Title: Default for 38.3 The Exponential Distribution
Course: HELM Workbooks 2023-2024

The exponential distribution

The exponential distribution is defined by\[  f(t)=\lambda e^{-\lambda t}\qquad t\ge 0\qquad \lambda \text{ a constant} \]or sometimes by\[ f(t)=\frac{1}{\mu} e^{- t/\mu}\qquad t\ge 0\qquad \mu \text{ a constant} \]The  advantage of this latter representation is that it may be shown that the mean of the distribution is  \(\mu\).

Example

The lifetime \(T\) (years) of an electronic component is a continuous random variable with a probability density function given by \[ f(t)=e^{-t}\qquad t\ge 0 \qquad (\text{i.e. } \lambda=1 \text{ or } \mu=1)\] Find the lifetime \(L\) which a typical component is \(60\%\) certain to exceed.  If five components are sold to a manufacturer, find the probability that at least one of them will have a lifetime less than \(L\) years.

Solution

We require  \(\text{P}(T>L)=0.6\).  We know that this probability is given by the relationship \[ \text{P}(T>L)=\int_L^{\infty} e^{-t}\,dt=\left[\phantom{\dfrac{1}{1}}\!\!\!-e^{-t}\right]_L^{\infty}=e^{-L}\] Solving \(e^{-L}=0.6\)  for the least value of \(L\) we obtain \(L=0.51\)   years.

Assuming that the lifetime of each component is independent we have \[\text{P} (\text{at least one component has a lifetime less than  0.51 years})\] \[ =1-\text{P} (\text{no component has a lifetime less than  0.51 years})\] \[ =1-0.6^5 =0.92\]

Exercise

Commonly, car cooling systems are controlled by electrically driven fans.  Assuming that the lifetime \(T\) in hours of a particular make of fan can be modelled by an exponential distribution with \(\displaystyle{\lambda_1 = [value]}\)  find the proportion of fans which will give at least 10000 hours service.

\(\text{P}(T > 10000) = \)

If the fan is redesigned so that its lifetime may be modelled by an exponential distribution with \(\displaystyle{\lambda_2 = [value]}\). Would you expect more or fewer fans to give at least \(10000\) hours service?
Feedback: Let \(T\) be the random variable for the lifetime in hours of a cooling fan, then \(T \sim \text{Exp}(\lambda)\). Under the original design, the rate parameter is \(\lambda = \lambda_1 = [value]\). The cumulative distribution function for this exponential distribution is \(F_T(t) = \text{P}(T \leq t) = 1 - e^{-\lambda_1 t}\). The probability that a fan will give at least 10000 hours of service is: \[\begin{align*}\text{P}(T>10000) &= 1 - \text{P}(T \leq 10000)\\ &= 1 - (1 - e^{[value] 10000})\\ &= e^{[value]}\\ &= [value] ~~(\text{to 4 d.p.})\end{align*}\] For the redesigned cooling fan, the new rate parameter is \(\lambda = \lambda_2 = [value]\). The probability that this type of fan will give at least 10000 hours of service can be calculated using the above method using the new rate parameter, giving \(\text{P}(T>10000) = [value] ~~(\text{to 3 s.f.})\). From comparing the two probabilities, we expect that \({\it fewer}\) of the redesigned cooling fans to give more than 10000 hours service.

Exercise

The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with mean \([value]\) minutes.

1. What is the probability, to 4 decimal places, that the time interval between two successive barges is less than \([value] \) minutes.
\(\text{P}(\text{less than $[value]$ minutes})\) =
2. Find the time interval \( t \), to 2 decimal places, so that we can be \([value]\)% sure that the time interval between two successive barges will be greater than \(t \).
\(t =\) minutes
Feedback: Let \(T\) denote the random variable between two successive barges so that \(T \sim \text{Exp}(\lambda\). The expectation of an exponentially distributed random variable is \(\text{E}(X) = \dfrac{1}{\lambda}\). With the mean interval time being \([value]\) minutes, it follows that the rate parameter is \(\lambda = \dfrac{1}{\text{E}(X)} = \displaystyle [value]\). 1. The probability for the time between successive barges being less than \([value]\) minutes is: \[\begin{align*} \text{P}(T \leq [value]) &= F_T([value])\\ &= 1 - e^{-[value] \times [value]}\\ & = [value] ~~(\text{to 4 d.p.}).\end{align*}\] 2. Using the complement rule, the probability that the time interval between successive barges exceeding \(t\) minutes is \(\text{P}(T>t) = 1 - \text{P}(T \leq t) = e^{-\lambda t}\). The value for \(t\) where this probability is \([value]\)% can be determined by: \[\begin{align*}e^{-[value]t} &= \dfrac{[value]}{100}\\ -[value]t &= \ln\left( \dfrac{[value]}{100} \right)\\ t & = [value] \ln\left( \dfrac{[value]}{100} \right) \end{align*}\] Hence, the required time interval is \(t = [value]\) minutes to 2 decimal places.

