Exponential distribution

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 next year. Then, $N_1\sim \text{Pois}(17)$ and so $\mathbb{E}(N_1)=17$.

Let $N_t$ be the number of major earthquakes in the next $t$ years. Then, $N_t\sim \text{Pois}(17t)$, where $t\in {\mathbb{R}}_{\ge 0}$. Let $T$ be the waiting time from now until the next major earthquake. Then, for all $t\ge 0$ in $N_t$, the CDF of $T$ is

$$F_T(t)=P(T\le t)=P(N_t>0)$$

We can further expand $P(N_t>0)=1-P(N_t\le 0)=1-P(N_t=0)$, 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 $\lambda >0$ if the PDF of $X$ is

$$f_X(x)=\lambda e^{-\lambda x}$$

for support $x>0$, and zero otherwise. The textbook writes $X\sim \text{Expo}(\lambda )$. Note that the $\lambda$ here is different from $\lambda$ in the Poisson.

The CDF is given by

$$F_X(x)=1-e^{-\lambda x}$$

Standard Exponential distribution

Just like the standard Normal and standard Uniform, we can let $\lambda =1$ and have $X\sim \text{Expo}(1)$.

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

Using a scale transformation

To go between $\text{Expo}(\lambda )$ and $\text{Expo(1)}$, we can simply divide by $\lambda$ and multiply by $\lambda$, respectively. If $X\sim \text{Expo}(1)$, then we have $\frac{X}{\lambda }\sim \text{Expo}(\lambda )$.

To prove, let $Y=\frac{X}{\lambda }$. Then, we have CDF $F_Y(y)=P\left(\frac{X}{\lambda }\le y\right)=P(X\le \lambda y)=F_X(\lambda y)=1-e^{-\lambda y}$. This will be true as long as $\lambda y>0$, the support.

• Then, to go to $\text{Expo(1)}$, multiply the r.v. by $\lambda$.
• To go to $\text{Expo}(\lambda )$, divide the r.v. by $\lambda$.

MGF

Standard Exponential

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

$$M_X(t)=\mathbb{E}(e^{tX})={\int }_0^{\infty }{e}^{xt}{e}^{-x}dx=\frac{1}{1-t}$$

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. $X\sim \text{Expo}(\lambda )$, we have $M_X(t)=\frac{\lambda }{\lambda -t}$.

General expectation and variance

Using the scale transformation property from above, we can derive the Expectation and Variance for a r.v. $Y\sim \text{Expo}(\lambda )$. We have $\mathbb{E}(Y)=\frac{1}{\lambda }$ and $\text{Var}(Y)=\frac{1}{{\lambda }^2}$.