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The time of deducting fees for variable annuities under the state-dependent fee structure
Insurance: Mathematics and Economics 61 (2015) 125–134
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Insurance: Mathematics and Economics journal homepage: www.elsevier.com/locate/ime
The time of deducting fees for variable annuities under the state-dependent fee structure Jiang Zhou a,∗ , Lan Wu a,b a
School of Mathematical Sciences, Peking University, Beijing 100871, PR China
b
Key Laboratory of Mathematical Economics and Quantitative Finance, Peking University, Beijing 100871, PR China
article
info
Article history: Received August 2014 Received in revised form December 2014 Accepted 28 December 2014 Available online 5 January 2015 Keywords: Variable annuities State-dependent fee Hyper-exponential jump diffusion process Laplace transform Refracted Lévy process
abstract We investigate the total time of deducting fees for variable annuities with state-dependent fee. This fee charging method is studied recently by Bernard et al. (2014) and Delong (2014) in which the fees deducted from the policyholder’s account depend on the account value. However, both of them have not considered the problem of analyzing probabilistic properties of the total time of deducting fees. We approximate the maturity of a general variable annuity contract by combinations of exponential distributions which are (weakly) dense in the space that is composed of all probability distributions on the positive axis. Working under general jump diffusion process, we derive analytic formulas for the expectation of the time of deducting fees as well as its Laplace transform. © 2015 Elsevier B.V. All rights reserved.
1. Introduction Variable Annuities (VAs) are generally issued with minimum guarantee on the death or maturity benefits and keep in high demand due to the embedded guarantees. Nowadays, a variety of guarantees are provided, such as Guaranteed Minimum Death Benefits (GMDBs), Guaranteed Minimum Maturity Benefits (GMMBs), Guaranteed Minimum Accumulation Benefits (GMABs) and so on, see Bauer et al. (2008). In the academic circle, VAs have received substantial attention, see among others, Lee (2003), Gerber and Shiu (2003), Ko et al. (2010), Bacinello et al. (2011) and Gerber et al. (2012, 2013). In general, insurers charge expenses for the provision of the guaranteed benefits from the policyholder’s account by a fixed rate, and most actuarial literatures assume that the fee rate is fixed as well, see e.g. Hyndman and Wenger (2014). However, this fixed fee structure has several disadvantages which have been noted by Bernard et al. (2014) and Delong (2014). For example, as the guaranteed benefits embedded in variable annuities are similar to put options, if the account value is high, the guarantees are (deep) outof-the-money. In this case, a higher deducted fee will yield incentives for policyholders to lapse the policy, see Bauer et al. (2008). Recently, Bernard et al. (2014) put forward a dynamic fee structure. In their paper, the fees are deducted at a fixed rate only if the account value is lower than a pre-specified level. They have derived formulas for calculating the fee rates under the Black–Scholes
∗
Corresponding author. E-mail addresses: [email protected] (J. Zhou), [email protected] (L. Wu).
http://dx.doi.org/10.1016/j.insmatheco.2014.12.008 0167-6687/© 2015 Elsevier B.V. All rights reserved.
model. Delong (2014) then extended their model to an incomplete financial market which is made up of two risk assets that are modeled by a two-dimensional Lévy process, and he considered a general state-dependent fee which is determined by a function of the account value. In his paper, the fee for the guaranteed benefit can be computed by solving an equation, and a strategy for hedging the guarantee is also characterized. Under a state-dependent fee structure, sometimes insurers do not have any fees income, e.g., when the account value is higher than a pre-specified level in Bernard et al. (2014). Therefore, insurers are interested in the problem that how long they can collect fees or alternatively, how long they cannot. In nature, insurers are interested in the probabilistic properties of the total time of collecting fee, for example its expectation. Besides, this total time will determine insurers’ income directly and have a significant impact on the account value indirectly. The above problem has not been considered in Bernard et al. (2014) and Delong (2014), and we are the first to investigate it to our knowledge. In this paper, we assume that the fees are deducted as in Bernard et al. (2014), which is one of the fee charging structures investigated in Delong (2014). Under the simple Black–Scholes model as well as the complex hyper-exponential jump diffusion process model, we obtain analytic formulas for the Laplace transform of the total time of deducting fees and its expectation as well. The rest of this paper is organized as follows. Our model is introduced in Section 2. In Section 3, under the hyper-exponential jump diffusion process, we derive our main results. In Section 4, three special cases of the general model are discussed in detail
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J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
where the formulas become simpler. Finally, some conclusions are given in Section 5.
We first consider the case that the maturity is Tz . Under the fee structure (2.3) and (2.4), the insurer is interested in the expectation of the total time of charging fees:
2. The model
X t = X 0 + µt + σ W t +
Nt
Zk ,
(2.2)
which can be explained as the value of one unity of the reference fund or stock underlying the variable annuity contract, and it is reasonable to assume that E [X1 ] > 0. According to (2.6) in Delong (2014), if we let Ft denote the policyholder’s account value at time t, then its dynamics are represented by the following stochastic differential equation (SDE) dSt − g (Ft − )dt , t > 0, (2.3) St − and F0 is the initial premium invested by the insured, and g is a function which represents a state-dependent fee charged by the insurer. In Bernard et al. (2014), they assumed that X is a Brownian motion and g is given by
dFt = Ft −
(2.4)
where α and B denote the deduction fee rate and a pre-specified level, respectively; and 1A is the indicator function of a set A. Function (2.4) means the expense is charged only when the account value is less than the pre-specified level B. In this paper, we consider the case that g satisfies Eq. (2.4) and the investigation of general g is left for future research. For the advantages of this fee deducting method, see Bernard et al. (2014) and Delong (2014). Let Ut satisfy the following SDE dUt = dXt − α 1{Ut
t > 0,
and U0 = X0 ,
=
1{Us
E
(2.7)
0
0
Remark 2.1. To evaluate (2.7), the value of α in (2.5) is required to be specified first. In principle, the value of α is determined by risk neutral pricing rule for some martingale measure. Here, we are more interested in obtaining the probability characteristic of time of deducting fee. Therefore, in this paper, we assume that the parameter α is exogenous and focus on the calculation of (2.7) with given α . From Ko and Ng (2007) or Dufresne (2007), we obtain the following result: in the space of all probability distribution on the positive axis, the subset of linear combinations of exponential distributions is (weakly) dense. Therefore, we can approximate fTz (t ) by i ci fTi (t ), where fTi (t ) is the density function for some exponential distributions. Hence, the problem of calculating (2.7) reduces to compute the following expectation: τ
E
1{Ut
(2.8)
0
where τ is an exponential random variable which is independent of the process U. For a contract with maturity date T, we first note that (2.8) is
T
related to the Laplace transform of E
0
1{Us
the contract expires at a fixed time T , the total time of deducting fees can also be obtained from (2.8) through taking inverse Laplace transform for instance. In addition, if the maturity is min{Tz , T }, then we have min{Tz ,T }
1{Us
E 0
Tz
T
1{Us
=E 0
1{Us
0
= P(Tz < T )
+∞
E
0
+ P(Tz ≥ T )E
t
1{Us
0 T
1{Us
(2.9)
0
(2.5)
B F0
1{Ut
t
where fTz (t ) denotes the probability density function of Tz . (2.1)
where µ and X0 are constants; {Wt ; t ≥ 0} is a standard Brownian motion with W0 = 0, and σ > 0 is the volatility of the diffusion; {Zk ; k = 1, 2, . . .} are independent and identically distributed random variables supported on R/{0}, and the common probability density function (pdf) of {Zk ; k = 1, 2, . . .} is denoted by fZ (z ); {Nt ; t ≥ 0} is a Poisson process with rate λ; moreover, {Wt }, {Nt } and {Zk ; k = 1, 2, . . .} are independent. The law of X starting from x is denoted by Px and Ex represents the corresponding expectation, and for the sake of brevity, we write P and E when x = 0. We define the process S = (St )t ≥0
g (x) = α x1{x
∞
0
k=1
St = S0 eXt −X0 ,
E
Suppose X = (Xt )t ≥0 is a jump diffusion process, i.e.,
where b = ln
Tz
. Then, when U0 = 0, we obtain the following
equation of Ft and Ut by Itô’s formula. Ft = F0 eUt , t ≥ 0. (2.6) Because the parameter σ > 0 and the jump part of X is a compound Poisson process, we have the following lemma of the existence of a unique strong solution of (2.5). For the proof of it, one can see Remarks 2 and 3 in Kyprianou and Loeffen (2010) and the proof of Lemma 2.1 in Delong (2014) for reference. Lemma 2.1. Eq. (2.5) exists a unique strong solution. Moreover the solution U is a strong Markov process. Consider a customer age z purchasing a VA contract. If the guarantee embedded is GMMBs, then the maturity of the contract is a given date T . For the guarantee of GMDBs, the maturity is given by min{Tz , T } for endowment policy or Tz for whole life (see Gerber et al., 2012 for example), where Tz is the random variable representing the time of death of the insured and independent of the account value process Ft . In this paper, we ignore other reasons of terminating the contract, e.g., the customer lapses the policy.
where gTz |Tz
ν e−ν t ,
t > 0,
(2.10)
where ν > 0 is a constant. Of course, it is very difficult (or even impossible) to calculate (2.8) for arbitrary pdf fZ (z ). In this paper, we assume that fZ (z ) is given by fZ (z ) =
m i=1
pi ηi e−ηi z 1{z >0} +
n
qj ϑj eϑj z 1{z <0} ,
(2.11)
j =1
where pi > 0, ηi > 0 for all i = 1, . . . , m; qj > 0, ϑj > 0 for all m n j = 1, . . . , n; i=1 pi + j=1 qj = 1, η1 < η2 < · · · < ηm and ϑ1 < ϑ2 < · · · < ϑn . Remark 2.2. The process X defined by (2.1) and (2.11) is called hyper-exponential jump diffusion process (HEP). The HEP has sev-
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
eral advantages and has been used to model the asset return in many papers, see e.g. Cai et al. (2009), here we just emphasize that the hyper-exponential distribution (2.11) can be used to approximate many complicated pdf fZ (z ). Therefore, for more general process X under which (2.8) cannot be solved, applying our results can obtain a very good approximation. Before deriving our results, we put up some notation. We set Y = {Yt := Xt − α t ; t ≥ 0} which is also a HEP. The law of Y start˜ y and the corresponding expectation is ing from y is denoted by P ˜ y ; and we write P˜ and E˜ when y = 0 for the sake of denoted by E brevity. For any h, H ∈ R, we define the following stopping times of X , Y and U.
τh−,X := inf{t ≥ 0 : Xt ≤ h},
τH+,X := inf{t ≥ 0 : Xt ≥ H }, (2.12)
τh−,Y := inf{t ≥ 0 : Yt ≤ h},
τH+,Y := inf{t ≥ 0 : Yt ≥ H }, (2.13)
κh− := inf{t ≥ 0 : Ut ≤ h},
κH+ := inf{t ≥ 0 : Ut ≥ H }. (2.14)
3. Main results For the HEP X given by (2.1) and (2.11), the Lévy exponent is given by
σ 2 2 z + µz ψ(z ) := ln E ezX1 = 2 m n pi η i q j ϑj +λ + −1 , ηi − z ϑj + z i =1 j=1
dUt = µdt + σ dWt + d
(3.1)
= ψ(z ) − α z. According
Zk
− α 1{Ut
Ex e
−qτh−,X
1BX
−γ (x−h) e 1,q −qτ − 0,h e−γ2,q (x−h) Ex e h,X 1BX 1,h = Qq−1 , .. .. . . −γn+1,q (x−h) − Ex e
−qτh,X
(3.5)
e
1BX
n,h
where
ϑ1 ϑ − γ1,q 1 ϑ 1 1 ϑ − γ2,q 1 Qq = . . .. .. ϑ1 1 ϑ1 − γn+1,q
ϑ2 ϑ2 − γ1,q ϑ2 ϑ2 − γ2,q .. . ϑ2 ϑ2 − γn+1,q
1
ϑn ϑn − γ1,q ϑn ··· ϑn − γ2,q , (3.6) .. .. . . ϑn ··· ϑn − γn+1,q ···
and BX0,h = {Xτ − = h}, BXj,h = {X crosses the boundary h by a jump h,X of an exp(ϑj )-distributed random variable }, j = 1, 2, . . . , n.
