# Probability that one Gaussian RV exceeds all others

Imagine we have $$k$$ Gaussian RVs $$X_i \sim N(\mu_i, \sigma_i^2) \text{ for } i=1, \ldots, k$$ and we sample from each of them independently to produce a vector, $$\vec{x} = (x_1, \ldots, x_k)$$.

For one of the Gaussian RVs, say $$X_j$$, I am interested in computing the probability that it exceeds all others, i.e. $$\Pr\left\{ \cap_{i\not= j} \, X_j > X_i \right\}.$$

I know I can use Monte Carlo sampling to estimate this probability. But are there any closed-form analytical methods or approximations?

• You can condition on the value of $X_j$ and then use the total probability formula. – Ian Jan 24 at 4:20
• @Math1000 What is written there is both deleted and incorrect. – Ian Jan 24 at 4:21
• @Ian - I see, but then I need to integrate over all possible conditional values for $X_j$, which doesn't seem nice. Is there an easy way to approximate this integral? – ted Jan 24 at 4:30
• A similar question has been asked before. It's not an exact duplicate -- in that case all the $X_i (i \neq j)$ are equi-distributed, which makes it potentially easier, but the question asked for the rank of $X_j$, which makes it potentially harder. Regardless, despite two years and a bounty, nobody was able to come up with anything beyond the obvious. :( In limit cases (e.g. $\sigma_j$ very large) some qualitative stmts can be made, but not in general... – antkam Jan 24 at 7:24

Given you have an expression for the distribution of the maximum, it is then relatively easy to write the probability as a two dimensional integral with the joint distribution (using appropriate integration limits corresponding to $$X_j > \max(\{X_i\}_{i\neq j})$$). Given the normal should be independent to the Gumbel the joint distribution should be separable and this should collapse nicely to just a single integral. This integral is likely quite ugly, but given the functions are all fairly nice it might be feasible to find where the integrand is maximal and just approximate this region to approximate the probability. While this doesn't solve the problem, it should hopefully reduce it to a simpler problem.