Topic 3-4 Exponential distribution
Definition: X is said to have an exponential distribution (or a negative exponential distribution) with parameter 
(positive) if its density function is
 e x when x  0
f ( x)  
when x  0
 0
Normalization check:



f(x)dx =


0
e-xdx =  e x

0
=1
Since actual problems usually involve P(X  x) or P(X  x), it is useful to derive the CDF of an exponentially
distributed r.v. once and for all, to avoid having to perform integration again:
F(x) =

x
0
e-xdx =  e x
x
0
= 1 – e-x for positive x, and F(x) = 0 otherwise.
The mean and variance of an exponentially distributed r.v. are calculated as follows:

1
E(X) =
xe-xdx =
(using integration by parts)


0
Check on units: since e is raised to -x which must be a dimensionless number,  must have units of (1/x), e.g.,
1
1
if x is in seconds,  would be in sec–1. Hence
must have the same units as x. Hence E(X) =
does have the


correct units.
To calculate Var(X), we calculate first
E(X2) =


0
2
x2e-xdx = 2/2, and then
Var(X) = E(X ) - (E(X))2 = 2/2 – 1/2 = 1/2
The exponential r.v. is useful in modeling something that is always positive and not likely to be very large, e.g.
the lifetime of light bulbs, length of cracks, etc.
The exponential distribution is closely related to the Poisson distribution: if a Poisson process has mean
occurrence rate , then the (random) time between two successive occurrences has an exponential distribution
with parameter . This can be established by asking what is the probability of zero occurrence during a duration
of length x. The answer is e-x by using  = x in the Poisson model. But the underlined event is equivalent to
“first occurrence happens at X > x”, where X is the instance when the first occurrence takes place. Hence
P(X  x) = FX(x) = 1 – e-x
 fX(x) = F’X(x) = e–x
See Ang and Tang, p.120-121 for a nice discussion on this.
Solved problem
Problem 3-4-1
The amount of delay time for a given flight is exponentially distributed with a mean of 0.5 hour. Ten passengers
on this flight need to take a subsequent connecting flight. The scheduled connection time is either 1 or 2 hours
depending on the final destination. Suppose 3 and 7 passengers are associated with these connection times,
respectively.
(a)
Suppose John is one of the 10 passengers needing a connection. What is the probability that he will miss
his connection? (ans. 0.053)
(b)
Suppose he met Mike on the plane, and Mike also needs to make a connection. However, Mike is going
to another destination and thus has a different connection time from John's. What is the probability that
both John and Mike will miss their connections? (ans. 0.018)
(c)
A friend of John’s, named Mary, happens to live close to the airport where John makes his connection.
She would like to take this opportunity to meet John at the airport. Suppose she has already waited for
30 minutes beyond John's scheduled arrival time. What is the probability that John will miss his
connection so that they could have a leisurely dinner together? Assume John’s scheduled connection
time is 1 hour in this part (c). (ans. 0.368)
Solution:
Let J1 and J2 denote the events that John’s scheduled connection time is 1 and 2 hours, respectively, where P(J 1)
= 0.3 and P(J2) = 0.7. Also, let X be the delay time of the flight in hours. Note that
P(X > x) = 1 – P(X  x) = 1 – F(x) = 1 – (1 – e-x/0.5)
= e-x/0.5
(a) Let M denote the event that John misses his connection, i.e. the flight delay time exceeded his scheduled
time for connection. Using the theorem of total probability,
P(M) = P(M | J1)P(J1) + P(M | J2)P(J2)
= P(X > 1)0.3 + P(X > 2)0.7
= e– 1/0.5  0.3 + e– 2/0.5  0.7
= 0.1353352830.3 + 0.0183156390.7  0.053
(b) Regardless of whether John has a connection time of 1 hour and Mike has 2, or the opposite, for them to
both miss their connections the flight must experience a delay of more than two hours, and the probability of
such an event is
P(X > 2) = e– 2/0.5  0.018
(c) Since Mary has already waited for 30 minutes, the flight will have a delay time of at least 0.5 hours when it
arrives. Hence the desired probability is
P(X > 1 | X > 0.5)
= P(X > 1 and X > 0.5) / P(X > 0.5)
= P(X > 1) / P(X > 0.5)
= e–1/0.5 / e–0.5/0.5 = e–2 / e–1 = 1/e  0.368
Exercises
Exercise 3-4-1
A city is subject to three types of natural hazards: (i) floods occur with mean rate of once in ten years; (ii)
earthquakes occur with a mean rate of once in twenty years; and (iii) tornadoes occur with a mean rate of once
in five years. Suppose the occurrence of floods, earthquakes and tornadoes can each be described by a Poisson
process.
(a)
What is the probability that the city will have two incidences of natural hazards (any combination of
floods / earthquakes / tornadoes) during the next year?
(b)
What is the probability that this city will be free from happenings of natural hazards in two of the next
three years?
(c)
What is the probability that the time between two successive earthquakes in this city will be at least ten
years apart?
Exercise 3-4-2
The traffic on the one way main street shown below may be satisfactorily described by a Poisson process with
an average rate of arrival  = 10 cars per minute. A driver (indicated by the box) on the side street is waiting to
cross the main street. He will cross as soon as he finds a gap of 15 seconds.
One way
(a)
Determine the probability, p, that a gap will be larger than 15 seconds. (ans. 0.082)
(b)
What is the probability that the driver will cross at the 4th gap?
(c)
Determine the mean number of gaps he has to wait until crossing the main street.
(d)
What is the probability that he will cross within the first 4 gaps?