# comparing Gaussian to order statistic of Gaussian

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?

If all the random variables $$Y,X_1,\dots,X_n$$ are independent, then $$P(Y>\max_iX_i)=\int_{-\infty}^\infty P(Y\in dy)P(\max_1^nX_i where $$\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.