The …nite time ruin probability in a risk
model with capital injections
Ciyu Nie, David C M Dickson and Shuanming Li
Abstract
We consider a risk model with capital injections as described in
Nie et al. (2011). We show that in the Sparre Andersen framework
the density of the time to ruin for the model with capital injections
can be expressed in terms of the density of the time to ruin in an
ordinary Sparre Andersen risk process. In the special case of Erlang
inter-claim times and exponential claims we show that there exists a
readily computable formula for the density of the time to ruin. When
the inter-claim time distribution is exponential, we obtain an explicit
solution for the density of the time to ruin when the individual claim
amount distribution is Erlang(2), and we explain techniques to …nd
the moments of the time to ruin. In the …nal section we consider the
related problem of the distribution of the duration of negative surplus
in the classical risk model, and we obtain explicit solutions for the
(defective) density of the total duration of negative surplus for two
individual claim amount distributions.
1
Introduction
In this paper we consider the …nite time ruin probability in the risk model
considered in Nie et al (2011). In that paper we considered a surplus process
into which capital was injected each time the surplus process, starting from
level u k, fell between 0 and k: The capital injections were assumed to be
secured by a reinsurance arrangement, and each capital injection restored the
surplus level to k. Ruin occurred if the surplus fell below 0. The focus of that
study was on the probability of ruin in in…nite time, and this measure was
used to determine quantities of interest, such as the value of k that minimised
the ultimate ruin probability for a given level of initial capital.
The scope of this study is more limited as we assume u and k are given
and we consider the calculation of the ruin probability in …nite time. We
1
derive general results for the density of the time to ruin in the framework
of a Sparre Andersen risk model, i.e. with a general claim inter-arrival time
distribution. In principle, these results lead to explicit solutions for the
density of the time to ruin in cases where explicit solutions exist for the
density of the time to ruin in standard Sparre Andersen surplus processes,
but these expressions can be complicated. However, in the special case when
claims are exponentially distributed we can exploit the fact that there is a
general expression for the density of the time to ruin for a standard Sparre
Andersen model, and we illustrate the use of this expression when claim interarrival times are Erlang(n) distributed. We make use of generalised binomial
functions to deal with convolutions of certain functions, and express solutions
in terms of generalised hypergeometric functions for computational e¢ ciency.
We also consider the situation when claims occur as a Poisson process.
We show how the density of the time to ruin can be found when the individual
claim amount distribution is Erlang(2), and we show that if the joint density
of the time to ruin and the de…cit at ruin (in the classical risk model) satis…es
a particular factorisation, then we can calculate moments of the time to
ruin. In the …nal section we discuss the related problem of the distribution
of the duration of negative surplus in the classical risk model, where these
techniques can also be applied.
2
Model and notation
In the standard Sparre Andersen surplus process, the surplus at time t is
U (t) = u + ct
N (t)
X
Xi
i=1
where u is the initial surplus, c is the rate of premium income per unit
time, fN (t)gt 0 is a counting process for the number of claims and fXi g1
i=1 is
a sequence of i.i.d. random variables representing individual claim amounts.
We assume that X1 has distribution function F , with F (0) = 0, density
function f , and mean m1 . We also assume that cE[V1 ] > m1 where fVi g1
i=1
is a sequence of i.i.d. random variables, independent of fXi g1
i=1 ; representing
the claim inter-arrival times.
We de…ne Tu to be the time to ruin from initial surplus u; and de…ne
(u; t) = Pr(Tu
t)
to be the …nite time ruin probability, with (defective) density wu (t) =
2
@
@t
(u; t).
De…ne Yu to be the de…cit at ruin from initial surplus u; and de…ne
W (u; y; t) = Pr(Tu
t; Yu
y)
to be the probability of ruin by time t with a de…cit of at most y at ruin,
and let
@2
wu (y; t) =
W (u; y; t)
@t @y
denote the (defective) joint density of Tu and Yu .
Further, let
(u) = 1
(u) = lim (u; t)
t!1
be the ultimate ruin probability, and let
G(u; y) = lim W (u; y; t)
t!1
be the probability of ultimate ruin with a de…cit of at most y at ruin.
