Proportionality between the expectation of the minimum of $N$ lognormal variables and the expectation of the lognormal distribution

Question

I am investigating the relationship between the expectation of the minimum of $N$ independent and identically distributed lognormal variables, $E[M]$, and the expectation of the lognormal distribution, $E[X]$. Based on my numerical simulations, it seems that there is an approximate proportionality between these expectations for fixed parameters $\mu$ and $\sigma$. However, I have not been able to find an analytical proof for this proportionality.

Let $X_1, X_2, \ldots, X_N$ be i.i.d. lognormal random variables with parameters $\mu$ and $\sigma$, and let $M = \min(X_1, X_2, \ldots, X_N)$. I am interested in the following ratio: \begin{gather*} \frac{E[M]}{E[X]} = \int_0^{\infty}\frac{1}{\exp\left(\mu + \frac{1}{2}\sigma^2\right)} m N \left(\frac{1}{2} \left( 1 + \operatorname{erf}\left(\frac{\ln m - \mu}{\sigma\sqrt{2}}\right) \right)\right)^{N-1}\\ \cdot \frac{1}{m\sigma\sqrt{2\pi}}\exp\left(-\frac{(\ln m - \mu)^2}{2\sigma^2}\right)\,dm \end{gather*}

My numerical simulations suggest that this ratio is approximately constant for any values of $\mu$, if $\sigma$ and $N$ are fixed.

In other words, it is saying that the following expression \begin{gather*} \frac{d}{d\mu} \frac{E[M]}{E[X]} = \int_0^{\infty}\frac{1}{\pi \sigma^2} 2^{-N} e^{-\mu - \frac{\sigma^2}{2} - \frac{(\mu - \ln m)^2}{\sigma^2}} N \left(1 + \operatorname{erf}\left(\frac{\mu - \ln m}{\sqrt{2}\sigma}\right)\right)^{-2 + N} \cdot\\ \left(2 (-1 + N) + \frac{1}{\sigma} \left(e^{\frac{(\mu - \ln m)^2}{2 \sigma^2}} \sqrt{2\pi} \left( -2 + \operatorname{erfc}\left(\frac{\mu - \ln m}{\sqrt{2}\sigma}\right) \right) (\mu + \sigma^2 - \ln m) \right)\right)\,dm \equiv 0 \end{gather*}

holds for all $\mu$, $\sigma$ and $N$.

Is there an analytical proof or explanation for this approximate proportionality? Or is there any literature discussing this phenomenon? I would appreciate any insights or references.

Thank you in advance!

Answer 1

$\def\d{\mathrm{d}}\def\e{\mathrm{e}}\DeclareMathOperator{\erf}{erf}\def\peq{\mathrel{\phantom=}}\def\paren#1{\left(#1\right)}$A change of variable for $\dfrac{E(M)}{E(X)}$ suffices:\begin{align*} &\peq \int_0^{+\infty} mN \paren{ \frac{1}{2} \paren{ 1 + \erf\paren{ \frac{ \ln m - \mu }{ \sqrt{2} \sigma } } } }^{ N - 1 }\\ &\peq \cdot \frac{1}{ \sqrt{2\pi} m\sigma } \exp\paren{ -\frac{1}{ 2\sigma^2 } (\ln m - \mu)^2 } \cdot \exp\paren{ -\mu - \frac{1}{2} \sigma^2 } \,\d m\\ &= \frac{N}{ \sqrt{2\pi} \sigma } \exp\paren{ -\frac{1}{2} \sigma^2 } \int_0^{+\infty} \paren{ \frac{1}{2} \paren{ 1 + \erf\paren{ \frac{ \ln m - \mu }{ \sqrt{2} \sigma } } } }^{ N - 1 }\\ &\peq \cdot \exp\paren{ -\frac{1}{ 2\sigma^2 } (\ln m - \mu)^2 } \cdot \e^{-\mu} \,\d m\\ &= \frac{N}{ \sqrt{2\pi} \sigma } \exp\paren{ -\frac{1}{2} \sigma^2 } \int_0^{+\infty} \paren{ \frac{1}{2} \paren{ 1 + \erf\paren{ \frac{ \ln n }{ \sqrt{2} \sigma } } } }^{ N - 1 } \exp\paren{ -\frac{ (\ln n)^2 }{ 2\sigma^2 } } \,\d n, \end{align*} where $n = m\e^{-\mu}$.

Answer 2

For a log-normal, changing $μ$ to $μ+c$ while keeping $σ$ and N constant simply multiplies the values (and their minimum and the expectations) by $e^c$.

So you have $\dfrac{E[e^cM]}{E[e^cX]}=\dfrac{E[M]}{E[X]}$ as you have observed.

You would get a similar result for any other random variable when multiplied by a positive constant $k$, providing the expectations exist. This multiplication would not change the scale-free coefficient of variation (standard deviation divided by mean) either.