Beta distribution

Properties

Measures of central tendency

Mode

The mode of a beta distributed random variable X with α, β > 1 is the most likely value of the distribution (corresponding to the peak in the PDF), and is given by the following expression:

$${\displaystyle \frac{\alpha -1}{\alpha +\beta -2}.}$$

When both parameters are less than one (α, β < 1), this is the anti-mode: the lowest point of the probability density curve.

Letting α = β, the expression for the mode simplifies to 1/2, showing that for α = β > 1 the mode (resp. anti-mode when α, β < 1), is at the center of the distribution: it is symmetric in those cases. See Shapes section in this article for a full list of mode cases, for arbitrary values of α and β. For several of these cases, the maximum value of the density function occurs at one or both ends. In some cases the (maximum) value of the density function occurring at the end is finite. For example, in the case of α = 2, β = 1 (or α = 1, β = 2), the density function becomes a right-triangle distribution which is finite at both ends. In several other cases there is a singularity at one end, where the value of the density function approaches infinity. For example, in the case α = β = 1/2, the beta distribution simplifies to become the arcsine distribution. There is debate among mathematicians about some of these cases and whether the ends (x = 0, and x = 1) can be called modes or not.

• Whether the ends are part of the domain of the density function
• Whether a singularity can ever be called a mode
• Whether cases with two maxima should be called bimodal

Median

The median of the beta distribution is the unique real number ${\displaystyle x=I_{1/2}^{[-1]}(\alpha ,\beta )}$ for which the regularized incomplete beta function ${\displaystyle I_x(\alpha ,\beta )=\frac{1}{2}}$. There is no general closed-form expression for the median of the beta distribution for arbitrary values of α and β. Closed-form expressions for particular values of the parameters α and β follow:

• For symmetric cases α = β, median = 1/2.
• For α = 1 and β > 0, median ${\displaystyle =1-2^{-1/\beta }}$ (this case is the mirror-image of the power function distribution)
• For α > 0 and β = 1, median = ${\displaystyle 2^{-1/\alpha }}$ (this case is the power function distribution)
• For α = 3 and β = 2, median = 0.6142724318676105..., the real solution to the quartic equation 1 − 8x³ + 6x⁴ = 0, which lies in [0,1].
• For α = 2 and β = 3, median = 0.38572756813238945... = 1−median(Beta(3, 2))

The following are the limits with one parameter finite (non-zero) and the other approaching these limits:

$${\displaystyle \begin{array}{c}\underset{\beta \to 0}{lim}\text{median}=\underset{\alpha \to \mathrm{\infty }}{lim}\text{median}=1,\\ \underset{\alpha \to 0}{lim}\text{median}=\underset{\beta \to \mathrm{\infty }}{lim}\text{median}=0.\end{array}}$$

A reasonable approximation of the value of the median of the beta distribution, for both α and β greater or equal to one, is given by the formula

$${\displaystyle \text{median}\approx \frac{\alpha -\frac{1}{3}}{\alpha +\beta -\frac{2}{3}}\text{ for }\alpha ,\beta \ge 1.}$$

When α, β ≥ 1, the relative error (the absolute error divided by the median) in this approximation is less than 4% and for both α ≥ 2 and β ≥ 2 it is less than 1%. The absolute error divided by the difference between the mean and the mode is similarly small:

Mean

The expected value (mean) (μ) of a beta distribution random variable X with two parameters α and β is a function of only the ratio β/α of these parameters:

$${\displaystyle \begin{array}{cc}\mu =\mathrm{E}[X]& ={\int }_0^{1}xf(x;\alpha ,\beta )dx\\ & ={\int }_0^{1}x\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{\mathrm{B}(\alpha ,\beta )}dx\\ & =\frac{\alpha }{\alpha +\beta }\\ & =\frac{1}{1+\frac{\beta }{\alpha }}\end{array}}$$

Letting α = β in the above expression one obtains μ = 1/2, showing that for α = β the mean is at the center of the distribution: it is symmetric. Also, the following limits can be obtained from the above expression:

$${\displaystyle \begin{array}{c}\underset{\frac{\beta }{\alpha }\to 0}{lim}\mu =1\\ \underset{\frac{\beta }{\alpha }\to \mathrm{\infty }}{lim}\mu =0\end{array}}$$

Therefore, for β/α → 0, or for α/β → ∞, the mean is located at the right end, x = 1. For these limit ratios, the beta distribution becomes a one-point degenerate distribution with a Dirac delta function spike at the right end, x = 1, with probability 1, and zero probability everywhere else. There is 100% probability (absolute certainty) concentrated at the right end, x = 1.

Similarly, for β/α → ∞, or for α/β → 0, the mean is located at the left end, x = 0. The beta distribution becomes a 1-point Degenerate distribution with a Dirac delta function spike at the left end, x = 0, with probability 1, and zero probability everywhere else. There is 100% probability (absolute certainty) concentrated at the left end, x = 0. Following are the limits with one parameter finite (non-zero) and the other approaching these limits:

$${\displaystyle \begin{array}{c}\underset{\beta \to 0}{lim}\mu =\underset{\alpha \to \mathrm{\infty }}{lim}\mu =1\\ \underset{\alpha \to 0}{lim}\mu =\underset{\beta \to \mathrm{\infty }}{lim}\mu =0\end{array}}$$

While for typical unimodal distributions (with centrally located modes, inflexion points at both sides of the mode, and longer tails) (with Beta(α, β) such that α, β > 2) it is known that the sample mean (as an estimate of location) is not as robust as the sample median, the opposite is the case for uniform or "U-shaped" bimodal distributions (with Beta(α, β) such that α, β ≤ 1), with the modes located at the ends of the distribution. As Mosteller and Tukey remark ( p. 207) "the average of the two extreme observations uses all the sample information. This illustrates how, for short-tailed distributions, the extreme observations should get more weight." By contrast, it follows that the median of "U-shaped" bimodal distributions with modes at the edge of the distribution (with Beta(α, β) such that α, β ≤ 1) is not robust, as the sample median drops the extreme sample observations from consideration. A practical application of this occurs for example for random walks, since the probability for the time of the last visit to the origin in a random walk is distributed as the arcsine distribution Beta(1/2, 1/2): the mean of a number of realizations of a random walk is a much more robust estimator than the median (which is an inappropriate sample measure estimate in this case).

