Editing Beta distribution (section)

====Mean absolute deviation around the mean====
[[File:Ratio of Mean Abs. Dev. to Std.Dev. Beta distribution with alpha and beta from 0 to 5 - J. Rodal.jpg|thumb|Ratio of ,ean abs.dev. to std.dev. for beta distribution with α and β ranging from 0 to 5]]
[[File:Ratio of Mean Abs. Dev. to Std.Dev. Beta distribution vs. nu from 0 to 10 and vs. mean - J. Rodal.jpg|thumb|Ratio of mean abs.dev. to std.dev. for beta distribution with mean 0 ≤ ''μ'' ≤ 1 and sample size 0 < ''ν'' ≤ 10]]
The [[mean absolute deviation]] around the mean for the beta distribution with shape parameters ''α'' and ''β'' is:<ref name="Handbook of Beta Distribution" />

:<math>\operatorname{E}[|X - E[X]|] = \frac{2 \alpha^\alpha \beta^\beta}{\Beta(\alpha,\beta)(\alpha + \beta)^{\alpha + \beta + 1}} </math>

The mean absolute deviation around the mean is a more [[Robust statistics|robust]] [[estimator]] of [[statistical dispersion]] than the standard deviation for beta distributions with tails and inflection points at each side of the mode, Beta(''α'',&nbsp;''β'') distributions with ''α'',''β'' > 2, as it depends on the linear (absolute) deviations rather than the square deviations from the mean.  Therefore, the effect of very large deviations from the mean are not as overly weighted.

Using [[Stirling's approximation]] to the [[Gamma function]], [[Norman Lloyd Johnson|N.L.Johnson]] and [[Samuel Kotz|S.Kotz]]<ref name=JKB /> derived the following approximation for values of the shape parameters greater than unity (the relative error for this approximation is only −3.5% for ''α'' = ''β'' = 1, and it decreases to zero as ''α'' → ∞, ''β'' → ∞):

:<math> \begin{align}
\frac{\text{mean abs. dev. from mean}}{\text{standard deviation}} &=\frac{\operatorname{E}[|X - E[X]|]}{\sqrt{\operatorname{var}(X)}}\\
&\approx \sqrt{\frac{2}{\pi}} \left(1+\frac{7}{12 (\alpha+\beta)}{}-\frac{1}{12 \alpha}-\frac{1}{12 \beta} \right), \text{ if } \alpha, \beta > 1.
\end{align}</math>

At the limit ''α'' → ∞, ''β'' → ∞, the ratio of the mean absolute deviation to the standard deviation (for the beta distribution) becomes equal to the ratio of the same measures for the normal distribution: <math>\sqrt{\frac{2}{\pi}}</math>.  For ''α'' = ''β'' = 1 this ratio equals <math>\frac{\sqrt{3}}{2}</math>, so that from ''α'' = ''β'' = 1 to ''α'', ''β'' → ∞ the ratio decreases by 8.5%.  For ''α'' = ''β'' = 0 the standard deviation is exactly equal to the mean absolute deviation around the mean. Therefore, this ratio decreases by 15% from ''α'' = ''β'' = 0 to ''α'' = ''β'' = 1, and by 25% from ''α'' = ''β'' = 0 to ''α'', ''β'' → ∞ . However, for skewed beta distributions such that ''α'' → 0 or ''β'' → 0, the ratio of the standard deviation to the mean absolute deviation approaches infinity (although each of them, individually, approaches zero) because the mean absolute deviation approaches zero faster than the standard deviation.

Using the [[Statistical parameter|parametrization]] in terms of mean ''μ'' and sample size ''ν'' = ''α'' + ''β'' > 0:

:''α'' = ''μν'', ''β'' = (1 − ''μ'')''ν''

one can express the mean [[absolute deviation]] around the mean in terms of the mean ''μ'' and the sample size ''ν'' as follows:

:<math>\operatorname{E}[| X - E[X]|] = \frac{2 \mu^{\mu\nu} (1-\mu)^{(1-\mu)\nu}}{\nu \Beta(\mu \nu,(1-\mu)\nu)}</math>

For a symmetric distribution, the mean is at the middle of the distribution, ''μ'' = 1/2, and therefore:

:<math> \begin{align}
\operatorname{E}[|X - E[X]|]  = \frac{2^{1-\nu}}{\nu \Beta(\tfrac{\nu}{2} ,\tfrac{\nu}{2})} &= \frac{2^{1-\nu}\Gamma(\nu)}{\nu (\Gamma(\tfrac{\nu}{2}))^2 } \\
\lim_{\nu \to 0} \left (\lim_{\mu \to \frac{1}{2}} \operatorname{E}[|X - E[X]|] \right ) &= \tfrac{1}{2}\\
\lim_{\nu \to \infty} \left (\lim_{\mu \to \frac{1}{2}} \operatorname{E}[| X - E[X]|] \right ) &= 0
\end{align}</math>

Also, the following limits (with only the noted variable approaching the limit) can be obtained from the above expressions:

:<math> \begin{align}
\lim_{\beta\to  0} \operatorname{E}[|X - E[X]|] &=\lim_{\alpha \to  0} \operatorname{E}[|X - E[X]|]= 0 \\
\lim_{\beta\to  \infty} \operatorname{E}[|X - E[X]|] &=\lim_{\alpha \to  \infty} \operatorname{E}[|X - E[X]|] = 0\\
\lim_{\mu \to  0} \operatorname{E}[|X - E[X]|]&=\lim_{\mu \to  1} \operatorname{E}[|X - E[X]|] = 0\\
\lim_{\nu \to  0} \operatorname{E}[|X - E[X]|] &= \sqrt{\mu (1-\mu)} \\
\lim_{\nu \to  \infty} \operatorname{E}[|X - E[X]|] &= 0
\end{align}</math>