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


 * $$ \sigma_g = \exp \left( \sqrt{ \sum_{i=1}^n ( \ln A_i - \ln \mu_g )^2 \over n } \right). \qquad \qquad (1) $$

Derivation
If the geometric mean is


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

then taking the natural logarithm of both sides results in


 * $$ \ln \mu_g = {1 \over n} \ln (A_1 A_2 \cdots A_n). $$

The logarithm of a product is a sum of logarithms (assuming $$A_i$$ is positive for all $$i$$), so


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

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


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

Thus


 * ln(geometric SD of A1, ..., An) = arithmetic (i.e. usual) SD of ln(A1), ..., ln(An).

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.