We now introduce notation for the Sparre Andersen process with capital
injections as described in the previous section. Let Tu;k denote the time to
ruin and let
k (u) = Pr(Tu;k < 1)
be the ultimate ruin probability. Nie et al (2011) give a general expression
for k (u). Next, let
Wu;k (t) = Pr(Tu;k t)
@
Wu;k (t).
be the …nite time ruin probability, with (defective) density wu;k (t) = @t
We use the notation MX to denote the moment generating function of a
random variable X, and we assume throughout that the required moment
generatingRfunctions exist. We denote the Laplace transform of a function
1
by ~ (s) = 0 e sx (x)dx:
3
General results
In this section we derive a general expression for the density of Tu;k , and,
as in problems considered in Nie et al (2011), we consider …rst the situation
when u = k; as this leads to solutions in the case u > k.
First, we observe that if ruin occurs on the nth (n = 1; 2; 3; : : :) occasion
that the surplus falls below k, then there must have been n 1 falls below
k to a level between 0 and k, resulting in a capital injection to restore the
surplus process to level k; and the nth fall below k results in a surplus below
0. Thus, the probability that ruin is caused by n falls in the surplus level
below k is G(0; k)n 1 ( (0) G(0; k)).
3
Second, we note that if ruin occurs on the nth occasion that the surplus
falls below k, the time to ruin is the sum of n 1 variables with distribution
function
W (0; k; t)
D0;1 (t) = Pr(T0 t j T0 < 1 and Y0 k) =
G(0; k)
and of a further random variable with distribution function
(0; t) W (0; k; t)
D0;2 (t) = Pr(T0 t j T0 < 1 and Y0 > k) =
:
(0) G(0; k)
These observations lead to
1
X
Wk;k (t) =
G(0; k)n 1 ( (0)
(n 1)
G(0; k)) D0;1
(3.1)
D0;2 (t);
n=1
0
D0;1
with
D0;2 (t) de…ned to be the same as D0;2 (t).
Next, we consider the case u > k. The situation is similar to the case
u = k, but now either
the time to ruin is the time to the …rst drop of the surplus process
below k if that drop results in ruin, or
the time to ruin is the sum of the time to the …rst drop below k, if that
drop does not result in ruin, and the time to ruin from surplus level k:
De…ne
Du
k;1 (t)
= Pr(Tu
=
k
t j Tu
k
< 1 and Yu
k
k)
W (u k; k; t)
G(u k; k)
to be the distribution of the time to the …rst drop of the surplus process
below k given that this drop does not result in ruin, and de…ne
Du
k;2 (t)
= Pr(Tu
=
k
t j Tu
(u k; t)
(u k)
k
W (u
G(u
< 1 and Yu
k
> k)
k; k; t)
k; k)
to be the distribution of the time to ruin given that ruin occurs on the …rst
drop of the surplus process below level k. Then, as the process re-starts from
k at the time of the …rst capital injection, we have
Wu;k (t) = ( (u
k)
G(u
k; k)) Du
4
k;2 (t)
+ G(u
k; k) Du
k;1
Wk;k (t) :
(3.2)
This is a general formula for Wu;k (t). Formulae for the distribution functions
Du;1 and Du;2 can be obtained for some claim amount distributions and claim
inter-arrival time distributions –see, for example, Dickson (2008) or Landriault and Willmot (2009) in the case of the classical risk model, or Dickson
and Li (2010) for the Erlang(2) risk model. In the case of exponentially distributed individual claims, the distribution functions Du;1 and Du;2 are easily
found, facilitating calculation of wu;k (t), as shown in Section 4.1.