Geometric mean

The logarithm of the geometric mean G_X of a distribution with random variable X is the arithmetic mean of ln(X), or, equivalently, its expected value:

$${\displaystyle \ln G_X=\mathrm{E}[\ln X]}$$

For a beta distribution, the expected value integral gives:

$${\displaystyle \begin{array}{cc}\mathrm{E}[\ln X]& ={\int }_0^1\ln xf(x;\alpha ,\beta )dx\\ & ={\int }_0^1\ln x\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{\mathrm{B}(\alpha ,\beta )}dx\\ & =\frac{1}{\mathrm{B}(\alpha ,\beta )}{\int }_0^1\frac{\mathrm{\partial }{x}^{\alpha -1}(1-x{)}^{\beta -1}}{\mathrm{\partial }\alpha }dx\\ & =\frac{1}{\mathrm{B}(\alpha ,\beta )}\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }{\int }_0^1{x}^{\alpha -1}(1-x{)}^{\beta -1}dx\\ & =\frac{1}{\mathrm{B}(\alpha ,\beta )}\frac{\mathrm{\partial }\mathrm{B}(\alpha ,\beta )}{\mathrm{\partial }\alpha }\\ & =\frac{\mathrm{\partial }\ln \mathrm{B}(\alpha ,\beta )}{\mathrm{\partial }\alpha }\\ & =\frac{\mathrm{\partial }\ln \mathrm{\Gamma }(\alpha )}{\mathrm{\partial }\alpha }-\frac{\mathrm{\partial }\ln \mathrm{\Gamma }(\alpha +\beta )}{\mathrm{\partial }\alpha }\\ & =\psi (\alpha )-\psi (\alpha +\beta )\end{array}}$$

where ψ is the digamma function.

Therefore, the geometric mean of a beta distribution with shape parameters α and β is the exponential of the digamma functions of α and β as follows:

$${\displaystyle G_X=e^{\mathrm{E}[\ln X]}=e^{\psi (\alpha )-\psi (\alpha +\beta )}}$$

While for a beta distribution with equal shape parameters α = β, it follows that skewness = 0 and mode = mean = median = 1/2, the geometric mean is less than 1/2: 0 < G_X < 1/2. The reason for this is that the logarithmic transformation strongly weights the values of X close to zero, as ln(X) strongly tends towards negative infinity as X approaches zero, while ln(X) flattens towards zero as X → 1.

Along a line α = β, the following limits apply:

$${\displaystyle \begin{array}{cc}& \underset{\alpha =\beta \to 0}{lim}{G}_X=0\\ & \underset{\alpha =\beta \to \mathrm{\infty }}{lim}{G}_X=\frac{1}{2}\end{array}}$$

Following are the limits with one parameter finite (non-zero) and the other approaching these limits:

$${\displaystyle \begin{array}{c}\underset{\beta \to 0}{lim}{G}_X=\underset{\alpha \to \mathrm{\infty }}{lim}{G}_X=1\\ \underset{\alpha \to 0}{lim}{G}_X=\underset{\beta \to \mathrm{\infty }}{lim}{G}_X=0\end{array}}$$

The accompanying plot shows the difference between the mean and the geometric mean for shape parameters α and β from zero to 2. Besides the fact that the difference between them approaches zero as α and β approach infinity and that the difference becomes large for values of α and β approaching zero, one can observe an evident asymmetry of the geometric mean with respect to the shape parameters α and β. The difference between the geometric mean and the mean is larger for small values of α in relation to β than when exchanging the magnitudes of β and α.

N. L.Johnson and S. Kotz suggest the logarithmic approximation to the digamma function ψ(α) ≈ ln(α − 1/2) which results in the following approximation to the geometric mean:

$${\displaystyle G_X\approx \frac{\alpha -\frac{1}{2}}{\alpha +\beta -\frac{1}{2}}\text{ if }\alpha ,\beta >1.}$$

Numerical values for the relative error in this approximation follow: [(α = β = 1): 9.39%]; [(α = β = 2): 1.29%]; [(α = 2, β = 3): 1.51%]; [(α = 3, β = 2): 0.44%]; [(α = β = 3): 0.51%]; [(α = β = 4): 0.26%]; [(α = 3, β = 4): 0.55%]; [(α = 4, β = 3): 0.24%].

Similarly, one can calculate the value of shape parameters required for the geometric mean to equal 1/2. Given the value of the parameter β, what would be the value of the other parameter, α, required for the geometric mean to equal 1/2?. The answer is that (for β > 1), the value of α required tends towards β + 1/2 as β → ∞. For example, all these couples have the same geometric mean of 1/2: [β = 1, α = 1.4427], [β = 2, α = 2.46958], [β = 3, α = 3.47943], [β = 4, α = 4.48449], [β = 5, α = 5.48756], [β = 10, α = 10.4938], [β = 100, α = 100.499].

