The exponential distribution

1 The exponential distribution

The exponential distribution is defined by

$f (t) = λ e^{- λ t} t \geq 0 λ$ a constant

or sometimes (see the Section on Reliability in HELM booklet 46) by

$f (t) = \frac{1}{μ} e^{- t ∕ μ} t \geq 0 μ$ a constant

The advantage of this latter representation is that it may be shown that the mean of the distribution is $μ$ .

Example 3

The lifetime $T$ (years) of an electronic component is a continuous random variable with a probability density function given by

$f (t) = e^{- t} t \geq 0$ (i.e. $λ = 1$ or $μ = 1$ )

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 $P (T > L) = 0.6$ . We know that this probability is given by the relationship

$P (T > L) = \int_{L}^{\infty} e^{- t} d t = {[- e^{- t}]}_{L}^{\infty} = e^{- L}$

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

$P$ (at least one component has a lifetime less than 0.51 years)

$= 1 - P$ (no component has a lifetime less than 0.51 years)

$= 1 - 0 . 6^{5}$

$= 0.92$

Task!

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 $λ = 0.0003$ find the proportion of fans which will give at least 10000 hours service. If the fan is redesigned so that its lifetime may be modelled by an exponential distribution with $λ = 0.00035$ , would you expect more fans or fewer to give at least 10000 hours service?

We know that $f (t) = 0.0003 e^{- 0.0003 t}$ so that the probability that a fan will give at least 10000 hours service is given by the expression

$P (T > 10000) = \int_{10000}^{\infty} f (t) d t = \int_{10000}^{\infty} 0.0003 e^{- 0.0003 t} d t = - {[e^{- 0.0003 t}]}_{10000}^{\infty} = e^{- 3} \approx 0.0498$

Hence about 5% of the fans may be expected to give at least 10000 hours service. After the redesign, the calculation becomes

$P (T > 10000) = \int_{10000}^{\infty} f (t) d t = \int_{10000}^{\infty} 0.00035 e^{- 0.00035 t} d t = - {[e^{- 0.00035 t}]}_{10000}^{\infty} = e^{- 3.5} \approx 0.0302$

and so only about 3% of the fans may be expected to give at least 10000 hours service.

Hence, after the redesign we expect fewer fans to give 10000 hours service.

Exercises

The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with mean 8 minutes.
1. Find the probability that the time interval between two successive barges is less than 5 minutes.
2. Find a time interval $t$ such that we can be 95% sure that the time interval between two successive barges will be greater than $t .$
It is believed that the time X for a worker to complete a certain task has probability density function f X ( x ) where

$f_{X} (x) = \{\begin{matrix} 0 & (x \leq 0) \\ k x^{2} e^{- λ x} & (x > 0) \end{matrix}$

where $λ$ is a parameter, the value of which is unknown, and $k$ is a constant which depends on $λ .$
1. Show that if $I_{n} = \int_{0}^{\infty} x^{n} e^{- λ x} d x$ then $I_{n} = \frac{n}{λ} I_{n - 1},$ where $n > 0$ and $λ > 0.$
  Evaluate $I_{0} = \int_{0}^{\infty} e^{- λ x} d x$ and hence find a general expression for $I_{n} .$
  
  This result can be used in the rest of this question.
2. Find, in terms of $λ,$ the value of $k .$
3. Find, in terms of $λ,$ the expected value of $X .$
4. Find, in terms of $λ,$ the variance of $X .$
5. Write down the expected value and variance of the sample mean of a sample of $n$ independent observations on $X .$
6. Find, in terms of $λ,$ the expected value of $X^{- 1} .$

We have μ = 8 so λ = 0 . 125 .
1. The probability is
  $P (T < 5) = \int_{0}^{5} 0.125 e^{- 0.125 t} d t = 1 - e^{- 0.125 \times 5} = 0.4647 .$
2. We require
  $\int_{t}^{\infty} 0.125 e^{- 0.125 x} d x = e^{- 0.125 t} = 0.95 .$
  
  So $- 0.125 t = log 0.95$ and
  
  $t = - \frac{log 0.95}{0.125} = 0.4103 .$
  
  That is, 24.6 s.
1. $I_{n} = \int_{0}^{\infty} x^{n} e^{- λ x} d x = {[- \frac{1}{λ} x^{n} e^{- λ x}]}_{0}^{\infty} + \frac{n}{λ} \int_{0}^{\infty} x^{n - 1} e^{- λ} d x = \frac{n}{λ} I_{n - 1}$
  $I_{0} = \int_{0}^{\infty} e^{- λ x} d x = {[- \frac{1}{λ} e^{- λ x}]}_{0}^{\infty} = \frac{1}{λ}$ hence $I_{n} = \frac{n!}{λ^{n + 1}} .$
2. $\int_{0}^{\infty} k x^{2} e^{- λ x} d x = 1 \Rightarrow k I_{2} = 1 \Rightarrow k = \frac{1}{I_{2}} = \frac{λ^{3}}{2}$
3. $E (X) = \int_{0}^{\infty} x f_{X} (x) d x = k I_{3} = \frac{λ^{3}}{2} \frac{6}{λ^{4}} = \frac{3}{λ}$
4. $E (X^{2}) = \int_{0}^{\infty} x^{2} f_{X} (x) d x = k I_{4} = \frac{λ^{3}}{2} \frac{24}{λ^{5}} = \frac{12}{λ^{2}}$
  so $V (X) = E (X^{2}) - {E (X)}^{2} = \frac{12}{λ^{2}} - \frac{9}{λ^{2}} = \frac{3}{λ^{2}}$
5. $E (\bar{X}) = \frac{3}{λ} V (\bar{X}) = \frac{3}{n λ^{2}}$
6. $E (\frac{1}{X}) = \int_{0}^{\infty} \frac{1}{x} f_{X} (x) d x - k I_{1} = \frac{λ^{3}}{2} \frac{1}{λ^{2}} = \frac{λ}{2}$