We can also derive moments of the time to ruin given that ruin occurs. Let
u;1 and u;2 be random variables with distribution functions Du;1 and Du;2
c
to be the time to ruin given that ruin occurs from
respectively. De…ne Tk;k
initial surplus k. Then from (3.1) and k (k) = ( (0) G(0; k))=(1 G(0; k))
(see Nie et al (2011)), we have
1
X
c (r)
MTk;k
=
G(0; k)n 1 (1
G(0; k)) M
0;1
(r)n
1
M
0;2
(r)
n=1
(1
1
=
G(0; k)) M 0;2 (r)
:
G(0; k) M 0;1 (r)
Similarly, we can write (3.2) as
c
Pr(Tu;k
(u
t) =
k)
G(u
k (u)
k; k)
+
Pr(
G(u
t)
u k;2
k; k)
k (u)
k (k)
Pr(
u k;1
c
+ Tk;k
t)
which gives
c (r) =
MTu;k
(u
k)
G(u
k (u)
k; k)
M
u k;2
(r)+
G(u
k; k)
k (u)
k (k)
M
u k;1
c (r):
(r)MTk;k
From this moment generating function we …nd
c
E[Tu;k
]=
(u
k)
and
h
i
2
c
E Tu;k
=
G(u
k (u)
(u
+
G(u
k)
k; k)
G(u
k (u)
k; k)
k (u)
E[
u k;2 ]+
k; k)
k (k)
E[
E[
5
G(u
k; k)
k (u)
k (k)
E[
u k;1 ]
c
+ E[Tk;k
]
(3.3)
2
u k;2 ]
2
u k;1 ]
+ 2E[
u
c
k;1 ]E[Tk;k ] + E
h
c
Tk;k
2
i
:
These give
c
] = E[
E[Tk;k
and
h
i
2
c
E Tk;k
= E[
2
0;2 ]+
G(0; k)
1 G(0; k)
0;2 ]
G(0; k) E[ 0;1 ]
1 G(0; k)
+
2E[
0;2 ]E[
0;1 ] +
(3.4)
2G(0; k) E[ 0;1 ]2
+ E[
1 G(0; k)
The calculation of moments of u;1 and u;2 depends on the density wu (y; t)
existing in a suitable form. We discuss this further in Section 4.3.
4
4.1
Some explicit results
Exponential claims
In this section we consider the situation when f (x) = e x , x > 0. Initially
we assume a general claim inter-arrival time distribution, before moving to a
speci…c assumption in Section 4.1.1. It is well known that by the memoryless
property of the exponential distribution
W (u; y; t) = (u; t) (1
with G(u; y) = (u) (1
e
y
);
). This leads to
(u; t)
(u)
Du;1 (t) = Du;2 (t) =
for all u
y
e
0. The density of Tk;k can then be found from (3.1) as
wk;k (t) =
1
X
(1
k n 1
e
)
k
w0n (t)
k
w0n
e
(j)
u
e
n=1
and using this and (3.2), the density of Tu;k is
wu;k (t) = e
k
wu k (t) +
1
X
(1
k n
e
) e
n=1
From Dickson and Li (2010) we know that
wu (t) =
1
X
w0j (t)
j=1
6
( u)j
1
wu k (t) :
(4.1)
2
0;1 ]
:
which means that we can write (4.1) as
wu;k (t) = e
u
1
X
(1
k n
e
)
n=0
1
X
(n+j)
w0
(t)
( (u
j=1
k))j
(j)
1
:
(4.2)
Thus, if we can evaluate convolutions w0n (t) then we can …nd wu;k (t). We
now show how this can be done for Erlang(n) claim inter-arrival times.
4.1.1
Erlang(n) claim inter-arrival times
Still assuming f (x) = e x , we now consider the situation when the distribution of claim inter-arrival times is Erlang(n; ) with density function
n n 1
t
en; (t) =
e
(n)
t
for t > 0. From Borovkov and Dickson (2008) we know that
w0 (t) = n
n
1
X
( c
n m (n+1)m+n 1
) t
e
m!(n(m + 1))!
m=0
( + c)t
and this function has Laplace transform
w~0 (s) = n
=
n
1
X
( c n )m ((n + 1)m + n 1)!
m!(n(m + 1))! ( + c + s)(n+1)m+n
m=0
+ c+s
1
nX
m=0
(n + 1)m + n
m
n Zm
(n + 1)m + n
where Z = c n =( + c + s)n+1 . We see that w~0 (s) can be written in terms
of the generalised binomial function de…ned as
Bt (z) =
1
X
tk + 1
1
zk ;
k
tk
+
1
k=0
with the property that
1
X
r
tk + r
Bt (z) =
zk :
k
tk
+
r
k=0
r
See Graham et al (1994) for details. Thus
n
w~0 (s) =
+ c+s
7
Bn+1 (Z)n :
Using properties of Laplace transforms and generalised binomial functions,
this gives
nr
w~0r
r
(s) = w
~0 (s) =
=
nr
Bn+1 (Z)nr
+ c+s
1
X
(n + 1)m + nr
nr
( c n )m
;
m
(n + 1)m + nr ( + c + s)m(n+1)+nr
m=0
and inversion yields
w0r
(t) =
nr nr 1
t
nr e
( + c)t
1
X
( c n tn+1 )m
.
m!(n(r + m))!
m=0
Finally, for ease of computation with software, we can apply techniques described in Graham et al (1994) to write w0r (t) in terms of generalised hypergeometric functions as
nr nr 1
w0r (t) =
t
e (
(nr)
+ c)t
0 Fn r +
2
n c n tn+1
1
;r + ;:::;r + ;
n
n
n
nn
(4.3)
where
p Fq (B1 ; B2 ; : : : ; Bp ; C1 ; C2 ; : : : ; Cq ; Z) =
1
X
(B1 )m (B2 )m : : : (Bp )m Z m
(C1 )m (C2 )m : : : (Cq )m m!
m=0
is the generalised hypergeometric function, with (a)m = (a + m)= (a).