The fundamental property of the geometric mean, which can be proven to be false for any other mean, is

$${\displaystyle G\left(\frac{X_i}{Y_i}\right)=\frac{G(X_i)}{G(Y_i)}}$$

This makes the geometric mean the only correct mean when averaging normalized results, that is results that are presented as ratios to reference values. This is relevant because the beta distribution is a suitable model for the random behavior of percentages and it is particularly suitable to the statistical modelling of proportions. The geometric mean plays a central role in maximum likelihood estimation, see section "Parameter estimation, maximum likelihood." Actually, when performing maximum likelihood estimation, besides the geometric mean G_X based on the random variable X, also another geometric mean appears naturally: the geometric mean based on the linear transformation ––(1 − X), the mirror-image of X, denoted by G_(1−X):

$${\displaystyle G_{1-X}=e^{\mathrm{E}[\ln (1-X)]}=e^{\psi (\beta )-\psi (\alpha +\beta )}}$$

Along a line α = β, the following limits apply:

$${\displaystyle \begin{array}{cc}& \underset{\alpha =\beta \to 0}{lim}{G}_{1-X}=0\\ & \underset{\alpha =\beta \to \mathrm{\infty }}{lim}{G}_{1-X}=\frac{1}{2}\end{array}}$$

Following are the limits with one parameter finite (non-zero) and the other approaching these limits:

$${\displaystyle \begin{array}{c}\underset{\beta \to 0}{lim}{G}_{(1-X)}=\underset{\alpha \to \mathrm{\infty }}{lim}{G}_{(1-X)}=0\\ \underset{\alpha \to 0}{lim}{G}_{(1-X)}=\underset{\beta \to \mathrm{\infty }}{lim}{G}_{(1-X)}=1\end{array}}$$

It has the following approximate value:

$${\displaystyle G_{(1-X)}\approx \frac{\beta -\frac{1}{2}}{\alpha +\beta -\frac{1}{2}}\text{ if }\alpha ,\beta >1.}$$

Although both G_X and G_1−X are asymmetric, in the case that both shape parameters are equal α = β, the geometric means are equal: G_X = G_(1−X). This equality follows from the following symmetry displayed between both geometric means:

$${\displaystyle G_X(\mathrm{B}(\alpha ,\beta ))=G_{1-X}(\mathrm{B}(\beta ,\alpha )).}$$

Harmonic mean

The inverse of the harmonic mean (H_X) of a distribution with random variable X is the arithmetic mean of 1/X, or, equivalently, its expected value. Therefore, the harmonic mean (H_X) of a beta distribution with shape parameters α and β is:

$${\displaystyle \begin{array}{cc}{H}_X& =\frac{1}{\mathrm{E}\left[\frac{1}{X}\right]}\\ & =\frac{1}{{\int }_0^1\frac{f(x;\alpha ,\beta )}{x}dx}\\ & =\frac{1}{{\int }_0^1\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{x\mathrm{B}(\alpha ,\beta )}dx}\\ & =\frac{\alpha -1}{\alpha +\beta -1}\text{ if }\alpha >1\text{ and }\beta >0\end{array}}$$

The harmonic mean (H_X) of a beta distribution with α < 1 is undefined, because its defining expression is not bounded in [0, 1] for shape parameter α less than unity.

Letting α = β in the above expression one obtains

$${\displaystyle H_X=\frac{\alpha -1}{2\alpha -1},}$$

showing that for α = β the harmonic mean ranges from 0, for α = β = 1, to 1/2, for α = β → ∞.

Following are the limits with one parameter finite (non-zero) and the other approaching these limits:

$${\displaystyle \begin{array}{cc}& \underset{\alpha \to 0}{lim}{H}_X\text{ is undefined}\\ & \underset{\alpha \to 1}{lim}{H}_X=\underset{\beta \to \mathrm{\infty }}{lim}{H}_X=0\\ & \underset{\beta \to 0}{lim}{H}_X=\underset{\alpha \to \mathrm{\infty }}{lim}{H}_X=1\end{array}}$$

The harmonic mean plays a role in maximum likelihood estimation for the four parameter case, in addition to the geometric mean. Actually, when performing maximum likelihood estimation for the four parameter case, besides the harmonic mean H_X based on the random variable X, also another harmonic mean appears naturally: the harmonic mean based on the linear transformation (1 − X), the mirror-image of X, denoted by H_{1 − X}:

$${\displaystyle H_{1-X}=\frac{1}{\mathrm{E}\left[\frac{1}{1-X}\right]}=\frac{\beta -1}{\alpha +\beta -1}\text{ if }\beta >1,\text{ and }\alpha >0.}$$

The harmonic mean (H_{(1 − X)}) of a beta distribution with β < 1 is undefined, because its defining expression is not bounded in [0, 1] for shape parameter β less than unity.

Letting α = β in the above expression one obtains

$${\displaystyle H_{(1-X)}=\frac{\beta -1}{2\beta -1},}$$

showing that for α = β the harmonic mean ranges from 0, for α = β = 1, to 1/2, for α = β → ∞.

Following are the limits with one parameter finite (non-zero) and the other approaching these limits:

$${\displaystyle \begin{array}{cc}& \underset{\beta \to 0}{lim}{H}_{1-X}\text{ is undefined}\\ & \underset{\beta \to 1}{lim}{H}_{1-X}=\underset{\alpha \to \mathrm{\infty }}{lim}{H}_{1-X}=0\\ & \underset{\alpha \to 0}{lim}{H}_{1-X}=\underset{\beta \to \mathrm{\infty }}{lim}{H}_{1-X}=1\end{array}}$$

Although both H_X and H_1−X are asymmetric, in the case that both shape parameters are equal α = β, the harmonic means are equal: H_X = H_1−X. This equality follows from the following symmetry displayed between both harmonic means:

$${\displaystyle H_X(\mathrm{B}(\alpha ,\beta ))=H_{1-X}(\mathrm{B}(\beta ,\alpha ))\text{ if }\alpha ,\beta >1.}$$

Measures of statistical dispersion

Variance

The variance (the second moment centered on the mean) of a beta distribution random variable X with parameters α and β is:

$${\displaystyle \mathrm{var}(X)=\mathrm{E}\left[(X-\mu {)}^2\right]=\frac{\alpha \beta }{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}}$$

Letting α = β in the above expression one obtains

$${\displaystyle \mathrm{var}(X)=\frac{1}{4(2\beta +1)},}$$

showing that for α = β the variance decreases monotonically as α = β increases. Setting α = β = 0 in this expression, one finds the maximum variance var(X) = 1/4 which only occurs approaching the limit, at α = β = 0.

