On optimal dividend problem for an insurance risk models
with surplus-dependent premiums
Zbigniew Palmowski
Joint work with E. Marciniak
ANZAPW 2015
Zbigniew Palmowski
1 / 17
Cramér-Lundberg model
Rt = x + pt −
Nt
X
Ck
k=1
Zbigniew Palmowski
2 / 17
Cramér-Lundberg model
Rt = x + pt −
Nt
X
Ck
k=1
where
x > 0 - initial capital
Zbigniew Palmowski
2 / 17
Cramér-Lundberg model
Rt = x + pt −
Nt
X
Ck
k=1
where
x > 0 - initial capital
{Ck }k∈N - i.i.d. claim sizes with d.f. F
Zbigniew Palmowski
2 / 17
Cramér-Lundberg model
Rt = x + pt −
Nt
X
Ck
k=1
where
x > 0 - initial capital
{Ck }k∈N - i.i.d. claim sizes with d.f. F
{Nt }t≥0 - independent Poisson process with intensity λ
Zbigniew Palmowski
2 / 17
Cramér-Lundberg model
Rt = x + pt −
Nt
X
Ck
k=1
where
x > 0 - initial capital
{Ck }k∈N - i.i.d. claim sizes with d.f. F
{Nt }t≥0 - independent Poisson process with intensity λ
p - premium rate
Zbigniew Palmowski
2 / 17
Cramér-Lundberg model
Rt = x + pt −
Nt
X
Ck
k=1
where
x > 0 - initial capital
{Ck }k∈N - i.i.d. claim sizes with d.f. F
{Nt }t≥0 - independent Poisson process with intensity λ
p - premium rate
Zbigniew Palmowski
2 / 17
Surplus-dependent premium
Risk process before dividend payments soleves SDE:
Z
0
Zbigniew Palmowski
t
p(Rs )ds −
Rt = x +
Nt
X
Ck
k=1
3 / 17
Surplus-dependent premium
Risk process before dividend payments soleves SDE:
Z
t
p(Rs )ds −
Rt = x +
0
Nt
X
Ck
k=1
where p is deterministic Lipschitz continuous positive function
We assume that:
Z ∞
e−qt p(rtx )dt ≤ Ax + B
0
for some A, B ≥ 0 where rtx = x +
and that
Rt → ∞
Zbigniew Palmowski
Rt
0
p(rsx )ds
P − a.s.
3 / 17
Dividend problem
Regulated risk process Xtπ satisfies:
Xtπ
Z
=x+
0
t
p(Xsπ )ds −
Nt
X
Ck − Lπs
k=1
for some π admissible strategy where Lπt denotes cumulative amount of
dividend payments up to time t
Zbigniew Palmowski
4 / 17
Dividend problem
Regulated risk process Xtπ satisfies:
Xtπ
Z
=x+
0
t
p(Xsπ )ds −
Nt
X
Ck − Lπs
k=1
for some π admissible strategy where Lπt denotes cumulative amount of
dividend payments up to time t
Remark !
In this model for p 6= const it is not true that:
Xtπ = Rt − Lπt
Zbigniew Palmowski
4 / 17
Value function
We want to maximize the mean of discounted cumulative dividend
payments collected up to ruin time taking into account penalty of at the
ruin:
Z T
−qs
π
e dLs + Ex e−qT w (XTπ ) ; T < ∞
vπ (x) = Ex
0
where:
T = T π := inf {t ≥ 0 : Xtπ < 0} - ruin time
q - discounted rate
w - penalty function
Ex - expectation w.r. Px (·) = P(·|X0 = x)
Zbigniew Palmowski
5 / 17
Value function
We want to maximize the mean of discounted cumulative dividend
payments collected up to ruin time taking into account penalty of at the
ruin:
Z T
−qs
π
e dLs + Ex e−qT w (XTπ ) ; T < ∞
vπ (x) = Ex
0
where:
T = T π := inf {t ≥ 0 : Xtπ < 0} - ruin time
q - discounted rate
w - penalty function
Ex - expectation w.r. Px (·) = P(·|X0 = x)
Goal: Find value fnction v and optimal strategy π ∗ s.t.
v(x) = sup vπ (x) = vπ∗ (x)
π∈Π
Zbigniew Palmowski
5 / 17
Barrier strategy
a
Zbigniew Palmowski
6 / 17
Verification theorem
HJB equation:
max Ak(x) − qf (x), 1 − k 0 (x) = 0 for x ≥ 0
k(x) = w(x) for x < 0
R∞
where Ak(x) = p(x)k 0 (x) + 0 (k(x − y) − k(x)) λdF (y) is full
generator of risk process R and k 0 is a density of k
Zbigniew Palmowski
7 / 17
Verification theorem
HJB equation:
max Ak(x) − qf (x), 1 − k 0 (x) = 0 for x ≥ 0
k(x) = w(x) for x < 0
R∞
where Ak(x) = p(x)k 0 (x) + 0 (k(x − y) − k(x)) λdF (y) is full
generator of risk process R and k 0 is a density of k
Verification theorem
Let π be an admissible dividend strategy such that vπ is absolutely
continuous ultimately dominated by some affine function. If HJB equation
holds for vπ then vπ (x) = v(x) for all x ≥ 0.
Zbigniew Palmowski
7 / 17
Idea of the proof
For any closed interval I ⊂ [0, ∞) we define:
TI := inf{t ≥ 0 : Rt ∈
/ I}.
By G we denote the family of function g such that
n
o
g,TI
−q(t∧TI )
M
:= e
g (Rt∧TI ) , t ≥ 0
is a supermartingale
and that g 0 (x) ≥ 1 for x > 0
g(x) ≤ w(x) for x < 0
and g is dominated by some linear function
Zbigniew Palmowski
8 / 17
Dual representation
Theorem
v(x) = min g(x)
g∈G
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9 / 17
Barrier strategy
a
Zbigniew Palmowski
10 / 17
Barrier strategy
We define
+
+
lim Ex [e−qτy , τy+ < τ0− ]/E0 [e−qτy , τy+ < τ0− ]
y→∞
h
i
−
Gq,w (x) = Ex e−qτ0 w(Rτ − )I{τ − <∞}
Wq (x) =
0
0
where τa+ = inf{t ≥ 0 : Rt > a}, τ0− = inf{t ≥ 0 : Rt < 0}
Zbigniew Palmowski
11 / 17
Barrier strategy
We define
+
+
lim Ex [e−qτy , τy+ < τ0− ]/E0 [e−qτy , τy+ < τ0− ]
y→∞
h
i
−
Gq,w (x) = Ex e−qτ0 w(Rτ − )I{τ − <∞}
Wq (x) =
0
0
where τa+ = inf{t ≥ 0 : Rt > a}, τ0− = inf{t ≥ 0 : Rt < 0}
Then for x ∈ (0, a):
h
i W (x)
+
q
Ex e−qτa I{τa+ <τ − } =
0
Wq (a)
Zbigniew Palmowski
11 / 17
Barrier strategy
Theorem
Let d.f. F of claim size has density. For a barrier strategy πa we have:

Wq (x)
0


 Wq0 (a) 1 − Gq,w (a) + Gq,w (x) x ≤ a
va (x) = vπa (x) =


x − a + v (a)
x>a
a
Moreover va is continuously differentiable for all x ≥ 0.
Zbigniew Palmowski
12 / 17
Denote:
Hq0 (y) :=
Wq0 (y)
1 − G0q,w (y)
Optimal barrier
Let
a∗ = sup a ≥ 0 : Hq0 (a) ≤ Hq0 (x) for all x ≥ 0
where Hq0 (0) = limx↓0 Hq0 (x).
Zbigniew Palmowski
13 / 17
Denote:
Hq0 (y) :=
Wq0 (y)
1 − G0q,w (y)
Optimal barrier
Let
a∗ = sup a ≥ 0 : Hq0 (a) ≤ Hq0 (x) for all x ≥ 0
where Hq0 (0) = limx↓0 Hq0 (x).
Theorem
Suppose that
Hq0 (a) ≤ Hq0 (b)
for all a∗ ≤ a ≤ b.
Then the barrier strategy at a∗ is optimal, that is v(x) = va∗ (x).
Zbigniew Palmowski
13 / 17
Example
Assume that claim size has exponential distribution with intensity µ
Let
p(x) = c + x
describing investments of the surplus into a bond with a fixed interest rate
>0
Functions Wq i Gq,w solve:
AWq (x) = qWq (x) for x ≥ 0
Wq (x) = 0 for x < 0
and
AGq,w = qGq,w (x) + ω(x) for x ≥ 0
Gq,w (x) = w(x)
where
Z
for x < 0
∞
w(x − z)dF (z)
ω(x) =
x
Zbigniew Palmowski
14 / 17
Example
We have: limx→∞ Wq (x) = +∞ and limx→∞ Gq,w (x) = 0
It means that optimal value function under mild conditions is a linear
combination of two Gerber-Shiu functions: unstable one that disappears on
negative half-line and tends to infinity at infinity (corresponding to dividend
payment, Wq in our notation) and stable one disappearing at infinity
(corresponding to penalty payment, Gq,w in our notation).
Zbigniew Palmowski
15 / 17
Example
Denote:
s1 (x) = U
q
+ 1,
λ+q
+ 1, µx +
µc (x
+ c)
λ+q
exp(−µx)
and
Γ(q/+1)
1 µ (λ+q)/
(x + c)(λ+q)/ exp(−µx − µc
Gu(x) = Γ((q+λ)/(1+))
)×
Z x
Z ∞
Z ∞
(0)
− U (x)
U
(x)
M (v) − M (x)
U (v) + M
U
(v)
u(v) dv,
U (0)
0
x
0
where U (u) and M (u) are Kummer functions and
∂
λ
( ∂x
+ µ) ω(x). Then
u(x) = − p(x)
Gq,w (x) = s1 (x) + Gu(x),
Moreover,
Wq (x) = M
Zbigniew Palmowski
q
+ 1,
λ+q
+ 1, µx +
µc (x
+ c)
λ+q
exp(−µx)
16 / 17
Example
We have:
Gq,w (x) = s1 (x) + Gu(x),
Wq (x) = M
q
+ 1,
λ+q
+ 1, µx +
Hence:
Hq0 (x) =
µc (x
+ c)
λ+q
exp(−µx)
Wq0 (x)
1 − G0q,w (x)
00
and we can identify optimal barrier a∗ solving Hq (a∗ ) = 0.
Zbigniew Palmowski
17 / 17
THANK YOU
for Your Attention !
Zbigniew Palmowski
18 / 17