Download Sums of exponential random variables and the Poisson distribution

3.4 Sums of exponential random variables and the Poisson distribution.
Sums of independent exponential random variables. Recall, a random variable is an exponential random
variable if it has density function of the form f(t) = e-t for t  0 and f(t) = 0 for t < 0. Exponential random
variables are important in stochastic processes since they are used to model the time between events, such
as the arrival of customers at a bank. Therefore, a sum of n exponential random variables is used to model
the time it takes for n occurrences of an event, such as the time it takes for n customers to arrive at a bank.
If the exponential random variables are independent and identically distributed the distribution of the sum
has an Erlang distribution.
Definition 1. A random variable has an Erlang distribution if it has a pdf of the form f(t) =
ntn-1e-t
(n-1)!
for
t  0 and f(t) = 0 for t < 0 where n is a positive integer and  is a positive real number.
Proposition 1. Let S = T1 + T2 + ... + Tn where T1, T2, ... , Tn are independent exponential random variables
each with density function f(t) = e-t for t  0 and f(t) = 0 for t < 0. Then S has density function h(t) of the
ntn-1e-t
form h(t) =
for t  0 and h(t) = 0 for t < 0. Furthermore, the mean of S is n/.
(n-1)!
Proof. This is proved by induction on n. It is clearly true for n = 1. We shall show it is true for n + 1 if it
is true for n. Let Sn = T1 + T2 + ... + Tn and Sn+1 = T1 + T2 + ... + Tn+1. We are assuming the pdf of Sn is
hn(t) =
ntn-1e-t
(n-1)!
. The pdf of Sn+1is
t
t
hn+1(t) =
 hn(r)f(t-r) dr =

0
=
t
n n-1 -r
n+1 -t
  r e e-(t-r) dr =  e  rn-1 dr
(n-1)! 
 (n-1)!
0
0
n+1tne-t
n!
which proves the first result. The fact the mean of S is n/ follows from the fact that the mean of a sum is
the sum of the means and the mean of an exponential random variable is 1/.
Example 1. Suppose customers arrive at an bank at a rate of 30 per hour and the times between arrivals are
exponentially distributed and independent. Let S2 = time (in min) until the second customer arrives. What
is the density function h(t) of S2?
Solution. S2 = T1+ T2 where T1 and T2 are independent and exponential with  = 1/2 min-1. So S2 has an
ntn-1e-t
Erlang distribution h(t) =
with n = 2 and  = 1/2. So h(t) = te-t/2/4 for t > 0.
(n-1)!
Poisson random variables. Let Sn = T1 + T2 + ... + Tn where T1, T2, ... , Tn are independent exponential
random variables each with density function f(t) = e-t for t  0 and f(t) = 0 for t < 0. To find the
ntn-1e-t
probability that S is between a and b one would have to integrate h(t) =
between a and b. This
(n-1)!
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integral can be done using integration by parts, but it takes some effort. It is usually easier to translate
questions about S into questions about a related Poisson random variable. We let
N(t) = max{n: Sn  t}
= the number of occurrences that have occurred by time t
For each t this is a random variable. We want to calculate Pr{N(t) = n} for n = 0, 1, 2. ..... One has
Theorem 2. Pr{N(t) = n} =
ntne-t
n!
Proof. Note that N(t) = n if and only if Sn  t and Sn+1 > t. So the event {N(t) = n} can be written as
{N(t) = n} = {Sn  t and Sn+1 > t} = {Sn  t} – {Sn+1  t}
where – is the difference between two sets, i.e A – B = {x: x  A and x  B} is the set of objects that are in
A but not in B. If B is contained in A then Pr{A - B } = Pr{A} - Pr{B} since A is the union of B and A - B
which are disjoint. The set {Sn+1  t} is contained in {Sn  t}, because if Sn+1  t then Sn  t. So
Pr{N(t) = n} = Pr{Sn  t} – Pr{Sn+1  t}
(1)
t
t
0
0
nsn-1e-s
n+1sne-s
= 
ds - 
ds
(n-1)!

 n!
In the second integral we integrate by parts letting u = sn/n! and dv = n+1e-s ds. Then du = sn-1/(n-1)! and
v = - ne-s, so that
t
t
0
0
n+1 n -s
n n - s t
n n-1 -s
  s e ds = -  s e  |
  s e ds
+
 n!  s = 0
 n!
 (n-1)!
= -
ntne-t
n!
t
+
  s e ds
 (n-1)!
n n-1 -s
0
Substituting into (1) proves the theorem. //
N(t) is an example of a Poisson random variable. Recall from section 1.2 that a random variable M is a
Poisson random variable if it only assumes values that are non-negative integers and there is a  > 0 such
ne-
that Pr{M = n} =
for n = 0, 1, 2, .....
n!
Example 2. Suppose customers arrive at an bank at a rate of 30 per hour. Suppose the times between
arrivals are exponentially distributed and independent.
a.
What is the probability that exactly 5 customers arrive in 10 minutes?
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b.
What is the probability that at least 5 customers arrive in 10 minutes.
c.
What is the probability that the 5th customer arrives between 9 and 10 minutes.
Solution. In this case  = 1/2 if we measure time in minutes. Let N(t) be the number of customers that
have arrived by time t. For part a we want
Pr{N(10) = 5} =
(1/2)5 105e-10/2
55e-5
=
= (54e-5)/24 = 625e-5/24  0.175
5!
120
For part b we want
50e-5
51e-5
52e-5
53e-5
54e-5
Pr{N(10)  5} = 1 - Pr{N(10)  4} = 1 - [ 0! + 1! + 2! + 3! + 4! ]
25
125
625
= 1 - [1 + 5 + 2 + 6 + 24 ] e-5  0.56
For part c, note that if the 5th customer arrives between the 9th and 10th minute then the customers who have
arrived by the 9th minute is 4 or less while the number of customers who have arrived th the 10 th minute is at
least 5. So the answer to part c is equal to
Pr{N(9)  4 and N(10)  5} = Pr{N(9)  4 but not N(10)  4}
Note that if N(10)  4 then N(9)  4. So
Pr{N(9)  4 but not N(10)  4} = Pr{N(9)  4} - Pr{N(10)  4}
4.50e-4.5
= [ 0!
+
4.51e-4.5
1!
+
4.52e-4.5
2!
+
4.53e-4.5
3!
+
4.54e-4.5
]
4!
50e-5
51e-5
52e-5
53e-5
54e-5
- [ 0! + 1! + 2! + 3! + 4! ]
 0.092
Problem 1. The dispatch at a central fire station has observed that the time between calls is an
exponential random variable with a mean of 32 minutes, i.e. the density function is f(t) = (1/32) e-t/32 for
t  0 and f(t) = 0 for t < 0.
a. What is the probability that the next call will arrive within the next half hour?
b. What is the probability that there will be more than two cars in the next hour?
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