The beta distribution may also be parametrized in terms of its mean μ (0 < μ < 1) and sample size ν = α + β (ν > 0) (see subsection Mean and sample size):

$${\displaystyle \begin{array}{ccc}\alpha & =\mu \nu ,& \text{ where }\nu =(\alpha +\beta )>0,\\ \beta & =(1-\mu )\nu ,& \text{ where }\nu =(\alpha +\beta )>0.\end{array}}$$

Using this parametrization, one can express the variance in terms of the mean μ and the sample size ν as follows:

$${\displaystyle \mathrm{var}(X)=\frac{\mu (1-\mu )}{1+\nu }}$$

Since ν = α + β > 0, it follows that var(X) < μ(1 − μ).

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

$${\displaystyle \mathrm{var}(X)=\frac{1}{4(1+\nu )}\text{ if }\mu =\frac{1}{2}}$$

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

$${\displaystyle \begin{array}{cc}& \underset{\beta \to 0}{lim}\mathrm{var}(X)=\underset{\alpha \to 0}{lim}\mathrm{var}(X)=\underset{\beta \to \mathrm{\infty }}{lim}\mathrm{var}(X)=\underset{\alpha \to \mathrm{\infty }}{lim}\mathrm{var}(X)=0\\ & \underset{\nu \to \mathrm{\infty }}{lim}\mathrm{var}(X)=\underset{\mu \to 0}{lim}\mathrm{var}(X)=\underset{\mu \to 1}{lim}\mathrm{var}(X)=0\\ & \underset{\nu \to 0}{lim}\mathrm{var}(X)=\mu (1-\mu )\end{array}}$$

Geometric variance and covariance

The logarithm of the geometric variance, ln(var_GX), of a distribution with random variable X is the second moment of the logarithm of X centered on the geometric mean of X, ln(G_X):

$${\displaystyle \begin{array}{cc}\ln {\mathrm{var}}_{GX}& =\mathrm{E}\left[{\left(\ln X-\ln G_X\right)}^2\right]\\ & =\mathrm{E}\left[{\left(\ln X-\mathrm{E}\left[\ln X\right]\right)}^2\right]\\ & =\mathrm{E}\left[{\left(\ln X\right)}^2\right]-{\left(\mathrm{E}[\ln X]\right)}^2\\ & =\mathrm{var}[\ln X]\end{array}}$$

and therefore, the geometric variance is:

$${\displaystyle {\mathrm{var}}_{GX}=e^{\mathrm{var}[\ln X]}}$$

In the Fisher information matrix, and the curvature of the log likelihood function, the logarithm of the geometric variance of the reflected variable 1 − X and the logarithm of the geometric covariance between X and 1 − X appear:

$${\displaystyle \begin{array}{cc}\ln \mathrm{v}\mathrm{a}{\mathrm{r}}_{\mathrm{G}(1\text{-}\mathrm{X})}& =\mathrm{E}\left[{\left(\ln (1-X)-\ln G_{1-X}\right)}^2\right]\\ & =\mathrm{E}\left[{\left(\ln (1-X)-\mathrm{E}[\ln (1-X)]\right)}^2\right]\\ & =\mathrm{E}\left[(\ln (1-X){)}^2\right]-{\left(\mathrm{E}[\ln (1-X)]\right)}^2\\ & =\mathrm{var}[\ln (1-X)]\\ & \\ \mathrm{v}\mathrm{a}{\mathrm{r}}_{\mathrm{G}(1\text{-}\mathrm{X})}& =e^{\mathrm{var}[\ln (1-X)]}\\ & \\ \ln \mathrm{c}\mathrm{o}{\mathrm{v}}_{\mathrm{G}\mathrm{X},1\text{-}\mathrm{X}}& =\mathrm{E}[(\ln X-\ln G_X)(\ln (1-X)-\ln G_{1-X})]\\ & =\mathrm{E}[(\ln X-\mathrm{E}[\ln X])(\ln (1-X)-\mathrm{E}[\ln (1-X)])]\\ & =\mathrm{E}\left[\ln X\ln (1-X)\right]-\mathrm{E}[\ln X]\mathrm{E}[\ln (1-X)]\\ & =\mathrm{cov}[\ln X,\ln (1-X)]\\ & \\ {\mathrm{cov}}_{GX,(1-X)}& =e^{\mathrm{cov}[\ln X,\ln (1-X)]}\end{array}}$$

For a beta distribution, higher order logarithmic moments can be derived by using the representation of a beta distribution as a proportion of two gamma distributions and differentiating through the integral. They can be expressed in terms of higher order poly-gamma functions. See the section § Moments of logarithmically transformed random variables. The variance of the logarithmic variables and covariance of ln X and ln(1−X) are:

$${\displaystyle \mathrm{var}[\ln X]={\psi }_1(\alpha )-{\psi }_1(\alpha +\beta )}$$ $${\displaystyle \mathrm{var}[\ln (1-X)]={\psi }_1(\beta )-{\psi }_1(\alpha +\beta )}$$ $${\displaystyle \mathrm{cov}[\ln X,\ln (1-X)]=-{\psi }_1(\alpha +\beta )}$$

where the trigamma function, denoted ψ₁(α), is the second of the polygamma functions, and is defined as the derivative of the digamma function:

$${\displaystyle {\psi }_1(\alpha )=\frac{d^2\ln \mathrm{\Gamma }(\alpha )}{d{\alpha }^2}=\frac{d\psi (\alpha )}{d\alpha }.}$$

Therefore,

$${\displaystyle \ln {\mathrm{var}}_{GX}=\mathrm{var}[\ln X]={\psi }_1(\alpha )-{\psi }_1(\alpha +\beta )}$$ $${\displaystyle \ln {\mathrm{var}}_{G(1-X)}=\mathrm{var}[\ln (1-X)]={\psi }_1(\beta )-{\psi }_1(\alpha +\beta )}$$ $${\displaystyle \ln {\mathrm{cov}}_{GX,1-X}=\mathrm{cov}[\ln X,\ln (1-X)]=-{\psi }_1(\alpha +\beta )}$$

The accompanying plots show the log geometric variances and log geometric covariance versus the shape parameters α and β. The plots show that the log geometric variances and log geometric covariance are close to zero for shape parameters α and β greater than 2, and that the log geometric variances rapidly rise in value for shape parameter values α and β less than unity. The log geometric variances are positive for all values of the shape parameters. The log geometric covariance is negative for all values of the shape parameters, and it reaches large negative values for α and β less than unity.

