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1
Poisson Processes
Chapter 3
2
Poisson Distribution and
the Poisson Process
Chapter 3.1
3
Poisson Distribution
z
The Poisson distribution with parameter
µ
> 0 is
given by
z
Let X be Poisson r.v. Then, E[X]=
µ
.
z
Moreover, E[X
2
] =
µ
2
+
µ
and
z
Var(X) =
µ
.
{}
!
Pr
k
e
p
k
X
k
k
µ
−
=
=
=
4
Theorem 3.1
z
Let X and Y be independent random
variables having Poisson distributions
with parameters
µ
and
ν
. Then the sum
X+Y has a Poisson distribution with
parameter
µ
+
ν
.
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Proof
{}
{
}
{
}
!
)
(
!
)!
(
!
Pr
Pr
,
Pr
Pr
)
(
0
)
(
0
0
0
n
e
k
n
n
e
k
n
e
k
e
k
n
Y
k
X
k
n
Y
k
X
n
Y
X
n
k
n
k
n
k
k
n
n
k
k
n
k
n
k
ν
µ
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎭
⎬
⎫
⎩
⎨
⎧
−
⎭
⎬
⎫
⎩
⎨
⎧
=
−
=
=
=
−
=
=
=
=
+
+
−
−
=
+
−
−
−
=
−
=
=
∑
∑
∑
∑
6
Theorem 3.2
z
Let N be a Poisson r.v. with parameter
µ
,
and conditional on N, let M have a
binomial distribution with parameters N
and p. Then the unconditional
distribution of M is Poisson with
parameter
µ
p.
7
Proof
{
}
{
}
[]
!
)
(
!
)
(
)!
(
)
1
(
!
)
(
!
)
1
(
)!
(
!
!
Pr

Pr
Pr
0
)
1
(
0
0
k
p
e
e
k
e
p
k
n
p
k
e
p
n
e
p
p
k
n
k
n
n
N
n
N
k
M
k
M
k
p
n
p
k
k
n
k
n
n
k
n
k
n
−
∞
=
−
−
−
−
−
∞
=
−
∞
=
=
=
−
−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎭
⎬
⎫
⎩
⎨
⎧
−
−
=
=
=
=
=
=
∑
∑
∑
8
The Poisson process
z
A Poisson process of intensity or rate
λ
> 0 is an
integervalued stochastic process {X(t); t
≥
0} for
which
•
For any time points t
0
=0< t
1
< … < t
n
, the process
increments,X(t
1
)X(t
0
),…, X(t
n
)X(t
n1
), are independent
r.v.
•
For s
≥
0 and t > 0, the r.v. X(s+t) – X(s) has the
Poisson distribution
•
And X(0) = 0.
,...
1
,
0
,
!
)
(
)
(
)
(
Pr
=
=
=
−
+
−
k
k
t
e
k
s
X
t
s
X
k
t
λ
9
Example 3.1
z
Defects occur along an undersea cable
according to a Poisson process of rate
λ
= 0.1
per mile.
z
A) What is the probability that no defects
appear in the first two miles of cable?
z
B) Given that there are no defects in the first
two miles of cable, what is the conditional
probability of no defect between points two and
three?
10
Example 3.1
z
To answer (A), observe that X(2) follows
Poisson with parameter (0.1)(2) = 0.2.
z
Hence, Pr{X(2) = 0} = e
0.2
= 0.8187.
z
For part (B), we use the independence of
X(3) – X(2) and X(2) – X(0) = X(2). Thus
the conditional prob is the same as the
unconditional prob.
Pr{X(3) – X(2) = 0}
= Pr{X(1) = 0} =e
0.1
= 0.9048.
11
Example 3.2
z
Customers arrive in a certain store
according to a Poisson process of rate
λ
=4 per hour.
z
Given that the store opens at 9:00 am,
what is the probability that exactly one
customer has arrived by 9:30 and a total
of five has arrived by 11:30am?
12
Example 3.2
z
Convert into probability language. We
are asked to find Pr{X(1/2) =1,X(5/2) =5}.
z
We use the independence property:
Pr{X(1/2) =1,X(5/2) =5}
= Pr{X(1/2) =1,X(5/2) – X(1/2) =4}
= e
4(1/2)
4(1/2){e
4(2)
[4(2)]
4
/4!}
=(2e
2
)[(512/3)e
8
]=0.0154965.
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Nonhomogeneous Processes
z
Nonhomogeneous Poisson process has all the
property of Poisson process except that the
second property is replaced by
z
Where
λ
(t) is a function of time t.
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