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Due to the inequality $2\sqrt{xy}\leq x+y$, the geometric mean is always closer to the smaller value than the arithmetic mean.

In my situation, I need a "mean" that is closer to the larger value, so I thought of simply "flipping" the geometric mean $g(x,y)$ at the arithmetic mean $a(x,y)$: $$a(x,y) + \Big(a(x,y)-g(x,y)\Big) = x+y-\sqrt{xy}$$ The result is always between the arithmetic mean and the larger value.

Is there some special name for this "mean"?

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    $\begingroup$ If you want a mean that emphasizes larger values, you could consider a a root-mean-square, or more generally $A(x^p)^{1/p}$, where $A()$ is the arithmetic mean, for $p > 1$ ... (RMS corresponds to $p=2$) $\endgroup$
    – Ben Bolker
    Apr 12, 2023 at 15:11
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    $\begingroup$ Ordinarily, to justify calling a function $f(\ ,\ )$ a "mean," you would like it to be increasing in both variables. With $f(x,y)=x+y-\sqrt{xy}$ (defined on non-negative values only), the partial derivative in the first variable is $1-\sqrt{y/(4x)},$ which is negative when $y \gt 4x.$ That's not a mean. Instead of guessing formulas and checking their properties, it would be more appropriate to develop a mean based on the characteristics of your underlying problem: what exactly is this mean intended to represent? $\endgroup$
    – whuber
    Apr 12, 2023 at 15:23
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    $\begingroup$ @BenBolker Wikipedia calls it generalized mean, power mean or Hölder mean. $\endgroup$ Apr 12, 2023 at 15:57
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    $\begingroup$ I'm not sure if I should answer or @whuber should - both comments seem valuable. It would be fine with me if you (cdalitz) composed an answer to your own question that incorporates whatever information from the comments is useful. $\endgroup$
    – Ben Bolker
    Apr 12, 2023 at 16:50
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    $\begingroup$ You can describe the geometric mean as $\exp\left (\frac1n \sum \log_e(x_i)\right)$ so one possible flipped version might be $\log_e\left (\frac1n \sum \exp(x_i)\right)$. This will be closer to the maximum value but will not exceed it. A curious effect is that while the geometric mean is in a sense scale-invariant but not location-invariant, this flipped version would in a similar sense be location-invariant but not scale-invariant (the arithmetic mean is both). $\endgroup$
    – Henry
    Apr 12, 2023 at 17:16

1 Answer 1

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Thanks for the valuable comments, which I try to summarize and put into a unifying framework in this answer.

As pointed out by @whuber, my suggested formula violated one plausible axiom for a mean, namely that it should increase in all arguments. @whuber also suggested to base the formula on a more rigorous axiomatic ground by postulating a number of reasonable properties of the desired "mean".

Hence I did a brief literature study, and it turned out that none less than Kolmogoroff (sic!) already did exactly that. In 1930, he postulated the following properties for a function $M:{\mathbb R}^n \to{\mathbb R}$ that represents a "regular mean":

  1. $M$ is continuous and increasing in each variable.
  2. $M$ is a symmetric function.
  3. The mean of repeated data equals the repeated value.
  4. The mean of a sample remains unchanged if a part of the sample is replaced by its corresponding mean

Kolmogoroff proved that, if these conditions hold, the mean must be of the form $$M(x_1,\ldots,x_n) = f^{-1}\left(\frac{1}{n}\sum_{i=1}^n f(x_i)\right)$$ which the Wikipedia page cited by @COOLSerdash calls "generalized f-mean", and the French Wikipedia calls it "quasi-arithmetic mean" or "Kolmogoroff mean".

A. Kolmogoroff: "Sur la notion de la moyenne." Atti Reale Accademia Nazionale dei Lincei, vol. 12,‎ 1930, p. 388–391

When applying this to my particular problem of defining a mean that is greater than the arithmetic mean, it is sufficient that $f$ is a convex function, because for a convex function we have $$\frac{1}{2}\Big(f(x)+f(y)\Big) \geq f\left(\frac{x+y}{2}\right) \Rightarrow f^{-1}\left(\frac{1}{2}(f(x)+f(y))\right) \geq \frac{x+y}{2}$$ The Hölder mean of order $p$, which is the root-mean-square for $p=2$, as suggested by @BenBolker, is the special case $f(x)=x^p$. The choice $f(x)=e^x$, as suggested by @Henry, is yet another special case.

For my use case, I have settled on the Hölder mean of order two.

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    $\begingroup$ +1 nice, authoritative summary. $\endgroup$
    – whuber
    Apr 13, 2023 at 13:48

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