Probability bound of the difference of order statistics for i.i.d. Gaussian random variables I have asked a related question before (with more stronger requirement):
Probability bound of the difference of order statistics for correlated and identical Gaussian random variables.
Now, I'm pretty sure that there doesn't exist any such bound or difficult to calculate such bound. However, in this question I'm relaxing one requirement to make it more simpler. So, my question is as follows:
Suppose, there are $n$ i.i.d. standard Gaussian random variables namely, $X_1, X_2, ..., X_n$ with $X_i\sim\mathcal{N}(0,1)$ for all $i\in\{1,2, ...n\}$. Then the order statistics of them can be denoted as $X_{(1)}\le X_{(2)} \le ...\le X_{(n)}$.
In this case, is it possible to calculate the upper bound of $Pr[X_{(k)}-X_{(l)}\le \tau]$ (or, is there any research paper exists which calculated such bound for i.i.d. Gaussian random variables) where, $X_{(k)}$ and $X_{(l)}$ are $k$-th and $l$-th order statistic with $l<k$. Even the bound for extreme case (i.e., $Pr[X_{(n)}-X_{(1)}\le\tau]$) would be helpful.

For large $n$, the difference $W=X_{(n)}-X_{(1)}$ has a distribution which is roughly the same as that for $2X_{(n)}$.
One standard way of writing that limit (as discussed in questions here or here) is
$$P\left[\frac{X_{(n)}-b_n}{a_n}<u\right] \to \exp(-e^{-u})$$
Using $W_n\sim 2X_{(n)}$ and $a_n\sim 1/b_n$, we get
$$P\left[\frac{Wb_n-2b_n^2}{2}<u\right] \to \exp(-e^{-u})$$
where $b_n\sim\sqrt{2\log n-\log\log n-\log(4\pi)}$.