˜x e E
−qτH+,Y
1B˜ Y
˜ 0,H eβ1,q (x−H ) ˜ −qτH+,Y β˜ 2,q (x−H ) Ex e 1B˜ Y e 1,H , = Q˜ q−1 . .. .. . β˜ m+1,q (x−H ) + e ˜ x e−qτH ,Y 1B˜ Y E
(3.7)
m,H
Nt
Lemma 3.2. For q > 0 and x > h,
Lemma 3.3. For q > 0 and x < H,
for z ∈ (−ϑ1 , η1 ). By analytic continuation, the function ψ(z ) can be extended to the whole complex plane except at ηi , i = 1, . . . , m and −ϑj , j = 1, . . . , n. Recalling the definition of Y , we have the
˜ z ) := ln E˜ ezY1 Lévy exponent of Y , ψ( to (2.5), the process (Ut )t ≥0 satisfies
127
t > 0,
k=1
where (3.2)
and U0 = X0 . The following lemma is taken from Lemmas 2.1 and 2.2 in Cai (2009), thus the proof is omitted. Lemma 3.1. For each given q ≥ 0, we have the following results. (1) For q > 0, the equation ψ(z ) = q has exactly (m + n + 2) real roots β1,q , β2,q , . . . , βm+1,q , −γ1,q , −γ2,q , . . . , −γn+1,q which satisfy 0 < β1,q < η1 < β2,q < · · · < ηm < βm+1,q < +∞, 0 < γ1,q < ϑ1 < γ2,q < · · · < ϑn < γn+1,q < +∞.
(3.3)
(2) For q > 0, the equation ψ(z ) − α z = q has exactly (m + n + 2) real roots β˜ 1,q , β˜ 2,q , . . . , β˜ m+1,q , −γ˜1,q , −γ˜2,q , . . . , −γ˜n+1,q which satisfy 0 < β˜ 1,q < η1 < β˜ 2,q < · · · < ηm < β˜ m+1,q < +∞, 0 < γ˜1,q < ϑ1 < γ˜2,q < · · · < ϑn < γ˜n+1,q < +∞.
(3.4)
(3) If q = 0, all the results in (1) hold except that β1,0 = 0 and γ1,0 > 0 (because we assume that E [X1 ] > 0); and all the results in (2) hold except that the signs of β˜ 1,0 and γ˜1,0 depend on the ˜ [Y1 ]. sign of E (4) For any i = 1, 2, . . . , m + 1, βi,q and β˜ i,q are continuous and strictly increasing functions of q on the interval [0, +∞). (5) For any j = 1, 2, . . . , n + 1, γj,q and γ˜j,q are continuous and strictly increasing functions of q on the interval [0, +∞). The following lemmas have been derived by Yin et al. (2013) (see Lemmas 2.2, 2.4 and 2.5), and one can refer to that paper for the details of the proofs.
1
1 Q˜ q = . ..
η1
η2
η1 − β˜ 1,q η1 η1 − β˜ 2,q
η2 − β˜ 1,q η2 η2 − β˜ 2,q
.. . η1 1 η1 − β˜ m+1,q
ηm
···
ηm − β˜ 1,q ηm ηm − β˜ 2,q
···
.. .. .. . . . η2 ηm ··· η2 − β˜ m+1,q ηm − β˜ m+1,q
, (3.8)
and B˜ Y0,H = {Yτ + = H }, B˜ Yi,H = {Y crosses the boundary H by a jump H ,Y of an exp(ηi )-distributed random variable }, i = 1, 2, . . . , m.
−qτh−,X
Remark 3.1. The expressions of Ex e −qτ − e h,X 1BX n,h
Ex
˜x and E
−qτ + e H ,Y 1B˜ Y 0,H
1BX
0,h
,..., +
, . . . , E˜ x e−qτH ,Y 1B˜ Y
m,H
can
be obtained by solving (3.5) and (3.7), see Theorems 2.1 and 2.2 in Yin et al. (2013). Lemma 3.4. (i) Consider any nonnegative measurable function g 0 such that −∞ g (h + y)eϑj y dy < ∞ for all j = 1, 2, . . . , n. For q > 0 and x > h, we have
Ex e
−qτh−,X
+
n j =1
g (Xτ − ) = g (h)Ex e h,X
−
ϑj Ex e−qτh,X 1BX
j,h
−qτh−,X
0
1BX
0,h
g (h + y)eϑj y dy.
(3.9)
−∞
(ii) Consider any nonnegative measurable function g such that ∞ g ( H + y)e−ηi y dy < ∞ for all i = 1, 2, . . . , m. For q > 0 0
128
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
and x < H, we have
˜x e E
−qτH+,Y
H ,Y
m
+
−qτH+,Y
˜x e g (Yτ + ) = g (H )E
˜x e E
−qτH+,Y
i,H
i=1
0,H
∞
1B˜ Y
1B˜ Y
g (H + y)ηi e−ηi y dy.
(3.10)
0
With the help of the above τ lemmas, we can derive formulas for the Laplace transform of 0 1{Us
τ
0 1{Ut
Ex e
τ
0 1{Ut
−θ
=
+1 ν m ˜ + Fi eβi,ξ (x−b) , ξ i=1
=1+
n+1
Gj e
−γj,ν (x−b)
x ≤ b, (3.11)
,
where ξ = θ +ν ; Fi , i = 1, 2, . . . , m+1, and Gj , j = 1, 2, . . . , n+1, are given by
β˜ j,ξ
Fi =
n +1
γj,ν
m
(β˜ i,ξ − ηj )
n
Gj = −
j=1,j̸=i
j =1
m+1
n +1
β˜ i,ξ
j =1
γi,ν
m
(γj,ν + ηi )
i=1,i̸=j
i =1
(ϑi − γj,ν )
0 1{Us
ν ν ν , M1 (b) − , . . . , Mm (b) − ξ ξ ξ Y = (y1 , y2 , . . . , yn+1 )
V (x; b) = Ex
(3.13)
νe
e
t
0 1{Us
= Ex
ν e−ν t e−θ t dt + Ex
0
dt
+
κb
ν e−ν t e−θ
t
0 1{Us
dt
P˜ x and {Ut , t < κb+ } under Px have the same law when x < b. Applying formula (3.10) with g (x) = V (x; b) to (3.14) yields
+ ν ν ˜ x e−ξ τb,Y 1B˜ Y V (x; b) = + E V (b; b) − 0,b ξ ξ m +∞ ν −ξ τb+,Y −ηi y ˜ + Ex e 1B˜ Y V (b + y; b)ηi e dy − , (3.15) i,b ξ 0 i =1 where ξ = θ + ν . For x > b, using again the strong Markov property, we can obtain the following equality from (3.9). +∞
ν e−ν t e
−θ
t
1 ds κb− {Us
.. .
β˜ m+1,ξ (x−b)
,
e
+∞
b,Y
κb−
˜
where the third equality follows from the strong Markov property; and in the fourth equality, we have used that {Yt , t < τb+,Y } under
V (x; b) = Ex
eβ1,ξ (x−b)
Eβ˜ =
+ + ν = 1 − Ex e−(ν+θ)κb + Ex e−(ν+θ)κb V (Uκ + ; b) b ν+θ + ν −(ν+θ)τ b,Y ˜x e 1−E = ν+θ + + E˜ x e−(ν+θ)τb,Y V (Yτ + ; b) , (3.14)
Q˜ ξ−1 ,
(3.19)
and Eβ˜ and Eγ are given by
0
κb+
V (b; b) −
−ν t −θ
(3.18)
:= (V (b; b) − 1, N1 (b) − 1, . . . , Nn (b) − 1) Qν−1 ,
.
+∞
j = 1 , 2 , . . . , n.
X = (x1 , x2 , . . . , xm+1 )
For x < b, we have
(3.17) V (y; b)ϑj eϑj (y−b) dy,
By using Lemmas 3.2, 3.3 and formulas (3.15), (3.16), we have
i=1
τ
i = 1, 2, . . . , m,
−∞
:=
Proof. (1) We define a function of x as V (x; b) = Ex e−θ
V (y; b)ηi e−ηi (y−b) dy,
where X and Y are defined as follows:
θ i =1 i =1 i =1 i=1,i̸=j . m n n +1 m +1 ξ ˜ η i ϑi (γi,ν − γj,ν ) (βi,ξ + γj,ν ) i=1
Nj (b) :=
b b
ν = X Eβ˜ , for x < b, ξ V (x; b) − 1 = Y Eγ , for x > b, (3.12)
n
−∞
V ( x; b) −
(β˜ i,ξ + ϑj )
θ j=1,j̸=i j =1 j =1 j =1 , m n m +1 n +1 ξ ηj ϑj (β˜ i,ξ − β˜ j,ξ ) (β˜ i,ξ + γj,ν ) j =1
+∞
Mi (b) :=
x > b,
j =1
m+1
j,b
j =1
where we have used the fact that {Xt , t < τb−,X } and {Ut , t < κb− } under Px have the same law for x > b in the third equality. To simplify the expressions, we denote
Theorem 3.1. For any θ > 0, we have
Ex e−θ
− − = Ex e−νκb V (Uκ − ; b) + Ex 1 − e−νκb b −ντ − −ντb−,X V (Xτ − ; b) + Ex 1 − e b,X = Ex e b,X −ντb−,X 1BX (V (b; b) − 1) = 1 + Ex e 0,b n 0 −ντ − V (b + y; b)ϑj eϑj y dy − 1 , (3.16) + Ex e b,X 1BX
dt
− + Ex 1 − e−νκb
e−γ1,ν (x−b)
.. .
Eγ =
e
−γn+1,ν (x−b)
.
(3.20)
Therefore, if we can derive the expressions of X and Y , then we obtain the expressions of V (x; b) from (3.18). (2) In this step, we derive formulas for X and Y . From the definitions of X and Y , we have
(N1 (b) − 1, N2 (b) − 1, . . . , Nn (b) − 1) = Y Bν , ν ν ν = X B˜ ξ , M1 (b) − , M2 (b) − , . . . , Mm (b) − ξ ξ ξ
(3.21)
where the matrixes Bν and B˜ ξ are obtained by deleting the first column of Qν and Q˜ ξ respectively. Moreover, m+1
xi +
i =1
n+1 ν = yj + 1, ξ j =1
(3.22)
which comes from the fact that both sides of (3.22) are equal to V (b; b); thus we have m+1
i =1
xi −
n +1 j =1
yj =
θ . ξ
(3.23)
We multiply both sides of (3.15) by ϑj eϑj (x−b) , for j = 1, 2, . . . , n, and integrate with respect to x from −∞ to b. From Lemma 3.3, we obtain the following equation after some calculations.
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
N1 (b) − N2 (b) − .. . Nn (b) −
ν
ν
ϑ1
ϑ1 + β˜ 1,ξ ϑ1 ϑ + β˜ 1 2,ξ P˜ ξ = .. . ϑ1 ϑ1 + β˜ m+1,ξ
ϑ2
(3.24)
ϑn
···
ϑ2 + β˜ 1,ξ ϑ2 ϑ2 + β˜ 2,ξ
.. . ϑ2 ϑ2 + β˜ m+1,ξ
ϑn + β˜ 1,ξ ϑn ϑn + β˜ 2,ξ
··· .. .
.. . ϑn ϑn + β˜ m+1,ξ
···
β˜ 1,ξ η1 − β˜ 1,ξ .. . ˜ β1,ξ ηm − β˜ 1,ξ ˜ β1,ξ ˜ M = ϑ1 + β1,ξ .. . β˜ 1,ξ ϑn + β˜ 1,ξ β˜ 1,ξ
. (3.25)
1
Similar to derive (3.24), using Lemma 3.2 and formula (3.16), we have V (b; b) − 1 M1 (b) − 1 N1 (b) − 1 M2 (b) − 1 − 1 ′ = (Q Pν ) , .. .. ν . . Nn (b) − 1 Mm (b) − 1
(3.26)
η1 η + 1 γ1,ν η1 η1 + γ2,ν Pν = .. . η1 η1 + γn+1,ν
··· ··· .. . ···
ηm ηm + γ1,ν ηm ηm + γ2,ν . (3.27) .. . ηm ηm + γn+1,ν ′
.. .
.. .
.. .
.. .
···
β˜ m+1,ξ ηm − β˜ m+1,ξ
γ1,ν ηm + γ1,ν
···
···
β˜ m+1,ξ ϑ1 + β˜ m+1,ξ
γ1,ν ϑ1 − γ1,ν
···
.. .
.. .
.. .
.. .
