# Geometric standard deviation

In probability theory and statistics, the geometric standard deviation describes how spread out are a set of numbers whose preferred average is the geometric mean. If the geometric mean of a set of numbers {A1, A2, ..., An} is denoted as μg, then the geometric standard deviation is

[itex] \sigma_g = \exp \left( \sqrt{ \sum_{i=1}^n ( \ln A_i - \ln \mu_g )^2 \over n } \right). \qquad \qquad (1) [itex]
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## Derivation

If the geometric mean is

[itex] \mu_g = \sqrt[n]{ A_1 A_2 \cdots A_n }.\, [itex]

then taking the natural logarithm of both sides results in

[itex] \ln \mu_g = {1 \over n} \ln (A_1 A_2 \cdots A_n). [itex]

The logarithm of a product is a sum of logarithms, so

[itex] \ln \mu_g = {1 \over n} [ \ln A_1 + \ln A_2 + \cdots + \ln A_n ].\, [itex]

It can now be seen that [itex] \ln \, \mu_g [itex] is the arithmetic mean of the set [itex] \{ \ln A_1, \ln A_2, \dots , \ln A_n \} [itex], therefore the arithmetic standard deviation of this same set should be

[itex] \ln \sigma_g = \sqrt{ \sum_{i=1}^n ( \ln A_i - \ln \mu_g )^2 \over n }. [itex]

'Correction: The arithmetic standard deviation of this set is what is shown on the right. It is not the log of the standard deviation. To prove it's wrong, compute the covariance of two sets in a like manner. Exponentiating the result produces a covariance that is strictly non-negative, which is not reasonable (covariance can be negative). Exponentiating both sides results in equation (1). Q.E.D.

## Geometric standard score

The geometric version of the standard score is

[itex] z = {\ln ( x/\mu_g ) \over \ln \sigma_g }.\, [itex]

If the geometric mean, standard deviation, and z-score of a datum are known, then the raw score can be reconstructed by

[itex] x = \mu_g \sigma_g^z. [itex]

## Relationship to log-normal distribution

The geometric standard deviation is related to the log-normal distribution. The log-normal distribution is a distribution which is normal for the logarithm transformed values. By a simple set of logarithm transformations we see that the geometric standard deviation is the exponentiated value of the standard deviation of the log transformed values (e.g. exp(stdev(ln(A))));

As such, the geometric mean and the geometric standard deviation of a sample of data from a log-normally distributed population may be used to find the bounds of confidence intervals analogously to the way the arithmetic mean and standard deviation are used to bound confidence intervals for a normal distribution. See discussion in log-normal distribution for details.

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