University of Illinois
Fall 2009
ECE 313:
Problem Set 11: Solutions
Poisson Processes; Function of a Random Variable; DecisionMaking
1.
[Conditional Probabilities in a Poisson Process]
From the deﬁnition of conditional probability,
P
{
N
(0
,τ
] =
k

N
(0
,t
] =
M
}
=
P
{
N
(0
,τ
] =
k
∩
N
(0
,t
] =
M
}
P
{
N
(0
,t
] =
M
}
.
(1)
Now,
{
N
(0
,τ
] =
k
} ∩ {
N
(0
,t
] =
M
}
=
{
N
(0
,τ
] =
k
} ∩ {
N
(
τ,t
] =
M

k
}
.
and observe that the intervals (0
,τ
] and (
τ,t
] are disjoint time intervals. So by the basic property of
Poisson processes, we have that
N
(0
,τ
] and
N
(
τ,t
] are independent random variables. This means that
P
{
N
(0
,τ
] =
k
∩
N
(0
,t
] =
M
}
=
P
{
N
(0
,τ
] =
k
} ·
P
{
N
(
τ,t
] =
M

k
}
.
Since the number of packets arriving in any interval is Poisson (with mean proportional to the time
interval length) we have
P
{
N
(0
,t
] =
M
}
=
e

λt
(
λt
)
M
M
!
P
{
N
(0
,τ
] =
k
}
=
e

λτ
(
λτ
)
k
k
!
P
{
N
(
τ,t
] =
M

k
}
=
e

λ
(
t

τ
)
(
λ
(
t

τ
))
M

k
M

k
!
.
Substituting these values in Equation (1) and after some straightforward simpliﬁcation, we get
P
{
N
(0
,τ
] =
k

N
(0
,t
] =
M
}
=
±
M
k
²
³
τ
t
´
k
±
t

τ
t
²
M

k
.
In other words, the conditional probability being asked is the same as that of a
binomial
random
variable with parameters (
M,
τ
t
) taking on the value
k
. The intuition is that any arrival in the time
interval (0
,t
] occurs uniformly in that interval.
2.