Introducing the beta function

The gamma distribution is mathematically defined from the gamma function. This post gives a brief introduction to the beta function. The goal is to establish one property that is the basis for defining the beta distribution.

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The Beta Function

For any positive constants a and b, the beta function is defined to be the following integral:

    \displaystyle B(a,b)=\int_0^1 t^{a-1} \ (1-t)^{b-1} \ dt \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (0)

The beta function can be evaluated directly if the parameters a and b are not too large. For example, B(3,2) is the integral \displaystyle \int_0^1 t^2 (1-t) \ dt, which is 1/12. Evaluating (0) in a case by case basis does not shed light on the beta function. Direct calculation can also be cumbersome (e.g. for large parameters that are integers) or challenging (e.g. for parameters a and b that are fractional). It turns out that the evaluation of the beta function B(a,b) is based on the gamma function.

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Connection to the Gamma Function

The remainder of the post is to establish the following value of the beta function:

    \displaystyle B(a,b)=\int_0^1 t^{a-1} \ (1-t)^{b-1} \ dt=\frac{\Gamma(a) \ \Gamma(b)}{\Gamma(a+b)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

To start the proof of (1), let X and Y be two independent random variables such that X follows a gamma distribution with shape parameter a and rate parameter \beta and that Y follows a gamma distribution with shape parameter b and rate parameter \beta. It does not matter what \beta is, as long as it is the rate parameter for both X and Y. Then the sum S=X+Y has a gamma distribution with shape parameter a+b and rate parameter \beta. The following is the density function for S=X+Y.

    \displaystyle f_S(s)=\frac{1}{\Gamma(a+b)} \ \beta^{a+b} \ s^{a+b-1} \ e^{-\beta s}  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

The density function of S=X+Y can also be derived from the convolution formula using the density functions of X and Y as follows:

    \displaystyle \begin{aligned} f_S(s)&=\int_0^s f_Y(s-x) \ f_X(x) \ dx \ \ \ \ \ \text{(convolution)} \\&=\int_0^s \frac{1}{\Gamma(b)} \ \beta^{b} \ (s-x)^{b-1} \ e^{-\beta (s-x)} \ \frac{1}{\Gamma(a)} \ \beta^{a} \ x^{a-1} \ e^{-\beta x} \ dx \\&=\frac{\beta^{a+b}}{\Gamma(a) \ \Gamma(b)} \ e^{-\beta s} \ \int_0^s x^{a-1} \ (s-x)^{b-1} \ dx \\&=\frac{\beta^{a+b}}{\Gamma(a) \ \Gamma(b)} \ e^{-\beta s} \ s^{a+b-1} \ \int_0^1 t^{a-1} \ (1-t)^{b-1} \ dt \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)  \end{aligned}

See here for more information on how to use the convolution formula. The last step in (3) is obtained by a change of variable in the integral from the step immediately above it by letting x=st. The last step in (3) must equal to (2). Setting the two equal would produce the equality in (1).

Note that if the function t^{a-1} \ (1-t)^{b-1} is normalized by the value B(a,b), it would be a density function, which is the beta distribution. The following is the density function of the beta distribution.

    \displaystyle f(x)=\frac{\Gamma(a+b)}{\Gamma(a) \ \Gamma(b)} \ x^{a-1} \ (1-x)^{b-1}; \ \ \ \ \ \ \ \ 0<x<1  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)

The beta distribution is further examined in the next post.

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\copyright \ 2016 - \text{Dan Ma}

2 thoughts on “Introducing the beta function

  1. Pingback: Introducing the beta distribution | Topics in Actuarial Modeling

  2. Pingback: Generalized beta distribution | Topics in Actuarial Modeling

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