Following are the limits with one parameter finite (non-zero) and the other approaching these limits:

$${\displaystyle \begin{array}{cc}& \underset{\alpha \to 0}{lim}\ln {\mathrm{var}}_{GX}=\underset{\beta \to 0}{lim}\ln {\mathrm{var}}_{G(1-X)}=\mathrm{\infty }\\ & \underset{\beta \to 0}{lim}\ln {\mathrm{var}}_{GX}=\underset{\alpha \to \mathrm{\infty }}{lim}\ln {\mathrm{var}}_{GX}=\underset{\alpha \to 0}{lim}\ln {\mathrm{var}}_{G(1-X)}=\underset{\beta \to \mathrm{\infty }}{lim}\ln {\mathrm{var}}_{G(1-X)}=0\\ & \underset{\alpha \to \mathrm{\infty }}{lim}\ln {\mathrm{cov}}_{GX,(1-X)}=\underset{\beta \to \mathrm{\infty }}{lim}\ln {\mathrm{cov}}_{GX,(1-X)}=0\\ & \underset{\beta \to \mathrm{\infty }}{lim}\ln {\mathrm{var}}_{GX}={\psi }_1(\alpha )\\ & \underset{\alpha \to \mathrm{\infty }}{lim}\ln {\mathrm{var}}_{G(1-X)}={\psi }_1(\beta )\\ & \underset{\alpha \to 0}{lim}\ln {\mathrm{cov}}_{GX,(1-X)}=-{\psi }_1(\beta )\\ & \underset{\beta \to 0}{lim}\ln {\mathrm{cov}}_{GX,(1-X)}=-{\psi }_1(\alpha )\end{array}}$$

Limits with two parameters varying:

$${\displaystyle \begin{array}{cc}& \underset{\alpha \to \mathrm{\infty }}{lim}(\underset{\beta \to \mathrm{\infty }}{lim}\ln {\mathrm{var}}_{GX})=\underset{\beta \to \mathrm{\infty }}{lim}(\underset{\alpha \to \mathrm{\infty }}{lim}\ln {\mathrm{var}}_{G(1-X)})=\underset{\alpha \to \mathrm{\infty }}{lim}(\underset{\beta \to 0}{lim}\ln {\mathrm{cov}}_{GX,(1-X)})=\underset{\beta \to \mathrm{\infty }}{lim}(\underset{\alpha \to 0}{lim}\ln {\mathrm{cov}}_{GX,(1-X)})=0\\ & \underset{\alpha \to \mathrm{\infty }}{lim}(\underset{\beta \to 0}{lim}\ln {\mathrm{var}}_{GX})=\underset{\beta \to \mathrm{\infty }}{lim}(\underset{\alpha \to 0}{lim}\ln {\mathrm{var}}_{G(1-X)})=\mathrm{\infty }\\ & \underset{\alpha \to 0}{lim}(\underset{\beta \to 0}{lim}\ln {\mathrm{cov}}_{GX,(1-X)})=\underset{\beta \to 0}{lim}(\underset{\alpha \to 0}{lim}\ln {\mathrm{cov}}_{GX,(1-X)})=-\mathrm{\infty }\end{array}}$$

Although both ln(var_GX) and ln(var_{G(1 − X)}) are asymmetric, when the shape parameters are equal, α = β, one has: ln(var_GX) = ln(var_G(1−X)). This equality follows from the following symmetry displayed between both log geometric variances:

$${\displaystyle \ln {\mathrm{var}}_{GX}(\mathrm{B}(\alpha ,\beta ))=\ln {\mathrm{var}}_{G(1-X)}(\mathrm{B}(\beta ,\alpha )).}$$

The log geometric covariance is symmetric:

$${\displaystyle \ln {\mathrm{cov}}_{GX,(1-X)}(\mathrm{B}(\alpha ,\beta ))=\ln {\mathrm{cov}}_{GX,(1-X)}(\mathrm{B}(\beta ,\alpha ))}$$

Mean absolute deviation around the mean

The mean absolute deviation around the mean for the beta distribution with shape parameters α and β is:

$${\displaystyle \mathrm{E}[|X-E[X]|]=\frac{2{\alpha }^{\alpha }{\beta }^{\beta }}{\mathrm{B}(\alpha ,\beta )(\alpha +\beta {)}^{\alpha +\beta +1}}}$$

The mean absolute deviation around the mean is a more 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, 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, N.L.Johnson and S.Kotz 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 → ∞):

$${\displaystyle \begin{array}{cc}\frac{\text{mean abs. dev. from mean}}{\text{standard deviation}}& =\frac{\mathrm{E}[|X-E[X]|]}{\sqrt{\mathrm{var}(X)}}\\ & \approx \sqrt{\frac{2}{\pi }}\left(1+\frac{7}{12(\alpha +\beta )}-\frac{1}{12\alpha }-\frac{1}{12\beta }\right),\text{ if }\alpha ,\beta >1.\end{array}}$$

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: ${\displaystyle \sqrt{\frac{2}{\pi }}}$. For α = β = 1 this ratio equals ${\displaystyle \frac{\sqrt{3}}{2}}$, 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 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:

$${\displaystyle \mathrm{E}[|X-E[X]|]=\frac{2{\mu }^{\mu \nu }(1-\mu {)}^{(1-\mu )\nu }}{\nu \mathrm{B}(\mu \nu ,(1-\mu )\nu )}}$$

