Properties, bounds and limits about difference of two inverse standard normal CDF variables and extreme value distribution

I'm interested in the variable: $$\sigma_n=\Phi^{-1}\left(1-{1\over n}e^{-1}\right)-\Phi^{-1}\left(1-{1\over n}\right),$$ where $$\Phi(\cdot)$$ is the CDF of standard normal distribution. I want to prove $$\sigma_n$$ is monotonically decreasing for $$n\ge 2$$, which is shown by my simulation, but I do not know how to prove it. If this property holds, then it implies that $$\sigma_n$$ also has a non-negative limit as $$\sigma_n>0$$ for all $$n\ge 2$$.

Or you can just tell me $$\sigma_n$$ is bounded or has a finite limit.

$$\sigma_n$$ comes from the distribution $$GEV(x;\mu_n,\sigma_n,0)$$, which is the distribution of the max of $$n$$ i.i.d. standard normal distribution (Extreme value Type I distribution). See https://en.wikipedia.org/wiki/Generalized_extreme_value_distribution.

My simulation shows:

• $\sigma_n$ is positive for $n\ge 2$. You might like to look at $\sigma_n \sqrt{\log(n)}$ - this does not provide bounds for all $n$, though for $n$ between $2$ and $10^{200}$ it is never greater than $0.78$ and never less than $0.70$, suggesting that $\sigma_n$ is likely to be monotonically decreasing towards $0$. Jan 9, 2021 at 1:43
• Hi, Henry, thanks very much! Do you have any reference for the value $\sigma_n\sqrt{\log(n)}$? I would like some theoretical guarantees. Jan 10, 2021 at 5:07
• It was just an empirical observation, but @ClaudeLeibovici's answer suggests to me that $\sigma_n\sqrt{\log(n)} \to \frac{1}{\sqrt{2}} \approx 0.7071$ Jan 10, 2021 at 16:54

We have to start $$\Phi^{-1}\left(1-\frac{x}{n}\right) = \sqrt{2}\ \text{erf}^{-1}\left(1-\frac{2x}{n}\right)$$ which makes $$\sigma_n=\sqrt{2}\Bigg[\text{erf}^{-1}\left(1-\frac{1}{en}\right)-\text{erf}^{-1}\left(1-\frac{1}{n}\right) \Bigg]$$
For small $$x$$ we have $$\text{erf}^{-1}\left(1-x\right)=\sqrt{\frac{1}{2} \left(\log \left(\frac{2}{\pi x^2}\right)-\log \left(\log \left(\frac{2}{\pi x^2}\right)\right)\right)}$$ (have a look here). It is very good for $$0\leq x \leq 0.1$$.
Using it, we have $$\sigma_n=\sqrt{\log \left(\frac{2 e^2 n^2}{\pi }\right)-\log \left(\log \left(\frac{2 e^2 n^2}{\pi }\right)\right)}-$$ $$\sqrt{\log \left(\frac{2 n^2}{\pi }\right)-\log \left(\log \left(\frac{2 n^2}{\pi }\right)\right)}$$ and the limit is $$0$$.
Making $$n=10^k$$, the table contains the approximate and exact values of $$\sigma_n$$ $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.430284 & 0.443256 \\ 2 & 0.328501 & 0.328637 \\ 3 & 0.272163 & 0.271588 \\ 4 & 0.236707 & 0.236187 \\ 5 & 0.211987 & 0.211577 \\ 6 & 0.193551 & 0.193231 \\ 7 & 0.179145 & 0.178892 \\ 8 & 0.167499 & 0.167296 \\ 9 & 0.157836 & 0.157671 \\ 10 & 0.149656 & 0.149518 \\ 11 & 0.142615 & 0.142498 \\ 12 & 0.136474 & 0.136354 \\ 13 & 0.131056 & 0.131152 \\ 14 & 0.126231 & 0.126569 \\ 15 & 0.121898 & 0.133751 \end{array} \right)$$