Exercise

It is believed that the time \( X \) for a worker to complete a certain  task  has probability density function \( f_{X}(x) \) where \[  f_{X}(x) = \left\{ \begin{array}{cc} 0 & (x \leq 0) \\ k x^{2}e^{-\lambda x} & (x>0) \end{array} \right. \]   where  \(\lambda \) is a parameter, the value of which is unknown, and \( k \) is a  constant which depends on \( \lambda. \)

1. Show that if \(\displaystyle{ I_{n} = \int_{0}^{\infty} x^{n} e^{-\lambda x}\,dx} \)  then \( \displaystyle{I_{n} = \frac{n}{\lambda} I_{n-1},}\) where \(n>0 \) and \(\lambda>0. \)
Evaluate \(\displaystyle{ I_{0} = \int_{0}^{\infty} e^{-\lambda x}\; dx}\)  and hence find a general expression for \( I_{n}. \)
This result can be used in the rest of this question.
\( I_0\) =
\( I_n \)=
2. Find, in terms of \( \lambda, \) the value of \( k. \)

3. Find, in terms of \(\lambda, \) the expected value of \( X. \)
4. Find, in terms of \( \lambda, \) the variance of \( X. \)
5. Write down the expected value and variance of the sample mean of a sample  of \( n \) independent observations on  \(X\).
\(\text{E}(\bar{X})\)= \(\text{V}(\bar{X})\)=
6. Find, in terms of \( \lambda, \) the expected value of \( X^{-1}. \)
Feedback: (a)\[ I_{n} = \int_{0}^{\infty} x^{n} e^{-\lambda x}\,dx = \left[ - \frac{1}{\lambda} x^{n} e^{-\lambda x} \right]_{0}^{\infty} + \frac{n}{\lambda} \qquad\quad \int_{0}^{\infty} x^{n-1} e^{-\lambda}\,dx = \frac{n}{\lambda} I_{n-1} \] \[ I_{0} = \int_{0}^{\infty} e^{-\lambda x}\,dx = \left[ - \frac{1}{\lambda} e^{-\lambda x} \right]_{0}^{\infty} = \frac{1}{\lambda} \text{ hence } \quad I_{n} = \frac{n!}{\lambda^{n+1} } .\] (b)\[ \int_{0}^{\infty} k x^{2} e^{-\lambda x} \,dx = 1 \Rightarrow k I_{2} = 1 \Rightarrow k = \frac{1}{I_{2}} = \frac{\lambda^{3}}{2} \] (c)\[ \text{E}(X) = \int_{0}^{\infty} x f_{X}(x)\,dx = kI_{3} = \frac{\lambda^{3}}{2} \frac{6}{\lambda^{4}} = \frac{3}{\lambda} \] (d)\[ \text{E}(X^{2}) = \int_{0}^{\infty} x^{2} f_{X}(x)\,dx = kI_{4} = \frac{\lambda^{3}}{2} \frac{24}{\lambda^{5}} = \frac{12}{\lambda^{2}}\] so\( \text{X} = \text{E}(X^{2}) - \{\text{E}(X)\}^{2} = \frac{12}{\lambda^{2}} - \frac{9}{\lambda^{2}} = \frac{3}{\lambda^{2}}\) (e)\[ \text{E}(\bar{X}) = \frac{3}{\lambda}\qquad\quad \text{\bar{X}} = \frac{3}{n \lambda^{2}} \] (f)\[ \text{E}\left(\frac{1}{X}\right) = \int_{0}^{\infty} \frac{1}{x} f_{X}(x)\,dx - kI_{1} = \frac{\lambda^{3}}{2} \frac{1}{\lambda^{2}} = \frac{\lambda}{2}\]

Exercise

Which of the following plots represents the probability distribution function for the exponential distribution?