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

$${\displaystyle \begin{array}{cc}\mathrm{E}[|X-E[X]|]=\frac{2^{1-\nu }}{\nu \mathrm{B}(\frac{\nu }{2},\frac{\nu }{2})}& =\frac{2^{1-\nu }\mathrm{\Gamma }(\nu )}{\nu (\mathrm{\Gamma }(\frac{\nu }{2}){)}^2}\\ \underset{\nu \to 0}{lim}\left(\underset{\mu \to \frac{1}{2}}{lim}\mathrm{E}[|X-E[X]|]\right)& =\frac{1}{2}\\ \underset{\nu \to \mathrm{\infty }}{lim}\left(\underset{\mu \to \frac{1}{2}}{lim}\mathrm{E}[|X-E[X]|]\right)& =0\end{array}}$$

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

$${\displaystyle \begin{array}{cc}\underset{\beta \to 0}{lim}\mathrm{E}[|X-E[X]|]& =\underset{\alpha \to 0}{lim}\mathrm{E}[|X-E[X]|]=0\\ \underset{\beta \to \mathrm{\infty }}{lim}\mathrm{E}[|X-E[X]|]& =\underset{\alpha \to \mathrm{\infty }}{lim}\mathrm{E}[|X-E[X]|]=0\\ \underset{\mu \to 0}{lim}\mathrm{E}[|X-E[X]|]& =\underset{\mu \to 1}{lim}\mathrm{E}[|X-E[X]|]=0\\ \underset{\nu \to 0}{lim}\mathrm{E}[|X-E[X]|]& =\sqrt{\mu (1-\mu )}\\ \underset{\nu \to \mathrm{\infty }}{lim}\mathrm{E}[|X-E[X]|]& =0\end{array}}$$

Mean absolute difference

The mean absolute difference for the beta distribution is:

$${\displaystyle \begin{array}{cc}\mathrm{M}\mathrm{D}& ={\int }_0^1{\int }_0^{1}f(x;\alpha ,\beta )f(y;\alpha ,\beta )\left|x-y\right|dxdy\\ & =\frac{4}{\alpha +\beta }\frac{B(\alpha +\beta ,\alpha +\beta )}{B(\alpha ,\alpha )B(\beta ,\beta )}\end{array}}$$

The Gini coefficient for the beta distribution is half of the relative mean absolute difference:

$${\displaystyle \mathrm{G}=\left(\frac{2}{\alpha }\right)\frac{B(\alpha +\beta ,\alpha +\beta )}{B(\alpha ,\alpha )B(\beta ,\beta )}}$$

Skewness

The skewness (the third moment centered on the mean, normalized by the 3/2 power of the variance) of the beta distribution is

$${\displaystyle {\gamma }_1=\frac{\mathrm{E}\left[{\left(X-\mu \right)}^3\right]}{{\left(\mathrm{var}(X)\right)}^{3/2}}=\frac{2\left(\beta -\alpha \right)\sqrt{\alpha +\beta +1}}{\left(\alpha +\beta +2\right)\sqrt{\alpha \beta }}.}$$

Letting α = β in the above expression one obtains γ₁ = 0, showing once again that for α = β the distribution is symmetric and hence the skewness is zero. Positive skew (right-tailed) for α < β, negative skew (left-tailed) for α > β.

Using the parametrization in terms of mean μ and sample size ν = α + β:

$${\displaystyle \begin{array}{ccc}\alpha & =\mu \nu ,& \text{ where }\nu =(\alpha +\beta )>0,\\ \beta & =(1-\mu )\nu ,& \text{ where }\nu =(\alpha +\beta )>0.\end{array}}$$

one can express the skewness in terms of the mean μ and the sample size ν as follows:

$${\displaystyle {\gamma }_1=\frac{\mathrm{E}[(X-\mu {)}^3]}{{\left(\mathrm{var}(X)\right)}^{3/2}}=\frac{2(1-2\mu )\sqrt{1+\nu }}{(2+\nu )\sqrt{\mu (1-\mu )}}.}$$

The skewness can also be expressed just in terms of the variance var and the mean μ as follows:

$${\displaystyle {\gamma }_1=\frac{\mathrm{E}[(X-\mu {)}^3]}{(\mathrm{var}(X){)}^{3/2}}=\frac{2(1-2\mu )\sqrt{\mathrm{var}}}{\mu (1-\mu )+\mathrm{var}}\text{ if }\mathrm{var}<\mu (1-\mu )}$$

The accompanying plot of skewness as a function of variance and mean shows that maximum variance (1/4) is coupled with zero skewness and the symmetry condition (μ = 1/2), and that maximum skewness (positive or negative infinity) occurs when the mean is located at one end or the other, so that the "mass" of the probability distribution is concentrated at the ends (minimum variance).

The following expression for the square of the skewness, in terms of the sample size ν = α + β and the variance var, is useful for the method of moments estimation of four parameters:

$${\displaystyle ({\gamma }_1{)}^2=\frac{{\left(\mathrm{E}[(X-\mu {)}^3]\right)}^2}{{\left(\mathrm{var}(X)\right)}^3}=\frac{4}{(2+\nu {)}^2}\left(\frac{1}{\mathrm{var}}-4(1+\nu )\right)}$$

This expression correctly gives a skewness of zero for α = β, since in that case (see § Variance): ${\displaystyle \mathrm{var}=\frac{1}{4(1+\nu )}}$.

