What Is the Moment-Generating Function of a Beta Distribution?

If $X\sim\operatorname{Beta}(\alpha,\beta)$, its moment-generating function can be written using the confluent hypergeometric function. In general, it does not simplify to a combination of elementary functions, but the formula gives a convergent series and makes moments easy to obtain.

The formula

Assume $\alpha>0$ and $\beta>0$, with the standard Beta distribution supported on $0<x<1$. Its density is

where $B(\alpha,\beta)$ is the beta function. By definition, the MGF is $M_X(t)=\mathbb{E}[e^{tX}]$. For a Beta random variable,

Here ${}_1F_1(a;c;t)$ is the confluent hypergeometric function, also called Kummer’s function. The semicolon separates its two parameters.

Deriving the MGF

Start with the expectation integral and substitute the density:

One way to identify the result is the integral representation

which applies when $\operatorname{Re}(c)>\operatorname{Re}(a)>0$. Set $a=\alpha$ and $c=\alpha+\beta$. The prefactor becomes the reciprocal beta-function normalization, giving the stated MGF.

You can also derive it from the power series for the exponential. The Beta moments are

where $(q)_n=q(q+1)\cdots(q+n-1)$ is the rising factorial, and $(q)_0=1$. Thus

Because $0\le X\le1$, the expectation of $e^{tX}$ is finite for every real $t$. In fact, the power series converges for every complex $t$ as well.

Worked example: $X\sim\operatorname{Beta}(2,3)$

For this distribution, $\alpha=2$ and $\alpha+\beta=5$, so

Expanding the series gives

The coefficient of $t^n$ is $\mathbb{E}[X^n]/n!$. For example, the coefficient of $t$ gives $\mathbb{E}[X]=2/5$. The coefficient of $t^2$ is $\mathbb{E}[X^2]/2!$, so $\mathbb{E}[X^2]=2(1/10)=1/5$. Therefore,

This agrees with the standard Beta variance formula $\alpha\beta/[(\alpha+\beta)^2(\alpha+\beta+1)]$.

Common mistakes

• Using the wrong second parameter. The second parameter of ${}_1F_1$ is $\alpha+\beta$, not just $\beta$.
• Forgetting the factorial in the series. The coefficient of $t^n$ is $\mathbb{E}[X^n]/n!$, not the moment itself.
• Confusing an MGF with a probability-generating function. The MGF uses $e^{tX}$ and a real parameter $t$; it is not obtained by substituting a probability value into a discrete generating function.
• Assuming every MGF has an elementary form. The hypergeometric expression is a standard closed form; for general shape parameters it need not reduce to elementary functions.

A useful check

At $t=0$, the formula must give $M_X(0)=1$, as every MGF does. Differentiating at zero gives $M_X'(0)=\alpha/(\alpha+\beta)$, the Beta mean. A formula that fails either check likely has an incorrect parameter or normalization.