ACTL2003&5103 Tutorial 4 Sample Solutions
1. Define
{
N
B
(
t
)
}
as a Poisson process with rate
λB
.
If we know what
B
is, then obviously for
x
=
0
,
1
,
2
, ...
Prob (
N
B
(
t
) =
x

B
=
b
) =
e

λbt
(
λbt
)
x
x
!
.
From standard conditional probability definitions, we have:
Prob (
B
=
b

N
B
(
t
) =
x
) =
Prob (
B
=
b, N
B
(
t
) =
x
)
Prob (
N
B
(
t
) =
x
)
The numerator represents the joint probability of
B
and
{
N
B
(
t
)
}
. For
x
= 0
,
1
,
2
, ...
and
b
∈ {
1
,
2
,
3
, ...
}
,
we have
Prob (
B
=
b, N
B
(
t
) =
x
)
=
Prob (
N
B
(
t
) =
x

B
=
b
) Prob (
B
=
b
)
=
e

λbt
(
λbt
)
x
x
!
π
(
b
)
,
from which we can also derive the denominator:
Prob (
N
B
(
t
) =
x
) =
X
b
Prob (
B
=
b, N
B
(
t
) =
x
) =
X
b
e

λbt
(
λbt
)
x
x
!
π
(
b
)
.
And finally we have:
Prob (
B
=
b

N
B
(
t
) =
x
) =
e

λbt
(
λbt
)
x
x
!
π
(
b
)
∑
a
e

λat
(
λat
)
x
x
!
π
(
a
)
.
To derive the probability that the specified individual will not have an accident in (
t, t
+ 1)
,
we can
do the following:
Prob (
N
B
(
t
+ 1)

N
B
(
t
) = 0

N
B
(
t
) =
x
)
=
X
b
Prob (
N
B
(
t
+ 1)

N
B
(
t
) = 0

B
=
b, N
B
(
t
) =
x
) Prob (
B
=
b

N
B
(
t
) =
x
)
=
X
b
(
e

λb
)
e

λbt
(
λbt
)
x
x
!
π
(
b
)
∑
a
e

λat
(
λat
)
x
x
!
π
(
a
)
=
∑
b
e

λb
(
t
+1)
b
x
π
(
b
)
∑
a
e

λat
a
x
π
(
a
)
.
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 Three '11
 kim
 Equations, pAA, initial condition, π, PROB, (λbt)

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