Combining (4.2) and (4.3) we have a readily computable formula for
wu;k (t). Of course, in…nite sums must be truncated at a suitable point.
Figure 4.1 gives the density w10;k (t)= k (10) for k = 1; 2; 3 under the classical
risk model (n = = 1) with = 1 and c = 1:2: The highest peak is the case
k = 1 whilst the lowest is the case k = 3. The shape of each density is not
surprising; these densities have the same sort of shape as in the classical risk
model (the case k = 0).
4.2
Classical risk model with Erlang(2) claims
In this section we assume that the claim arrival process is a Poisson process
with parameter and we consider the situation when the claim size distribution is Erlang(2, ). From Dickson (2008) we know that wu (y; t) is of the
form
wu (y; t) = l1 (u; t) 2 ye y + l2 (u; t) e y .
8
Figure 4.1: w10;k (t)=
In particular,
10 (k)
1
X
(
l1 (0; t) =
2 2 n
c ) t3n e ( +
n!(2n + 1)!
n=0
and
l2 (0; t) = 2
c
1
X
(
2 2 n
c ) t3n+1 e (
n!(2n + 2)!
n=0
so that
~l1 (0; s) =
1
X
n=0
=
for k = 1; 2; 3
c)t
+ c)t
n
( 2 c2 ) (3n + 1)
n!(2n + 1)! ( + c + s)3n+1
+ c+s
B3 Z~
where
Z~ =
and
~l2 (0; s) =
2 2
c =( + c + s)3 ;
c
~ 2
2 B3 (Z) :
( + c + s)
9
(4.4)
Now consider wk;k (t): We have
d0;1 (t) =
=
@
W (0; k; t)
@t
G(0; k)
=
l1 (0; t) 1
e
k
ke k + l2 (0; t) 1
G(0; k)
k
e
ak l1 (0; t) + bk l2 (0; t)
G(0; k)
where ak = 1
e
k
ke
w0 (t)
d0;2 (t) =
(0)
=
(1
k
and bk = 1
@
W (0; k; t)
@t
G(0; k)
k
e
k
l1 (0; t)(e
=
, and
(1 + k)) + l2 (0; t)e
(0) G(0; k)
k
ak ) l1 (0; t) + (1 bk ) l2 (0; t)
:
(0) G(0; k)
It then follows from (3.1) that
w~k;k (s) =
1
X
n 1
ak ~l1 (0; s) + bk ~l2 (0; s)
(1
ak ) ~l1 (0; s) + (1
bk ) ~l2 (0; s) :
1 j
~l1 (0; s)j+1 ~l2 (0; s)n
n=1
From this we obtain
w~k;k (s) = (1
ak )
1 X
n 1
X
n
n=1 j=0
+(1
bk )
1
j
1 X
n 1
X
n
n=1 j=0
= (1
ak )
1 X
n 1
X
n
n=1 j=0
+(1
bk )
ak )
n=1 j=0
bk )
n=1 j=0
j
(4.5)
( c)n 1 j
( + c + s)2n 1
j
B3 Z~
1 j
1 j
( c)n j
( + c + s)2n
n
( c)n
j
B3 Z~
2n 1 j
2n j
1 j
2n 1 j
( 2 c2 )i
3i + 2n 1 j ( + c + s)3i+2n
j
1
j
~l1 (0; s)j ~l2 (0; s)n
n
ajk bkn
j
1 j
ajk bkn
j
1
n 1
1 X
X
n
1 j
1 j
n
ajk bkn
1
1
X
3i + 2n 1
i
i=0
+(1
ajk bkn
j
j
1 X
n 1
X
n
1 X
n 1
X
n
1
1
n=1 j=0
= (1
ajk bkn
ajk bkn
10
1 j
n
( c)n
j
1 j
1
X
3i + 2n
i
i=0
( 2 c2 )i
2n j
3i + 2n j ( + c + s)3i+2n
j
j
:
The inverse of
1
X
3i + 2n 1
i
i=0
( 2 c2 )i
2n 1 j
3i + 2n 1 j ( + c + s)3i+2n
j
1 j
is
1
X
3i + 2n 1
i
i=0
= e
( + c)t 2n 2 j
t
e
( + c)t 2n 2 j
t
2 j)!
