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The geometric mean is a multiplicative alternative to the arithmetic mean, which we could call additive mean, thereby calling the geometric mean multiplicative mean. My question is the following: what comes after the multiplicative mean (aka geometric mean)?

This is interesting in the context of the log-log-normal distribution, which is a continuous probability distribution of a random variable whose logarithm logarithm $\ln(\ln(x))$ is normally distributed. Its PDF is:

$f(x) = \displaystyle \frac{1}{\sqrt{2\pi\sigma}x\ln(x)}\exp\Bigg({\frac{-\big(\ln(\ln(x)) - \mu\big)^2}{2\sigma^2}}\Bigg) \quad x \geq 1$

Now, we could rewrite this equation as:

$f(x) = \displaystyle \frac{1}{\sqrt{2\pi\sigma}x\ln(x)}\exp\Bigg({\frac{-\big(\ln(\ln(x^{\frac{1}{e^{\mu}}}))\big)^2}{2\sigma^2}}\Bigg) \quad x \geq 1$

In this equation, $\frac{1}{e^{\mu}}$ is the parameter that I am interested in. This is a level 2 mean if we consider the additive mean (aka arithmetic mean) $\mu$ (for a normal distribution) to be a level 0 mean and the multiplicative mean (aka geometric mean) $e^{\mu}$ (for the log-normal distribution) to be a level 1 mean.

Is there a standard definition for this type of mean?

And did anyone study the sequence of means introduced here?

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    $\begingroup$ I don't know of anyone looking at the specific thing you're asking about, but there are a lot of different kinds of "mean". $\endgroup$ – Glen_b -Reinstate Monica Jan 14 '19 at 23:57
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And did anyone study the sequence of means introduced here?

Many. Check the Generalized Mean

$$M_p(x_1,\dots,x_n) = \left( \frac{1}{n} \sum_{i=1}^n x_i^p \right)^{\frac{1}{p}} $$

in

https://en.wikipedia.org/wiki/Generalized_mean

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  • $\begingroup$ This is off topic in this thread, I am just pinging you. Given your expertise in related subjects, perhaps you could be interested in my question with a bounty on the Probabilistic Reduction approach of Aris Spanos. $\endgroup$ – Richard Hardy Jan 18 '19 at 9:39
  • $\begingroup$ @RichardHardy Richard, thanks for the pointer. I won't be able to respond on time for the bounty, things are frantic right now. But I am indeed interested in posting an answer there and will do eventually. $\endgroup$ – Alecos Papadopoulos Jan 18 '19 at 11:05
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I think all you need is to get the arithmetic mean of log-log of your series. Here's why.

The geometric mean is basically an exponent of arithmetic mean of log series: $$\bar y=\frac 1 n\sum_{i=1}^n y_i$$ $$\frac 1 n\sum_{i=1}^n\log x_i=\frac 1 n\log\prod_{i=1}^nx_i= \log\left(\prod_{i=1}^nx_i\right)^{\frac 1 n}$$ $$\exp\left(\frac 1 n\sum_{i=1}^n\log x_i\right)\equiv e^{\bar x}= \left(\prod_{i=1}^nx_i\right)^{\frac 1 n}$$

If you have the base normal distribution $\mathcal N(\mu,\sigma^2)$, which forms a lognormal distribution then the geometric mean of the lognormal series is going to give you a decent estimator $\hat\mu$. In my opinion this is the motivation for using the geometric mean. It's a good supparization of multiplicative processes such as lognormal, geometric Brownian motion etc.

Using the same logic, you can try the arithmetic mean of log-log series: $$\hat\mu'=\frac 1 n\sum_{i=1}^n\log\log x_i$$

This will give a decent estimator $\hat\mu'$ of the mean of the base normal of your log-lognormal distribution. I won't say that it's the most optimal estimator, not sure about that, but it will be a sensible one.

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The mean used for the log-log-normal distribution remains the multiplicative mean (geometric mean), like the one used for the log-normal distribution. But in the case of the log-log-normal distribution, the variable is raised to the power of the inverse of that mean, while it is simply divided by that mean in the case of the log-normal distribution.

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