Exponential, and Gamma

If \(T \sim \Exp(\text{rate}=\lambda)\), then

\[\begin{aligned} \P\{ T \in dt \} = \lambda e^{-\lambda t} dt . \end{aligned}\]

1. \(T\) can be any nonnegative real number.

2. \(T\) is memoryless: \[\begin{aligned} \P\{ T > x + y \given T > x \} = \P\{ T > y \} . \end{aligned}\]

3. A machine produces \(n\) widgets per second; each widget has probability \(\lambda/n\) of being broken. The time until the first broken widget appears (in seconds) is approximately \(\sim \Exp(\lambda)\).

If \(S \sim \Gam(\text{shape}=\alpha, \text{rate}=\lambda)\), then

\[\begin{aligned} \P\{ S \in dt \} = \frac{\alpha^\lambda}{\Gamma(\alpha)} t^{\alpha - 1} e^{-\lambda t} dt . \end{aligned}\]

1. \(S\) has mean \(\alpha \lambda\) and variance \(\alpha/\lambda^2\).

2. If \(T_1, \ldots, T_k\) are independent \(\Exp(\lambda)\), then \(S = T_1 + \cdots + T_k\) is \(\Gam(k, \lambda)\).

3. A machine produces \(n\) widgets per second; each widget has probability \(\lambda/n\) of being broken. The time until the \(k^\text{th}\) broken widget appears (in seconds) is approximately \(\sim \Gam(k, \lambda)\).