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I read this post How is the minimum of a set of IID random variables distributed? where I can find how to compute the density distribution of the minimum between $N$ positive random variables. If the cumulative of these random variables is $F(x)$ the cumulative of the minimum between $N$ of these random variables is:

$$ \mathbb{P}(\text{min}\le x)=1-\big(1-F(x)\big)^N$$

From this result it is easy to compute the mean value of the minimum (simply using integration by parts): $$ \overline{\text{min}}=\int_{0}^{\infty}x\,\frac{d \mathbb{P}(\text{min}\le x)}{dx}dx=\int_{0}^{\infty}\big(1-F(x)\big)^Ndx $$ At this point I can compute $\overline{\text{min}}$ for a general cumulative distribution $F(x)$.

I want now to get the scaling of $\overline{\text{min}}$ for $N\rightarrow\infty$ when $F(x)=\frac {1}{r!}\int_{0}^{x}l^re^{-l}dl$ for $r\in\mathbb{Z}$ and $r>-1$. (A simple case can be the one for $r=0$, that is the density distribution for the variables is exponential, in this case the average min value scale as $1/N$). In the general case I can arrive to: $$ \overline{\text{min}}=\int_{0}^{\infty}\left(1-\frac{1}{r!}\int_0^{x}l^re^{-l}dl\right)^Ndx $$ but I don't know how to approximate it. For general $r$ it can be useful to approximate $\frac{l^re^{-l}}{r!}\simeq \frac{l^r}{r!}\propto l^r$, since the important contribution for the average minimum value will be $\ll1$. In this approximation my result is: $$ \overline{\text{min}}=\int_{0}^{\infty}\left(1-\int_0^{x}l^rdl\right)^Ndx $$ but still can't figure out which is the scaling with large $N$.

The result should be: $$ \overline{\text{min}}\simeq N^{-\frac{1}{r+1}} $$

I am studying the following paper: https://hal.science/jpa-00232897/document .

Here the authors find the scaling of the mean value of what I suppose is the minimum of the random variables, which in this case are $l_{ij}$, called “distances”. I cannot find the scaling found after Eq.(5), can anyone help me?

Ruth Murphy
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  • Can you clarify the exact mathematical meaning of the operator "$\simeq$": does $a_N \simeq b_N$ means $a_N/b_N \to 1$ as $N \to \infty$? – Zhanxiong May 28 '23 at 15:38
  • By "$a_N\simeq b_N$" I mean $a_N/b_N\rightarrow c$ as $N\rightarrow\infty$ where $c$ is a constant (independent on $N$). – Ruth Murphy May 28 '23 at 17:19
  • The result is a simple one and wholly unrelated to how you frame the question. The energy in (2) is a sum over neighbors of $2N$ terms. The "short links" assumption implies there are $O(N)$ such terms altogether. Each term $l_{ij}$ is approximately $N^{-1/(r+1)}$ following equation (5), whence upon multiplication by $O(N)$ you obtain the stated result. – whuber May 28 '23 at 18:29
  • My question is why equation (5) implies that scaling ($N^{-1/(r+1)}$) – Ruth Murphy May 29 '23 at 12:58
  • And the answer is that you multiply by $N$. – whuber May 29 '23 at 15:35
  • I understand that, given the scaling of the "minimal length", the energy scales like $N^{1-1/(r+1)}$, but why does the minimal length scale like $N^{-1/(r+1)}? – Ruth Murphy May 29 '23 at 15:43

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Here is a way to arrive at the result

  • With $f(l) = \frac{l^re^{-l}}{r!}\simeq \frac{l^r}{r!}\propto l^r$ we have cdf $F(l) \propto l^{r+1}$ and quantile function $Q(p) \propto p^{\frac{1}{r+1}}$
  • The minimum statistic can be expressed with by the transform of a beta distributed variable $$min \sim Q(p) \quad \text{where $p \sim Beta(1,n)$ or $f_P(p) = n(1-p)^{n-1}$}$$
  • The mean can then be approximated with an integral $$\begin{array}{} mean(min)& \approx &\int_0^1 Q(p) f_P(p)\, \text{d}p \\&\approx& n \int_0^1 p^{\frac{1}{r+1}} (1-p)^{n-1}\, \text{d}p\\ &\approx&n Beta(1+\frac{1}{r+1},n) \\&\approx& n \Gamma\left(1+\frac{1}{r+1}\right) n^{-(1+\frac{1}{r+1})}\\& \approx &\Gamma\left(1+\frac{1}{r+1}\right) n^{-\frac{1}{r+1}} \end{array} $$
Sextus Empiricus
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