···
β˜ m+1,ξ ϑn + β˜ m+1,ξ
γ1,ν ϑn − γ1,ν
···
···
β˜ m+1,ξ
γ1,ν
···
···
1
−1
···
γn+1,ν η1 + γn+1,ν .. . γn+1,ν ηm + γn+1,ν γn+1,ν ϑ1 − γn+1,ν . .. . γn+1,ν ϑn − γn+1,ν γn+1,ν −1
= 1, 2, . . . , m + 1 and j =
j =1
β˜ j,ξ
n+1
γj,ν
m
(β˜ i,ξ − ηj )
j=1,j̸=i
j =1 m+1
n
i =1
β˜ i,ξ
j =1
(3.33)
n +1
m
n
γi,ν (γj,ν + ηi ) (ϑi − γj,ν ) θ i =1 i=1 i =1 i =1 , m n n +1 m +1 ξ ˜ η i ϑi (γi,ν − γj,ν ) (βi,ξ + γj,ν )
yj γj,ν = −
i=1,i̸=j
i =1
thus we complete the proof.
i=1
1
˜ ξ = Y Dν , XD
Remark 3.2. For all i = 1, 2, . . . , m + 1 and j = 1, 2, . . . , n + 1, we have Fi > 0 and Gj < 0 in Theorem 3.1. Moreover,
(3.28) m+1
˜ ξ and Dν are given by where D β˜ 1,ξ β˜ 2,ξ ˜ξ = D . , .. β˜ m+1,ξ
···
(β˜ i,ξ + ϑj ) θ j =1 j =1 j=1 j=1 = m , n m +1 n +1 ξ ˜ ˜ ˜ ηj ϑj (βi,ξ − βj,ξ ) (βi,ξ + γj,ν )
It follows from the condition V (b−; b) = V (b+; b) (the proof will be given in the Appendix) and formula (3.18) that ′
γ1,ν η1 + γ1,ν
Solving Eq. (3.31), for all i 1, 2, . . . , n + 1, we have
xi β˜ i,ξ
η2 η2 + γ1,ν η2 η2 + γ2,ν .. . η2 η2 + γn+1,ν
β˜ m+1,ξ η1 − β˜ m+1,ξ
(3.32)
m+1
where Pν is given by
···
where P˜ ξ is given by
129
where
V (b; b) − ξ ξ ν ν M1 (b) − ξ ξ , ˜ −1 ˜ ′ = (Qξ Pξ ) .. . ν ν Mm (b) − ξ ξ
Fi <
i=1
−γ1,ν −γ2,ν Dν = .. . .
(3.29)
Using (3.19), we can write (3.24) and (3.26) as follows: Y Bν +
(3.30)
M (x1 , . . . , xm+1 , y1 , . . . , yn+1 )′ =
0, 0, . . . , 0,
θ ξ
′
n+1
Gj > −1,
(3.34)
j =1
which can be proved by using the fact that 0 < Eb e−θ < 1.
+∞ 0
1{Ut
m+1
=
Fi0 eβi,θ (x−b) , ˜
0 1{Ut
x ≤ b,
i=1
,
τ
Corollary 3.1. For any θ > 0,
Ex e−θ
Combining equalities (3.23), (3.28) and (3.30), we obtain
and
From Lemma 3.1, for θ > 0 and i = 1, . . . , m + 1, j = 1, . . . , n + 1, we have limν↓0 β˜ i,ξ = β˜ i,θ > 0 and limν↓0 γj,ν = γj,0 > 0. Therefore, if we let ν ↓ 0 in Theorem 3.1, then the Laplace +∞ transform of 0 1{Us
−γn+1,ν
θ (1, 1, . . . , 1) = X P˜ξ , ξ θ X B˜ ξ − (1, 1, . . . , 1) = Y Pν . ξ
θ ξ
(3.31)
Ex e
−θ 0+∞ 1{Ut
=1+
n +1
(3.35) G0j e−γj,0 (x−b)
,
x > b,
j=1
1 This equality means that the right derivative of V at b equals its left derivative.
where Fi0 , i = 1, 2, . . . , m + 1, and G0j , j = 1, 2, . . . , n + 1, are given by
130
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134 m+1
β˜ j,θ
j=1,j̸=i
Fi0 =
m
β˜ i,θ
i =1
G0j = −
m
ηi
i=1
i=1,i̸=j
m
γi,0
(γj,0 + ηi )
(3.36)
(ϑi − γj,0 )
i=1
(γi,0 − γj,0 )
i=1,i̸=j
(β˜ i,θ + γj,0 )
i =1
following corollary. Corollary 3.2. 1{Ut
Ex
1
=
τ
1{Ut
=
ν
0
m+1
1−
Hi e
β˜ i,ν (x−b)
,
(3.38)
Ij e−γj,ν (x−b) ,
x > b,
j =1
where Hi , i = 1, 2, . . . , m + 1, and Ij , j = 1, 2, . . . , n + 1, are given by m+1
j=1,j̸=i
Hi =
m
ηj
j =1 m+1
m i=1
n +1
n
β˜ i,ν
ηi
γj,ν
j =1
ϑj
j =1
i=1
Ij =
β˜ j,ν
j=1,j̸=i
n+1
i=1,i̸=j
n i=1
ϑi
m
γi,ν
i=1,i̸=j
,
n +1
(β˜ i,ν + γj,ν )
j =1
m
(γj,ν + ηi )
(γi,ν − γj,ν )
n
(3.39)
(ϑi − γj,ν )
i=1
τ
1{Ut ≥b} dt
Ex 0
m+1
Remark 3.5. As functions of x, Ex
Ex e
τ
(3.41)
τ 0
t
0 1{Us
≈ (−1)n−1
∂ n−1 (V (x; b)/ν) n ν /Γ (n) |ν=n/t . ∂ν n−1
(3.42)
Similarly, t
1{Us
Ex 0
≈ (−1)
n −1
τ ∂ n−1 Ex 0 1{Us
which can be computed from Corollary 3.2. The calculations of (3.42) and (3.43) are complicated when the value of n is large, whereas this method becomes very easy if X is a Brownian motion, see Section 4. Alternatively, one can take inverse Laplace transform to obtain the left-hand side of (3.42) and (3.43).
In this section, we consider three special cases. The first one is that X is a double exponential jump diffusion process, then X is simplified to have only negative jumps, and finally, X is assumed to be a Brownian motion. 4.1. Double exponential jump diffusion process case
ψ(z ) =
Ex e−θ
σ2 2
z + µz + λ 2
p1 η1
η1 − z
+
q1 ϑ1
ϑ1 + z
(3.40)
1{Ut
0 1{Us
−1
(4.1)
0
˜
=
˜
θ ,0 γ1,0 (b−x)
1 − A3 e
τ
x ≤ b, x > b.
1{Us
−Aθ1,0 eβ1,θ (x−b) − Aθ2,0 eβ2,θ (x−b) ,
x > b.
τ
ν − Aξ ,ν eβ˜ 1,ξ (x−b) − Aξ ,ν eβ˜ 2,ξ (x−b) , 1 2 = ξ ξ ,ν γ1,ν (b−x) ξ ,ν γ2,ν (b−x) 1 − A3 e − A4 e , +∞
Ex e−θ
x ≤ b,
are continuous at b, which can be proved
easily by using (3.23).
ν n t n−1 −ν t −θ t 1{U
(2005). Under this case, results in the previous section can be simplified as follows.
= Ex [τ ] − Ex 1{Ut
−θ 0+∞ 1{Ut
=
β2,q (β˜ 2,q ), γ1,q (γ˜1,q ) and γ2,q (γ˜2,q ), see Appendix B in Kou et al.
(β˜ i,ν + γj,ν )
i=1
Ex e−θ
Ex
+∞
for z ∈ (−ϑ1 , η1 ). For the solutions in Lemma 3.1 of (4.1), here we only add that one can obtain the analytical forms for β1,q (β˜ 1,q ),
.
Remark 3.4. Note that Hi > 0 for all i = 1, 2, . . . , m + 1. Hence, m+1 β˜ (x−b) i,ν we have > 0 in (3.38). More importantly, we i=1 Hi e have the following result.
Consider the process X with double exponential jump distribution, i.e., m = n = 1 in (2.11). Now the Lévy exponent of X reduces to
(β˜ i,ν + ϑj )
j =1
(β˜ i,ν − β˜ j,ν )
i=1
n+1
n
(β˜ i,ν − ηj )
j =1
m+1
1{Us
4. Special cases
x ≤ b,
i=1
n +1 1
Ex
ν
0
0
n ; t
Therefore, for large n,
1 1 ∂ β˜ i,ξ = lim = , (3.37) lim ′ ′ ˜ (β˜ i,ξ ) θ→0 ψ θ→0 ∂θ ψ˜ (β˜ i,ν ) ˜ z ) = ψ(z ) − α z which can be obtained where the function ψ( from (3.1). Therefore, taking a derivative with respect to θ in Theτ orem 3.1 and then letting θ equal to 0 yield Ex 0 1{Us
Γ (n,ν)
0
˜ β˜ i,ξ ) = ξ for i = 1, . . . , m + 1, we Note that ξ = ν + θ and ψ( have
τ
−θ
∂ n−1 (V (x; b)/ν) n = (−1)n−1 ν /Γ (n). ∂ν n−1
Remark 3.3. For the process U 1 (see Appendix), Renaud (2013) +∞ has derived the Laplace transform of 0 1{U 1
Ex e
.
m+1
. In fact, by the strong law of large numbers, we
could construct an Erlang random Γ (n, ν) ≈ t for large n if ν = and for the independent Erlang random Γ (n, ν), we have
(β˜ i,θ + γj,0 )
i=1
i =1
,
n
0 1{Us
Ex e−θ
j =1
n +1
ϑi
(β˜ i,θ − β˜ j,θ )
n+1
constant t, we can use Theorem 3.1 to approximate For a given t
(β˜ i,θ + ϑj )
j =1
j=1,j̸=i n +1
n
n
(β˜ i,θ − ηj )
j =1
j=1
m+1
m
γj,0
m+1
ϑj
j =1
j =1
n
ηj
n+1
θ ,0 γ2,0 (b−x)
− A4 e
x ≤ b,
,
x > b.
1{Us
0 1 + Aν,ν eβ˜ 1,ν (x−b) + Aν,ν eβ˜ 2,ν (x−b) , 1 2 = ν ν,ν γ1,ν (b−x) ν,ν γ2,ν (b−x) A3 e + A4 e ,
x ≤ b, x > b.
(4.2)
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
ξ ,ν
In the above formulas, the constants Ai , i = 1, . . . , 4, are given by ξ ,ν
A1
ξ ,ν
A2
ξ ,ν
A3
ξ ,ν
A4
β˜ 2,ξ γ2,ν γ1,ν (β˜ 1,ξ − η1 )(ϑ1 + β˜ 1,ξ )θ , (β˜ 1,ξ + γ1,ν )(β˜ 1,ξ + γ2,ν )(β˜ 2,ξ − β˜ 1,ξ )η1 ξ ϑ1 β˜ 1,ξ γ2,ν γ1,ν (β˜ 2,ξ − η1 )(ϑ1 + β˜ 2,ξ )θ = , (β˜ 2,ξ + γ1,ν )(β˜ 2,ξ + γ2,ν )(β˜ 1,ξ − β˜ 2,ξ )η1 ξ ϑ1 β˜ 1,ξ β˜ 2,ξ γ2,ν (γ1,ν + η1 )(ϑ1 − γ1,ν )θ = , (β˜ 1,ξ + γ1,ν )(β˜ 2,ξ + γ1,ν )(γ2,ν − γ1,ν )η1 ξ ϑ1 β˜ 1,ξ β˜ 2,ξ γ1,ν (γ2,ν + η1 )(γ2,ν − ϑ1 )θ = , (β˜ 1,ξ + γ2,ν )(β˜ 2,ξ + γ2,ν )(γ2,ν − γ1,ν )η1 ξ ϑ1
=
θ,0
and Ai
ν,ν
is defined similarly and Ai
Ex e−θ (4.3)
σ
2
z 2 + µz + λq1
ξ ,ν
ξ ,ν
:= lim A1 = − η1 ↑∞
ξ ,ν
lim A2
η1 ↑∞ ξ ,ν
B3
ξ ,ν
ξ ,ν
B4
ξ ,ν
η1 ↑∞
z > −ϑ1 .
(4.4)
γ2,ν γ1,ν (ϑ1 + β˜ 1,ξ )θ , ˜ (β1,ξ + γ1,ν )(β˜ 1,ξ + γ2,ν )ξ ϑ1
τ
β˜ 1,ξ γ2,ν (ϑ1 − γ1,ν )θ , ˜ (β1,ξ + γ1,ν )(γ2,ν − γ1,ν )ξ ϑ1 β˜ 1,ξ γ1,ν (γ2,ν − ϑ1 )θ = , (β˜ 1,ξ + γ2,ν )(γ2,ν − γ1,ν )ξ ϑ1
=
1
ν
ν,ν
β1,ν (x−b) + Bν,ν , 1 e ˜
x ≤ b,
ν,ν
B3 eγ1,ν (b−x) + B4 eγ2,ν (b−x) ,
θ ,0
where Bi
ξ ,ν
= limν↓0 Bi
ν,ν
and Bi
(4.8)
x > b, ξ ,ν
= limθ↓0 Bi /θ , for i = 1, 3, 4.
Remark 4.3. Note that X is a Lévy process without positive jump in this subsection and the later one. When x = b, formulas (4.7) and (4.14) can be found from Kyprianou and Loeffen (2014) (see Corollary 2). In addition, note that the process U reduces to X if α = 0. Thus we remark that results (4.7) and (4.14) with α = 0 have been established by other authors, see, e.g., Corollary 1 in Landriault et al. (2011).
In this subsection, we consider the simplest case Xt = X0 +µt +
σ Wt with µ > 0. Thus the process (Ut )t ≥0 satisfies dUt = dXt − α 1{Ut
= µdt + σ dWt − α 1{Ut
(4.9)
(4.5)
σ2
z 2 + µz , (4.10) 2 and for the sake of brevity, we denote the roots of ψ(z ) = q and ψ(z ) − α z = q for q > 0 by βq > 0, γq < 0 and β˜ q > 0, γ˜q < 0 respectively, see also (A.9) and (A.10). Noting that (4.10) can be obtained from (4.4) by letting ϑ1 ↑ ∞, we can derive
βq = lim β1,q , ϑ1 ↑∞
x > b.
