Is “geometric mean” the same as “the first moment of the lognormal distribution”?

I would like to compare the results of two studies, one reporting "geometric mean diameter" and the other one reporting "the first moment of the lognormal size distribution". I am not sure whether the reported values are the same or not. Thank you.

• Thank you very much Xi'an.That answers my question. – Mehrdad Jan 2 at 18:51
• apologies, I forgot a $1/n$ term, see the answer below. – Xi'an Jan 2 at 21:22

The geometric mean $$(X_1\ldots X_n)^{1/n}$$ of a sample $$X_1,\ldots, X_n$$ is an estimator while the first moment $$\mathbb{E}[X]$$ of the lognormal distribution is a theoretical quantity.

If you are asking whether or not the geometric mean $$(X_1\ldots X_n)^{1/n}$$ of a lognormal sample is an unbiased estimator of the first moment $$\mathbb{E}[X]$$ of the lognormal distribution, $$\mathcal{LN}(\mu,\sigma)$$ the answer is no, as a simple query to Wikipedia can show: indeed $$\mathbb{E}[X]=\int_0^\infty x \frac{1}{\sqrt{2\pi}\sigma x}\exp\{-(\ln(x)-\mu)^2/2\sigma^2\}\,\text{d}x=\overbrace{\mathbb{E}[e^Y]}^{Y=\ln X}=\overbrace{\exp\{\mu+\sigma^2/2\}}^\text{Laplace transform}$$ and $$\mathbb{E}[(X_1\ldots X_n)^{1/n}]=\mathbb{E}[X_1^{1/n}]^n=\underbrace{\mathbb{E}[e^{Y/n}]^n}_{Y=\ln X_1}=\underbrace{\exp\{n(\mu/n+\sigma^2/2n^2)\}}_{\text{var}(Y/n)=\sigma^2/n^2}=\exp\{\mu+\sigma^2/2n\}$$

• But a relatively straight forward transformation gives a distribution whose first moment has an easy straightforward estimator :) – AdamO Jan 2 at 22:36

They're not the same quantity. One is geometric mean and the other is arithmetic, but in an empirical setup the difference can be indistinguishable.

Important! - since we're talking about the values I assume we're dealing with estimator of the first moment, not the first moment itself. The first moment is a characteristic of the distribution (population), it's unknown. We can estimate it though. Its common estimator is the arithmetic average $$\bar x=(1/n)\sum_{i=1}^n \ln x_i$$

The geometric mean is given by formula $$\mu=(\prod_{i=1}^n x_i)^{1/n}=\exp \left[(1/n)\sum_{i=1}^n \ln x_i\right]$$

So, clearly $$\mu\ne\bar x$$, hence geometric mean is not equal to the arithmetic average.

I've been careful so far to not make a statement about relation of geometric mean and first moment because the latter cannot be known, only its estimator can be calculator. Moreover, there usually are multiple estimators of the same population parameter, that are optimal under different conditions.

So, here's the punchline. In empirical setup the geometric mean can be as good of an estimator of the first moment as a more common arithmetic average. Particularly, when there is high degree of uncertainty about the variance of the distribution.