Class 18
\(\displaystyle f(x) = \begin{cases} 0 & \text{ if } x < 0\\\lambda e^{-\lambda x} & \text{ if } x \ge 0\end{cases}\)
\(\displaystyle F(x) = \begin{cases} 0 & \text{ if } x < 0\\1 - e^{-\lambda x} & \text{ if } x \ge 0\end{cases}\)
\(\displaystyle \operatorname{P}(X > x) = G(x) = 1 - F(x) = \begin{cases} 1 & \text{ if } x < 0\\ e^{-\lambda x} & \text{ if } x \ge 0\end{cases}\)
\(\operatorname{E} X = \frac{1}{\lambda}\)
For each $s , and \(t \ge 0\)
\[\operatorname{P}\left(X > t + s\mid X > t\right) = \operatorname{P}\left(X > s\right)\]
\(r(t) = \lambda\)
Randomly choosing one of \(n\) exponentially distributed variables with rates \(\lambda_1, \lambda_2, \dots, \lambda_n\), with probabilities \(P_1, P_2, \dots, P_n\).
Parameters \(\lambda_1, \lambda_2, \dots, \lambda_n\), \(P_1, P_2, \dots, P_n\), where \(\displaystyle \sum_{j=1}^n P_n = 1\).
CDF: \(\displaystyle F(t) = 1 - \sum_{j=1}^n P_je^{-\lambda_j t}\)
pdf: \(\displaystyle f(t) = \sum_{j=1}^n \lambda_jP_je^{-\lambda_j t}\)
\(\displaystyle r(t) = \frac{\sum_{j=1}^n P_j\lambda_je^{-\lambda_j t}}{\sum_{j-1}^n P_j e^{-\lambda_j t}} \class{fragment}{{}= \sum_{j=1}^n \lambda_j\operatorname{P}\left(T = j\mid X > t\right)} \class{fragment}{\rightarrow \min_j \lambda_j \text{ as } t \rightarrow \infty}\)
Let \(X_1, X_2, \dots, X_n\) are independent indentically distributed exponential random variables with rate \(\lambda\).
\[X_j ~ \operatorname{Exp}(\lambda) \text{ for } j = 1, 2, \dots, n\]
What is the distribution of \(X = X_1 + X_2 + \cdots + X_n\)?
(Waiting time for \(n\)-th occurrence.)
\(X \sim \operatorname{Gamma}(n,\lambda)\), or \(\displaystyle f(t) = \lambda e^{-\lambda t}\frac{(\lambda t)^{n-1}}{(n-1)!}\)
Let \(X_1\) and \(X_2\) be independent exponentially distributed with rates \(\lambda_1\) and \(\lambda_2\). What is \(\operatorname{P}(X_1 < X_2)\)?
Condition on \(X_1\)
Suppose \(X_i\) is exponential with rate \(\lambda_i\) for \(i = 1, 2, \ldots, n\) and \(X_i\)’s are independent. Then
\[ \begin{aligned} \operatorname{P}(\min(X_1, X_2, \ldots, X_n) > x) &= \operatorname{P}(X_i > x \text{ for each } i = 1, 2, \ldots, n) \\ &= \prod_{i=1}^n \operatorname{P}(X_i > x)\\ &= \prod_{i=1}^n e^{-\lambda_i} x\\ &= \exp\left(-\left(\sum_{i=1}^n \lambda_i\right) x\right) \end{aligned} \]
In other words, \(\min(X_1, X_2, \ldots, X_n)\) has exponential distribution with the rate \(\lambda = \lambda_1 + \lambda_2 + \cdots + \lambda_n\).
Suppose you arrive at a post office where there are two clerks that are both busy at the time, but there is no one waiting in line in front of you. At the moment one of the two clerks is free, it will be your turn.
Assume that the service time for clerk \(i\) is exponential with rate \(\lambda_i\), for \(i = 1, 2\). Let \(T\) be the total time you spend at the post office. Find \(\operatorname{E} T\).