ϑ1 ↑∞
(4.11)
ϑ1 ↑∞
ϑ1 ↑∞
Formula (4.5) and the fact that ϑ1 < γ2,ν yield ξ ,ν
=
ξ ,ν
=
lim B3
(4.6)
β˜ q = lim β˜ 1,q ,
γq = − lim γ1,q and γ˜q = − lim γ˜1,q .
lim B1
x ≤ b,
t > 0,
and U0 = X0 ,
ϑ1 ↑∞
ν ξ ,ν ˜ − B1 eβ1,ξ (x−b) , ξ = ξ ,ν ξ ,ν 1 − B3 eγ1,ν (b−x) − B4 eγ2,ν (b−x) ,
(4.7)
x > b,
1{Us
ψ(z ) =
τ e−θ 0 1{Us
,
and the process (Yt )t ≥0 is given by Yt = X0 + (µ − α)t + σ Wt . The Lévy exponent of X is
Theorem 4.1. For θ > 0,
Ex
− B4 e
4.3. Brownian motion case
where −γ1,ν < 0 and −γ2,ν < 0 are the roots of ψ(z ) = ν with ψ(z ) given by (4.4); and β˜ 1,ξ > 0 is the unique positive root of ψ(z ) − α z = ξ . Then, from the first equation in (4.2), we obtain Theorem 4.1.
x ≤ b,
θ ,0 γ2,0 (b−x)
0
:= lim A3 = := lim A4
˜
θ ,0 γ1,0 (b−x)
= 0,
η1 ↑∞
−Bθ1,0 eβ1,θ (x−b) ,
Ex
The reason why we denote the Poisson intensity in (4.4) by λq1 not λ is that now (4.4) can be obtained from (4.1) by letting η1 ↑ ∞. Applying the fact that η1 < β˜ 2,ξ , for fixed ϑ1 > 0, we have B1
and
ξ ,ν
ϑ1 −1 , ϑ1 + z
1{Us
1 − B3 e
In this case, the Lévy exponent is given by
ψ(z ) =
0
=
= limθ↓0 Ai /θ .
+∞
4.2. The process X only has negative jumps with exponential distribution
that is similar to the proof of Theorem 3.1. This verification process is simple because there is only one-sided jump, and we omit the details here. It follows from (4.6) that
Remark 4.1. When α = 0, the two processes U and X are equal. We comment that the first result in (4.2) with α = 0 have been derived in Cai et al. (2010) (see Theorem 3.2). Here, we make a remark that the approach used in Cai et al. (2010) is completely different from ours. We have to say that our method is simple relatively, because in this paper it is unnecessary to take too much effort as in Cai et al. (2010) to establish an integro-differential equation. For more details, we refer the reader to Cai et al. (2010).
2
131
ϑ1 ↑∞
θ γν , ˜ ξ (βξ − γν ) θ β˜ ξ ξ (β˜ ξ − γν )
(4.12) and
ν,ξ
lim B4
ϑ1 ↑∞
= 0.
Applying (4.6), for θ > 0, we can deduce the following theorem. Theorem 4.2.
Remark 4.2. Strictly speaking, to derive the above theorem, we still need to show that the left-hand side of (4.6) is equal to
Ex e−θ
τ
0 1{Us
in (4.2) after letting η1 ↑ ∞. Even though some-
one is likely to prove this conclusion though some efforts, we recommend the readers to verify formula (4.6) by using an approach
ν θ γν β˜ (x−b) , −e ξ τ ξ ξ (β˜ ξ − γν ) −θ 0 1{Us
x ≤ b, (4.13) x > b,
132
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
γ0 ˜ − eβθ (x−b) , (β˜ θ − γ0 ) +∞ Ex e−θ 0 1{Us
x ≤ b, (4.14) x > b,
τ
Ex 0
γν ˜ 1/ν + eβν (x−b) , ν(β˜ ν − γν ) 1{Us
dUt1 = dX˜ t − α 1{U 1 >b} dt ,
t > 0,
t
(A.3)
and U01 = X˜ 0 .
From 22 in Kyprianou and Loeffen (2010), we have Corollary Px Ut1 = b = 0 for Lebesgue almost every t ≥ 0. Therefore, from (A.3), we have
2µ
where γ0 = − σ 2 , and
First, for any given b, let the process U 1 = Ut1 t ≥0 satisfy the following SDE:
from which we have
x ≤ b, (4.15) x > b.
t
Ut1 = Y˜t + α
1{U 1
0
t ≥ 0,
(A.4)
where Y˜t = X˜ t − α t for t ≥ 0.
Remark 4.4. Similar to Remark 4.2, we have to verify that equality (4.13) is correct. Compared with Remark 4.2, the verification process is much simpler because now there is no jumps. In addition, the computations of (3.42) and (3.43) from (4.13) and (4.15) become easier as well, especially when x = b. Remark 4.5. For α = 0 and σ = 1, formula (4.13) has been derived by other author, see formula 1.5.1 on p. 256 of Borodin and Salminen (2002) for instance.
Remark A.1. When X˜ in (A.3) is a Lévy process without positive jump, Kyprianou and Loeffen (2010) (see Theorem 1) have showed that there exists a unique strong solution U 1 = Ut1 t ≥0 to Eq. (A.3). In addition, U 1 is a strong Markov process. The process U 1 is called a refracted Lévy process, and for more details about this process, we refer to Kyprianou and Loeffen (2010). For any given ε > 0, we define the stopping time τε1 as τε1 := inf{t ≥ 0, Ut1 > b + ε or Ut1 < b − ε}. In the following, we give a technical lemma which has been obtained by Renaud (2013), see also Theorem 4 of Kyprianou and Loeffen (2010). Lemma A.1 (Theorem 2 in Renaud, 2013). For q > 0 and α ≥ 0,
5. Conclusion
In this paper, we investigate the problem of computing the time of deducting fees for variable annuities with a state-dependent fee deducting method which is proposed recently. Under a general jump diffusion model, we have derived closed-form formulas for the time of charging expenses. In addition, some of our outcomes extend the results obtained by previous authors like Renaud (2013) to the process which has two-sided jumps. From the derivation of Theorem 3.1, we claim that our results can be extended to a general jump diffusion process X only if we know how to compute (3.9) and (3.10). For example, the process X can be assumed to have jumps with phase-type distributions, see Proposition 2 in Asmussen et al. (2004) for the calculations of (3.9) and (3.10). Acknowledgments
W (q) (ε)
Eb e−qτε 1{U 1 1
=
=b+ε} 1
τε
ω(q) (b
+ ε; b − ε)
,
(A.5)
and
Eb e−qτε 1{U 1
1
τε1
= Z (q) (ε) −
=b−ε}
z (q) (b + ε; b − ε)
ω(q) (b + ε; b − ε)
W (q) (ε). (A.6)
Here ω(q) (b + ε; b − ε) and z (q) (b + ε; b − ε) are given as follows.
ω(q) (b + ε; b − ε) = W (q) (2ε) + α
b+ε
˜ (q) (b + ε − y)W (q)′ (y − b + ε)dy, W
b
(A.7)
z (q) (b + ε; b − ε)
= Z (q) (2ε) + α q
b+ε
˜ (q) (b + ε − y)W (q) (y − b + ε)dy, W
b
The authors are grateful to an anonymous reviewer for various useful comments and suggestions on an earlier version, which help to improve the structure and text of the paper. They also thank the editor Elias S. Shiu for helpful comments and suggestions and kindly sending an article about writing.
where W (q) (x) =
Recalling the process X given by (2.1), we denote the first jump time of the Poisson process Nt by T1 which is an exponential random variable with parameter λ. We denote the continuous part of X by X˜ = (X˜ t )t ≥0 which is given by X˜ t = X0 + µt + σ Wt ,
t ≥ 0.
(A.1)
The process U˜ related to X˜ is defined by dU˜ t = dX˜ t − α 1{U˜ t
t > 0,
σ 2 (βq − γq )
,
x > 0,
2(eβq x − eγ˜q x ) ˜
˜ (q) (x) = W
Appendix
2(eβq x − eγq x )
, x > 0, σ 2 (β˜ q − γ˜q ) ˜ (q) (x) = 0, x ≤ 0, W (q) (x) = W x (q) Z (x) = 1 + q W (q) (y)dy, x ∈ R.
(A.8)
0
In (A.8), βq > 0 and γq < 0 are the two roots of the equation:
µx +
σ2 2
x2 = q,
(A.9)
and β˜ q > 0 and γ˜q < 0 are the two roots of the equation: (A.2)
(µ − α)x +
σ2 2
x2 = q.
(A.10)
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
Another function Z˜ q (x) defined as Z˜ (q) (x) = 1 + q
x
˜ (q) (y)dy, W
133
Remark A.4. Note that the limitations: limε↓0
x ∈ R,
1/2−Eb e−qτ˜ε 1{U˜
(A.11) and limε↓0
0
ε
1/2−Eb e−qτ˜ε 1{U˜
ε
τ˜ε =b−ε}
τ˜ε =b+ε}
do not depend on q.
will be used in the following.
are known as the q-scale functions of X˜ Y˜ , one can refer to
Proof. {The proof of V ′ (b−; b) = V ′ (b+; b)} From (3.18), it is obvious that both V ′ (b−; b) and V ′ (b+; b) exist. Therefore, to prove they are equal, we only need to show that
Chapter 8 of Kyprianou (2006) for more details about these functions.
lim
˜ (q) (x) and Z (q) (x) Z˜ (q) (x) Remark A.2. The functions W (q) (x) W
From Renaud (2013) and Kyprianou and Loeffen (2010), it is easy to show that Lemma A.1 holds for α < 0 as well. Actually, as Renaud stated, Theorem 2 in his paper is a restatement of Theorem 4 in Kyprianou and Loeffen (2010) which holds for α ∈ R. Therefore, recalling (A.2), we have U˜ t = X˜ t − α
t
1{U˜ s
0
ε↓0
(V (b + ε; b) + V (b − ε; b)) /2 − V (b; b) = 0. ε
From the definition of V (x; b), we have V (b; b) = Eb
0
1{U˜ s
(A.12)
Eb e
(A.13)
(A.14)
where the stopping time τ˜ε is defined as τ˜ε := inf{t ≥ 0, U˜ t > b + ε or U˜ t < b − ε} and
˜ =W
(2ε) + α
dt
t
0 1{Us
dt
ν e−ν t e−θ
t
ν e−ν t e−θ
t
0 1{Us
+∞ τ˜ε
0 1{Us
dt1{τ˜ε ≥T1 }
dt1{τ˜ε
(A.18)
τ˜ε
Eb
νe
−ν t −θ
e
t
0 1{Us
0
W
(q)
˜ (b + ε − y)W
(q)′
b
(y − b + ε)dy,
≤ Eb τ˜ε
Eb
νe
= 1 − Eb e−ν τ˜ε
ν e−ν t e−θ
t
0 1{Us
dt
≥
0
˜ (q) (y − b + ε)dy. W (q) (b + ε − y)W
b
−ν t
(A.19)
0
(A.15) b+ε
τ˜ε
b+ε
z˜ (q) (b + ε; b − ε)
= Z˜ (q) (2ε) + α q
0 1{Us
Because θ > 0, thus
ω˜ (q) (b + ε; b − ε) (q)
τ˜ε
+ Eb ˜ (q) (ε), W
e
t
+∞
ω˜ (q) (b + ε; b − ε)
−ν t −θ
ν e−ν t e−θ
+ Eb
τ˜ε =b−ε}
= Z˜ (q) (ε) −
dt
0
z˜ (q) (b + ε; b − ε)
0 1{Us
+∞ t ν e−ν t e−θ 0 1{Us
and
Eb e−qτ˜ε 1{U˜
νe
τ˜ε
1{U˜ =b+ε} = (q) , τ˜ε ω˜ (b + ε; b − ε)
t
0
= Eb
˜ (q) (ε) W
τ˜ε
= Eb
Corollary A.1. For q > 0, −qτ˜ε
ν e−ν t e−θ 0
Then, comparing (A.4) and (A.12), we obtain the following corollary from Lemma A.1.
+∞
t
(A.17)
ν 1 − Eb e−(ν+θ )τ˜ε . ν+θ
Then applying (A.16) yields
Remark A.3. Corollary A.1 can be extended to a Lévy process without positive jump, i.e., the process X˜ in (A.12) can be assumed to be a general Lévy process with only negative jump except that X˜ t − α t has monotone paths. In fact, in this case, we still have (A.4)–(A.6) and (A.12), from which this extension is derived. Alternatively, one can show this conclusion directly by using almost the same method as in Renaud (2013), for interested readers, see that paper for the details. Using L’Hospital’s rule twice, we can derive the following results from Corollary A.1 after some calculations.
