# The moment generating function of the gamma

The moment-generating function of the gamma distribution is $M(t) = (1-\theta t)^{-\alpha}$. Such a friendly little guy. Not what you would expect when you start with this: $M(t) = \int_{0}^{\infty}e^{tx}\frac{1}{\Gamma (\alpha)\theta^{\alpha}}x^{\alpha - 1}e^{-x/\theta}dx$

How do we get there? First let’s combine the two exponential terms and move the gamma fraction out of the integral: $M(t) = \frac{1}{\Gamma (\alpha)\theta^{\alpha}}\int_{0}^{\infty}x^{\alpha - 1}e^{tx-x/\theta}dx$

Multiply $tx$ in the exponential by $\frac{\theta}{\theta} = 1$, add the two terms together and factor out $-x$: $M(t) = \frac{1}{\Gamma (\alpha)\theta^{\alpha}}\int_{0}^{\infty}x^{\alpha - 1}e^{-x(1-\theta t)/\theta}dx$

Now we’re ready to do so substitution in the integral. Let $u = \frac{x(1-\theta t)}{\theta}$. That means we have $du = \frac{1-\theta t}{\theta}dx$. Now solve those for x and dx, respectively. We get $x = \frac{\theta u}{1-\theta t}$ and $dx = \frac{\theta}{1-\theta t}du$. Now substitute those in the integral to see something that looks truly awful: $M(t) = \frac{1}{\Gamma (\alpha)\theta^{\alpha}}\int_{0}^{\infty}\frac{\theta u}{1-\theta t}^{\alpha - 1}e^{-u}\frac{\theta}{1-\theta t}du$

If re-write everything inside the integral with exponents we can do some cancellation and make this integral more attractive: $M(t) = \frac{1}{\Gamma (\alpha)\theta^{\alpha}}\int_{0}^{\infty}(1-\theta t)^{-\alpha + 1}(1-\theta t)^{-1}(\theta u)^{\alpha-1}\theta e^{-u}du$

Using the rules of exponents we see that $(1-\theta t)^{-\alpha + 1}(1-\theta t)^{-1} = (1- \theta t)^{-\alpha}$ which can be moved out of the integral since it doesn’t have $u$ in it: $M(t) = \frac{(1- \theta t)^{-\alpha}}{\Gamma (\alpha)\theta^{\alpha}}\int_{0}^{\infty}(\theta u)^{\alpha-1}\theta e^{-u}du$

Using the rules of exponents again we see that $\theta^{\alpha -1}\theta = \theta^{\alpha}$: $M(t) = \frac{(1- \theta t)^{-\alpha}}{\Gamma (\alpha)\theta^{\alpha}}\int_{0}^{\infty}\theta^{\alpha}u^{\alpha - 1} e^{-u}du$

Now notice we can cancel out the two $\theta^{\alpha}$ terms: $M(t) = \frac{(1- \theta t)^{-\alpha}}{\Gamma (\alpha)}\int_{0}^{\infty}u^{\alpha - 1} e^{-u}du$

The integral is now the gamma function: $\Gamma (\alpha) = \int_{0}^{\infty}u^{\alpha - 1} e^{-u}du$. Make that substitution: $M(t) = \frac{(1- \theta t)^{-\alpha}}{\Gamma (\alpha)}\Gamma(\alpha)$

Cancel out the $\Gamma(\alpha)$ terms and we have our nice-looking moment-generating function: $M(t) = (1- \theta t)^{-\alpha}$

If we take the derivative of this function and evaluate at 0 we get the mean of the gamma distribution: $M'(t) = -\alpha (1-\theta t)^{-\alpha -1}(-\theta)$ $M'(0) = -\alpha(1 - 0)^{-\alpha -1}(-\theta) = \alpha\theta = E(X)$

Recall that $\theta$ is the mean time between events and $\alpha$ is the number of events. Multiply them together and you have the mean. This makes sense. If the mean wait time between events is 2 minutes ( $\theta = 2$) and we’re interested in the distribution of times until 4 events occur ( $\alpha = 4$), it seems reasonable to think the mean wait time to observe the 4th event would be 8 minutes ( $4(2)$).

If we take the second derivative of the moment-generating function and evaluate at 0, we get the second moment about the origin which we can use to find the variance: $M''(t) = (-\alpha - 1)\theta\alpha(1-\theta t)^{-\alpha - 2}(-\theta)$ $M''(0) = (-\alpha - 1)\theta\alpha(1-0)^{-\alpha - 2}(-\theta) = (-\alpha - 1)(-\theta^{2})\alpha$

Now find the variance: $Var(X) = (-\alpha - 1)(-\theta^{2})\alpha - (\alpha\theta)^{2}$ $Var(X) = \alpha^{2}\theta^{2} + \alpha\theta^{2} - \alpha^{2}\theta^{2}$ $Var(X) = \alpha\theta^{2}$

Going back to our example with $\alpha = 4$ (number of events) and $\theta=2$ (mean time between events), we have as our variance $4(2^{2})=16$. The square root of that gives our standard deviation as 4.

Plugging the mean and standard deviation into the dgamma function in R, we can plot this particular gamma distribution:

xval <- seq(0,30, by = 0.01)
ga.4 <- dgamma(xval, shape=4, scale=2)
plot(xval,ga.4,col=1, type = "l", ylab="f(x)", xlab="x")
title(main="Gamma Distribution, scale = 2, shape = 4") This site uses Akismet to reduce spam. Learn how your comment data is processed.