The Exponential distribution models the *waiting time* until the next “arrival” (occurrence) of some event.

- The next text message
*arriving*on your phone - The next car accident
*arriving*in Philly

This distribution is closely related to the Poisson distribution, which models the *number* of arrivals (successes) in a fixed time period. It is also Memoryless.

Example: motivation

Assume there are 17 major earthquakes per year. Let $N_{1}$ be the number of major earthquakes in the

nextyear. Then, $N_{1}∼Pois(17)$ and so $E(N_{1})=17$.Let $N_{t}$ be the number of major earthquakes in the next $t$ years. Then, $N_{t}∼Pois(17t)$, where $t∈R_{≥0}$. Let $T$ be the waiting time from now until the next major earthquake. Then, for all $t≥0$ in $N_{t}$, the CDF of $T$ is

$F_{T}(t)=P(T≤t)=P(N_{t}>0)$We can further expand $P(N_{t}>0)=1−P(N_{t}≤0)=1−P(N_{t}=0)$, and taking the derivative we can find the PDF to be $f_{T}(t)=17e_{−17t}$, for support $t>0$.

## PDF and CDF

A Random variable $X$ has the **Exponential distribution** with **rate** parameter $λ>0$ if the PDF of $X$ is

for support $x>0$, and zero otherwise. The textbook writes $X∼Expo(λ)$. Note that the $λ$ here is different from $λ$ in the Poisson.

The CDF is given by

$F_{X}(x)=1−e_{−λx}$## Standard Exponential distribution

Just like the standard Normal and standard Uniform, we can let $λ=1$ and have $X∼Expo(1)$.

We have $E(X)=1$ and $Var(X)=1$. For a proof of the variance, see the example for the first theorem under Usefulness.

### Using a scale transformation

To go between $Expo(λ)$ and $Expo(1)$, we can simply divide by $λ$ and multiply by $λ$, respectively. If $X∼Expo(1)$, then we have $λX ∼Expo(λ)$.

To prove, let $Y=λX $. Then, we have CDF $F_{Y}(y)=P(λX ≤y)=P(X≤λy)=F_{X}(λy)=1−e_{−λy}$. This will be true as long as $λy>0$, the support.

- Then, to go to $Expo(1)$, multiply the r.v. by $λ$.
- To go to $Expo(λ)$, divide the r.v. by $λ$.

## MGF

### Standard Exponential

The MGF of the standard exponential distribution can be found using LOTUS, where $g(X)=e_{tX}$ and $X∼Expo(1)$. We have

$M_{X}(t)=E(e_{tX})=∫_{0}e_{xt}e_{−x}dx=1−t1 $To guarantee that the integral remains finite, we set $t<1$. So the open interval is given by $(−1,1)$.

### General

For a r.v. $X∼Expo(λ)$, we have $M_{X}(t)=λ−tλ $.

## General expectation and variance

Using the scale transformation property from above, we can derive the Expectation and Variance for a r.v. $Y∼Expo(λ)$. We have $E(Y)=λ1 $ and $Var(Y)=λ_{2}1 $.