Download Question Describe what is meant by a compound Poisson process

Question
Describe what is meant by a compound Poisson process. Show that if A(z)
is the probability generating function for the number of events occurring at
each point of the process, then the random variable X(t) - the total number
of events occurring in a time interval of length t - has probability generating
function
Gt (z) = exp(λtA(z) − λt)
Points in time occur in a Poisson process with rate λ. At each point two
fair coins are tossed. Find the probability generating function for the total
number of heads occurring in a time interval of length t. Find the mean
number of heads occurring in a time interval of length t.
Let W denote the waiting time before any heads occur.
Show that
Ã
−3λt
P (W > t) = exp
4
!
Answer
Suppose that
(i) points occur in a Poisson process {N (t) : t ≥ 0 with rate λ
(ii) at the ith point Yi events occur, where Y1 , Y2 , · · · are i.i.d random variables.
(iii) Yi and {N (t) : t ≥ 0} are independent.
The total number of events occurring in a time interval of length t is
N (t)
X(t) =
X
Yi
i=1
{X(t) : t ≥ 0} is said to be a compound Poisson process.
Let the p.g.f of each Yi be A(z). Then X(t) has p.g.f
∞
X
j=0
z j P (X(t) = j) =
∞
∞ X
X
j=0 n=0
∞ X
∞
X
xj P (X(t) = j | N (t) = n)P (N (t) = n)
(λt)n e−λt
z P (Y1 + · · · + Yn = j)
n!
j=0 n=0
j
1
=
∞
X
j=0
∞
X
(
∞
X
j
z P (Y1 + · · · + Yn = j)
n=0
)
(λt)n e−λt
n!
(λt)n e−λt
n!
n=0
since the Yi are independent
= exp(λtA(z) − λt)
=
[A(z)]n
1
1 1
Now A(x) = + x + z 2 in this case.
4 2
4
So the number of
heads
in a time interval
¶
¶of length t has p.g.f.
µ µ
1 2
1 1
+ z + z − λt
Gt (z) = exp λt
2
4 ¶¶
µ µ4
1 2 1
3
= exp λt z + z −
4
2
4
The mean number of heads is G0 (1).
1
1
3
1
1
· λt z +
G (z) = exp λt z 2 + z −
4
2
4
2
2
so G0 (1) = e0 λt = λt
Let W be the waiting time before a head is recorded.
µ
0
¶¶
µ
µ
EITHER
P (W > t) = P(no events µ
in compound
process in (0, t])
¶
3
= Gt (0) = exp − λt
4
OR
P (W > t) = P(no events in Poisson process)
+
∞
X
n=1
= e
−λt
−λt
+
(n events and 2 tails at each event)
∞ µ ¶n
X
1 (λt)n e−λt
n=1
−λt
= e +e
3λt
= e− 4
h4
e
λ 4t
−1
in!
2
¶