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Exponential distribution and the
Poisson process
• Many useful applications, especially in queueing
systems, inventory management, and reliability
analysis.
• A connection between discrete time Markov chains
and continuous time Markov chains.
The exponential distribution
A continuous random variable X is said to have an exponential
distribution with parameter  ,   0, if its probability density
function is given by
  e x
f ( x)  
0,
x0
x0

F ( x)  
x

1  e   x
f ( y ) dy  
 0,
x0
x0
E( X )  



xf ( x)dx   x e   x dx
0
Integrating by parts leads to
E ( X )   xe
 x 
0

 e
0
 x
dx 
1

The moment generating function  (t ) is given by

 (t )  E[e ]   e  e
tX
tx
 x
0

dx    e  (   t ) x dx
0

d (t )
 E[ X ] 
dt
2
d
 (t )
2
E[ X ] 
dt 2

t 0
t 0
 t
1

2

(  t ) 3
Var ( X )  E[ X ]  E[ X ] 
2

2

t 0
1
2
2
2
, for t  
The memoryless property
A random variable is said to be memoryless if
P ( X  s  t | X  t )  P ( X  s ) for all s, t  0

P( X  s  t , X  t ) P( X  s  t )

 P( X  s)
P( X  t )
P( X  t )

P( X  s  t )  P( X  t ) P( X  s)
If X has the exponential distrbution, then
P ( X  s  t )  e   ( s  t )  e   s e  t  P ( X  t ) P ( X  s )
 Exponentially distributed random variables are
memoryless.
It can be shown that the exponential distribution
is the only distribution that has the memoryless
property.
Example 1: The amount of time one spends in a bank is
exponentially distributed with mean ten minutes ( = 1/10).
What is the probability that a customer spends more than 15
minutes? What is the probability that the customer spends
more than 15 minutes given that she is in the bank after 10
minutes?
Example 1: The amount of time one spends in a bank is
exponentially distributed with mean ten minutes ( = 1/10).
What is the probability that a customer spends more than 15
minutes? What is the probability that the customer spends
more than 15 minutes given that she is in the bank after 10
minutes?
P ( X  15)  e 15  e 1.5  0.220
P ( X  15 | X  10)  P( X  5)  e 5  e 0.5  0.604
Example 2: Consider a branch of a bank with two agents
serving customers. The time an agent takes with a customer
is exponentially distributed with mean 1/. A customer
enters and finds the two agents busy serving two other
customers. What is the probability that the customer that just
entered would be last to leave?
The minimum of n exponentially distributed
random variables
Suppose that X1, X2, ..., Xn are independent exponential
random variables, with Xi having rate mi, i=1, ..., n.
What is P(min(X1, X2, ..., Xn )>x)?
The minimum of n exponentially distributed
random variables
Suppose that X1, X2, ..., Xn are independent exponential
random variables, with Xi having rate mi, i=1, ..., n.
What is P(min(X1, X2, ..., Xn )>x)?
P(min( X 1 , X 2 , ..., X n )  x)  P{( X 1  x), ( X 2  x), ..., (X n  x)}
 P{( X 1  x)}P{( X 2  x)} ...P{(X n  x )}
 e  1 x e  2 x ...e  N x
 e  ( 1  2 ... N ) x
 The distribution of the random variable P(min( X 1 , X 2 , ..., X n )
1
is exponentially distributed with mean
.
1  2  ...  n
Comparing two exponentially distributed
random variables
Suppose that X1 and X2 are independent exponentially
distributed random variables with rates 1 and 2.
What is P(X1 < X2)?
Comparing two exponentially distributed
random variables
Suppose that X1 and X2 are independent exponentially
distributed random variables with rates 1 and 2.
What is P(X1 < X2)?

P ( X 1  X 2 )   P ( X 1  X 2 | X 1  x) f X1 ( x )dx
0

  P ( X 1  X 2 | X 1  x )1e
0

 e
 2 x
0

1
1  2
1e
.
 1 x

 1 x
dx   P ( x  X 2 )1e  1 x dx
dx   1e  ( 1  2 ) x dx
0

0
The sum of 2 identical exponentially distributed
random variables
Suppose that X1 and X2 are independent exponential
random variables with rates 1=2 =. What is the
distribution of fX1+X2 (x)?)?
The sum of 2 identical exponentially distributed
random variables
Suppose that X1 and X2 are independent exponential
random variables with rates 1=2 =. What is the
distribution of fX1+X2 (x)?
x
f X1  X 2 ( x)   f X1 ( s ) f X 2 ( x  s )ds
0
x
   e   s  e   ( x  s ) ds
0
2  x
 e

x
0
ds   2 xe   x
The sum of n identical exponentially distributed
random variables
Suppose that X1,..., XN are independent exponential random
variables with rates i= for i =1, ..., n. What is the
distribution of fX1+...+Xn (x)?
f X1  X 2 ... X n ( x)   e
 x
( x)n 1
(n  1)!
 X1,..., XN has the gamma distribution with parameters n
and .
The sum of 2 exponentially distributed random
variables
Suppose that X1 and X2 are independent exponential
random variables with rates 1≠2. What is the distribution
of fX1+X2 (x)?
The sum of 2 exponentially distributed random
variables
Suppose that X1 and X2 are independent exponential
random variables with rates 1≠2. What is the distribution
of fX1+X2 (x)?
x
f X1  X 2 ( x)   f X1 ( s ) f X 2 ( x  s )ds
0
x
  1e  1s 2 e  2 ( x  s ) ds
0
 12 e

