Let $n \ge 3$ be an integer and let $X=(X_1,\ldots,X_n)$ be random vector with iid coordinates from $N(0,1)$. For $1 \le k \le n$, let $X_{(k)}$ be the value of the $k$th largest coordinate of $X$.

Question.What are good (anti-)concentration inequalities for $X_{(1)} - X_{(2)}$ ?

References welcome!

## A crude concentration inequality

Note that $X_{(1)} - X_{(2)} \le \Delta := \max_{i,j} X_i - X_j = X_{(1)} - X_{(n)}$. Moreover, $E[\Delta] \le 2\max_i X_i \le 2\sqrt{2\log n},$ and so using the result from this post, we have

$$ \begin{split} P(X_{(1)} - X_{(2)} &\ge 2\sqrt{2 \log n} + t) \le P(\Delta \ge E[\Delta] + t)\\ &\le 2 P(|\max_i X_i - \mathbb E[\max_i X_i]| \ge t/2) \le 2e^{-t^2/8}. \end{split} $$

I wonder if my above somewhat naive bounds can be improved.

## Edit: $P(X_{(1)} - X_{(2)} > t)$ when $n \ge 3$ and $t \ge 4 \sqrt{2 \log n}$

Inspired by the posted answers and the above Borell-TIS inequality, one may compute $$ \begin{split} P(X_{(1)} - X_{(2)} > t) &\le P(X_{(1)} > E X_{(1)} + t/2) + P(X_{(2)} < E X_{(1)} - t/2)\\ &\le e^{-t^2/8} + P(X_{(2)} < \sqrt{2 \log n} - t/2)\\ &= e^{-t^2/8} + P(X_{(2)} \le -t/4), \text{ if }t \ge 4\sqrt{2 \log n}\\ &= e^{-t^2/8} + (\Phi(-t/2))^n + n(1 - \Phi(-t/4))\cdot (\Phi(-t/4))^{n-1}\\ &=e^{-t^2/8} + (\Phi^c(t/4))^n + n\Phi(-t/4)\cdot (\Phi^c(t/4))^{n-1} \\ &\le e^{-t^2/8} + (\Phi(t/4)+n\Phi(-t/4))\cdot e^{-(n-1)t^2/16}\\ &\le e^{-t^2/8}+(1+n/2)e^{-(n-1)t^2/16} \le 2e^{-t^2/8}, \text{ if }n \ge 3. \end{split} $$

We deduce that

If $n \ge 3$ and $ t \ge 4\sqrt{2 \log n}$, then we have the concentration inequality $$ P(X_{(1)} - X_{(2)} \ge t) \le 2e^{-Ct^2}. $$

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