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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.

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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

$$\epsilon_n\xrightarrow[n\to\infty]{}\begin{cases}+\infty,&x<0,\\0,&x\geq0.\end{cases}$$

Moreover,

$$\mathbf{P}\left(X_{(n-1)}\leq\sqrt{2\log n}+x\right) = \left(1-\frac{\epsilon_n}{n}\right)^n+\epsilon_n\left(1-\frac{\epsilon_n}{n}\right)^{n-1} = (1 + o(1)) (1 - \epsilon_n) e^{-\epsilon_n}$$

This allows to show that $X_{(n-1)}-\sqrt{2\log n}$ also converges in distribution to $0$.

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