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Given a set of $n$ observations $x_1,\cdots,x_n$, the power mean is defined as $$M_p = \Big(\frac{x_1^p+\cdots + x_n^p}{n}\Big)^{1/p}.$$ Likewise, the exponential mean is defined as $$m_p = \log_p\Big(\frac{p^{x_1}+\cdots + p^{x_n}}{n}\Big).$$ The logarithm is in base $p> 0$. If $p\rightarrow 0$, then $m_p$ tends to the minimum observation. If $p\rightarrow \infty$, then $m_p$ tends to the maximum observation. If $p\rightarrow 1$, then $m_p$ tends to the arithmetic mean. Note that $m_p$ (unlike $M_p$) is defined even if any of the observations is negative. For details, see here.

I am wondering if the following results are correct:

  • If $p < q$, then $m_p \leq m_q$, and $m_p = m_q$ if and only if $x_1 = ... = x_n$.
  • If $p > 1$, then $m_p \leq M_p$, and $m_p = M_p$ if and only if $x_1 = ... = x_n$.

Hint for the first inequality

The second one is not true, see Michael's answer below. The first one may or may not be true, but the following could help settle the question. I used WolframAlpha to compute the derivative of $m_p$ with respect to $p$, see here.

$$m'_p =\frac{dm_p}{dp} =\frac{1}{p \log p}\cdot\Big[\frac{x_1p^{x_1}+\cdots+x_np^{x_n}}{p_1^{x_1}+\cdots+p_n^{x_n}}-m_p\Big].$$

The next step is to prove (or disprove) that $m'_p \geq 0$. Since $\log p > 0$ if $p>1$ and $\log p < 0$ if $p<1$, it suffices to prove that

$$\frac{x_1p^{x_1}+\cdots+x_np^{x_n}}{p_1^{x_1}+\cdots+p_n^{x_n}} \geq m_p \mbox{ if } p>1, $$ $$\frac{x_1p^{x_1}+\cdots+x_np^{x_n}}{p_1^{x_1}+\cdots+p_n^{x_n}} \leq m_p \mbox{ if } p<1. $$

The derivative $m'_p$ is well approximated by a power function, positive everywhere for $p\geq0$, suggesting that the first inequality is correct. A remarkable fact is the following:

$$\lim_{p\rightarrow 1} m'_p=\frac{1}{2n^2}\sum_{1\leq i<j\leq n}(x_i-x_j)^2 \geq 0.$$ I haven't explicitly computed that limit in general, instead I used WolfFramAlpha (see here) to compute it for $n=2,3,4,5$. I've found a pattern, and generalized it to any $n$; see also this MO question. Since $m_\infty = \max(x_1,\cdots,x_n)$ we also have $\lim_{p\rightarrow\infty}m'_p = 0$. It also seems that if the $x_i's$ are not all identical, then $\lim_{p\rightarrow 0^+} m'_p=+\infty$. To complete the proof of the first inequality, one may proceed recursively on $n$.

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  • $\begingroup$ I hope we work with positive numbers. Or maybe not? $\endgroup$ – Michael Rozenberg Aug 23 '20 at 14:41
  • $\begingroup$ We can restrict ourselves to positive numbers, I thought about that. We have too anyway when making comparisons between $m_p$ and $M_p$. $\endgroup$ – Vincent Granville Aug 23 '20 at 14:49
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The second inequality is wrong.

Try $n=3$, $p=e$, $x_1=x_2=\frac{1+\sqrt5}{2}$ and $x_3=2$.

In this case $$M_e-m_e=-0.000719...<0$$

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    $\begingroup$ Thank you Michael. $\endgroup$ – Vincent Granville Aug 23 '20 at 19:38

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