23  Exponential and Gamma Distribution

23.1 Exponential Distribution

An exponential random variable \(X\) with parameter \(\lambda > 0\) is a random variable with pdf \[f(x) = \begin{cases}\lambda e^{-\lambda x} \quad &\text{if } x > 0 \\ 0 \quad &\text{otherwise} \end{cases}\]

where the parameter \(\lambda\) is called the rate parameter. We denote \(X\sim \text{Expo}(\lambda)\). Exponential random variables are often used to model arrival times, waiting times, and equipment failure times.

The cdf of \(X\) is: \[F(x) = 1 - e^{-\lambda x}, \qquad x \ge 0\] and the sdf of \(X\) is: \[S(x) = e^{-\lambda x}, \qquad x \ge 0\] Properties: * \(\text E[X] = \frac{1}{\lambda}\) * \(\text{Var}(X) = \frac{1}{\lambda^2}\) * Memorylessness: \(P(X > s + t | X > s) = P(X > t)\) for \(s, t \ge 0\). * \(\varphi(t) = \frac{\lambda}{\lambda -it}\) * \(\text E[X^k] = \frac{k!}{\lambda^k}\) * \(\text E[X\wedge x] = \frac{F(x)}{\lambda}\)

Example: Suppose that the time \(X\) (in hours) required to repair a car has an exponential distribution with mean \(4\). Find
* The cumulative distribution function of \(X\) * \(P(X > 4)\) * \(P(X > 10|X > 8)\)

Solution: The rate parameter \(\lambda\) is equal to \(\frac{1}{4}\). Hence, the cdf of \(X\) is \[F(x) = 1 - e^{-\frac{x}{4}}, \qquad x \ge 0\] and \(0\) otherwise. The survival function is \(S(x) = e^{-\frac{x}{4}}\) for \(x \ge 0\).

Then, \(P(X > 4) = S(4) = e^{-1}\) and \(P(X > 10 | X > 8) = S(2) = e^{-0.5}\). ## Gamma Distribution We introduce the Gamma function: \[\Gamma(\alpha) = \int_0^\infty e^{-t} t^{\alpha - 1}dt\] Properties: * \(\Gamma(1) = 1\) * \(\Gamma(\frac{1}{2}) = \sqrt{\pi}\) * \(\Gamma(n) = (n-1)!\) for \(n \in \mathbb Z^+\) * \(\Gamma(\alpha) = (\alpha-1)\Gamma(\alpha-1)\)

A Gamma random variable with parameters \(\alpha > 0\) and \(\lambda > 0\) has a pdf \[f(x) = \frac{\lambda e^{-\lambda x}(\lambda x)^{\alpha - 1}}{\Gamma(\alpha)}, \qquad x \ge 0\] The cdf of the Gamma distribution is \[F(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^x t^{\alpha-1}e^{-\lambda t} dt\] and in the case of \(\alpha \in \mathbb Z^+\) is equal to: \[F(x)= 1 - \sum_{k=0}^{\alpha - 1}P(Y = k)\] for \(Y \sim \text{Pois}(\lambda x)\).

Proposition: Let \(X_i \sim \text{Expo}(\lambda)\) be independent exponential random variables. Then \[X_1 + ... + X_m \sim \text{Gamma}(m, \lambda )\] Properties: * \(\text E[X] = \frac{\alpha}{\lambda}\) * \(\text{Var}(X) = \frac{\alpha}{\lambda^2}\) * \(\varphi_X(t) = (1-\frac{it}{\lambda})^{-\alpha}\)