Thanks in advance for patience.

I am trying to find an approximation for geometric mean. The one that is often mentioned is

geometric mean = arithmetic mean - 1/2 * variance.

I also read in multiple sources that there are ways to make it more accurate but so far I have not been able to find any. Hence I am here asking.

Q: What are some ways of making geometric mean approximation more accurate?

Why am I asking?: I am looking into implementing the Kelly criterion for more complex portfolios where the geometric mean is very difficult to calculate. Therefore variance in this case is the typical covariance matrix $\Sigma$ with the invested fractions vector $f$.

$f^T\Sigma f$

Thank you so much.

  • $\begingroup$ What does variance mean here? $\endgroup$ May 19, 2018 at 11:08
  • $\begingroup$ @LordSharktheUnknown Thanks for your reply, I edited the question according to your question. $\endgroup$
    – no6
    May 19, 2018 at 11:14

1 Answer 1


Approximations for the geometric mean are found by expressing it in terms of a series of central moments (variance, skewness, kurtosis and higher) and eliminating various terms by arguing that they are small for the case concerned. By including those terms which are not small for your application you can improve your estimate.

In summary, the geometric mean is given by,

$$ \left(\prod_{i=1}^n X_i\right)^{1/n} = \prod_{i=1}^n X_i^{1/n} = \prod_{i=1}^n \exp \left[ \frac{1}{n}\ln X_i \right] = \exp \left[ \frac{1}{n} \sum_{i=1}^n \ln X_i \right] $$ Using the arithmetic mean, $\mu_1$, we define $$ x_i=\frac{X_i - \mu_1}{\mu_1} $$ and use the binomial expansion for $\ln (1 + x)$ $$ \ln X_i - \ln \mu_1 = \ln (1+x_i) = x_i - \frac{x_i^2}{2} + \frac{x_i^3}{3} - \frac{x_i^4}{4} + \frac{x_i^5}{5} - \ldots $$ $$ \frac{1}{n}\sum_{i=1}^n \ln X_i = \ln \mu_1 - \frac{1}{n}\sum_{i=1}^n \frac{x_i^2}{2} + \frac{1}{n}\sum_{i=1}^n\frac{x_i^3}{3} - \frac{1}{n}\sum_{i=1}^n\frac{x_i^4}{4} + \frac{1}{n}\sum_{i=1}^n\frac{x_i^5}{5} - \ldots $$

So, without approximation we can write

$$ \left(\prod_{i=1}^n X_i\right)^{1/n} = \exp \left[ \frac{1}{n} \sum_{i=1}^n \ln X_i \right] = \mu_1 \exp\left[{ -\frac{\mu_2}{2 \mu_1^2} +\frac{\mu_3}{3 \mu_1^3} -\frac{\mu_4}{4 \mu_1^4} +\frac{\mu_5}{5 \mu_1^5} - \ldots }\right] $$

Where the $\mu_i$ are the population$^\dagger$ central moments (arithmetic mean, variance, skewness, kurtosis and higher order)

$$ \mu_1 = \frac{1}{n} \sum_{i=1}^n X_i \hspace{1cm} \mu_{j>1} = \frac{1}{n} \sum_{i=1}^n(X_i-\mu_1)^j $$ At this point we may choose how many moments to include and how many terms to retain in the expansion of $\exp x = 1 + x + \frac{x^2}{2!} + \ldots$

For example if we retain only $\mu_2$ and truncate $\exp x = 1 + x$ we arrive at:

$$ \left(\prod_{i=1}^n X_i\right)^{1/n} \simeq \mu_1 -\frac{\mu_2}{2 \mu_1} $$ Squaring this and noting that we have implicitly ordered $\left(\frac{\mu_2}{\mu_1}\right)^2$ negligible gives the Latané approximation $$ \text{Geometric Mean}^2 = \text{Arithmetic Mean}^2 - \text{Population Standard Deviation}^2 $$

Higher order approximations may be found by including higher moments in $\ln (1 + x_i)$ and retaining more terms in the exponential.

$\dagger$ The sample central moments are defined analogously with $\frac{1}{n} \rightarrow \frac{1}{n-1}$. This is Bessel's correction aimed at the reduction of finite population bias.

The reference you probably want is:

Journal of Financial and Quantitative Analysis 1969 / 06 Vol. 4; Iss. 2
Geometric Mean Approximations of Individual Security and Portfolio Performance

William E. Young and Robert H. Trent

screnshot of calculation

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