15.2 - Exponential Properties

The exponential probability density function:

\(f(x)=\dfrac{1}{\theta} e^{-x/\theta}\)

for \(x\ge 0\) and \(\theta>0\) is a valid probability density function.

Proof

The moment generating function of an exponential random variable \(X\) with parameter \(\theta\) is:

\(M(t)=\dfrac{1}{1-\theta t}\)

for \(t<\frac{1}{\theta}\).

Proof

\(M(t)=E(e^{tX})=\int_0^\infty e^{tx} \left(\dfrac{1}{\theta}\right) e^{-x/\theta} dx\)

Simplifying and rewriting the integral as a limit, we have:

\(M(t)=\dfrac{1}{\theta}\lim\limits_{b \to \infty} \int_0^b e^{x(t-1/\theta)} dx\)

Integrating, we have:

\(M(t)=\dfrac{1}{\theta}\lim\limits_{b \to \infty} \left[ \dfrac{1}{t-1/\theta} e^{x(t-1/\theta)} \right]^{x=b}_{x=0}\)

Evaluating at \(x=0\) and \(x=b\), we have:

\(M(t)=\dfrac{1}{\theta}\lim\limits_{b \to \infty} \left[ \dfrac{1}{t-1/\theta} e^{b(t-1/\theta)} - \dfrac{1}{t-1/\theta} \right]=\dfrac{1}{\theta}\lim\limits_{b \to \infty} \left\{ \left(\dfrac{1}{t-1/\theta}\right) e^{b(t-1/\theta)} \right\}-\dfrac{1}{t-1/\theta}\)

Now, the limit approaches 0 provided \(t-\frac{1}{\theta}<0\), that is, provided \(t<\frac{1}{\theta}\), and so we have:

\(M(t)=\dfrac{1}{\theta} \left(0-\dfrac{1}{t-1/\theta}\right)\)

Simplifying more:

\(M(t)=\dfrac{1}{\theta} \left(-\dfrac{1}{\dfrac{\theta t-1}{\theta}}\right)=\dfrac{1}{\theta}\left(-\dfrac{\theta}{\theta t-1}\right)=-\dfrac{1}{\theta t-1}\)

and finally:

\(M(t)=\dfrac{1}{1-\theta t}\)

provided \(t<\frac{1}{\theta}\), as was to be proved.

The mean of an exponential random variable \(X\) with parameter \(\theta\) is:

\(\mu=E(X)=\theta\)

Proof

The variance of an exponential random variable \(X\) with parameter \(\theta\) is:

\(\sigma^2=Var(X)=\theta^2\)

Proof