Let $X_1,..., X_n$ be an i.i.d. standard normal.
Let $X_{(1)}= \min \{ X_1,X_2,\ldots, X_n \}$ and $X_{(n)}= \max \{ X_1,X_2,\ldots, X_n \}$ and define the range as $R=|X_{(n)}-X_{(1)}|$.
Are there any concentration results for the range? That is can we find bound on \begin{align} \mathbb{P}[R \ge t ] \le ???? \end{align}
I know that there are concentration results for the maximum like this is on \begin{align} \mathbb{P}[ X_{(n)} \ge \sqrt{ 2 \log (2n) }+t ] \le 2e^{- \frac{t^2}{2}}. \end{align}
I had a feeling that because of symmetry the range can be estimated by $R$ approximate by $2 | X_{(n)}|$. Not sure how to do this though.
Maybe you want a tighter bound than the straightforward one in the following? $$ \mathbb{P}[ R_n \ge 2\sqrt{2 \log(2n)}+2t ] \le 4e^{-t^2/2} $$ Which I doubt to have any kind of better estimation.
Update
You see , by symmetry, $-X_{(1)}$ also has the same distribution as $X_{(n)}$ .
Then we have this straight forword probability relation for all $T>0$ $$ \mathbb{P}( R_n > 2T ) = \underbrace{\mathbb{P}( R_n > 2T, X_{(n)}>T )}_{ \le \mathbb{P}( X_{(n)}>T )}+\underbrace{\mathbb{P}( R_n > 2T, X_{(n)}\le T, -X_{(1)} > T )}_{\le \mathbb{P}(-X_{(1)} > T) } + \underbrace{\mathbb{P}( R_n > 2T, X_{(n)}\le T, -X_{(1)} \le T )}_{=0}$$ Thus $$ \mathbb{P}( R_n > 2T )\le 2\mathbb{P}( X_{(n)}>T )$$
By choosing $T:= \sqrt{2 \log(2n)}+t$, we'll the above result.