Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Gamma distribution ▷ Moment-generating function

Theorem: Let $X$ follow a gamma distribution:

\[\label{eq:gam} X \sim \mathrm{Gam}(a, b) \; .\]

Then, the moment-generating function of $X$ is

\[\label{eq:gam-mgf} M_X(t) = \left( 1 - \frac{t}{b} \right)^{-a} \; .\]

Proof: The moment-generating function of a random variable $X$ is defined as:

\[\label{eq:mgf} M_X(t) = \mathrm{E} \left[ e^{tX} \right], \quad t \in \mathbb{R} \; .\]

Applying the law of the unconscious statistician, we have:

\[\label{eq:gam-mgf-s1} M_X(t) = \int_{\mathcal{X}} e^{tx} \cdot f_X(x) \, \mathrm{d}x \; .\]

With the probability density function of the gamma distribution, we have:

\[\label{eq:gam-mgf-s2} M_X(t) = \int_{\mathbb{R}} \exp[t x] \cdot \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \, \mathrm{d}x \; .\]

Now we summarize the two exponential functions inside the integral:

\[\label{eq:gam-mgf-s3} \begin{split} M_X(t) &= \int_{\mathbb{R}} \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-(b-t) x] \, \mathrm{d}x \\ &= \int_{\mathbb{R}} \frac{(b-t)^a}{(b-t)^a} \cdot \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-(b-t) x] \, \mathrm{d}x \\ &= \int_{\mathbb{R}} \frac{b^a}{(b-t)^a} \cdot \frac{(b-t)^a}{\Gamma(a)} x^{a-1} \exp[-(b-t) x] \, \mathrm{d}x \\ &= \left( \frac{b}{b-t} \right)^a \int_{\mathbb{R}} \frac{(b-t)^a}{\Gamma(a)} x^{a-1} \exp[-(b-t) x] \, \mathrm{d}x \; . \end{split}\]

The integrand is equal to the probability density function of a gamma distribution:

\[\label{eq:gam-mgf-s4} M_X(t) = \left( \frac{b}{b-t} \right)^a \int_{\mathbb{R}} \mathrm{Gam}(x; a, b-t) \, \mathrm{d}x \; .\]

Because the entire probability density integrates to one, we finally have:

\[\label{eq:gam-mgf-s5} M_X(t) = \left( \frac{b}{b-t} \right)^a = \left( \frac{b-t}{b} \right)^{-a} = \left( \frac{b}{b} - \frac{t}{b} \right)^{-a} = \left( 1 - \frac{t}{b} \right)^{-a} \; .\]
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Metadata: ID: P437 | shortcut: gam-mgf | author: JoramSoch | date: 2024-02-23, 14:31.