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The density of beta distribution is given by the following

$$f(x\mid \alpha ,\beta ) = \frac 1 {\operatorname{B}(\alpha,\beta)} x^{\alpha - 1} (1 - x)^{\beta - 1}$$

where

$$ \operatorname{B}(\alpha,\beta) = \int_0^1 x^{\alpha - 1} (1 - x)^{\beta - 1} \,dx $$

is the Beta function as the normalizing constant.

I understand one use of Beta distribution is to draw random probabilities from it. I am interested in what gives rise to this distribution? In particular, why it is "$x^{\alpha - 1} (1 - x)^{\beta - 1}$", not something else? What kinds of desired properties uniquely determine this density? And how the density formula is derived?

I have googled but could not find a good explanation. Many text books just throw this in front of you like it comes from nowhere.

Michael Hardy
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Ralph B.
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1 Answers1

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  • One way in which the Beta distribution arises is as the distribution of the order statistics of a random sample from the uniform distribution. Suppose $X_1,\ldots, X_n$ are i.i.d. and uniformly distributed in $[0,1]$. Let $X_{(1)} < \cdots < X_{(n)}$ be the corresponding order statistics, i.e. the observations sorted into increasing order. Then $X_{(k)} \sim\operatorname{Beta}(k,n-k+1).$

  • Another way in which the Beta distribution arises is as the arrival times in a Poisson process, conditioned on the number of arrivals in a specified time interval. For example, suppose it is given that there were exactly $n$ arrivals between times $0$ and $T$. Let $S$ be the time of the $k$th arrival. Then $S/T \sim\operatorname{Beta}(k,n-k+1).$

  • Suppose $R\sim\operatorname{Uniform}(0,1)$ and $X_1,\ldots,X_n\mid R \sim \operatorname{i.i.d.} \operatorname{Bernoulli}(R),$ i.e. $$\begin{cases} \Pr(X_1=1\mid R) = R, \\ \Pr(X_1=0\mid R) = 1-R. \end{cases}$$ Then $$ R \mid (X_1+\cdots +X_n = k) \sim \operatorname{Beta}(k+1,n-k+1).$$

  • Suppose $X$ has the Gamma distribution $\displaystyle \frac 1 {\Gamma(\alpha)} (x/\mu)^{\alpha-1} e^{-(x/\mu)} (dx/\mu)$ and $Y$ is independent of $X$ and has the Gamma distribution $\displaystyle \frac 1 {\Gamma(\beta)} (x/\mu)^{\beta-1} e^{-(x/\mu)} (dx/\mu). \vphantom{\frac {\displaystyle\int} {\displaystyle\int}} $ The $X/(X+Y)$ is independent of $X+Y$ and $X/(X+Y) \sim\operatorname{Beta}(\alpha,\beta).$

Michael Hardy
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