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The Exponential
Distribution
Introduction
If an engineer is responsible for the quality of, say, copper wire for use in domestic wiring systems, he or she might be interested in knowing both the number of faults in a given length of wire and also the distances between such faults. While the number of faults may be analysed by using the Poisson distribution, the distances between faults along the wire may be shown to give rise to the exponential distribution defined and used in this Section.
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Prerequisites
Before starting this Section you should...
- understand the concepts of probability
- be familiar with the concepts of expectation and variance
- be familiar with the concepts of continuous distributions, in particular the Poisson distribution.
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Learning Outcomes
On completion you should be able to...
- understand what is meant by the term exponential distribution
- calculate the mean and variance of an exponential distribution
- use the exponential distribution to solve simple practical problems
HELM (2008): Section 38.3: The Exponential Distribution
1. The exponential distribution
The exponential distribution is defined by
f (t) = λe−λt^ t ≥ 0 λ a constant
or sometimes (see the Section on Reliability in 46) by
f (t) =
μ
e−t/μ^ t ≥ 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 ≥ 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) =
L
e−t^ dt =
[
− e−t
]∞
L
= 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. 65
= 0. 92
24 HELM (2008):
Workbook 38: Continuous Probability Distributions
Exercises
- The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with mean 8 minutes.
(a) Find the probability that the time interval between two successive barges is less than 5 minutes. (b) 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 fX (x) where
fX (x) =
0 (x ≤ 0) kx^2 e−λx^ (x > 0)
where λ is a parameter, the value of which is unknown, and k is a constant which depends on λ.
(a) Show that if In =
0
xne−λx^ dx then In =
n λ
In− 1 , where n > 0 and λ > 0.
Evaluate I 0 =
0
e−λx^ dx and hence find a general expression for In.
This result can be used in the rest of this question. (b) Find, in terms of λ, the value of k. (c) Find, in terms of λ, the expected value of X. (d) Find, in terms of λ, the variance of X. (e) Write down the expected value and variance of the sample mean of a sample of n inde- pendent observations on X. (f) Find, in terms of λ, the expected value of X−^1.
26 HELM (2008):
Workbook 38: Continuous Probability Distributions
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Answers
- We have μ = 8 so λ = 0. 125.
(a) The probability is
P(T < 5) =
0
- 125 e−^0.^125 t^ dt = 1 − e−^0.^125 ×^5 = 0. 4647.
(b) We require ∫ (^) ∞
t
- 125 e−^0.^125 x^ dx = e−^0.^125 t^ = 0. 95.
So − 0. 125 t = log 0. 95 and
t = −
log 0. 95
- 125
That is, 24.6 s.
(a) In =
0
xne−λx^ dx =
[
λ
xne−λx
]∞
0
n λ
0
xn−^1 e−λ^ dx =
n λ
In− 1
I 0 =
0
e−λx^ dx =
[
λ
e−λx
]∞
0
λ
hence In =
n! λn+^
(b)
0
kx^2 e−λx^ dx = 1 ⇒ kI 2 = 1 ⇒ k =
I 2
λ^3 2
(c) E(X) =
0
xfX (x) dx = kI 3 =
λ^3 2
λ^4
λ
(d) E(X^2 ) =
0
x^2 fX (x) dx = kI 4 =
λ^3 2
λ^5
λ^2
so V(X) = E(X^2 ) − {E(X)}^2 =
λ^2
λ^2
λ^2
(e) E( X¯) =
λ
V( X¯) =
nλ^2
(f) E
X
0
x
fX (x) dx − kI 1 =
λ^3 2
λ^2
λ 2
HELM (2008): Section 38.3: The Exponential Distribution
