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It is easy to prove that $\lim_{p\rightarrow 1} m_p = (x_1 + \cdots + x_n)/n$. The following fact about the derivative of $m_p$ with respect to $p$ is also elementary: $$m'_p =\frac{dm_p}{dp} =\frac{1}{p \log p}\cdot\Big[\frac{x_1p^{x_1}+\cdots+x_np^{x_n}}{p_1^{x_1}+\cdots+p_n^{x_n}}-m_p\Big].$$ My interest in this is to create an alternative to the power mean, called the exponential mean: see here and here. The limit I am interested in is $\lim_{p\rightarrow 1} m'_p$. Using WolframAlpha, I computed the limit for $n=2,3,4,5$ (see here) and the following remarkable pattern emerges: $$\lim_{p\rightarrow 1} m'_p=\frac{1}{2n^2}\sum_{1\leq i<j\leq n}(x_i-x_j)^2.$$ How do you go about formally proving this fact? It does not sound elementary to me. Also, it sounds like $m_p$ is a strictly increasing function of $p$ (its derivative beeing positive everywhere, with $m'_0 =+\infty$ and $m'_\infty =0$) unless all the $x_i$'s are identical.

Update

In short, $m_1$ is the arithmetic mean and $m'_1$ is half the empirical variance of $x_1,\cdots,x_n$. I tried to see if such simple formulas existed for the power mean $M_p$, but I could not find anything interesting other than the well known fact that $M_1=m_1$ is the arithmetic mean. It would be interesting to see how the second and third derivatives of $m_p$ at $p=1$ are linked to the higher empirical moments of $x_1,\cdots,x_n$.

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    $\begingroup$ The 5th example of the quasi-arithmetic mean wikipedia page appears to be your expression $m_p$, which may be useful. $\endgroup$
    – Mark
    Aug 25, 2020 at 17:45
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    $\begingroup$ Thanks for asking a well written and very polite question. I think, though, that this is going to just yield to L'Hospital's rule. $\endgroup$ Aug 25, 2020 at 17:54
  • $\begingroup$ Thanks Mark. I actually researched the mean in question in more details before posting, but I could not find anything leading to either a statement or proof of my result regarding $m'_1$. $\endgroup$ Aug 25, 2020 at 17:56
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    $\begingroup$ I found it a bit confusing when reading this question that it started with a computation involving an apparently undefined quantity $m_p$. I eventually noticed that it was defined in the title, but, if you ever have occasion to edit, it may be appropriate to reproduce the definition in the body. $\endgroup$
    – LSpice
    Jan 16 at 15:57

1 Answer 1

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$\newcommand\bar\overline$ Letting $t:=\ln p$, we see that the limit in question is the limit of $$d(t):=\frac1t\Big(\sum_1^n x_j e^{tx_j}\Big/\sum_1^n e^{tx_j}-m_{e^t}\Big)$$ as $t\to0$. Next, letting $\bar x:=\frac1n\,\sum_1^n x_j$, $\bar{x^2}:=\frac1n\,\sum_1^n x_j^2$, and $s^2=\bar{x^2}-\bar x^2$, we have $$\sum_1^n x_j e^{tx_j}=\sum_1^n x_j (1+tx_j+o(t)) =n(\bar x+t\bar{x^2})+o(t),$$ $$\sum_1^n e^{tx_j}=\sum_1^n (1+tx_j+o(t)) =n(1+t\bar x)+o(t),$$ $$m_{e^t}=\log_{e^t}\Big(\frac1n\,\sum_1^n e^{tx_j}\Big) \\ =\log_{e^t}(1+t\bar x+t^2\bar{x^2}/2+o(t^2)) \\ =\tfrac1t\,\ln(1+t\bar x+t^2\bar{x^2}/2+o(t^2)) \\ =\bar x+ts^2/2+o(t).$$ So, $$d(t)=\frac1t\Big(\frac{\bar x+t\bar{x^2}}{1+t\bar x}+o(t)-\bar x-ts^2/2\Big) \\ =\frac1t\Big((\bar x+t\bar{x^2})(1-t\bar x)+o(t)-\bar x-ts^2/2\Big) \\ =s^2/2+o(1). $$ So, the limit in question is $$s^2/2 =\frac1{4n^2}\sum_{1\le i,j\le n}(x_i-x_j)^2 \\ =\frac1{4n^2}\sum_{1\le i,j\le n,\ i\ne j}(x_i-x_j)^2 \\ =\frac1{2n^2}\sum_{1\le i<j\le n}(x_i-x_j)^2,$$ as conjectured.


Details on the first equality in the last three-line display: The left-hand side of that equality is $\frac12\,Var\,X$, where $X$ is any random variable whose distribution is $\frac1n\,\sum_1^n\delta_{x_j}$, where $\delta_a$ is the Dirac probability measure at point $a$. The right-hand side of that equality is $$\frac14\,E(X-X')^2=\frac14\,Var(X-X')=\frac12\,Var\,X,$$ where $X'$ is an independent copy of $X$.

Of course, that equality can also be checked by straightforward algebraic calculations.

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  • $\begingroup$ Thanks Losif. Still trying to figure out why WA gives half your value, see dsc.news/2FXbHpX when $n=2$. $\endgroup$ Aug 25, 2020 at 18:55
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    $\begingroup$ @VincentGranville : Sorry, I thought $m_p$ was defined as $\bar x$. The necessary correction has now been made, and your conjecture holds. $\endgroup$ Aug 25, 2020 at 19:52
  • $\begingroup$ @Losif: thank you. One of my other problems is to prove (or disprove) that $m_p \leq m_q$ if $p<q$. This inequality holds for the power mean $M_p$, wondering if it is true too for the expo mean $m_p$. I'm wondering if I should open another question regarding this, or maybe you have a sense that it might be trivial. $\endgroup$ Aug 25, 2020 at 20:26
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    $\begingroup$ @VincentGranville : Yes, $m_p$ is nondecreasing in $p>0$, because $L(t):=\ln Ee^{tX}$ is convex in $t$, with $L(0)=0$. (The first letter in my first name is, not L, but the upper case of i.) $\endgroup$ Aug 25, 2020 at 20:45
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    $\begingroup$ @LSpice : Thank you for the reminder. $\endgroup$ Jan 16 at 16:16

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