2.2 The Beta Function

The gamma function of Chapter 2 interpolated the factorial. The beta function does the same for the binomial coefficient, and the two are so closely related that everything about the second reduces to the first — a relation established in Theorem 2.2.2 below and used constantly thereafter.

Two reasons make the beta function worth its own chapter. It converts a large class of definite integrals into gamma values, so that integrals with no elementary antiderivative can nonetheless be evaluated exactly; the trigonometric form of Theorem 2.2.1 is the workhorse here. And it is the normalising constant of the beta distribution, which is why the function appears throughout statistics — in the distribution of order statistics met in any course on non-parametric methods, in Bayesian inference as the conjugate prior for a probability, and in the \(F\) and \(t\) distributions.

We define the Beta function of \(x\) and \(y\), by \[\beta (x,y) = \int ^1_0t^{x-1}\, (1 - t)^{y-1}\, dt\] for \(x>0\) and \(y>0\).

Theorem 2.2.1. \[\beta (x,y) = 2\int _0^{\frac {\pi }{2}}\sin ^{2x-1}\theta \cos ^{2y-1}\theta \, d\theta .\]

Proof. Let \(\, t = \sin ^2\theta \), \(\, \hspace {0.3cm} dt = 2\sin \theta \cos \theta \, d\theta \) \begin {align*} \beta (x,y) & = \int _0^1t^{x-1}\, (1 - t)^{y - 1}\, dt\\ & = \int _0^{\frac {\pi }{2}}\sin ^{2(x-1)}\theta \, (1 - \sin ^2\theta )^{y - 1}\cdot 2\sin \theta \cos \theta \, d\theta \\ & = 2\int _0^{\frac {\pi }{2}}\sin ^{2x - 1}\theta \, \cos ^{2y - 1}\theta \, d\theta . \end {align*} □

Theorem 2.2.2. \[\beta (x,y) = \frac {\Gamma (x)\, \Gamma (y)}{\Gamma (x+y)}.\]

Proof. \begin {align*} \beta (x,y) & = 2\int _0^{\frac {\pi }{2}} \sin ^{2x - 1}\theta \, \cos ^{2y-1}\theta \, d\theta \\ & = \frac {2\, \Gamma (x)\, \Gamma (y)}{2\Gamma (x+y)}\hspace {0.5cm} \text {by theorem 2.4}\\ & = \frac {\Gamma (x)\, \Gamma (y)}{\Gamma (x + y)}. \end {align*} □

Note 2.2.3. This is the result the chapter is built around. Every beta integral is now a gamma computation, and the symmetry \(\beta (x,y)=\beta (y,x)\), which is obvious from the defining integral under \(t\mapsto 1-t\), becomes obvious again from the right-hand side. It also explains the appearance of the beta function in statistics: the density of the \(\operatorname {Beta}(x,y)\) distribution is \(t^{x-1}(1-t)^{y-1}\) divided by exactly this constant, and the moments of that distribution are ratios of gamma values in consequence.

Theorem 2.2.4.

(a).: \(\beta (x,y) = \beta (y,x)\)
(b).: \(\beta (x+1,y) = \frac {x}{x+y}\, \beta (x,y)\)
(c).: \(\beta (x, y+1) = \frac {y}{x + y}\, \beta (x,y)\)

Working.

(a).: \(\displaystyle {\beta (x,y) = \frac {\Gamma (x)\, \Gamma (y)}{\Gamma (x+y)}= \frac {\Gamma (y)\, \Gamma (x)}{\Gamma (y + x)} = \beta (y,x).}\)
(b).: \(\begin {aligned}[t] \beta (x+1, y) & = \frac {\Gamma (x + 1)\, \Gamma (y)}{\Gamma (x + 1 + y)}\\\\ & = \frac {x\, \Gamma (x) \, \Gamma (y)}{(x + y)\, \Gamma (x + y)}\, , \hspace {0.6cm} \Gamma (x + 1 ) = x\Gamma (x)\\\\ & = \frac {x}{x + y}\, \frac {\Gamma (x)\, \Gamma (y)}{\Gamma (x + y)}. \end {aligned}\)
(c).: follows similarly.