Eb
τ˜ε 0
lim
ν e−ν t e−θ
t
0 1{Us
dt
= 0.
ε
ε↓0
(A.20)
Note that T1 is independent of X˜ and thus independent of τ˜ε . We have +∞
0 ≤ Eb
τ˜ε
ν e−ν t e−θ
t
0 1{Us
dt1{τ˜ε ≥T1 }
≤ Pb (τ˜ε ≥ T1 ) = 1 − Eb e−λτ˜ε ,
(A.21)
Corollary A.2.
lim
1/2 − Eb e−qτ˜ε 1{U˜ =b+ε} τ˜ε
1/2 − Eb e lim ε↓0
=
ε
ε↓0
−qτ˜ε
ε
1{U˜ =b−ε} τ˜ε
and using (A.16) again, we obtain
=
3α − 2µ 4σ
2
2µ − 3α 4σ 2
,
(A.16)
.
lim Eb ε↓0
+∞ τ˜ε
νe
−ν t −θ
e
t
0 1{Us
dt1{τ˜ε ≥T1 } /ε = 0.
(A.22)
Using the strong Markov property and that {Xt , t < T1 } and {X˜ t , t < T1 } have the same law under Pb , we have
134
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
Eb
+∞
ν e−ν t e−θ
τ˜ε
−ν τ˜ε −θ
τ˜ε
t
0 1{Us
1{U˜
References
dt1{τ˜ε
˜ {τ˜ε
= Eb e τ˜ −(ν+λ)τ˜ε −θ 0 ε 1{U˜
ε
−(ν+λ)τ˜ε −θ
+ Eb e
τ˜ε 0
1{U˜
{U˜ τ˜ε =b−ε} V (b − ε; b) ,
(A.23)
where the second equality comes from the fact that T1 is indepen˜ Similar to (A.19), we can show that dent of the process U.
1/2 − Eb e−(ν+λ)τ˜ε e lim
−θ
τ˜ε 0
1{U˜
1{U˜ =b+ε} τ˜ε
=
ε
ε↓0
1/2 − Eb e−(ν+λ)τ˜ε e lim ε↓0
−θ
τ˜ε 0
1{U˜
{U˜ τ˜ε =b−ε}
ε
3α − 2µ 4σ 2
=
,
(A.24)
2µ − 3α 4σ 2
.
Therefore, it follows from (A.18), (A.20), (A.22) and (A.24) that
(V (b + ε; b) + V (b − ε; b)) /2 − V (b; b) ε τ˜ −θ 0 ε 1{U˜
lim ε↓0
= V (b; b) × 0 = 0, thus we complete the proof.
(A.25)
Remark A.5. Note that the distribution of Z1 (see (2.1)) does not appear in the above proof. Thus the conclusion V ′ (b−; b) = V ′ (b+; b) holds for general jump diffusion processes.
Asmussen, S., Avram, F., Pistorius, M.R., 2004. Russian and American put options under exponential phase-type Lévy models. Stochastic Process. Appl. 109, 79–111. Bacinello, A.R., Millossovich, P., Olivieri, A., Pitacco, E., 2011. Variable annuities: a unifying valuation approach. Insurance Math. Econom. 49, 285–297. Bauer, D., Kling, A., Russ, J., 2008. A universal pricing framework for guaranteed minimum benefits in variable annuities. Astin Bull. 38, 621–651. Bernard, C., Hardy, M., Mackay, A., 2014. State-dependent fees for variable annuity guarantees. Astin Bull. 44, 559–585. Borodin, A.N., Salminen, P., 2002. Handbook of Brownian Motion Facts and Formulae, second ed. Birkhäuser Verlag, Basel, Boston, Berlin. Cai, N., 2009. On the first passage times of a hyper-exponential jump diffusion process. Oper. Res. Lett. 37, 127–134. Cai, N., Chen, N., Wan, X., 2009. Pricing double-barrier options under a flexible jump diffusion model. Oper. Res. Lett. 37, 163–167. Cai, N., Chen, N., Wan, X., 2010. Occupation times of jump-diffusion processes with double exponential jumps and the pricing of options. Math. Oper. Res. 35, 412–437. Delong, Ł., 2014. Pricing and hedging of variable annuities with state-dependent fees. Insurance Math. Econom. 58, 24–33. Dufresne, D., 2007. Fitting combinations of exponentials to probability distributions. Appl. Stoch. Models Bus. Ind. 23, 23–48. Gerber, H.U., Shiu, E.S.W., 2003. Pricing perpetual fund protection with withdrawal option. N. Am. Actuar. J. 7 (2), 60–77. Discussion 77–92. Gerber, H.U., Shiu, E.S.W., Yang, H., 2012. Valuing equity-linked death benefits and other contingent options: a discounted density approach. Insurance Math. Econom. 51, 73–92. Gerber, H.U., Shiu, E.S.W., Yang, H., 2013. Valuing equity-linked death benefits in jump diffusion models. Insurance Math. Econom. 53, 615–623. Hyndman, C.B., Wenger, M., 2014. Valuation perspectives and decompositions for variable annuities with GMWB riders. Insurance Math. Econom. 55, 283–290. Ko, B., Ng, A.C.Y., 2007. Discussion on stochastic life annuities by D. Dufresne. N. Am. Actuar. J. 11 (3), 170–171. Ko, B., Shiu, E.S.W., Wei, L., 2010. Pricing maturity guarantee with dynamic withdrawal benefit. Insurance Math. Econom. 47, 216–223. Kou, S.G., Petrella, G., Wang, H., 2005. Pricing path-dependent options with jump risk via Laplace transforms. Kyoto Econ. Rev. 74 (1), 1–23. Kyprianou, A.E., 2006. Introductory Lectures on Fluctuations of Lévy Processes with Applications. In: Universitext, Springer-Verlag, Berlin, pp. 210–238. Kyprianou, A.E., Loeffen, R.L., 2010. Refracted Lévy processes. Ann. Inst. H. Poincare Probab. Statist. 46, 24–44. Kyprianou, A.E., Loeffen, R.L., 2014. Occupation times of refracted Lévy process. J. Theoret. Probab. 27, 1292–1315. Landriault, D., Renaud, J.F., Zhou, X., 2011. Occupation times of spectrally negative Lévy processes with applications. Stochastic Process. Appl. 121, 2629–2641. Lee, H., 2003. Pricing equity-indexed annuities with path-dependent options. Insurance Math. Econom. 33, 677–690. Renaud, J.F., 2013. On the time spent in the red by a refracted Lévy risk process. arXiv:1306.4619v1 [math.PR]. Yin, C., Shen, Y., Wen, Y., 2013. Exit problems for jump process with applications to dividend problems. J. Comput. Appl. Math. 245, 30–52.
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Insurance: Mathematics and Economics journal homepage: www.elsevier.com/locate/ime
The time of deducting fees for variable annuities under the state-dependent fee structure Jiang Zhou a,∗ , Lan Wu a,b a
School of Mathematical Sciences, Peking University, Beijing 100871, PR China
b
Key Laboratory of Mathematical Economics and Quantitative Finance, Peking University, Beijing 100871, PR China
article
info
Article history: Received August 2014 Received in revised form December 2014 Accepted 28 December 2014 Available online 5 January 2015 Keywords: Variable annuities State-dependent fee Hyper-exponential jump diffusion process Laplace transform Refracted Lévy process
abstract We investigate the total time of deducting fees for variable annuities with state-dependent fee. This fee charging method is studied recently by Bernard et al. (2014) and Delong (2014) in which the fees deducted from the policyholder’s account depend on the account value. However, both of them have not considered the problem of analyzing probabilistic properties of the total time of deducting fees. We approximate the maturity of a general variable annuity contract by combinations of exponential distributions which are (weakly) dense in the space that is composed of all probability distributions on the positive axis. Working under general jump diffusion process, we derive analytic formulas for the expectation of the time of deducting fees as well as its Laplace transform. © 2015 Elsevier B.V. All rights reserved.
1. Introduction Variable Annuities (VAs) are generally issued with minimum guarantee on the death or maturity benefits and keep in high demand due to the embedded guarantees. Nowadays, a variety of guarantees are provided, such as Guaranteed Minimum Death Benefits (GMDBs), Guaranteed Minimum Maturity Benefits (GMMBs), Guaranteed Minimum Accumulation Benefits (GMABs) and so on, see Bauer et al. (2008). In the academic circle, VAs have received substantial attention, see among others, Lee (2003), Gerber and Shiu (2003), Ko et al. (2010), Bacinello et al. (2011) and Gerber et al. (2012, 2013). In general, insurers charge expenses for the provision of the guaranteed benefits from the policyholder’s account by a fixed rate, and most actuarial literatures assume that the fee rate is fixed as well, see e.g. Hyndman and Wenger (2014). However, this fixed fee structure has several disadvantages which have been noted by Bernard et al. (2014) and Delong (2014). For example, as the guaranteed benefits embedded in variable annuities are similar to put options, if the account value is high, the guarantees are (deep) outof-the-money. In this case, a higher deducted fee will yield incentives for policyholders to lapse the policy, see Bauer et al. (2008). Recently, Bernard et al. (2014) put forward a dynamic fee structure. In their paper, the fees are deducted at a fixed rate only if the account value is lower than a pre-specified level. They have derived formulas for calculating the fee rates under the Black–Scholes
∗
Corresponding author. E-mail addresses: [email protected] (J. Zhou), [email protected] (L. Wu).
http://dx.doi.org/10.1016/j.insmatheco.2014.12.008 0167-6687/© 2015 Elsevier B.V. All rights reserved.
model. Delong (2014) then extended their model to an incomplete financial market which is made up of two risk assets that are modeled by a two-dimensional Lévy process, and he considered a general state-dependent fee which is determined by a function of the account value. In his paper, the fee for the guaranteed benefit can be computed by solving an equation, and a strategy for hedging the guarantee is also characterized. Under a state-dependent fee structure, sometimes insurers do not have any fees income, e.g., when the account value is higher than a pre-specified level in Bernard et al. (2014). Therefore, insurers are interested in the problem that how long they can collect fees or alternatively, how long they cannot. In nature, insurers are interested in the probabilistic properties of the total time of collecting fee, for example its expectation. Besides, this total time will determine insurers’ income directly and have a significant impact on the account value indirectly. The above problem has not been considered in Bernard et al. (2014) and Delong (2014), and we are the first to investigate it to our knowledge. In this paper, we assume that the fees are deducted as in Bernard et al. (2014), which is one of the fee charging structures investigated in Delong (2014). Under the simple Black–Scholes model as well as the complex hyper-exponential jump diffusion process model, we obtain analytic formulas for the Laplace transform of the total time of deducting fees and its expectation as well. The rest of this paper is organized as follows. Our model is introduced in Section 2. In Section 3, under the hyper-exponential jump diffusion process, we derive our main results. In Section 4, three special cases of the general model are discussed in detail
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J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
where the formulas become simpler. Finally, some conclusions are given in Section 5.
