# Covariance & conditional probability: given $x_1>x_2$, what is the probability that $x_1>x_3$?

Suppose there are three normal variables, and we know their means, variances, and correlations/covariances. How would I go about calculating the probability that $$x_1>x_3$$, given that $$x_1>x_2$$?

The true problem I'm trying to solve is a ranking problem: if one were to sample a single value from each of $$n$$ random distributions (not independent), what are the probabilities for each value being the largest (larger than the other n-1 values)? My original/title question is trying to figure this out using Bayes Theorem.

• Are they all jointly normal, in addition to being marginally normal? – Kodiologist Oct 12 '16 at 5:21
• No, just marginally normal – Alex W Oct 12 '16 at 19:59

calculating the probability that x1>X3, given that x1>x2

Assuming the joint distribution of $(X_1,X_2,X_3)$ is normal, the joint distribution of $(X_1-X_3,X_1-X_2)$ is normal as well and the probability is thus $$\mathfrak{p}=\dfrac{\mathbb{P}(X_1-X_3\ge 0,X_1-X_2\ge 0)}{\mathbb{P}(X_1-X_2\ge 0)}$$ The denominator is of the form $\Phi(\mu_{12}/\sigma_{12})$ when $X_1-X_2\sim\text{N}(\mu_{12},\sigma_{12}^2)$ [which one could consider as a closed-form expression] but as far as I know there is no generic formula for computing the numerator.

what are the probabilities for each value being the largest (larger than the other n-1 values)

There is no generic (useful) formula for computing this probability when the random variables are not iid. In the iid case, this is an order statistics result.