□

Theorem 2.2.5 (Legendre Duplication formula). \[\Gamma (2x) = \frac {2^{2x - 1}}{\sqrt {\pi }}\, \Gamma (x)\, \Gamma \left (x + \frac {1}{2}\right ).\]

Proof. \[\frac {\Gamma (x)\, \Gamma (x)}{\Gamma (x + x)} = \beta (x,x) = \int _0^1t^{x-1}\, (1 - t)^{x - 1}\, dt,\]

let \(t = \frac {1}{2}(1 + s)\), then \(dt = \frac {1}{2}ds\)

\begin {align*} \frac {\Gamma (x)\, \Gamma (x)}{\Gamma (x + x)} & = \int _{-1}^1 \frac {1}{2^{x-1}}\, (1 + s)^{x - 1}\, \frac {1}{2^{x - 1}}(1 - s)^{x - 1}\, \frac {1}{2}\, ds\\ & = \frac {1}{2^{2x - 1}}\int _{-1}^1 (1 - s^2 )^{x-1}\, ds\\ & = \frac {2}{2^{2x - 1}}\int _0^1(1 - s^2)^{x-1}\, ds\\ & = 2^{-2x +2}\int _0^1(1 - u)^{x-1}\, \frac {1}{2}\, u^{-\frac {1}{2}}\, du\, , \hspace {0.3cm} \text {letting}\, \, u = s^2 \implies \,\, du = 2sds\\ & = 2^{-2x+1}\, \beta \left (\frac {1}{2}, x\right )\\ & = 2^{-2x + 1}\, \frac {\Gamma \left (\frac {1}{2}\right ) \, \Gamma (x)}{\Gamma \left (x + \frac {1}{2}\right )}. \end {align*}

Hence, \[\frac {\Gamma (x)}{\Gamma (2x)} = \frac {2^{-2x + 1}\, \sqrt {\pi }}{\Gamma \left (x + \frac {1}{2}\right )}.\] i.e \[\Gamma (2x) = \frac {2^{2x - 1} \, \Gamma (x)\, \Gamma \left (x + \frac {1}{2}\right )}{\sqrt {\pi }}.\] □

Exercise 2.2.6.

(a).

Express in terms of Gamma or Beta functions

(i).: \(\displaystyle {\int _0^1 \frac {1}{\sqrt [3]{1 - x^3}}\, dx}\hspace {0.5cm}\) let \(t = x^3.\)
(ii).: \(\displaystyle {\int _{-1}^1 \left (\frac {1 + x}{1 - x}\right )^{\frac {1}{2}}\, dx = \int _{-1}^1(1 + x)^{\frac {1}{2}}\, (1 - x)^{-\frac {1}{2}}\, dx = \pi }\).
(iii).: \(\displaystyle {\int _0^1\left (\frac {1}{x} - 1\right )^{\frac {1}{4}}\, dx = \int _0^1 \left (\frac {1 - x}{x}\right )\, dx}\).
(iv).: \(\displaystyle {\int _0^1\frac {1}{\sqrt {1 - x^n}}\, dx\hspace {0.5cm}}, n > 0.\)

(b).

Show that the area enclosed by the curve \(x^4 + y^4 = 1\) is give by \[A = \frac {\left [\Gamma \left (\frac {1}{4}\right )\right ]^2}{2\sqrt {\pi }}.\]

(c).

Show that \[\int _0^{\frac {\pi }{2}} \sin ^{2n-1}\theta \, d\theta = \frac {1}{2}\frac {\Gamma (n)\, \Gamma \left (\frac {1}{2}\right )}{\Gamma \left (n+\frac {1}{2}\right )}.\] Hence, deduce that \[\int _0^{\frac {\pi }{2}} \sin ^{2n-1}\theta \, d\theta = \frac {2^{n-1}\, \Gamma (n)}{1\cdot 3\cdot 5\cdots (2n-1)} = \frac {2\cdot 4\, \cdots \, (2n-2)}{1\cdot 3\cdot 5 \, \cdots \, (2n-1)}.\]

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