comparing Gaussian to order statistic of Gaussian

Question

I would like to compute the probability of $\mathbb{P}[Y > max(X_i)], Y\sim N(0, 1), X_i \sim N(0, \sigma_i)$

All the random variables have zero mean, but the variances are different.

My approaches so far were unsuccessful. I tried looking at events $A_i = P(Y>X_i)$ and their intersections and unions. But that didn't work out.

Then I took a step back and tried to search for related results online. If the $X_i$ were IID, this would be comparing Y to the n-th order statistic of a normal random variable. This question seems related: https://stats.stackexchange.com/questions/9001/approximate-order-statistics-for-normal-random-variables It seems as if even for the simplified case, there is no closed form solution for the order statistic.

Am I missing a trick here. Is it possible to calculate this in closed form?

Answer 1

If all the random variables $Y,X_1,\dots,X_n$ are independent, then $$P(Y>\max_iX_i)=\int_{-\infty}^\infty P(Y\in dy)P(\max_1^nX_i<y) \\ =\int_{-\infty}^\infty P(Y\in dy)\prod_1^n P(X_i<y) \\ =\int_{-\infty}^\infty dy\, \varphi(y)\prod_1^n \Phi(y/\sigma_i), $$ where $\Phi$ and $\varphi$ are the standard normal cdf and pdf, respectively.

Mathematica can do nothing with the latter integral even when $n=2$ and $\sigma_1=\sigma_2$. So, it is highly unlikely that the probability $P(Y>\max_iX_i)$ can be computed in closed form in general.

However, when $\sigma_1=\cdots=\sigma_n=1$, then the integral equals $\frac1{n+1}$, which can be easily obtained by the substitution $u=\Phi(y)$. The same result is also obvious by symmetry, because then the random variables $Y,X_1,\dots,X_n$ are identically distributed and hence exchangeable.