# Limit distribution of the second maximum for standard normal random variates

Suppose $$X_1, \dots, X_n$$ are independent, standard normal random variables, and let $$X_{(1)} \leq \dots \leq X_{(n)}$$ denote their order statistics. I am interested in the joint distribution of $$X_{(n-1)}$$ and $$X_{(n)}$$ (or equivalently, $$X_{(1)}$$ and $$X_{(2)}$$), and in particular I would like to show that for large $$n$$, with overwhelming probability both maximum values are very close to their expected values, which I believe are equal to $$\sqrt{2 \log n}$$.

For the first minimum/maximum, from the Fisher–Tippett–Gnedenko theorem we know exactly what the limiting distribution of $$X_{(n)}$$ looks like, and so we can then use standard tail bounds on the Gumbel distribution to get the required bounds. For the second maximum however, I cannot find a similar limit theorem, and my intuition is not helping me find a solution.

Looking at e.g. Exercise 8.5 of "A First Course in Order Statistics" by Arnold, Balakrishnan, Nagaraja, this seems like it should somehow be an easy exercise:

When $$n \to \infty$$ but $$i$$ is held fixed, how are the asymptotic distributions of $$X_{(1)}$$ and $$X_{(i)}$$ related? Express the limiting cdf and pdf of $$X_{(i)}$$ in terms of those of $$X_{(1)}$$.

Writing out the formulas, I get $$f_{X_{(2)}}(x) = f_{X_{(1)}}(x) \cdot (n-1) F(x) / (1 - F(x))$$ where $$f$$ and $$F$$ are the standard normal cdf and pdf, but I am not sure how to (conveniently) continue from here to show that $$X_{(2)}$$ is also tightly concentrated around $$\sqrt{2 \log n}$$ for large $$n$$.

Any feedback is appreciated.

For each $$x\in\mathbb{R}$$, define $$\epsilon_n$$ so that $$F(\sqrt{2\log n}+x)=1-\frac{\epsilon_n}{n}$$ holds. Then from the asymptotic formula $$1 - F(x) \sim \frac{1}{\sqrt{2\pi}x}e^{-x^2/2}$$ as $$x\to+\infty$$, we can check that $$\epsilon_n=n^{o(1)}$$ and
$$\epsilon_n\xrightarrow[n\to\infty]{}\begin{cases}+\infty,&x<0,\\0,&x\geq0.\end{cases}$$
$$\mathbf{P}\left(X_{(n-1)}\leq\sqrt{2\log n}+x\right) = \left(1-\frac{\epsilon_n}{n}\right)^n+\epsilon_n\left(1-\frac{\epsilon_n}{n}\right)^{n-1} = (1 + o(1)) (1 - \epsilon_n) e^{-\epsilon_n}$$
This allows to show that $$X_{(n-1)}-\sqrt{2\log n}$$ also converges in distribution to $$0$$.