(2n
c )e ( +
(3i + 2n
1
X
(2n
c)t 3i+2n 2 j
t
1
j)
i
1 j) ( 2 c2 t3 )
i!(2i + 2n 1 j)!
i=0
=
2 2 i
2n 1 j (
3i + 2n 1 j
j
0 F2
j
;n
2
n
2 2 3
j 1
+ ;
2 2
ct
4
:
Similarly, the inverse of
1
X
3i + 2n
i
i=0
is
e
j
( 2 c2 )i
2n j
3i + 2n j ( + c + s)3i+2n
( + c)t 2n 1 j
(2n
t
1 j)!
0 F2
j 1
+ ;n
2 2
n
j
2 2 3
j
+ 1;
2
ct
4
:
These results give
wk;k (t) = (1
ak )
1 X
n 1
X
n
n=1 j=0
e
bk )
j
t
2 j)!
n 1
1 X
X
n
n=1 j=0
e
ajk bkn
1 j
( + c)t 2n 2 j
(2n
+(1
1
0 F2 n
1
j
ajk bkn
( + c)t 2n 1 j
(2n
t
1 j)!
0 F2
11
n
j
;n
2
1 j
n
( c)n
1 j
2 2 3
j 1
+ ;
2 2
n
( c)n
j 1
+ ;n
2 2
ct
4
j
j
+ 1;
2
2 2 3
ct
4
;
a formula which is easy to compute with appropriate truncation of the in…nite
sums. The derivation of this result has a further application, as shown in
Section 5.
We now look at the case where u k. We have
Z t
Z t
Z kZ t
l2 (u; ) d
l1 (u; ) d + bk
wu (y; ) d dy = ak
W (u; k; t) =
0
0
0
0
and hence
(u; t) =
Z
t
(l1 (u; ) + l2 (u; )) d :
0
Then from equation (3.2) we have
Wu;k (t) = (u
k; t)
W (u
k; k; t) + Wk;k
W (u
k; k; t);
and hence, by di¤erentiating and taking the Laplace transform, we have
ak )le1 (u k; s) + (1 bk )le2 (u k; s)
+[ak le1 (u k; s) + bk le2 (u k; s)]w
ek;k (s):
w
eu;k (s) = (1
(4.6)
Dickson (2008) gives formulae for the functions l1 (u; t) and l2 (u; t) and shows
that their Laplace transforms are
Z 1
e
l1 (u; s) =
e st l1 (u; t)dt
0
p
1
1 XX p e
=
l1 (0; s)q+1 le2 (0; s)p q ep+q+1; (u);
q
p=0 q=0
Z 1
e st l2 (u; t)dt
le2 (u; s) =
(4.7)
0
=
p
1
1 XX p e
l1 (0; s)q+1 le2 (0; s)p q ep+q+2; (u)
q
p=0 q=0
p
1
1 XX p e
+
l1 (0; s)q le2 (0; s)p
q
p=0 q=0
q+1
ep+q+1; (u):
(4.8)
Substituting these results and equation (4.5) for w
ek;k ( ) into equation (4.6),
and by inverting the Laplace transform using the property of the generalised
binomial function and the same method used in deriving wk;k (t), we obtain
our expression for wu;k (t) as
12
wu;k (t)
= (1 ak ) l1 (u k; t) + (1 bk ) l2 (u k; t)
p
1
1 n 1
e t XXXX p n 1 i n
+
ak b k
q
i
p=0 q=0 n=1 i=0
n+p+1
1 i
(1
ak )
c
[ak ep+q+1; (u
k) + bk ep+q+2; (u k)]
2n + 2p i q + 1 2n + 2p i q + 2
;
;
e2n+2p i q; c (t) 0 F2
2
2
p
1
1 n 1
n+p+1
e t XXXX p n 1 i n 1 i
+
ak b k
q
i
c
p=0 q=0 n=1 i=0
2 2 3
ct
4
[(ak + bk
2 ak bk ) ep+q+1; (u k) + bk (1 bk ) ep+q+2 (u k)]
2 2 3
ct
2n + 2p i q + 2 2n + 2p i q + 3
;
;
e2n+2p i q+1; c (t) 0 F2
2
2
4
p
1
1
n
1
n+p+1
e t XXXX p n 1 i n i
+ (1 bk )
ak b k
ep+q+1; (u
q
i
c
p=0 q=0 n=1 i=0
e2n+2p
i q+2; c (t) 0 F2
2n + 2p
i
2
q + 3 2n + 2p
;
i
2
q+4
2 2 3
;
ct
4
This formula can easily be implemented by appropriate truncation of the
in…nite sums.