We first consider the case that the maturity is Tz . Under the fee structure (2.3) and (2.4), the insurer is interested in the expectation of the total time of charging fees:
2. The model
X t = X 0 + µt + σ W t +
Nt
Zk ,
(2.2)
which can be explained as the value of one unity of the reference fund or stock underlying the variable annuity contract, and it is reasonable to assume that E [X1 ] > 0. According to (2.6) in Delong (2014), if we let Ft denote the policyholder’s account value at time t, then its dynamics are represented by the following stochastic differential equation (SDE) dSt − g (Ft − )dt , t > 0, (2.3) St − and F0 is the initial premium invested by the insured, and g is a function which represents a state-dependent fee charged by the insurer. In Bernard et al. (2014), they assumed that X is a Brownian motion and g is given by
dFt = Ft −
(2.4)
where α and B denote the deduction fee rate and a pre-specified level, respectively; and 1A is the indicator function of a set A. Function (2.4) means the expense is charged only when the account value is less than the pre-specified level B. In this paper, we consider the case that g satisfies Eq. (2.4) and the investigation of general g is left for future research. For the advantages of this fee deducting method, see Bernard et al. (2014) and Delong (2014). Let Ut satisfy the following SDE dUt = dXt − α 1{Ut
t > 0,
and U0 = X0 ,
=
1{Us
E
(2.7)
0
0
Remark 2.1. To evaluate (2.7), the value of α in (2.5) is required to be specified first. In principle, the value of α is determined by risk neutral pricing rule for some martingale measure. Here, we are more interested in obtaining the probability characteristic of time of deducting fee. Therefore, in this paper, we assume that the parameter α is exogenous and focus on the calculation of (2.7) with given α . From Ko and Ng (2007) or Dufresne (2007), we obtain the following result: in the space of all probability distribution on the positive axis, the subset of linear combinations of exponential distributions is (weakly) dense. Therefore, we can approximate fTz (t ) by i ci fTi (t ), where fTi (t ) is the density function for some exponential distributions. Hence, the problem of calculating (2.7) reduces to compute the following expectation: τ
E
1{Ut
(2.8)
0
where τ is an exponential random variable which is independent of the process U. For a contract with maturity date T, we first note that (2.8) is
T
related to the Laplace transform of E
0
1{Us
the contract expires at a fixed time T , the total time of deducting fees can also be obtained from (2.8) through taking inverse Laplace transform for instance. In addition, if the maturity is min{Tz , T }, then we have min{Tz ,T }
1{Us
E 0
Tz
T
1{Us
=E 0
1{Us
0
= P(Tz < T )
+∞
E
0
+ P(Tz ≥ T )E
t
1{Us
0 T
1{Us
(2.9)
0
(2.5)
B F0
1{Ut
t
where fTz (t ) denotes the probability density function of Tz . (2.1)
where µ and X0 are constants; {Wt ; t ≥ 0} is a standard Brownian motion with W0 = 0, and σ > 0 is the volatility of the diffusion; {Zk ; k = 1, 2, . . .} are independent and identically distributed random variables supported on R/{0}, and the common probability density function (pdf) of {Zk ; k = 1, 2, . . .} is denoted by fZ (z ); {Nt ; t ≥ 0} is a Poisson process with rate λ; moreover, {Wt }, {Nt } and {Zk ; k = 1, 2, . . .} are independent. The law of X starting from x is denoted by Px and Ex represents the corresponding expectation, and for the sake of brevity, we write P and E when x = 0. We define the process S = (St )t ≥0
g (x) = α x1{x
∞
0
k=1
St = S0 eXt −X0 ,
E
Suppose X = (Xt )t ≥0 is a jump diffusion process, i.e.,
where b = ln
Tz
. Then, when U0 = 0, we obtain the following
equation of Ft and Ut by Itô’s formula. Ft = F0 eUt , t ≥ 0. (2.6) Because the parameter σ > 0 and the jump part of X is a compound Poisson process, we have the following lemma of the existence of a unique strong solution of (2.5). For the proof of it, one can see Remarks 2 and 3 in Kyprianou and Loeffen (2010) and the proof of Lemma 2.1 in Delong (2014) for reference. Lemma 2.1. Eq. (2.5) exists a unique strong solution. Moreover the solution U is a strong Markov process. Consider a customer age z purchasing a VA contract. If the guarantee embedded is GMMBs, then the maturity of the contract is a given date T . For the guarantee of GMDBs, the maturity is given by min{Tz , T } for endowment policy or Tz for whole life (see Gerber et al., 2012 for example), where Tz is the random variable representing the time of death of the insured and independent of the account value process Ft . In this paper, we ignore other reasons of terminating the contract, e.g., the customer lapses the policy.
where gTz |Tz
ν e−ν t ,
t > 0,
(2.10)
where ν > 0 is a constant. Of course, it is very difficult (or even impossible) to calculate (2.8) for arbitrary pdf fZ (z ). In this paper, we assume that fZ (z ) is given by fZ (z ) =
m i=1
pi ηi e−ηi z 1{z >0} +
n
qj ϑj eϑj z 1{z <0} ,
(2.11)
j =1
where pi > 0, ηi > 0 for all i = 1, . . . , m; qj > 0, ϑj > 0 for all m n j = 1, . . . , n; i=1 pi + j=1 qj = 1, η1 < η2 < · · · < ηm and ϑ1 < ϑ2 < · · · < ϑn . Remark 2.2. The process X defined by (2.1) and (2.11) is called hyper-exponential jump diffusion process (HEP). The HEP has sev-
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
eral advantages and has been used to model the asset return in many papers, see e.g. Cai et al. (2009), here we just emphasize that the hyper-exponential distribution (2.11) can be used to approximate many complicated pdf fZ (z ). Therefore, for more general process X under which (2.8) cannot be solved, applying our results can obtain a very good approximation. Before deriving our results, we put up some notation. We set Y = {Yt := Xt − α t ; t ≥ 0} which is also a HEP. The law of Y start˜ y and the corresponding expectation is ing from y is denoted by P ˜ y ; and we write P˜ and E˜ when y = 0 for the sake of denoted by E brevity. For any h, H ∈ R, we define the following stopping times of X , Y and U.
τh−,X := inf{t ≥ 0 : Xt ≤ h},
τH+,X := inf{t ≥ 0 : Xt ≥ H }, (2.12)
τh−,Y := inf{t ≥ 0 : Yt ≤ h},
τH+,Y := inf{t ≥ 0 : Yt ≥ H }, (2.13)
κh− := inf{t ≥ 0 : Ut ≤ h},
κH+ := inf{t ≥ 0 : Ut ≥ H }. (2.14)
3. Main results For the HEP X given by (2.1) and (2.11), the Lévy exponent is given by
σ 2 2 z + µz ψ(z ) := ln E ezX1 = 2 m n pi η i q j ϑj +λ + −1 , ηi − z ϑj + z i =1 j=1
dUt = µdt + σ dWt + d
(3.1)
= ψ(z ) − α z. According
Zk
− α 1{Ut
Ex e
−qτh−,X
1BX
−γ (x−h) e 1,q −qτ − 0,h e−γ2,q (x−h) Ex e h,X 1BX 1,h = Qq−1 , .. .. . . −γn+1,q (x−h) − Ex e
−qτh,X
(3.5)
e
1BX
n,h
where
ϑ1 ϑ − γ1,q 1 ϑ 1 1 ϑ − γ2,q 1 Qq = . . .. .. ϑ1 1 ϑ1 − γn+1,q
ϑ2 ϑ2 − γ1,q ϑ2 ϑ2 − γ2,q .. . ϑ2 ϑ2 − γn+1,q
1
ϑn ϑn − γ1,q ϑn ··· ϑn − γ2,q , (3.6) .. .. . . ϑn ··· ϑn − γn+1,q ···
and BX0,h = {Xτ − = h}, BXj,h = {X crosses the boundary h by a jump h,X of an exp(ϑj )-distributed random variable }, j = 1, 2, . . . , n.
˜x e E
−qτH+,Y
1B˜ Y
˜ 0,H eβ1,q (x−H ) ˜ −qτH+,Y β˜ 2,q (x−H ) Ex e 1B˜ Y e 1,H , = Q˜ q−1 . .. .. . β˜ m+1,q (x−H ) + e ˜ x e−qτH ,Y 1B˜ Y E
(3.7)
m,H
Nt
Lemma 3.2. For q > 0 and x > h,
Lemma 3.3. For q > 0 and x < H,
for z ∈ (−ϑ1 , η1 ). By analytic continuation, the function ψ(z ) can be extended to the whole complex plane except at ηi , i = 1, . . . , m and −ϑj , j = 1, . . . , n. Recalling the definition of Y , we have the
˜ z ) := ln E˜ ezY1 Lévy exponent of Y , ψ( to (2.5), the process (Ut )t ≥0 satisfies
127
t > 0,
k=1
where (3.2)
and U0 = X0 . The following lemma is taken from Lemmas 2.1 and 2.2 in Cai (2009), thus the proof is omitted. Lemma 3.1. For each given q ≥ 0, we have the following results. (1) For q > 0, the equation ψ(z ) = q has exactly (m + n + 2) real roots β1,q , β2,q , . . . , βm+1,q , −γ1,q , −γ2,q , . . . , −γn+1,q which satisfy 0 < β1,q < η1 < β2,q < · · · < ηm < βm+1,q < +∞, 0 < γ1,q < ϑ1 < γ2,q < · · · < ϑn < γn+1,q < +∞.
(3.3)
(2) For q > 0, the equation ψ(z ) − α z = q has exactly (m + n + 2) real roots β˜ 1,q , β˜ 2,q , . . . , β˜ m+1,q , −γ˜1,q , −γ˜2,q , . . . , −γ˜n+1,q which satisfy 0 < β˜ 1,q < η1 < β˜ 2,q < · · · < ηm < β˜ m+1,q < +∞, 0 < γ˜1,q < ϑ1 < γ˜2,q < · · · < ϑn < γ˜n+1,q < +∞.
(3.4)
(3) If q = 0, all the results in (1) hold except that β1,0 = 0 and γ1,0 > 0 (because we assume that E [X1 ] > 0); and all the results in (2) hold except that the signs of β˜ 1,0 and γ˜1,0 depend on the ˜ [Y1 ]. sign of E (4) For any i = 1, 2, . . . , m + 1, βi,q and β˜ i,q are continuous and strictly increasing functions of q on the interval [0, +∞). (5) For any j = 1, 2, . . . , n + 1, γj,q and γ˜j,q are continuous and strictly increasing functions of q on the interval [0, +∞). The following lemmas have been derived by Yin et al. (2013) (see Lemmas 2.2, 2.4 and 2.5), and one can refer to that paper for the details of the proofs.
1
1 Q˜ q = . ..
η1
η2
η1 − β˜ 1,q η1 η1 − β˜ 2,q
η2 − β˜ 1,q η2 η2 − β˜ 2,q
.. . η1 1 η1 − β˜ m+1,q
ηm
···
ηm − β˜ 1,q ηm ηm − β˜ 2,q
···
.. .. .. . . . η2 ηm ··· η2 − β˜ m+1,q ηm − β˜ m+1,q
, (3.8)
and B˜ Y0,H = {Yτ + = H }, B˜ Yi,H = {Y crosses the boundary H by a jump H ,Y of an exp(ηi )-distributed random variable }, i = 1, 2, . . . , m.
−qτh−,X
Remark 3.1. The expressions of Ex e −qτ − e h,X 1BX n,h
Ex
˜x and E
−qτ + e H ,Y 1B˜ Y 0,H
1BX
0,h
,..., +
, . . . , E˜ x e−qτH ,Y 1B˜ Y
m,H
can
be obtained by solving (3.5) and (3.7), see Theorems 2.1 and 2.2 in Yin et al. (2013). Lemma 3.4. (i) Consider any nonnegative measurable function g 0 such that −∞ g (h + y)eϑj y dy < ∞ for all j = 1, 2, . . . , n. For q > 0 and x > h, we have
Ex e
−qτh−,X
+
n j =1
g (Xτ − ) = g (h)Ex e h,X
−
ϑj Ex e−qτh,X 1BX
j,h
−qτh−,X
0
1BX
0,h
g (h + y)eϑj y dy.
(3.9)
−∞
(ii) Consider any nonnegative measurable function g such that ∞ g ( H + y)e−ηi y dy < ∞ for all i = 1, 2, . . . , m. For q > 0 0
128
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
and x < H, we have
˜x e E
−qτH+,Y
H ,Y
m
+
−qτH+,Y
˜x e g (Yτ + ) = g (H )E
˜x e E
−qτH+,Y
i,H
i=1
0,H
∞
1B˜ Y
1B˜ Y
g (H + y)ηi e−ηi y dy.
(3.10)
0
With the help of the above τ lemmas, we can derive formulas for the Laplace transform of 0 1{Us
τ
0 1{Ut
Ex e
τ
0 1{Ut
−θ
=
+1 ν m ˜ + Fi eβi,ξ (x−b) , ξ i=1
=1+
n+1
Gj e
−γj,ν (x−b)
x ≤ b, (3.11)
,
where ξ = θ +ν ; Fi , i = 1, 2, . . . , m+1, and Gj , j = 1, 2, . . . , n+1, are given by
β˜ j,ξ
Fi =
n +1
γj,ν
m
(β˜ i,ξ − ηj )
n
Gj = −
j=1,j̸=i
j =1
m+1
n +1
β˜ i,ξ
j =1
γi,ν
m
(γj,ν + ηi )
i=1,i̸=j
i =1
(ϑi − γj,ν )
0 1{Us
ν ν ν , M1 (b) − , . . . , Mm (b) − ξ ξ ξ Y = (y1 , y2 , . . . , yn+1 )
V (x; b) = Ex
(3.13)
νe
e
t
0 1{Us
= Ex
ν e−ν t e−θ t dt + Ex
0
dt
+
κb
ν e−ν t e−θ
t
0 1{Us
dt
P˜ x and {Ut , t < κb+ } under Px have the same law when x < b. Applying formula (3.10) with g (x) = V (x; b) to (3.14) yields
+ ν ν ˜ x e−ξ τb,Y 1B˜ Y V (x; b) = + E V (b; b) − 0,b ξ ξ m +∞ ν −ξ τb+,Y −ηi y ˜ + Ex e 1B˜ Y V (b + y; b)ηi e dy − , (3.15) i,b ξ 0 i =1 where ξ = θ + ν . For x > b, using again the strong Markov property, we can obtain the following equality from (3.9). +∞
ν e−ν t e
−θ
t
1 ds κb− {Us
.. .
β˜ m+1,ξ (x−b)
,
e
+∞
b,Y
κb−
˜
where the third equality follows from the strong Markov property; and in the fourth equality, we have used that {Yt , t < τb+,Y } under
V (x; b) = Ex
eβ1,ξ (x−b)
Eβ˜ =
+ + ν = 1 − Ex e−(ν+θ)κb + Ex e−(ν+θ)κb V (Uκ + ; b) b ν+θ + ν −(ν+θ)τ b,Y ˜x e 1−E = ν+θ + + E˜ x e−(ν+θ)τb,Y V (Yτ + ; b) , (3.14)
Q˜ ξ−1 ,
(3.19)
and Eβ˜ and Eγ are given by
0
κb+
V (b; b) −
−ν t −θ
(3.18)
:= (V (b; b) − 1, N1 (b) − 1, . . . , Nn (b) − 1) Qν−1 ,
.
