Risk Analysis in Finance and Insurance, 2nd Edition

240 Risk Analysis in Finance and Insurance

Since f(0) = 0, then (8.16) also holds for k = 0. Now we can extend (8.16)

to all integers k:

φ(k) −(1 − q) φ(0) 1

+

(k)=q





φ ∗ 1

+



(k) −



φ ∗ 1

+

∗ f



(k)



. (8.17)

Introduce function

δ(j):=



1 ,j=0

0 ,j=0

.

Then (8.17) can be written in the form

φ(k) ∗



δ(k) − q



1

+

(k) ∗



δ(k) −f (k)





= c 1

+

(k) ,

where c =(1− q) φ(0).

A solution to this equation can be written in the form of the following

Neumann series:

φ(k)=c

∞



n=0

q

n





δ(k) −f (k)



∗n

∗ 1

∗(n+k)

+

(k)



,

where g

∗0

= δ, g

∗n

= g

∗(n−1)

∗ g, n =1, 2,... .

If k →∞, then (8.16) gives

1 − (1 −q) φ(0) = q

∞



j=−∞



1

+

(j) −



1

+

∗ f



(j)



= q

∞



j=0



1 − P



{ω : X

1

≤ j}





= qμ.

Hence,

φ(0) =

1 − qμ

1 − q

.

Mathematical ...