4.3
Moments
c
We now illustrate how E[Tu;k
] can be found. The techniques hinge on being
able to decompose the joint density wu (y; t) as
wu (y; t) =
n
X
hi (u; t) pi (y)
i=1
for some functions fhi (u; t)gni=1 and densities fpi (y)gni=1 . The key point is
that we do not need to know the functions fhi (u; t)gni=1 : Our techniques
readily extend to higher moments, and so we omit details. We work through
the case u = k in detail, then illustrate the key points for the case u > k.
We consider the situation when
f (x) = p e
13
x
+q e
x
k)
(4.9)
:
c
< . To …nd E[Tk;k
] we require E[
where p + q = 1; 0 < p < 1 and
and E[ 0;2 ], given by
Z
1
E[ 0;1 ] =
G(0; k)
and
E[
0;2 ]
1
(0) G(0; k)
=
Z
1
@
W (0; k; t) dt
@t
t
0
1
t
0
0;1 ]
@
( (0; t)
@t
W (0; k; t)) dt:
From Dickson (2008) we know that
y
w0 (y; t) = (0; t) e
+ (0; t) e
y
where
~(0; ) =
Z
1
t
e
(0; t)dt =
p
+
c
0
and
~ (0; ) =
q
+
c
;
where (= ( )) is the unique positive solution of Lundberg’s fundamental
equation for the classical risk model, so that
c = f~( ):
+
See Gerber and Shiu (1998). Thus,
@
W (0; k; t) = (0; t) 1
@t
and
@
( (0; t)
@t
e
k
e
k
+ (0; t) e
k
+ (0; t) 1
W (0; k; t)) = (0; t) e
k
:
Hence
1
E[ 0;1 ] =
G(0; k)
Now
Z 1
t (0; t) dt =
0
since
= 0 when
Z
1
t
(0; t) 1
e
k
+ (0; t) 1
e
k
dt:
0
d
~(0; )
d
=
=0
p
d
2
c( + ) d
=
=0
p
c
= 0 and dd =0 = 1=(c
m1 ). Similarly,
Z 1
q
t (0; t) dt =
.
2
c (c
m1 )
0
14
2 (c
m1 )
Hence
1
E[ 0;1 ] =
G(0; k)
p 1
c 2 (c
e
k
q 1
+
m1 ) c 2 (c
k
e
m1 )
!
and it is straightforward to show that
G(0; k) =
p
1
c
e
k
q
1
c
+
e
k
:
Similarly,
E[
0;2 ]
=
1
(0) G(0; k)
k
pe
2
c (c
qe
+
2
m1 ) c (c
k
;
m1 )
c
giving everything required to …nd E[Tk;k
] from formula (3.4).
c
We now consider E[Tu;k ]. From formula (3.3) we see that we require
E[ u k;1 ] and E[ u k;2 ]. We again know from Dickson (2008) that wu (y; t) is
of the form
wu (y; t) = (u; t) e y + (u; t) e y :
(4.10)
The key quantities that we need to evaluate to obtain E[ u k;1 ] and E[
are
Z 1
Z 1
t (u k; t) dt and
t (u k; t) dt:
0
u k;2 ]
0
We start by deriving formulae for these, then indicate an alternative approach.
Rt
A rather complex formula for 0 (u; s) ds is givenR in Dickson (2008),
1
from which we could obtain (u; t). However, to obtain 0 t (u k; t) dt it
is considerably easier to note from Dickson (2008) that
Z 1
e t (u k; t) dt
0
=
1
n
1 XX n
~(0; )j+1 ~ (0; )n
j
n=0 j=0
j
e
(u k)
j
( (u k))n
1 F1 (j + 1; n + 1; (
n!