+∞
j = 1 , 2 , . . . , n.
X = (x1 , x2 , . . . , xm+1 )
For x < b, we have
(3.17) V (y; b)ϑj eϑj (y−b) dy,
By using Lemmas 3.2, 3.3 and formulas (3.15), (3.16), we have
i=1
τ
i = 1, 2, . . . , m,
−∞
:=
Proof. (1) We define a function of x as V (x; b) = Ex e−θ
V (y; b)ηi e−ηi (y−b) dy,
where X and Y are defined as follows:
θ i =1 i =1 i =1 i=1,i̸=j . m n n +1 m +1 ξ ˜ η i ϑi (γi,ν − γj,ν ) (βi,ξ + γj,ν ) i=1
Nj (b) :=
b b
ν = X Eβ˜ , for x < b, ξ V (x; b) − 1 = Y Eγ , for x > b, (3.12)
n
−∞
V ( x; b) −
(β˜ i,ξ + ϑj )
θ j=1,j̸=i j =1 j =1 j =1 , m n m +1 n +1 ξ ηj ϑj (β˜ i,ξ − β˜ j,ξ ) (β˜ i,ξ + γj,ν ) j =1
+∞
Mi (b) :=
x > b,
j =1
m+1
j,b
j =1
where we have used the fact that {Xt , t < τb−,X } and {Ut , t < κb− } under Px have the same law for x > b in the third equality. To simplify the expressions, we denote
Theorem 3.1. For any θ > 0, we have
Ex e−θ
− − = Ex e−νκb V (Uκ − ; b) + Ex 1 − e−νκb b −ντ − −ντb−,X V (Xτ − ; b) + Ex 1 − e b,X = Ex e b,X −ντb−,X 1BX (V (b; b) − 1) = 1 + Ex e 0,b n 0 −ντ − V (b + y; b)ϑj eϑj y dy − 1 , (3.16) + Ex e b,X 1BX
dt
− + Ex 1 − e−νκb
e−γ1,ν (x−b)
.. .
Eγ =
e
−γn+1,ν (x−b)
.
(3.20)
Therefore, if we can derive the expressions of X and Y , then we obtain the expressions of V (x; b) from (3.18). (2) In this step, we derive formulas for X and Y . From the definitions of X and Y , we have
(N1 (b) − 1, N2 (b) − 1, . . . , Nn (b) − 1) = Y Bν , ν ν ν = X B˜ ξ , M1 (b) − , M2 (b) − , . . . , Mm (b) − ξ ξ ξ
(3.21)
where the matrixes Bν and B˜ ξ are obtained by deleting the first column of Qν and Q˜ ξ respectively. Moreover, m+1
xi +
i =1
n+1 ν = yj + 1, ξ j =1
(3.22)
which comes from the fact that both sides of (3.22) are equal to V (b; b); thus we have m+1
i =1
xi −
n +1 j =1
yj =
θ . ξ
(3.23)
We multiply both sides of (3.15) by ϑj eϑj (x−b) , for j = 1, 2, . . . , n, and integrate with respect to x from −∞ to b. From Lemma 3.3, we obtain the following equation after some calculations.
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
N1 (b) − N2 (b) − .. . Nn (b) −
ν
ν
ϑ1
ϑ1 + β˜ 1,ξ ϑ1 ϑ + β˜ 1 2,ξ P˜ ξ = .. . ϑ1 ϑ1 + β˜ m+1,ξ
ϑ2
(3.24)
ϑn
···
ϑ2 + β˜ 1,ξ ϑ2 ϑ2 + β˜ 2,ξ
.. . ϑ2 ϑ2 + β˜ m+1,ξ
ϑn + β˜ 1,ξ ϑn ϑn + β˜ 2,ξ
··· .. .
.. . ϑn ϑn + β˜ m+1,ξ
···
β˜ 1,ξ η1 − β˜ 1,ξ .. . ˜ β1,ξ ηm − β˜ 1,ξ ˜ β1,ξ ˜ M = ϑ1 + β1,ξ .. . β˜ 1,ξ ϑn + β˜ 1,ξ β˜ 1,ξ
. (3.25)
1
Similar to derive (3.24), using Lemma 3.2 and formula (3.16), we have V (b; b) − 1 M1 (b) − 1 N1 (b) − 1 M2 (b) − 1 − 1 ′ = (Q Pν ) , .. .. ν . . Nn (b) − 1 Mm (b) − 1
(3.26)
η1 η + 1 γ1,ν η1 η1 + γ2,ν Pν = .. . η1 η1 + γn+1,ν
··· ··· .. . ···
ηm ηm + γ1,ν ηm ηm + γ2,ν . (3.27) .. . ηm ηm + γn+1,ν ′
.. .
.. .
.. .
.. .
···
β˜ m+1,ξ ηm − β˜ m+1,ξ
γ1,ν ηm + γ1,ν
···
···
β˜ m+1,ξ ϑ1 + β˜ m+1,ξ
γ1,ν ϑ1 − γ1,ν
···
.. .
.. .
.. .
.. .
···
β˜ m+1,ξ ϑn + β˜ m+1,ξ
γ1,ν ϑn − γ1,ν
···
···
β˜ m+1,ξ
γ1,ν
···
···
1
−1
···
γn+1,ν η1 + γn+1,ν .. . γn+1,ν ηm + γn+1,ν γn+1,ν ϑ1 − γn+1,ν . .. . γn+1,ν ϑn − γn+1,ν γn+1,ν −1
= 1, 2, . . . , m + 1 and j =
j =1
β˜ j,ξ
n+1
γj,ν
m
(β˜ i,ξ − ηj )
j=1,j̸=i
j =1 m+1
n
i =1
β˜ i,ξ
j =1
(3.33)
n +1
m
n
γi,ν (γj,ν + ηi ) (ϑi − γj,ν ) θ i =1 i=1 i =1 i =1 , m n n +1 m +1 ξ ˜ η i ϑi (γi,ν − γj,ν ) (βi,ξ + γj,ν )
yj γj,ν = −
i=1,i̸=j
i =1
thus we complete the proof.
i=1
1
˜ ξ = Y Dν , XD
Remark 3.2. For all i = 1, 2, . . . , m + 1 and j = 1, 2, . . . , n + 1, we have Fi > 0 and Gj < 0 in Theorem 3.1. Moreover,
(3.28) m+1
˜ ξ and Dν are given by where D β˜ 1,ξ β˜ 2,ξ ˜ξ = D . , .. β˜ m+1,ξ
···
(β˜ i,ξ + ϑj ) θ j =1 j =1 j=1 j=1 = m , n m +1 n +1 ξ ˜ ˜ ˜ ηj ϑj (βi,ξ − βj,ξ ) (βi,ξ + γj,ν )
It follows from the condition V (b−; b) = V (b+; b) (the proof will be given in the Appendix) and formula (3.18) that ′
γ1,ν η1 + γ1,ν
Solving Eq. (3.31), for all i 1, 2, . . . , n + 1, we have
xi β˜ i,ξ
η2 η2 + γ1,ν η2 η2 + γ2,ν .. . η2 η2 + γn+1,ν
β˜ m+1,ξ η1 − β˜ m+1,ξ
(3.32)
m+1
where Pν is given by
···
where P˜ ξ is given by
129
where
V (b; b) − ξ ξ ν ν M1 (b) − ξ ξ , ˜ −1 ˜ ′ = (Qξ Pξ ) .. . ν ν Mm (b) − ξ ξ
Fi <
i=1
−γ1,ν −γ2,ν Dν = .. . .
(3.29)
Using (3.19), we can write (3.24) and (3.26) as follows: Y Bν +
(3.30)
M (x1 , . . . , xm+1 , y1 , . . . , yn+1 )′ =
0, 0, . . . , 0,
θ ξ
′
n+1
Gj > −1,
(3.34)
j =1
which can be proved by using the fact that 0 < Eb e−θ < 1.
+∞ 0
1{Ut
m+1
=
Fi0 eβi,θ (x−b) , ˜
0 1{Ut
x ≤ b,
i=1
,
τ
Corollary 3.1. For any θ > 0,
Ex e−θ
Combining equalities (3.23), (3.28) and (3.30), we obtain
and
From Lemma 3.1, for θ > 0 and i = 1, . . . , m + 1, j = 1, . . . , n + 1, we have limν↓0 β˜ i,ξ = β˜ i,θ > 0 and limν↓0 γj,ν = γj,0 > 0. Therefore, if we let ν ↓ 0 in Theorem 3.1, then the Laplace +∞ transform of 0 1{Us
−γn+1,ν
θ (1, 1, . . . , 1) = X P˜ξ , ξ θ X B˜ ξ − (1, 1, . . . , 1) = Y Pν . ξ
θ ξ
(3.31)
Ex e
−θ 0+∞ 1{Ut
=1+
n +1
(3.35) G0j e−γj,0 (x−b)
,
x > b,
j=1
1 This equality means that the right derivative of V at b equals its left derivative.
where Fi0 , i = 1, 2, . . . , m + 1, and G0j , j = 1, 2, . . . , n + 1, are given by
130
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134 m+1
β˜ j,θ
j=1,j̸=i
Fi0 =
m
β˜ i,θ
i =1
G0j = −
m
ηi
i=1
i=1,i̸=j
m
γi,0
(γj,0 + ηi )
(3.36)
(ϑi − γj,0 )
i=1
(γi,0 − γj,0 )
i=1,i̸=j
(β˜ i,θ + γj,0 )
i =1
following corollary. Corollary 3.2. 1{Ut
Ex
1
=
τ
1{Ut
=
ν
0
m+1
1−
Hi e
β˜ i,ν (x−b)
,
(3.38)
Ij e−γj,ν (x−b) ,
x > b,
j =1
where Hi , i = 1, 2, . . . , m + 1, and Ij , j = 1, 2, . . . , n + 1, are given by m+1
j=1,j̸=i
Hi =
m
ηj
j =1 m+1
m i=1
n +1
n
β˜ i,ν
ηi
γj,ν
j =1
ϑj
j =1
i=1
Ij =
β˜ j,ν
j=1,j̸=i
n+1
i=1,i̸=j
n i=1
ϑi
m
γi,ν
i=1,i̸=j
,
n +1
(β˜ i,ν + γj,ν )
j =1
m
(γj,ν + ηi )
(γi,ν − γj,ν )
n
(3.39)
(ϑi − γj,ν )
i=1
τ
1{Ut ≥b} dt
Ex 0
m+1
Remark 3.5. As functions of x, Ex
Ex e
τ
(3.41)
τ 0
t
0 1{Us
≈ (−1)n−1
∂ n−1 (V (x; b)/ν) n ν /Γ (n) |ν=n/t . ∂ν n−1
(3.42)
Similarly, t
1{Us
Ex 0
≈ (−1)
n −1
τ ∂ n−1 Ex 0 1{Us
which can be computed from Corollary 3.2. The calculations of (3.42) and (3.43) are complicated when the value of n is large, whereas this method becomes very easy if X is a Brownian motion, see Section 4. Alternatively, one can take inverse Laplace transform to obtain the left-hand side of (3.42) and (3.43).
In this section, we consider three special cases. The first one is that X is a double exponential jump diffusion process, then X is simplified to have only negative jumps, and finally, X is assumed to be a Brownian motion. 4.1. Double exponential jump diffusion process case
ψ(z ) =
Ex e−θ
σ2 2
z + µz + λ 2
p1 η1
η1 − z
+
q1 ϑ1
ϑ1 + z
(3.40)
1{Ut
0 1{Us
−1
(4.1)
0
˜
=
˜
θ ,0 γ1,0 (b−x)
1 − A3 e
τ
x ≤ b, x > b.
1{Us
−Aθ1,0 eβ1,θ (x−b) − Aθ2,0 eβ2,θ (x−b) ,
x > b.
τ
ν − Aξ ,ν eβ˜ 1,ξ (x−b) − Aξ ,ν eβ˜ 2,ξ (x−b) , 1 2 = ξ ξ ,ν γ1,ν (b−x) ξ ,ν γ2,ν (b−x) 1 − A3 e − A4 e , +∞
Ex e−θ
x ≤ b,
are continuous at b, which can be proved
easily by using (3.23).
ν n t n−1 −ν t −θ t 1{U
(2005). Under this case, results in the previous section can be simplified as follows.
= Ex [τ ] − Ex 1{Ut
−θ 0+∞ 1{Ut
=
β2,q (β˜ 2,q ), γ1,q (γ˜1,q ) and γ2,q (γ˜2,q ), see Appendix B in Kou et al.
(β˜ i,ν + γj,ν )
i=1
Ex e−θ
Ex
+∞
for z ∈ (−ϑ1 , η1 ). For the solutions in Lemma 3.1 of (4.1), here we only add that one can obtain the analytical forms for β1,q (β˜ 1,q ),
.
