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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 $\hat\mu$.

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.

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.

I think all you need is to get exponent of the arithmetic mean of log-log of your series.

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

Using the same logic idea, you can try the arithmetic mean of log-log series: $$\frac 1 n\sum_{i=1}^n\log\log x_i=\frac 1 n\log\prod_{i=1}^n\log x_i= \log\left(\prod_{i=1}^n\log x_i\right)^{\frac 1 n}$$

The last term is linked to something called a "nested exponent": $$\prod_{i=1}^n\log x_i=\log x_1^{{x_2}^{{\dots}^{x_n}}}$$

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 $\hat\mu$.

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.

The geometric means is basically arithmetic mean on log series: $$\frac 1 n\sum_{i=1}^n\log x_i=\frac 1 n\log\prod_{i=1}^nx_i= \log\left(\prod_{i=1}^nx_i\right)^{\frac 1 n}$$ $$\exp\left(\frac 1 n\sum_{i=1}^n\log x_i\right)= \left(\prod_{i=1}^nx_i\right)^{\frac 1 n}$$

I think all you need is to get exponent of the arithmetic mean of log-log of your series.

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

Using the same logic idea, you can try the arithmetic mean of log-log series: $$\frac 1 n\sum_{i=1}^n\log\log x_i=\frac 1 n\log\prod_{i=1}^n\log x_i= \log\left(\prod_{i=1}^n\log x_i\right)^{\frac 1 n}$$

The last term is linked to something called a "nested exponent": $$\prod_{i=1}^n\log x_i=\log x_1^{{x_2}^{{\dots}^{x_n}}}$$