)(u
k)):
(4.11)
Consider
d
~(0; )j+1 ~ (0; )n
d
j
= (j + 1)~(0; )j
15
d
~(0; ) ~ (0; )n
d
j
+~(0; )j+1 (n
j)~ (0; )n
d
~ (0; )
d
j 1
giving
d
~(0; )j+1 ~ (0; )n
d
j
= (j + 1)
=0
p
c
+
=
This gives
Z
1
t (u
p
c
2
1
c
j+1
1
c
j
p
c
m1
(n
p
c
j)
m1
q
c
n j 1
c
j+1
q
n j
q
q
2
c
n j
j+1
c
1
c
+
m1
n
k; t)dt
0
=
1
(c
1 X
n
X
n
m1 ) n=0 j=0 j
j
e
(u k)
p
c
j+1
q
n j
j+1
c
+
( (u k))n
1 F1 (j + 1; n + 1; (
n!
n
j
)(u
k)):
A similar argument yields
Z 1
t (u k; t)dt
0
=
1
(c
1 X
n
X
n
m1 ) n=0 j=0 j
j
e
(u k)
q
j+1
c
( (u k))n
1 F1 (n
n!
p
c
n j
j+1
j; n + 1; (
+
n
)(u
j
k))
and we are thus able to compute E[ u k;1 ] and E[ Ru k;2 ]. It is clear that if
1
we di¤erentiate (4.11) a second time we can obtain 0 t2 (u k; t)dt. This
is a straightforward task that involves no new ideas.
A second approach uses results given in Lin and Willmot (2000). They
provide formulae for both E[Tu I(Tu < 1)] and E[Tu jU (Tu )j I(Tu < 1)],
and their methodology can be applied to …nd quantities such as
16
j
:
E[Tu2 jU (Tu )j I(Tu < 1)]. It follows from formula (4.10) that
Z 1
Z 1
E[Tu I(Tu < 1)] =
t (u; t) dt +
t (u; t) dt
0
0
and
E[Tu jU (Tu )j I(Tu < 1)] =
1
Z
1
t (u; t) dt +
0
1
Z
1
t (u; t) dt:
0
R1
R1
These two identities can then be used to …nd 0 t (u; t) dt and 0 t (u; t) dt.
Again, the decomposition of wu (y; t) is the key. If the individual claim
amount distribution was a mixture of three exponentials, say
f (x) =
3
X
wi
i
expf
i xg;
i=1
then we would have
wu (y; t) =
3
X
hi (u; t)
i
expf
i xg;
i=1
and in this case we would need to apply formulae given by LinR and Willmot
1
(2000) to …nd E[Tu jU (Tu )j2 I(Tu < 1)] in order to identify 0 t hi (u; t)dt
for i = 1; 2; 3.
c
Figure 4.2 shows E[Tu;k
] for k = 0; 1; 2 and 3 when f is given by (4.9)
with p = 1=3, = 1=2 and = 2; with = 1 and c = 1:2: Some features
of this plot are not particularly surprising –as u increases for a given value
c
c
of k, E[Tu;k
] increases and as k increases, E[Tu;k
] increases for a given value
of u. An interesting feature of the plot is that the lines are almost parallel
from around u = 5.
5
The duration of negative surplus
The distribution of the total duration of negative surplus in the classical
risk model was studied by Dickson and Egídio dos Reis (1996). In that
paper a recursion formula is given for the distribution function of the total
duration of negative surplus, and expressions are given for the density of the
duration of individual periods of negative surplus when the individual claim
amount distribution is exponential and Erlang(2). We now revisit each of
these examples and show that there is an explicit solution for the density of
the total duration of negative surplus.
17
c
Figure 4.2: E[Tu;k
] for k = 0; 1; 2 and 3
We follow the notation of Dickson and Egídio dos Reis (1996): a denotes
the density function of the …rst period of negative surplus given that ruin
occurs, d denotes the density function of any subsequent period of negative
surplus given that the surplus falls below 0 after the last recovery, and k
denotes the (defective) density function of the total duration of negative
surplus, which we denote by T T . (Note that Pr(T T = 0) = (u).) As in
Section 4.2, claims occur as a Poisson process with parameter .
5.1
Exponential claims
When f (x) = e x , x > 0, we have a(t) = d(t) = w0 (t)= (0): Thus, from
formula (3.1) of Dickson and Egídio dos Reis (1996) we have
k(t) =
(u) (0)
1
X
(0)n a(n+1) (t)
n=0
=
=
1
(u) (0) X n
w0 (t)
(0)
n=1
1
(u) (0) X
(0)
n=1
n n 1
t
18
e (
(n)
+ c)t
0 F1
n + 1; c t2 ;
(5.1)
where the …nal line follows from formula (4.3). It is straightforward to implement this formula numerically by truncating the in…nite sum. We can see
from formula (5.1) that the conditional distribution of T T given that T T > 0
is independent of u, which is a consequence of the distribution of the de…cit
at ruin, given that ruin occurs, being independent of u.
