If you have data that are sampled from a normal distribution, what is the relationship between the arithmetic and geometric means? Would it ever make sense to report the geometric mean instead of the arithmetic mean? (Assume that all the values are positive; no zeros, no negative values)
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4$\begingroup$ Would you be happy to have a geometric mean $\sqrt{x_1x_2}$ have imaginary value when $x_1$ and $x_2$ have opposite sign? $\endgroup$– Dilip SarwateCommented Feb 15, 2016 at 23:38
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3$\begingroup$ Any normal distribution, no matter what $\mu$ & $\sigma^2$, will include $0$ & negative values. It is of course possible that your particular sample includes only strictly positive values, but the population must include negative values. $\endgroup$– gung - Reinstate MonicaCommented Feb 16, 2016 at 15:11
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$\begingroup$ The geometric mean is sort of built into the Gaussian pdf in its denominator term: $\sqrt{2\pi * \sigma^2}$. The geometric mean is defined as $\left(\prod _{i=1}^{n}x_{i}\right)^{\frac {1}{n}}={\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}$. And so the denominator term is really just the geometric mean of $2\pi$ and the variance. I left this as a comment not an answer because it doesn't address your question, it's just an interesting way to look at the Gaussian pdf. $\endgroup$– jbuddy_13Commented Mar 30, 2021 at 17:27
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3 Answers
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Why do you want a geometric mean of normal distributed observations? I can see no good reason. Applications of the Geometric Mean gives several good examples of use of geometric mean. A typical case is return on investment. Returns combine multiplicatively, so If you want one "typical" return that would result in same winning, if the return was held constant over years, you get that from the geometric mean of the yearly returns.
Common to all such examples is that the random variable in question cannot be negative, and, since every normal distributed variable is negative with some (maybe very small) positive probability, geometric means do not look natural to use. So, again, why do you want to use a geometric mean?
See also the related Estimating with the geometric mean, Which "mean" to use and when?
answered Apr 27, 2017 at 16:44
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Edited because I am an idiot:
If you have normal samples you might have negatives depending on your mean and variance which would flip the sign. It feels unstable to flip a sign irrespective of the magnitude. Bummer.
More interesting are distributions with support on the positive or negative reals.
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$\begingroup$ Can you explain how you get a geometric mean of zero when negative values is included? $\endgroup$ Commented Sep 16, 2021 at 0:53
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1$\begingroup$ Damn you are absolutely right! It would just endlessly change the sign. Feels unstable. $\endgroup$– ChrisCommented Sep 16, 2021 at 0:57
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2$\begingroup$ (+1) I hope you won't be offended at me upvoting an answer where you self-describe as an idiot --- I just love a good self-deprecating sense of humour. $\endgroup$– BenCommented Sep 16, 2021 at 1:04
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For a normal distribution then, approximately: $\mu_g=\mu_a-0.5\sigma^2$, where $\sigma$ is the standard deviation of the normal distribution. I'm defining $\mu_g=\sqrt[n]{(1+r_1)(1+r_2)...(1+r_T)}-1$ where $r_j$ are drawn from a normal distribution. I'm thinking here of the case of finance where r are asset returns, it does make sense to consider geometric returns as they represent growth rates.
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$\begingroup$ As pointed out in comments and other answers, the underlying Normal distribution does not have a geometric mean. Moreover, there are plenty of cases where your formula will give an impossible negative value! $\endgroup$– whuber ♦Commented Mar 13, 2023 at 15:37