15.6  Gamma Properties
15.6  Gamma PropertiesHere, after formally defining the gamma distribution (we haven't done that yet?!), we present and prove (well, sort of!) three key properties of the gamma distribution.
 Gamma Distribution

A continuous random variable \(X\) follows a gamma distribution with parameters \(\theta>0\) and \(\alpha>0\) if its probability density function is:
\(f(x)=\dfrac{1}{\Gamma(\alpha)\theta^\alpha} x^{\alpha1} e^{x/\theta}\)
for \(x>0\).
Before we get to the three theorems and proofs, two notes:

We consider \(\alpha>0\) a positive integer if the derivation of the p.d.f. is motivated by waiting times until α events. But the p.d.f. is actually a valid p.d.f. for any \(\alpha>0\) (since \(\Gamma(\alpha)\) is defined for all positive \(\alpha\)).

The gamma p.d.f. reaffirms that the exponential distribution is just a special case of the gamma distribution. That is, when you put \(\alpha=1\) into the gamma p.d.f., you get the exponential p.d.f.
Theorem
The moment generating function of a gamma random variable is:
\(M(t)=\dfrac{1}{(1\theta t)^\alpha}\)
for \(t<\frac{1}{\theta}\).
Proof
By definition, the moment generating function \(M(t)\) of a gamma random variable is:
\(M(t)=E(e^{tX})=\int_0^\infty \dfrac{1}{\Gamma(\alpha)\theta^\alpha}e^{x/\theta} x^{\alpha1} e^{tx}dx\)
Collecting like terms, we get:
\(M(t)=E(e^{tX})=\int_0^\infty \dfrac{1}{\Gamma(\alpha)\theta^\alpha}e^{x\left(\frac{1}{\theta}t\right)} x^{\alpha1} dx\)
Now, let's use the change of variable technique with:
\(y=x\left(\dfrac{1}{\theta}t\right)\)
Rearranging, we get:
\(x=\dfrac{\theta}{1\theta t}y\) and therefore \(dx=\dfrac{\theta}{1\theta t}dy\)
Now, making the substitutions for \(x\) and \(dx\) into our integral, we get:
Theorem
The mean of a gamma random variable is:
\(\mu=E(X)=\alpha \theta\)
Proof
The proof is left for you as an exercise.
Theorem
The variance of a gamma random variable is:
\(\sigma^2=Var(X)=\alpha \theta^2\)
Proof
This proof is also left for you as an exercise.