5.2
Erlang(2) claims
We now consider the case when f (x) = 2 xe x , x > 0. Dickson and Egídio
dos Reis (1996) give a formula for a(t) (which contains a typographical error
–the powers of t should be 3n) from which we obtain
a
~(s) = (1
=
c
c
( c)2
~ 2
~ + (u)
B3 (Z)
B3 (Z)
+ c+s
( + c + s)2
(u))
(u)) ~l1 (0; s) + (u) ~l2 (0; s)
(1
where Z~ is given by (4.4) and the function (u) is given in Dickson and Egídio
dos Reis (1996), and is such that (0) = 21 ; meaning that
~ =
d(s)
c
2
~l1 (0; s) + ~l2 (0; s) :
~ = ~l1 (0; s) + ~l2 (0; s) = (0),
We remark that d(t) = w0 (t)= (0), and so d(s)
with (0) = 2 =( c). It is straightforward to show (e.g. from formula (3.2)
of Dickson and Egídio dos Reis (1996) that
~
k(s)
= (u) (0)
1
X
~ n
(0)n a
~(s) d(s)
n=0
and so we get
~
k(s)
=
(u) (0)
1
X
c
(1
(u)) ~l1 (0; s) + (u) ~l2 (0; s)
~l1 (0; s) + ~l2 (0; s)
n=0
=
(u) (0)
+ (u) (0)
c
(1
c
1 X
n
X
n ~
(u))
l1 (0; s)i+1 ~l2 (0; s)n
i
n=0 i=0
1 X
n
X
n ~
(u)
l1 (0; s)i ~l2 (0; s)n
i
n=0 i=0
19
i+1
:
i
n
From our inversion of w~k;k (s) in Section 4.2 we see that this yields
k(t) =
c
(u) (0)
0 F2
(1
n+1
+ (u) (0)
c
0 F2 n +
3
2
1 X
n
X
n
(u))
i
n=0 i=0
i
3
;n +
2
2
i
;
2
1 X
n
X
n
(u)
i
n=0 i=0
i
;n + 2
2
i
;
2
n+1
( c)n i t2n i e
(2n i)!
( + c)t
2 2 3
ct
4
n+1
( c)n+1 i t2n+1 i e
(2n + 1 i)!
( + c)t
2 2 3
ct
4
:
Again it is straightforward to implement this formula numerically by truncating the in…nite sums.
References
[1] Borovkov, K.A. and Dickson, D.C.M. (2008) On the ruin time distribution
for a Sparre Andersen process with exponential claim sizes. Insurance:
Mathematics & Economics 42, 1104–1108.
[2] Dickson, D.C.M. (2008) Some explicit solutions for the joint density of
the time of ruin and the de…cit at ruin. ASTIN Bulletin 38, 259–276.
[3] Dickson, D.C.M. and Egídio dos Reis, A.D. (1996) On the distribution
of the duration of negative surplus. Scandinavian Actuarial Journal 1996,
148–164.
[4] Dickson, D.C.M. and Li, S. (2010) Finite time ruin problems for the
Erlang(2) risk model. Insurance: Mathematics & Economics 46, 12–18.
[5] Gerber, H.U. and Shiu, E.S.W. (1998). On the time value of ruin. North
American Actuarial Journal 2, 1, 48–78.
[6] Graham, R.L., Knuth, D.E. and Patashnik, O. (1994) Concrete Mathematics, 2nd edition. Addison-Wesley, Upper Saddle River, NJ.
[7] Landriault, D. and Willmot, G.E. (2009) On the joint distributions of
the time to ruin, the surplus prior to ruin, and the de…cit at ruin in the
classical risk model. North American Actuarial Journal 13, 2, 252–279.
20
[8] Lin, X.S. and Willmot, G.E. (2000) The moments of the time of ruin,
the surplus before ruin, and the de…cit at ruin. Insurance: Mathematics
& Economics 27, 19–44.
[9] Nie, C., Dickson, D.C.M. and Li, S. (2011) Minimizing the ruin probability
through capital injections. Annals of Actuarial Science 5, 2, 195–209.
21