|
|
Line 1: |
Line 1: |
| 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]]. For such data, it may be preferred to the more usual [[standard deviation]]. Note that unlike the usual ''arithmetic'' standard deviation, the ''geometric'' standard deviation is a multiplicative factor, and thus is [[dimensionless]], rather than having the same [[Dimensional analysis | dimension]] as the input values.
| | I am 33 years old and my name is Xavier Mordaunt. I life in Kierspe (Germany).<br><br>My weblog - [http://www.merrittislandvacationrentals.com/ForumRetrieve.aspx?ForumID=2825&TopicID=1072492&NoTemplate=False how to get free castle clash gems] |
| | |
| | |
| ==Definition==
| |
| If the geometric mean of a set of numbers {''A''<sub>1</sub>, ''A''<sub>2</sub>, ..., ''A''<sub>''n''</sub>} is denoted as μ<sub>''g''</sub>, then the geometric standard deviation is
| |
| | |
| :<math> \sigma_g = \exp \left( \sqrt{ \sum_{i=1}^n ( \ln { A_i \over \mu_g } )^2 \over n } \right). \qquad \qquad (1) </math>
| |
| | |
| ==Derivation==
| |
| | |
| If the geometric mean is
| |
| | |
| :<math> \mu_g = \sqrt[n]{ A_1 A_2 \cdots A_n }.\, </math>
| |
| | |
| then taking the [[natural logarithm]] of both sides results in
| |
| | |
| :<math> \ln \mu_g = {1 \over n} \ln (A_1 A_2 \cdots A_n). </math> | |
| | |
| The logarithm of a product is a sum of logarithms (assuming <math>A_i</math> is positive for all <math>i</math>), so
| |
| | |
| :<math> \ln \mu_g = {1 \over n} [ \ln A_1 + \ln A_2 + \cdots + \ln A_n ].\, </math>
| |
| | |
| It can now be seen that <math> \ln \, \mu_g </math> is the [[arithmetic mean]] of the set <math> \{ \ln A_1, \ln A_2, \dots , \ln A_n \} </math>, therefore the arithmetic standard deviation of this same set should be
| |
| | |
| :<math> \ln \sigma_g = \sqrt{ \sum_{i=1}^n ( \ln A_i - \ln \mu_g )^2 \over n }.</math>
| |
| | |
| This simplifies to
| |
| | |
| :<math> \sigma_g = \exp{\sqrt{ \sum_{i=1}^n ( \ln { A_i \over \mu_g } )^2 \over n }}. </math>
| |
| | |
| ==Geometric standard score== | |
| | |
| The geometric version of the [[standard score]] is
| |
| | |
| :<math> z = {{\ln ( x ) - \ln ( \mu_g )} \over \ln \sigma_g } = {\log _{\sigma_g} (x / \mu_g)}.\, </math>
| |
| | |
| If the geometric mean, standard deviation, and z-score of a datum are known, then the [[raw score]] can be reconstructed by
| |
| | |
| :<math> x = \mu_g {\sigma_g}^z. </math>
| |
| | |
| ==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 interval]]s 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.
| |
| | |
| {{Unreferenced|date=November 2010}}
| |
| | |
| | |
| {{DEFAULTSORT:Geometric Standard Deviation}}
| |
| [[Category:Scale statistics]]
| |
| [[Category:Statistical terminology]]
| |
I am 33 years old and my name is Xavier Mordaunt. I life in Kierspe (Germany).
My weblog - how to get free castle clash gems