[Image: Plot of the probability distribution function]

Select answer from drop-down options

Exercise

Let \(X\) be a random variable that follows an exponential distribution with rate parameter \(\lambda = [value]\). What is the expectation, variance, median and mode of \(X\). Give all answers as either a whole number or a fraction, except for the median which should be calculated to 3 decimal places.

- \(\text{E}(X) = \)
- \(\text{V}(X) = \)
- \(\text{median}(X) = \)
- \(\text{mode}(X) = \)
Feedback: If \(X \sim \text{Exp}(\lambda)\) with \(\lambda = [value]\), then: - \(\text{E}(X) = \dfrac{1}{\lambda} = [value]\) - \(\text{V}(X) = \dfrac{1}{\lambda^2} = [value]\) - The median \(m\) for \(X\) is the solution of \(\text{P}(X \leq m) = \dfrac{1}{2}\): \[\begin{align*} 1 - e^{-\lambda m} & = \dfrac{1}{2}\\ e^{-\lambda m} & = \dfrac{1}{2}\\ -\lambda m& = -\ln 2\\ m = \dfrac{1}{\lambda} \ln 2\end{align*}\] With \(\lambda = [value]\), the median value for \(X\) is \([value]\) to 3 decimal places. - The probability density function for the exponential distribution is an exponentially decaying function on the range \(x \geq 0\), therefore the modal value for \(X\) is at \([value]\).

Exercise

You arrive at the bus stop and the time (in minutes) it takes for the next bus to arrive follows an Exponential distribution with rate \(\displaystyle[value]\). Calculate the following probabilities to \(4\) decimal places.

1. You wait less than \(5\) minutes?
2. You wait longer than \(7\) minutes?
3. You wait for between \(5\) and \(7\) minutes?
Feedback: Let \(T\) be the random variable for the wait time at the bus stop, then \(T \sim \text{Exp}(\lambda = [value])\). 1. The probability of waiting less than \(5\) minutes is: \[\text{P}(T \leq 5) = F_T(5) = 1 - e^{-\lambda 5} = [value]~ (\text{to 4 d.p.}).\] 2. The probability of waiting more than \(7\) minutes is: \[\text{P}(T > 7) = 1 - F_T(7) = e^{-\lambda 7} = [value]~ (\text{to 4 d.p.}).\] 3. The probability of waiting between \(5\) and \(7\) minutes is: \[\begin{align*}\text{P}(5 \leq T \leq 7) &= F_T(7) - F_T(5)\\ &= \left[1 - e^{-\lambda 7}\right] - \left[1 - e^{-\lambda 5}\right]\\ &= e^{-\lambda 5} - e^{-\lambda 7}\\ &= [value]~ (\text{to 4 d.p.}).\end{align*}\] An alternative approach to evaluating this probability can be achieved by noting that \(\text{P}(5 \leq T \leq 7) = 1 - \text{P}(T \leq 5) - \text{P}(T > 7).\)

Exercise

1. Let \(X\) be an Exponentially distributed random variable with rate parameter \(\lambda\). You are told that the probability that random variable takes the value \(x\) or less is \(p\), \(\mathbb{P}(X<x) = p\). Find an expression for \(x\) in terms of \(p\).
\(x=\)
2. The half-life for Radon-222 is \(3.8235\). If the time it takes for an atom of Radon-222 to decay follows an Exponential distribution, what is the value of its rate parameter? Give your answer to 4 d.p.
\( \lambda\)=
3. What is the probability that it takes longer than \verb+days+ days for an atom of Radon-222 to decay? Give your answer to 3 d.p.
Feedback: a) \(x = \frac{-\ln(1-p)}{\lambda}\) \begin{eqnarray*} 1-exp(-\lambda x) & = & p \\ exp(-\lambda x) & = & 1-p \\ -\lambda x & = & \ln(1-p) \\ x & = & \frac{-\ln(1-p)}{\lambda} \end{eqnarray*} b) From part a, \begin{eqnarray*} \mathtt{[value]} & = & \frac{-\ln(1 - 0.5)}{\lambda} \\ \lambda & = & \frac{-\ln(1 - 0.5)}{\mathtt{[value]}} \end{eqnarray*} which gives \(\lambda = [value]\) to 4pd. c) \(\mathbb{P}(X > \mathtt{[value]}) = \exp(- \mathtt{[value]} \lambda) = \mathtt{[value]}\) \end{itemize}