Notes Review for Loss Models 1 - ACTSC 431/831, FALL 2011
Part 6 The Classical Continuous Time Ruin Model
In this part, times are measured in years, unless stated otherwise.
1. Poisson Process: Let Nt be the number of claims up to time t or the number of
claims occurring in the time interval (0, t], t > 0. The process {Nt , t 0} is said to
be a Poisson process with rate > 0 if the following three conditions hold:
(a) N0 = 0.
(b) The number of claims in any time interval of length t has a Poisson distribution
with mean t, namely, for all s 0 and t > 0,
Pr{there are n claims in (s, s + t]} = Pr{Nt+s Ns = n} =
(t)n et
.
n!
(c) The process {Nt , t 0} has stationary and independent increments.
Stationary increments mean that for all n = 1, 2, ..., 0 t0 < t1 < < tn and
h 0, the distribution of the random vector (Nt1 +h Nt0 +h , Nt2 +h Nt1 +h , , Ntn +h
Ntn1 +h ) does not depend on h.
In particular, Nt+h Nh has the same distribution as Nt .
Independent increments mean that for all n = 1, 2, ..., 0 t0 < t1 < < tn ,
random variables Nt1 Nt0 , Nt2 Nt1 , , Ntn Ntn1 are independent.
For the Poisson process {Nt , t 0},
(a) For any t > 0, Nt has a Poisson distribution with mean t.
(b) For any 0 < s < t, Ns and Nt Ns are independent. Note that Ns and Nt are
not independent.
2. Inter-claim times and waiting times: Let Tn be the time of the nth claim in a
Poisson process {Nt , t 0} with rate > 0. Then 0 < T1 < T2 < T3 < .
Let Tn Tn1 = Wn be the time between the (n 1)th claim and the nth claim for
n = 1, 2, ... with T0 = 0. Then W1 , W2 , W3 ,... are called inter-claim times or waiting
times.
(a) An important result about the inter-claim times: The inter-claim times
W1 , W2 , W3 ,... are independent and identically distributed exponential random
variables with the same mean 1/.
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3. Compound Poisson Process: Assume that the number of claims up to time t is a
Poisson process Nt with rate > 0. The size of the j th claim is Xj , j = 1, 2, .... Then
the aggregate claims up to time t are
Nt
St =
Xj
j =1
with St = 0 if Nt = 0. The process {St , t 0} is called a compound Poisson process.
Unless stated otherwise, for a compound Poisson process {St , t 0}, we assume that
the Poisson process {Nt , t 0} is independent of the claim sizes {X1 , X2 , } and
the claim sizes {X1 , X2 , } are independent and identically distributed nonnegative
random variables and have the same distributions as X .
In particular, for any t > 0,
E (St ) = tE (X ),
V ar(St ) = tE (X 2 ).
4. Surplus Process: Assume that the initial surplus at time 0 is u 0, the premium
rate per year is c > 0 collected continuously, the aggregative claims up to time t is a
compound Poisson process Nt Xj . Then surplus the at time t is given by
j =1
Nt
Ut = u + ct
Xj
j =1
with U0 = u, where {X1 , X2 , ...} have the same distribution function F (x) and the
same mean > 0 and the Poisson process {Nt , t 0} has rate > 0.
(a) Ruin probabilities of the surplus process: Ruin is said to occur when the
surplus Ut is negative.
Note that ruin can occur only at claim times in the surplus process.
Let (u) be the innite-time ruin probability or
(u) = Pr{ruin occurs at some time t > 0 | U0 = u}
and let n (u) be the probability that ruin occurs on or before the nth claim,
n = 1, 2, ..., namely,
n (u) = Pr{ruin occurs on or before the nth claim | U0 = u}
2
(b) We assume that c > in the surplus process. This condition guarantees that
the ruin probabilities are less than one.
(c) The relative security loading factor > 0 of a surplus process Ut = u + ct
Nt
j =1 Xj is dened by
=
c E (Xj )
.
E (Xj )
(d) The adjustment coecient R > 0 of a surplus process Ut = u + ct
is the smallest positive solution to the following Lundberg equation:
Nt
j =1
1 + (1 + )t = E (etX1 ) = MX1 (t).
Note that the adjustment coecient R > 0 satises
E (eR(X1 c T1 ) ) = E (eR X1 ) E (eR c T1 ) = 1,
where X1 and T1 are the amount and the time of the rst claim, respectively.
(e) Some important results about ruin probabilities:
i. For any u 0,
1 (u) 2 (u) 3 (u)
and
lim n (u) = (u).
n
ii. Recursion formula for n (u): for any u 0 and n = 1, 2, ...,
n+1 (u) =
0
0
n (u + ct x)f (x)dx +
0
=
u+ct
et
f (x)dx dt
u+ct
u+ct
et
n (u + ct x)f (x)dx + S (u + ct) dt
0
with
1 (u) =
et S (u + ct)dt,
0
where f (x) and S (x) are the pdf and the survival function of X1 .
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Xj
iii. Inequalities for the survival function S (x): for any x 0,
S (x) eR x E (eRX1 ),
where R > 0 is the adjustment coecient.
iv. The Lundberg upper bounds for the ruin probabilities: Let R > 0 be the
adjustment coecient. Then for any u 0,
1 (u) eR u ,
n (u) eR u , for all n = 2, 3, ...,
and hence
(u) eR u .
v. If the initial surplus u = 0, then
1
.
1+
(0) =
vi. The ruin probability (u) satises the following integro-dierential equation
u
(u x)f (x)dx S (x) , u 0.
(u)
c
0
vii. The ruin probability (u) satises the following defective renewal equation
(u) =
(u) =
Fe (x)
1
+
1+
1+
u
(u t)dFe (t), u 0,
0
where
Fe (x) =
1
x
S (t)dt, x 0
0
is the equilibrium distribution of F (x) and
1
Fe (x) = 1 Fe (x) =
S (t)dt, x 0,
x
is the survival function of Fe (x).
(f) The general solution to the rst order linear dierential equation of
d
y (t) = (t)y (t) + (t)
dt
can be expresses as
t
y (t) = e
0
t
(s)ds
e
s
0
(u)du
(s) ds + C ,
0
where C is an arbitrary constant and can be determined from the boundary
conditions on y (t).
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