# Handbook of Descriptive Statistics/Measures of Statistical Variability/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

${\displaystyle \sigma _{g}=\exp \left({\sqrt {\sum _{i=1}^{n}(\ln A_{i}-\ln \mu _{g})^{2} \over n}}\right).\qquad \qquad (1)}$

## Derivation

If the geometric mean is

${\displaystyle \mu _{g}={\sqrt[{n}]{A_{1}A_{2}\cdots A_{n}}}.\,}$

then taking the natural logarithm of both sides results in

${\displaystyle \ln \mu _{g}={1 \over n}\ln(A_{1}A_{2}\cdots A_{n}).}$

The logarithm of a product is a sum of logarithms (assuming ${\displaystyle A_{i}}$ is positive for all ${\displaystyle i}$), so

${\displaystyle \ln \mu _{g}={1 \over n}[\ln A_{1}+\ln A_{2}+\cdots +\ln A_{n}].\,}$

It can now be seen that ${\displaystyle \ln \,\mu _{g}}$ is the arithmetic mean of the set ${\displaystyle \{\ln A_{1},\ln A_{2},\dots ,\ln A_{n}\}}$, therefore the arithmetic standard deviation of this same set should be

${\displaystyle \ln \sigma _{g}={\sqrt {\sum _{i=1}^{n}(\ln A_{i}-\ln \mu _{g})^{2} \over n}}.}$

Thus

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