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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.

More info:

$\sigma_n$ 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:

  1. when $n$ increases from 2 to 2000, $\sigma_n$ decreases from 0.9 to 0.27

  2. $n=10, 10^2, ..., 10^7$, $\sigma_n=0.90,0.51,0.35,0.29,0.25,0.22,0.20,0.18$, respectively

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  • $\begingroup$ $\sigma_n$ 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 $0$. $\endgroup$
    – Henry
    Jan 9, 2021 at 1:43
  • $\begingroup$ Hi, Henry, thanks very much! Do you have any reference for the value $\sigma_n\sqrt{\log(n)}$? I would like some theoretical guarantees. $\endgroup$ Jan 10, 2021 at 5:07
  • $\begingroup$ It was just an empirical observation, but @ClaudeLeibovici's answer suggests to me that $\sigma_n\sqrt{\log(n)} \to \frac{1}{\sqrt{2}} \approx 0.7071$ $\endgroup$
    – Henry
    Jan 10, 2021 at 16:54

1 Answer 1

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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 $\sigma_n$ $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.430284 & 0.443256 \\ 2 & 0.328501 & 0.328637 \\ 3 & 0.272163 & 0.271588 \\ 4 & 0.236707 & 0.236187 \\ 5 & 0.211987 & 0.211577 \\ 6 & 0.193551 & 0.193231 \\ 7 & 0.179145 & 0.178892 \\ 8 & 0.167499 & 0.167296 \\ 9 & 0.157836 & 0.157671 \\ 10 & 0.149656 & 0.149518 \\ 11 & 0.142615 & 0.142498 \\ 12 & 0.136474 & 0.136354 \\ 13 & 0.131056 & 0.131152 \\ 14 & 0.126231 & 0.126569 \\ 15 & 0.121898 & 0.133751 \end{array} \right)$$

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  • $\begingroup$ Hi, Claude, thanks very much! Could you give me more information or reference about the expansion of erf, as I could not find this expansion of inverse erf containing log and square root. If this is an asymptotic expansion, then we probably could not ignore any terms bigger than o(1) in the expansion when taking the difference. $\endgroup$ Jan 10, 2021 at 5:03
  • $\begingroup$ @JuntaoWang. Have a look at reference.wolfram.com/language/ref/InverseErf.html $\endgroup$ Jan 10, 2021 at 5:29

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