Page 29 - A First Course In Stochastic Models

20 THE POISSON PROCESS AND RELATED PROCESSES

Since the D i are independent of each other, it follows that
∞
j
P {D 0 + · · · + D n = j}z = E(z D 0 +···+D n )
j=0
n
= E(z D 0 ) · · · E(z D n ) = [A(z)] .
Thus
∞ n
−λt (λt) n −λt[1−A(z)]
R(z, t) = e [A(z)] = e
n!
n=0
which proves (1.2.2). To prove part (b) for fixed t, we write R(z) = R(z, t) for ease
of notation. It follows immediately from the definition of the generating function
that the probability r j (t) is given by

j
1 d R(z)
r j (t) = .
j! dz j
z=0
It is not possible to obtain (1.2.3) directly from this relation and (1.2.2). The
following intermediate step is needed. By differentiation of (1.2.2), we find
′
′
R (z) = λtA (z)R(z), |z| ≤ 1.
This gives

∞ ∞ ∞
j−1 k−1 ℓ
jr j (t)z = λt ka k z r ℓ (t)z
j=1 k=1 ℓ=0
∞ ∞
k+ℓ−1
= λtka k r ℓ (t)z .
k=1 ℓ=0
Replacing k + l by j and interchanging the order of summation yields
∞ ∞ ∞
j−1 j−1
jr j (t)z = λtka k r j−k (t)z
j=1 k=1 j=k
 
j
∞
j−1
=  λtka k r j−k (t)  z .
j=1 k=1
Next equating coefficients gives the recurrence relation (1.2.3).
The recursion scheme for the r j (t) is easy to program and is numerically stable.
It is often called Adelson’s recursion scheme after Adelson (1966). In the insurance
literature the recursive scheme is known as Panjer’s algorithm. Note that for the
special case of a 1 = 1 the recursion (1.2.3) reduces to the familiar recursion