For the symmetric case (α = β), skewness = 0 over the whole range, and the following limits apply:

$${\displaystyle \underset{\alpha =\beta \to 0}{lim}{\gamma }_1=\underset{\alpha =\beta \to \mathrm{\infty }}{lim}{\gamma }_1=\underset{\nu \to 0}{lim}{\gamma }_1=\underset{\nu \to \mathrm{\infty }}{lim}{\gamma }_1=\underset{\mu \to \frac{1}{2}}{lim}{\gamma }_1=0}$$

For the asymmetric cases (α ≠ β) the following limits (with only the noted variable approaching the limit) can be obtained from the above expressions:

$${\displaystyle \begin{array}{cc}& \underset{\alpha \to 0}{lim}{\gamma }_1=\underset{\mu \to 0}{lim}{\gamma }_1=\mathrm{\infty }\\ & \underset{\beta \to 0}{lim}{\gamma }_1=\underset{\mu \to 1}{lim}{\gamma }_1=-\mathrm{\infty }\\ & \underset{\alpha \to \mathrm{\infty }}{lim}{\gamma }_1=-\frac{2}{\sqrt{\beta }},\underset{\beta \to 0}{lim}(\underset{\alpha \to \mathrm{\infty }}{lim}{\gamma }_1)=-\mathrm{\infty },\underset{\beta \to \mathrm{\infty }}{lim}(\underset{\alpha \to \mathrm{\infty }}{lim}{\gamma }_1)=0\\ & \underset{\beta \to \mathrm{\infty }}{lim}{\gamma }_1=\frac{2}{\sqrt{\alpha }},\underset{\alpha \to 0}{lim}(\underset{\beta \to \mathrm{\infty }}{lim}{\gamma }_1)=\mathrm{\infty },\underset{\alpha \to \mathrm{\infty }}{lim}(\underset{\beta \to \mathrm{\infty }}{lim}{\gamma }_1)=0\\ & \underset{\nu \to 0}{lim}{\gamma }_1=\frac{1-2\mu }{\sqrt{\mu (1-\mu )}},\underset{\mu \to 0}{lim}(\underset{\nu \to 0}{lim}{\gamma }_1)=\mathrm{\infty },\underset{\mu \to 1}{lim}(\underset{\nu \to 0}{lim}{\gamma }_1)=-\mathrm{\infty }\end{array}}$$

Kurtosis

The beta distribution has been applied in acoustic analysis to assess damage to gears, as the kurtosis of the beta distribution has been reported to be a good indicator of the condition of a gear. Kurtosis has also been used to distinguish the seismic signal generated by a person's footsteps from other signals. As persons or other targets moving on the ground generate continuous signals in the form of seismic waves, one can separate different targets based on the seismic waves they generate. Kurtosis is sensitive to impulsive signals, so it's much more sensitive to the signal generated by human footsteps than other signals generated by vehicles, winds, noise, etc. Unfortunately, the notation for kurtosis has not been standardized. Kenney and Keeping use the symbol γ₂ for the excess kurtosis, but Abramowitz and Stegun use different terminology. To prevent confusion between kurtosis (the fourth moment centered on the mean, normalized by the square of the variance) and excess kurtosis, when using symbols, they will be spelled out as follows:

$${\displaystyle \begin{array}{cc}\text{excess kurtosis}& =\text{kurtosis}-3\\ & =\frac{\mathrm{E}[(X-\mu {)}^4]}{(\mathrm{var}(X){)}^2}-3\\ & =\frac{6[{\alpha }^3-{\alpha }^2(2\beta -1)+{\beta }^2(\beta +1)-2\alpha \beta (\beta +2)]}{\alpha \beta (\alpha +\beta +2)(\alpha +\beta +3)}\\ & =\frac{6[(\alpha -\beta {)}^2(\alpha +\beta +1)-\alpha \beta (\alpha +\beta +2)]}{\alpha \beta (\alpha +\beta +2)(\alpha +\beta +3)}.\end{array}}$$

Letting α = β in the above expression one obtains

$${\displaystyle \text{excess kurtosis}=-\frac{6}{3+2\alpha }\text{ if }\alpha =\beta .}$$

Therefore, for symmetric beta distributions, the excess kurtosis is negative, increasing from a minimum value of −2 at the limit as {α = β} → 0, and approaching a maximum value of zero as {α = β} → ∞. The value of −2 is the minimum value of excess kurtosis that any distribution (not just beta distributions, but any distribution of any possible kind) can ever achieve. This minimum value is reached when all the probability density is entirely concentrated at each end x = 0 and x = 1, with nothing in between: a 2-point Bernoulli distribution with equal probability 1/2 at each end (a coin toss: see section below "Kurtosis bounded by the square of the skewness" for further discussion). The description of kurtosis as a measure of the "potential outliers" (or "potential rare, extreme values") of the probability distribution, is correct for all distributions including the beta distribution. When rare, extreme values can occur in the beta distribution, the higher its kurtosis; otherwise, the kurtosis is lower. For α ≠ β, skewed beta distributions, the excess kurtosis can reach unlimited positive values (particularly for α → 0 for finite β, or for β → 0 for finite α) because the side away from the mode will produce occasional extreme values. Minimum kurtosis takes place when the mass density is concentrated equally at each end (and therefore the mean is at the center), and there is no probability mass density in between the ends.

Using the parametrization in terms of mean μ and sample size ν = α + β:

$${\displaystyle \begin{array}{cc}\alpha & =\mu \nu ,\text{ where }\nu =(\alpha +\beta )>0\\ \beta & =(1-\mu )\nu ,\text{ where }\nu =(\alpha +\beta )>0.\end{array}}$$

one can express the excess kurtosis in terms of the mean μ and the sample size ν as follows:

$${\displaystyle \text{excess kurtosis}=\frac{6}{3+\nu }(\frac{(1-2\mu {)}^2(1+\nu )}{\mu (1-\mu )(2+\nu )}-1)}$$

The excess kurtosis can also be expressed in terms of just the following two parameters: the variance var, and the sample size ν as follows:

$${\displaystyle \text{excess kurtosis}=\frac{6}{(3+\nu )(2+\nu )}\left(\frac{1}{\text{ var }}-6-5\nu \right)\text{ if }\text{var}<\mu (1-\mu )}$$

and, in terms of the variance var and the mean μ as follows:

$${\displaystyle \text{excess kurtosis}=\frac{6\text{ var }(1-\text{ var }-5\mu (1-\mu ))}{(\text{var }+\mu (1-\mu ))(2\text{ var }+\mu (1-\mu ))}\text{ if }\text{var}<\mu (1-\mu )}$$

