8.3 The Exponential Density:
The exponential random variable can be used to describe the life time of a machine,
industrial product and Human being. Also, it can be used to describe the waiting time
of a customer for some service.
Exponetial Probability Density Function:
A random variable X taking values in [0,  ] has the exponential
probability density function f (x) if
f ( x) 
where
1

x
e  , 0 x ,
  0. .
f(x)
The graph of f (x) is
0
x
Properties of Exponential Density Function:
Let X be the random variable with the exponential density function
f (x ) and the parameter  . Then
1
1.
P( X  x0 )  1  e
 x0

,
x0  0 .
for any
2.

E( X )   x 
0
x
1
 e  dx  

and

Var ( X )   x    
2
0
1

x
 e  dx  2 .
[derivation:]
x
x0
x
x0

x
0
1 
x
P( X  x0 )    e dx   e  d   e  y dy

 0
0
0
 e
x0
y 
0
 e
 x0

 
  e0  1  e
x
(y  )

 x0

The derivation of 2 is left as exercise.
Note: S ( x0 )  P( X  x0 )  1  P( X  x0 )  e
 x0

is called the
survival function.
Example:
Let X represent the life time of a washing machine. Suppose the average lifetime for
this type of washing machine is 15 years. What is the probability that this washing
2
machine can be used for less than 6 years? Also, what is the probability that this
washing machine can be used for more than 18 years?
[solution:]
X has the exponential density function with   15 (years). Then,
P( X  6)  1  e
6
15
 0.3297 and P( X  18)  e
18
15
 0.3012
Thus, for this washing machine, it is about 30% chance that it can be used for quite a
long time or a short time.
Relationship Between Poisson and Exponential Random
Variable:
Let Y be a Poisson random variable representing the number of
occurrences in an time interval of length t with the probability
distribution
where

e u  i
P(Y  i) 
,
i!
is the mean number of occurrences in this time interval.
Then, if X represent the time of one occurrence, X has the exponential
density function with mean E ( X )   
1
 (t).
The intuition of the above result is as follows. Suppose the time interval is [0,1] (in
hour) and   4 . Then, on the average, there are 4 occurrences during 1 hour period.
Thus, the mean time for one occurrence is  
1


1
(hour). The number of
4
occurrences can be described by a Poisson random variable (discrete) with mean 4
while the time of one occurrence can be described by an exponential random variable
(continuous) with mean
1
.
4
3
Example:
Suppose the average number of car accidents on the highway in two days is 8. What is
the probability of no accident for more than 3 days?
[solutions:]
The average number of car accidents on the highway in one day is
mean time of one occurrence is
8
 4 . Thus, the
2
1
(day) .
4
Let Y be the Poisson random variable with mean 4 representing the number of car
accidents in one day while X be the exponential random variable with mean
1
(day)
4
representing the time of one accident occurrence. Thus,
P(No accident for more than 3 days)  P( the time of one occurrence larger tha n 3)
 P( X  3)  e
Online Exercise:
Exercise 8.3.1
4
3
1
4
 e 12  0