Exponential Distribution

A random variable $X$ has a exponential distribution $X \sim E x p (θ)$ if and only if it has the pdf

f (x; θ) = \frac{1}{θ} e^{- x / θ} for x > 0

Mean $μ = θ$

Variance $σ^{2} = θ^{2}$

Equivalent to the pdf of a Gamma Distribution with $α = 1$

\frac{1}{θ^{1} Γ (1)} x^{1 - 1} e^{- x / θ}

Recall that if recurrent events happen according to a Poisson process with mean number of events per unit time is λ, then the number of events that occurs during an unit of time has a Poisson distribution with mean λ. The waiting time until the first event occurs has an exponential distribution with parameter θ = 1 λ and the mean waiting time is 1 λ . In addition, the waiting time between two successive events has the same exponential distribution

CDF

F_{X} (x) = 1 - e^{- x / θ}, x > 0, θ > 0

Order Statistics

\begin{aligned} f_{X_{(1)}} & = \frac{1}{θ / n} e^{- x / (θ / n)} \\ \sim E x p o n e n t i a l (\frac{θ}{n}) \end{aligned}

\begin{aligned} f_{X_{(n)}} & = n [F_{X} (x)]^{n - 1} f_{X} (x) \\ = \frac{n}{θ} e^{- x / θ} [1 - e^{- x / θ}]^{n - 1} \end{aligned}

Let $n = 2 m + 1$

f_{\tilde{X}} (x) = \frac{(2 m + 1)!}{m! m!} \frac{1}{θ} e^{- (m + 1) x / θ} (1 - e^{- x / θ})^{m}