The plot of excess kurtosis as a function of the variance and the mean shows that the minimum value of the excess kurtosis (−2, which is the minimum possible value for excess kurtosis for any distribution) is intimately coupled with the maximum value of variance (1/4) and the symmetry condition: the mean occurring at the midpoint (μ = 1/2). This occurs for the symmetric case of α = β = 0, with zero skewness. At the limit, this is the 2 point Bernoulli distribution with equal probability 1/2 at each Dirac delta function end x = 0 and x = 1 and zero probability everywhere else. (A coin toss: one face of the coin being x = 0 and the other face being x = 1.) Variance is maximum because the distribution is bimodal with nothing in between the two modes (spikes) at each end. Excess kurtosis is minimum: the probability density "mass" is zero at the mean and it is concentrated at the two peaks at each end. Excess kurtosis reaches the minimum possible value (for any distribution) when the probability density function has two spikes at each end: it is bi-"peaky" with nothing in between them.

On the other hand, the plot shows that for extreme skewed cases, where the mean is located near one or the other end (μ = 0 or μ = 1), the variance is close to zero, and the excess kurtosis rapidly approaches infinity when the mean of the distribution approaches either end.

Alternatively, the excess kurtosis can also be expressed in terms of just the following two parameters: the square of the skewness, and the sample size ν as follows:

$${\displaystyle \text{excess kurtosis}=\frac{6}{3+\nu }(\frac{(2+\nu )}{4}(\text{skewness}{)}^2-1){\text{ if (skewness)}}^2-2<\text{excess kurtosis}<\frac{3}{2}(\text{skewness}{)}^2}$$

From this last expression, one can obtain the same limits published over a century ago by Karl Pearson for the beta distribution (see section below titled "Kurtosis bounded by the square of the skewness"). Setting α + β = ν = 0 in the above expression, one obtains Pearson's lower boundary (values for the skewness and excess kurtosis below the boundary (excess kurtosis + 2 − skewness² = 0) cannot occur for any distribution, and hence Karl Pearson appropriately called the region below this boundary the "impossible region"). The limit of α + β = ν → ∞ determines Pearson's upper boundary.

$${\displaystyle \begin{array}{cc}& \underset{\nu \to 0}{lim}\text{excess kurtosis}=(\text{skewness}{)}^2-2\\ & \underset{\nu \to \mathrm{\infty }}{lim}\text{excess kurtosis}=\frac{3}{2}(\text{skewness}{)}^2\end{array}}$$

therefore:

$${\displaystyle (\text{skewness}{)}^2-2<\text{excess kurtosis}<\frac{3}{2}(\text{skewness}{)}^2}$$

Values of ν = α + β such that ν ranges from zero to infinity, 0 < ν < ∞, span the whole region of the beta distribution in the plane of excess kurtosis versus squared skewness.

For the symmetric case (α = β), the following limits apply:

$${\displaystyle \begin{array}{cc}& \underset{\alpha =\beta \to 0}{lim}\text{excess kurtosis}=-2\\ & \underset{\alpha =\beta \to \mathrm{\infty }}{lim}\text{excess kurtosis}=0\\ & \underset{\mu \to \frac{1}{2}}{lim}\text{excess kurtosis}=-\frac{6}{3+\nu }\end{array}}$$

For the unsymmetric cases (α ≠ β) the following limits (with only the noted variable approaching the limit) can be obtained from the above expressions:

$${\displaystyle \begin{array}{cc}& \underset{\alpha \to 0}{lim}\text{excess kurtosis}=\underset{\beta \to 0}{lim}\text{excess kurtosis}=\underset{\mu \to 0}{lim}\text{excess kurtosis}=\underset{\mu \to 1}{lim}\text{excess kurtosis}=\mathrm{\infty }\\ & \underset{\alpha \to \mathrm{\infty }}{lim}\text{excess kurtosis}=\frac{6}{\beta },\underset{\beta \to 0}{lim}(\underset{\alpha \to \mathrm{\infty }}{lim}\text{excess kurtosis})=\mathrm{\infty },\underset{\beta \to \mathrm{\infty }}{lim}(\underset{\alpha \to \mathrm{\infty }}{lim}\text{excess kurtosis})=0\\ & \underset{\beta \to \mathrm{\infty }}{lim}\text{excess kurtosis}=\frac{6}{\alpha },\underset{\alpha \to 0}{lim}(\underset{\beta \to \mathrm{\infty }}{lim}\text{excess kurtosis})=\mathrm{\infty },\underset{\alpha \to \mathrm{\infty }}{lim}(\underset{\beta \to \mathrm{\infty }}{lim}\text{excess kurtosis})=0\\ & \underset{\nu \to 0}{lim}\text{excess kurtosis}=-6+\frac{1}{\mu (1-\mu )},\underset{\mu \to 0}{lim}(\underset{\nu \to 0}{lim}\text{excess kurtosis})=\mathrm{\infty },\underset{\mu \to 1}{lim}(\underset{\nu \to 0}{lim}\text{excess kurtosis})=\mathrm{\infty }\end{array}}$$

Characteristic function

The characteristic function is the Fourier transform of the probability density function. The characteristic function of the beta distribution is Kummer's confluent hypergeometric function (of the first kind):

${\displaystyle \begin{array}{cc}{\phi }_X(\alpha ;\beta ;t)& =\mathrm{E}\left[e^{itX}\right]\\ & ={\int }_0^1{e}^{itx}f(x;\alpha ,\beta )dx\\ & ={}_1{F}_1(\alpha ;\alpha +\beta ;it)\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{{\alpha }^{\overline{n}}(it{)}^n}{(\alpha +\beta {)}^{\overline{n}}n!}\\ & =1+\sum _{k=1}^{\mathrm{\infty }}\left(\prod _{r=0}^{k-1}\frac{\alpha +r}{\alpha +\beta +r}\right)\frac{(it{)}^k}{k!}\end{array}}$