11.3: The Exponential Distribution
Here, we will discuss another brand name random variable, the Exponential random variable which is related to the Poisson random variable discussed a few sections ago.
Definition: Let \( \theta > 0 \). Suppose \( X \) is a random variable who has the following density function:
\[
f(x) =
\begin{cases}
\frac{1}{\theta}e^{-x/\theta} & \text{if} ~ x \geq 0 \ \\
0 , & \text{otherwise} \\
\end{cases}
\nonumber\ \]
Then we say \( X \) is an Exponential random variable with parameter \( \theta \) and we write \( X \sim Exp(\theta) \).
Notice that the Exponential random variable is something which is not totally new for us since we actually studied the \(Exp(\frac{1}{10}) \) random variable in the last section! Additionally, recall that the point of a random variable is to model some sort of quantity under certain assumptions and so we can ask, "what does the Exponential random variable model?"
Recall that the Poisson random variable models the number of occurrences in a given interval (for instance, the number of earthquakes per week). This number was a discrete-type of random variable that follows the Poisson distribution with parameter or rate or mean \( \lambda \). However, suppose we were interested in modeling the waiting time between successive occurrences (for instance, we wished to model the time elapsed between two successive earthquakes). Then it can be shown that the waiting time between the next occurrence is an Exponential random variable with parameter \( \theta = \frac{1}{\lambda} \).
Theorem: If \( X \sim Exp(\theta) \) then
\begin{align*}
\mathbb{E}[X] = \theta ~~\text{and} ~~ \mathbb{V}ar[X] = \theta^2
\end{align*}
Theorem: If \( X \sim Exp(\theta) \) then the cumulative distribution of \(X\) is given by:
\[
F(x) =
\begin{cases}
0 & \text{if} ~ x < 0 \\
1 - e^{-x/\theta}, & \text{if} ~ x \geq 0 \\
\end{cases}
\nonumber\ \]
Customer arrive in a certain shop according to a Poisson process with a mean rate of \( \lambda = 1/3 \) customers per minute. What is the probability that the shopkeeper will have to wait more than five minutes for the arrival of the first customer?
- Answer
-
According to the interpretation of the Exponential random variable, if we have a Poisson process with rate \( \lambda \), then the waiting time for the next customer is going to be an exponential random variable with parameter \( \theta = \frac{1}{\lambda}\). Since customers arrive to the store with rate \(\lambda = 1/3 \), then the waiting time for the next customer is going to be distributed as an Exponential random variable with parameter \( \theta = \frac{1}{\lambda} = \frac{1}{1/3} = 3 \). So if \( X \) models the waiting time for the next customer, then it follows that \( X \sim Exp(3) \). That is, \(X\) has the following density:
\[
f(x) =
\begin{cases}
\frac{1}{3}e^{-x/3} & \text{if} ~ x \geq 0 \ \\
0 , & \text{otherwise} \\
\end{cases}
\nonumber\ \]Hence
\begin{align*}
P(\text{the waiting time is more than} ~ 5 ~ \text{minutes} ) &= P(X > 5) \\
&= \int_{5}^{\infty} \frac{1}{3}e^{-x/3} ~ dx \\
&= e^{-5/3} \\
& \approx 0.1889
\end{align*}Note that we can achieve the same answer by utilizing the cdf:
\begin{align*}
P(\text{the waiting time is more than} ~ 5 ~ \text{minutes} ) &= P(X > 5) \\
&= 1 - P(X \leq 5) \\
&= 1 - F(5) \\
&= 1 - (1 - e^{-5/3}) \\
&= e^{-5/3} \\
& \approx 0.1889
\end{align*}
Example: Suppose that a certain type of electronic component has an exponential distribution with a mean life of 500 hours. If $X$ denotes the lifetime of this component then find \( P(X > 900 | X > 300) \).
- Answer 1
-
Let \( X \) model the lifetime of the component. From the problem, we know that \( X \sim Exp(\theta)\). The problem tells us the mean is 500 hours and for an Exponential random variable, the mean and parameter are the same thing! So we know that \( X \sim Exp(500)\). That is, the density is given by
\[
f(x) =
\begin{cases}
\frac{1}{500}e^{-x/500} & \text{if} ~ x \geq 0 \ \\
0 , & \text{otherwise} \\
\end{cases}
\nonumber\ \]Thus we have
\begin{align*}
P(X > 900 | X > 300) &= \frac{P\bigg( \{X > 900 \} \cap \{X > 300 \} \bigg)}{P(X > 300)} \\
&= \frac{P(X > 900)}{P(X>300)} \\
&= \frac{\int_{900}^{\infty} \frac{1}{500}e^{-x/500} ~ dx }{\int_{300}^{\infty} \frac{1}{500}e^{-x/500} ~ dx } \\
&= \frac{e^{-900/500}}{e^{-300/500}} \\
& = e^{-6/5} \\
& \approx 0.3012
\end{align*}
- Answer 2
-
Or, as noted in the last problem, we can use the cdf to our advantage to avoid the integration.
Alternative Solution:
\begin{align*}
P(X > 900 | X > 300) &= \frac{P\bigg( \{X > 900 \} \cap \{X > 300 \} \bigg)}{P(X > 300)} \\
&= \frac{P(X > 900)}{P(X>300)} \\
&= \frac{1 - P(X \leq 900)}{1 - P(X \leq 300)} \\
&= \frac{1 - F(900)}{1 - F(300)} \\
&= \frac{1 - (1 - e^{-900/500})}{1 - (1 - e^{-300/500})} \\
&= \frac{e^{-900/500}}{e^{-300/500}} \\
& = e^{-6/5} \\
& \approx 0.3012
\end{align*}