}$

So now we need to calculate two expressions of the form $P(m<X|Y=m)$. Such conditional probabilities of the bivariate normal distribution always have a normal distribution $\mathcal{N}\left(\mu_{X|Y=m}, s^2_{X|Y=m}\right)$ with parameters:

$\mu_{X|Y=m} = \mu_1+\frac{\displaystyle \sigma_{12}}{\displaystyle \sigma_{22}}({m}-\mu_2) \tag{2}$

$s^2_{X|Y=m} = \sigma_{11}-\frac{\displaystyle \sigma_{12}^2}{\displaystyle \sigma_{22}} \tag{3} $

Note that in the original problem definition, $\sigma_{ij}$ referred to elements of the covariance matrix, contrary to the more common convention of using $\sigma$ for standard deviation. Below, we will find it more convenient to use $s^2$ for the variance and $s$ for the standard deviation of the conditional probability distribution.

Knowing these two parameters, we can calculate the probability than $m<X$ from the cumulative distribution function.

$P(m<X|Y=m) = \Phi \left(\frac{\displaystyle \mu_{X;Y=m} -m}{\displaystyle s_{X;Y=m}} \right) \tag{4}$

mutatis mutandis, we have a similar expression for $P(Y>m|X=m)$. Let

$ z_{X|Y=m} = \frac{\displaystyle \mu_{X;Y=m} - m}{\displaystyle s_{X;Y=m}} \tag{5} $


$ z_{Y|X=m} = \frac{\displaystyle \mu_{Y;X=m} -m}{\displaystyle s_{Y;X=m}} \tag{6} $

Then we can write the complete solution compactly in terms of these two $z$ scores:

$ P(X<Y|\min(X, Y)=m) = 1 - \frac{ \displaystyle \Phi(z_{X|Y=m}) }{ \displaystyle \Phi(z_{X|Y=m})+\Phi(z_{Y|X=m}) } \tag{7}$

Based on simulation code provided by the question author, we can compare this theoretical result to the simulated results:

enter image description here

  • $\begingroup$ In (3) I think that the left hand side should have a square, because it is the conditional variance while the standard deviation is used later. $\endgroup$
    – Yves
    Jul 8, 2018 at 7:09
  • $\begingroup$ You are quite right @Yves, and I believe my recent edits have fixed the issue. Thank you. $\endgroup$
    – olooney
    Jul 8, 2018 at 15:23
  • $\begingroup$ @olooney, thank you for this reply. I can follow the derivation and it seems correct. However, I tried verifying (1) and (7) in a simulation and the results were pretty different. You can see my R code here gist.github.com/mikeguggis/d041df05565f63f8be2c6c51f5cf8961 $\endgroup$
    – mike
    Jul 9, 2018 at 14:02
  • $\begingroup$ @mike, I think I had a sign error. After fixing that, the theoretical result seems to agree with the results of the simulation. gist.github.com/olooney/e88a66d2d2fa7f2f0cd0d0dd6b708739 $\endgroup$
    – olooney
    Jul 9, 2018 at 19:01
  • $\begingroup$ @olooney, good catch. I am still unable to understand why the two simulation based estimates do not match (lines 30-32 in my code). $\endgroup$
    – mike
    Jul 9, 2018 at 20:42

The question can be rewritten using a modified version of Bayes theorem (and an abuse of notion for $Pr$)

\begin{align} Pr(X<Y|min(X,Y) = m) &= \frac{Pr(min(X,Y)=m|X<Y)Pr(X<Y)}{Pr(min(X,Y)=m|X<Y)Pr(X<Y)+Pr(min(X,Y)=m|X\geq Y)Pr(X\geq Y)}\\ &= \frac{Pr(X<Y,min(X,Y)=m)}{Pr(X<Y,min(X,Y)=m)+Pr(X\geq Y,min(X,Y)=m)}. \end{align}

Define $f_{X,Y}$ to be the bivariate PDF of $X$ and $Y$, $\phi(x) = \frac{1}{\sqrt{2\pi}}exp(-\frac{1}{2}x^2)$ and $\Phi(x) = \int_{-\infty}^x\phi(t)dt$. Then

\begin{align} Pr(X<Y,min(X,Y)=m) &=Pr(X=m,Y>m) \\ &= \int_m^\infty f_{X,Y}(m,t)dt \end{align}


\begin{align} Pr(X\geq Y,min(X,Y)=m) &=Pr(X\geq m,Y=m) \\ &= \int_m^\infty f_{X,Y}(t,m)dt \end{align}

Using normality and the definition of conditional probability the integrands can be rewritten as

$$f_{X,Y}(m,t) = f_{Y|X}(t)f_X(m) = \frac{1}{\sqrt{\sigma_{Y|X}}}\phi\left(\frac{t-\mu_{Y|X}}{\sqrt{\sigma_{Y|X}}}\right)\frac{1}{\sqrt{\sigma_{11}}}\phi\left(\frac{m-\mu_1}{\sqrt{\sigma_{11}}}\right)$$


$$f_{X,Y}(t,m) = f_{X|Y}(t)f_Y(m) = \frac{1}{\sqrt{\sigma_{X|Y}}}\phi\left(\frac{t-\mu_{X|Y}}{\sqrt{\sigma_{X|Y}}}\right)\frac{1}{\sqrt{\sigma_{22}}}\phi\left(\frac{m-\mu_2}{\sqrt{\sigma_{22}}}\right).$$

Where $$\mu_{X|Y} = \mu_1 + \frac{\sigma_{12}}{\sigma_{22}}(m-\mu_2),$$

$$\mu_{Y|X} = \mu_2 + \frac{\sigma_{12}}{\sigma_{11}}(m-\mu_1),$$

$$\sigma_{X|Y} = \left(1-\frac{\sigma_{12}^2}{\sigma_{11}\sigma_{22}}\right)\sigma_{11}$$


$$\sigma_{Y|X} = \left(1-\frac{\sigma_{12}^2}{\sigma_{11}\sigma_{22}}\right)\sigma_{22}.$$


\begin{equation} Pr(X<Y|min(X,Y) = m) = \frac{\left(1-\Phi\left(\frac{m-\mu_{Y|X}}{\sqrt{\sigma_{Y|X}}}\right)\right)\frac{1}{\sqrt{\sigma_{11}}}\phi\left(\frac{m-\mu_1}{\sqrt{\sigma_{11}}}\right)}{\left(1-\Phi\left(\frac{m-\mu_{Y|X}}{\sqrt{\sigma_{Y|X}}}\right)\right)\frac{1}{\sqrt{\sigma_{11}}}\phi\left(\frac{m-\mu_1}{\sqrt{\sigma_{11}}}\right)+\left(1-\Phi\left(\frac{m-\mu_{X|Y}}{\sqrt{\sigma_{X|Y}}}\right)\right)\frac{1}{\sqrt{\sigma_{22}}}\phi\left(\frac{m-\mu_2}{\sqrt{\sigma_{22}}}\right)}. \end{equation}

This final form is very similar to the result @olooney arrived at. The difference is his probabilities are not weighted by the normal densities.

An R script for numerical verification can be found here


Your Answer

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

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