Given a Gaussian random vector of length N:
$$ X \sim N(\mu,\Sigma) $$
where $\Sigma$ is non-diagonal, what is the probability that a chosen component of $X$ is larger than any other? i.e.
$$P(X_n > max(X_i)) \; \text{ for } \; i = [1,2, ..., n-1,n+1,...,N]$$ For a diagonal covariance matrix, i.e. vector with independent components, we can solve the problem as shown here. I'm not sure how to approach this for the general case, though.
Furthermore -- how can the answer to be extended to include a group of components? e.g.:
$$P(max(X_{n}, X_{n+1}, X_{n+2}) > max(X_i)) \; \text{ for } \; i = [1,2, ..., n-1,n+3,...,N]$$
One idea is to form a new random Gaussian vector where we subtract the component of interest from all of the others, then determine the probability that all the elements of this new vector are positive (similar to the approach in the link I shared). I'm not too sure how to go about doing that, though. You could evaluate the CDF at 0, but for larger dimensions this becomes intractable.
– Jake Oct 14 '21 at 05:46