1
 2 x
1  2

x
0
2 e
e  ( 1  2 ) s ds
 2 x

2
1  2
1e  x
1
The Poisson Process
Counting Processes
A stochastic process {N(t), t ≥ 0} is said to be a counting
process if N(t) represents the total number of “events” that
occur by time t (i.e., in the time interval [0, t]).
Counting Processes
A stochastic process {N(t), t ≥ 0} is said to be a counting
process if N(t) represents the total number of “events” that
occur by time t (i.e., in the time interval [0, t]).
Example 1: N(t) is the number of customers that enter a store
at or prior to time t. An event corresponds to a person
entering the store.
Counting Processes
A stochastic process {N(t), t ≥ 0} is said to be a counting
process if N(t) represents the total number of “events” that
occur by time t (i.e., in the time interval [0, t]).
Example 1: N(t) is the number of customers that enter a store
at or prior to time t. An event corresponds to a person
entering the store.
Example 2: N(t) is the number of individuals born at or prior
to time t. An event occurs whenever a child is born.
Counting Processes
A stochastic process {N(t), t ≥ 0} is said to be a counting
process if N(t) represents the total number of “events” that
occur by time t (i.e., in the time interval [0, t]).
Example 1: N(t) is the number of customers that enter a store
at or prior to time t. An event corresponds to a person
entering the store.
Example 2: N(t) is the number of individuals born at or prior
to time t. An event occurs whenever a child is born.
Example 3: N(t) is the number of calls made to a technical
help line at or prior to time t. An event occurs whenever a
call is placed.
Properties of counting processes
A counting process satisfies the following properties.
(i) N(t) ≥ 0.
(ii) N(t) is integer valued.
(iii) If s < t, then N(s) ≤ N(t).
(iv) For s < t, N(t) – N(s) equals the number of events that
occurs in the time interval (s, t].
A counting process is said to possess independent increments
if the number of events that occur in disjoint intervals are
independent.
A counting process is said to possess independent increments
if the number of events that occur in disjoint intervals are
independent.
Example 1: The number of customers N(10) that enter the
store in the interval [0, 10] is independent from the number
of customers N(15) – N(10) that enter the store in the interval
(10, 15].
A counting process is said to possess independent increments
if the number of events that occur in disjoint intervals are
independent.
Example 1: The number of customers N(10) that enter the
store in the interval [0, 10] is independent from the number
of customers N(15) – N(10) that enter the store in the interval
(10, 15].
Example 2: The number of individuals N(10) born in the
interval [1996, 2000] is not independent from the number of
individuals N(2004) – N(2000) that enter the store in the
interval (2000, 2004].
A counting process is said to possess stationary increments
if the distribution of the number of events that occur in an
interval depend only on the length of the interval and not the
starting time of the interval.
A counting process is said to possess stationary increments
if the distribution of the number of events that occur in an
interval depend only on the length of the interval and not the
starting time of the interval.
Example 1: The number of customers N(t) – N(s) that enter
the store in the interval (s, t] does not depend on s (this is
true if there is not a particular time of day where more
customers enter the store).
The Poisson processes
The counting process {N(t) t ≥ 0} is said to be a Poisson process
having rate ,  > 0, if
(i) N(0) = 0.
(ii) The process has independent increments.
(iii) The number of events in any interval of length t is Poisson
distributed with mean t. That is for all s, t ≥ 0
n
(

t
)
P{N (t  s)  N (t )}  e t
,
for n  0,1,...
n!
The distribution of interarrival times
• Let Tn describe the time that elapses between (n-1)th event and
the nth event for n > 1 and let T1 be the time of the first event.
• The sequence {Tn , n = 1, 2, ...} is called the sequence of
interarrival times.
The distribution of interarrival times
• Let Tn describe the time that elapses between (n-1)th event and
the nth event for n > 1 and let T1 be the time of the first event.
• The sequence {Tn , n = 1, 2, ...} is called the sequence of
interarrival times.
Example: if T1 = 5 and T2 = 10  the first event arrives at time t
= 5 and event 2 occurs at time t = 15.
The distribution of interarrival times
• P(T1 > t) = P(N(t) = 0) = e-t  T1 has the exponential
distribution.
The distribution of interarrival times
• P(T1 > t) = P(N(t) = 0) = e-t  T1 has the exponential
distribution.
• P(T2 > t) = E[P(T2 > t|T1) ]
Since P(T2 > t|T1=s) = P(0 events in (s, s+t]|T1=s)
= P(0 events in (s, s+t])
= e-t
Then, P(T2 > t) = E[P(T2 > t|T1) ] = e-t  T2 has the exponential
The distribution of interarrival times
• P(T1 > t) = P(N(t) = 0) = e-t  T1 has the exponential
distribution.
• P(T2 > t) = E[P(T2 > t|T1) ]
Since P(T2 > t|T1=s) = P(0 events in (s, s+t]|T1=s)
= P(0 events in (s, s+t])
= e-t
Then, P(T2 > t) = E[P(T2 > t|T1) ] = e-t  T2 has the exponential
•The same applies to other values of n  Tn has the exponential
distribution.
The distribution of interarrival times
Let Tn denote the inter-arrival time between the (n-1)th event
and the nth event of a Poisson process, then the Tn (n=1, 2, ...)
are independent, identically distributed exponential random
variables having mean 1/.