Remark 3.4. Note that Hi > 0 for all i = 1, 2, . . . , m + 1. Hence, m+1 β˜ (x−b) i,ν we have > 0 in (3.38). More importantly, we i=1 Hi e have the following result.
Consider the process X with double exponential jump distribution, i.e., m = n = 1 in (2.11). Now the Lévy exponent of X reduces to
(β˜ i,ν + ϑj )
j =1
(β˜ i,ν − β˜ j,ν )
i=1
n+1
n
(β˜ i,ν − ηj )
j =1
m+1
1{Us
4. Special cases
x ≤ b,
i=1
n +1 1
Ex
ν
0
0
n ; t
Therefore, for large n,
1 1 ∂ β˜ i,ξ = lim = , (3.37) lim ′ ′ ˜ (β˜ i,ξ ) θ→0 ψ θ→0 ∂θ ψ˜ (β˜ i,ν ) ˜ z ) = ψ(z ) − α z which can be obtained where the function ψ( from (3.1). Therefore, taking a derivative with respect to θ in Theτ orem 3.1 and then letting θ equal to 0 yield Ex 0 1{Us
Γ (n,ν)
0
˜ β˜ i,ξ ) = ξ for i = 1, . . . , m + 1, we Note that ξ = ν + θ and ψ( have
τ
−θ
∂ n−1 (V (x; b)/ν) n = (−1)n−1 ν /Γ (n). ∂ν n−1
Remark 3.3. For the process U 1 (see Appendix), Renaud (2013) +∞ has derived the Laplace transform of 0 1{U 1
Ex e
.
m+1
. In fact, by the strong law of large numbers, we
could construct an Erlang random Γ (n, ν) ≈ t for large n if ν = and for the independent Erlang random Γ (n, ν), we have
(β˜ i,θ + γj,0 )
i=1
i =1
,
n
0 1{Us
Ex e−θ
j =1
n +1
ϑi
(β˜ i,θ − β˜ j,θ )
n+1
constant t, we can use Theorem 3.1 to approximate For a given t
(β˜ i,θ + ϑj )
j =1
j=1,j̸=i n +1
n
n
(β˜ i,θ − ηj )
j =1
j=1
m+1
m
γj,0
m+1
ϑj
j =1
j =1
n
ηj
n+1
θ ,0 γ2,0 (b−x)
− A4 e
x ≤ b,
,
x > b.
1{Us
0 1 + Aν,ν eβ˜ 1,ν (x−b) + Aν,ν eβ˜ 2,ν (x−b) , 1 2 = ν ν,ν γ1,ν (b−x) ν,ν γ2,ν (b−x) A3 e + A4 e ,
x ≤ b, x > b.
(4.2)
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
ξ ,ν
In the above formulas, the constants Ai , i = 1, . . . , 4, are given by ξ ,ν
A1
ξ ,ν
A2
ξ ,ν
A3
ξ ,ν
A4
β˜ 2,ξ γ2,ν γ1,ν (β˜ 1,ξ − η1 )(ϑ1 + β˜ 1,ξ )θ , (β˜ 1,ξ + γ1,ν )(β˜ 1,ξ + γ2,ν )(β˜ 2,ξ − β˜ 1,ξ )η1 ξ ϑ1 β˜ 1,ξ γ2,ν γ1,ν (β˜ 2,ξ − η1 )(ϑ1 + β˜ 2,ξ )θ = , (β˜ 2,ξ + γ1,ν )(β˜ 2,ξ + γ2,ν )(β˜ 1,ξ − β˜ 2,ξ )η1 ξ ϑ1 β˜ 1,ξ β˜ 2,ξ γ2,ν (γ1,ν + η1 )(ϑ1 − γ1,ν )θ = , (β˜ 1,ξ + γ1,ν )(β˜ 2,ξ + γ1,ν )(γ2,ν − γ1,ν )η1 ξ ϑ1 β˜ 1,ξ β˜ 2,ξ γ1,ν (γ2,ν + η1 )(γ2,ν − ϑ1 )θ = , (β˜ 1,ξ + γ2,ν )(β˜ 2,ξ + γ2,ν )(γ2,ν − γ1,ν )η1 ξ ϑ1
=
θ,0
and Ai
ν,ν
is defined similarly and Ai
Ex e−θ (4.3)
σ
2
z 2 + µz + λq1
ξ ,ν
ξ ,ν
:= lim A1 = − η1 ↑∞
ξ ,ν
lim A2
η1 ↑∞ ξ ,ν
B3
ξ ,ν
ξ ,ν
B4
ξ ,ν
η1 ↑∞
z > −ϑ1 .
(4.4)
γ2,ν γ1,ν (ϑ1 + β˜ 1,ξ )θ , ˜ (β1,ξ + γ1,ν )(β˜ 1,ξ + γ2,ν )ξ ϑ1
τ
β˜ 1,ξ γ2,ν (ϑ1 − γ1,ν )θ , ˜ (β1,ξ + γ1,ν )(γ2,ν − γ1,ν )ξ ϑ1 β˜ 1,ξ γ1,ν (γ2,ν − ϑ1 )θ = , (β˜ 1,ξ + γ2,ν )(γ2,ν − γ1,ν )ξ ϑ1
=
1
ν
ν,ν
β1,ν (x−b) + Bν,ν , 1 e ˜
x ≤ b,
ν,ν
B3 eγ1,ν (b−x) + B4 eγ2,ν (b−x) ,
θ ,0
where Bi
ξ ,ν
= limν↓0 Bi
ν,ν
and Bi
(4.8)
x > b, ξ ,ν
= limθ↓0 Bi /θ , for i = 1, 3, 4.
Remark 4.3. Note that X is a Lévy process without positive jump in this subsection and the later one. When x = b, formulas (4.7) and (4.14) can be found from Kyprianou and Loeffen (2014) (see Corollary 2). In addition, note that the process U reduces to X if α = 0. Thus we remark that results (4.7) and (4.14) with α = 0 have been established by other authors, see, e.g., Corollary 1 in Landriault et al. (2011).
In this subsection, we consider the simplest case Xt = X0 +µt +
σ Wt with µ > 0. Thus the process (Ut )t ≥0 satisfies dUt = dXt − α 1{Ut
= µdt + σ dWt − α 1{Ut
(4.9)
(4.5)
σ2
z 2 + µz , (4.10) 2 and for the sake of brevity, we denote the roots of ψ(z ) = q and ψ(z ) − α z = q for q > 0 by βq > 0, γq < 0 and β˜ q > 0, γ˜q < 0 respectively, see also (A.9) and (A.10). Noting that (4.10) can be obtained from (4.4) by letting ϑ1 ↑ ∞, we can derive
βq = lim β1,q , ϑ1 ↑∞
x > b.
ϑ1 ↑∞
(4.11)
ϑ1 ↑∞
ϑ1 ↑∞
Formula (4.5) and the fact that ϑ1 < γ2,ν yield ξ ,ν
=
ξ ,ν
=
lim B3
(4.6)
β˜ q = lim β˜ 1,q ,
γq = − lim γ1,q and γ˜q = − lim γ˜1,q .
lim B1
x ≤ b,
t > 0,
and U0 = X0 ,
ϑ1 ↑∞
ν ξ ,ν ˜ − B1 eβ1,ξ (x−b) , ξ = ξ ,ν ξ ,ν 1 − B3 eγ1,ν (b−x) − B4 eγ2,ν (b−x) ,
(4.7)
x > b,
1{Us
ψ(z ) =
τ e−θ 0 1{Us
,
and the process (Yt )t ≥0 is given by Yt = X0 + (µ − α)t + σ Wt . The Lévy exponent of X is
Theorem 4.1. For θ > 0,
Ex
− B4 e
4.3. Brownian motion case
where −γ1,ν < 0 and −γ2,ν < 0 are the roots of ψ(z ) = ν with ψ(z ) given by (4.4); and β˜ 1,ξ > 0 is the unique positive root of ψ(z ) − α z = ξ . Then, from the first equation in (4.2), we obtain Theorem 4.1.
x ≤ b,
θ ,0 γ2,0 (b−x)
0
:= lim A3 = := lim A4
˜
θ ,0 γ1,0 (b−x)
= 0,
η1 ↑∞
−Bθ1,0 eβ1,θ (x−b) ,
Ex
The reason why we denote the Poisson intensity in (4.4) by λq1 not λ is that now (4.4) can be obtained from (4.1) by letting η1 ↑ ∞. Applying the fact that η1 < β˜ 2,ξ , for fixed ϑ1 > 0, we have B1
and
ξ ,ν
ϑ1 −1 , ϑ1 + z
1{Us
1 − B3 e
In this case, the Lévy exponent is given by
ψ(z ) =
0
=
= limθ↓0 Ai /θ .
+∞
4.2. The process X only has negative jumps with exponential distribution
that is similar to the proof of Theorem 3.1. This verification process is simple because there is only one-sided jump, and we omit the details here. It follows from (4.6) that
Remark 4.1. When α = 0, the two processes U and X are equal. We comment that the first result in (4.2) with α = 0 have been derived in Cai et al. (2010) (see Theorem 3.2). Here, we make a remark that the approach used in Cai et al. (2010) is completely different from ours. We have to say that our method is simple relatively, because in this paper it is unnecessary to take too much effort as in Cai et al. (2010) to establish an integro-differential equation. For more details, we refer the reader to Cai et al. (2010).
2
131
ϑ1 ↑∞
θ γν , ˜ ξ (βξ − γν ) θ β˜ ξ ξ (β˜ ξ − γν )
(4.12) and
ν,ξ
lim B4
ϑ1 ↑∞
= 0.
Applying (4.6), for θ > 0, we can deduce the following theorem. Theorem 4.2.
Remark 4.2. Strictly speaking, to derive the above theorem, we still need to show that the left-hand side of (4.6) is equal to
Ex e−θ
τ
0 1{Us
in (4.2) after letting η1 ↑ ∞. Even though some-
one is likely to prove this conclusion though some efforts, we recommend the readers to verify formula (4.6) by using an approach
ν θ γν β˜ (x−b) , −e ξ τ ξ ξ (β˜ ξ − γν ) −θ 0 1{Us
x ≤ b, (4.13) x > b,
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J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
γ0 ˜ − eβθ (x−b) , (β˜ θ − γ0 ) +∞ Ex e−θ 0 1{Us
x ≤ b, (4.14) x > b,
τ
Ex 0
γν ˜ 1/ν + eβν (x−b) , ν(β˜ ν − γν ) 1{Us
dUt1 = dX˜ t − α 1{U 1 >b} dt ,
t > 0,
t
(A.3)
and U01 = X˜ 0 .
From 22 in Kyprianou and Loeffen (2010), we have Corollary Px Ut1 = b = 0 for Lebesgue almost every t ≥ 0. Therefore, from (A.3), we have
2µ
where γ0 = − σ 2 , and
First, for any given b, let the process U 1 = Ut1 t ≥0 satisfy the following SDE:
from which we have
x ≤ b, (4.15) x > b.
t
Ut1 = Y˜t + α
1{U 1
0
t ≥ 0,
(A.4)
where Y˜t = X˜ t − α t for t ≥ 0.
Remark 4.4. Similar to Remark 4.2, we have to verify that equality (4.13) is correct. Compared with Remark 4.2, the verification process is much simpler because now there is no jumps. In addition, the computations of (3.42) and (3.43) from (4.13) and (4.15) become easier as well, especially when x = b. Remark 4.5. For α = 0 and σ = 1, formula (4.13) has been derived by other author, see formula 1.5.1 on p. 256 of Borodin and Salminen (2002) for instance.
Remark A.1. When X˜ in (A.3) is a Lévy process without positive jump, Kyprianou and Loeffen (2010) (see Theorem 1) have showed that there exists a unique strong solution U 1 = Ut1 t ≥0 to Eq. (A.3). In addition, U 1 is a strong Markov process. The process U 1 is called a refracted Lévy process, and for more details about this process, we refer to Kyprianou and Loeffen (2010). For any given ε > 0, we define the stopping time τε1 as τε1 := inf{t ≥ 0, Ut1 > b + ε or Ut1 < b − ε}. In the following, we give a technical lemma which has been obtained by Renaud (2013), see also Theorem 4 of Kyprianou and Loeffen (2010). Lemma A.1 (Theorem 2 in Renaud, 2013). For q > 0 and α ≥ 0,
5. Conclusion
In this paper, we investigate the problem of computing the time of deducting fees for variable annuities with a state-dependent fee deducting method which is proposed recently. Under a general jump diffusion model, we have derived closed-form formulas for the time of charging expenses. In addition, some of our outcomes extend the results obtained by previous authors like Renaud (2013) to the process which has two-sided jumps. From the derivation of Theorem 3.1, we claim that our results can be extended to a general jump diffusion process X only if we know how to compute (3.9) and (3.10). For example, the process X can be assumed to have jumps with phase-type distributions, see Proposition 2 in Asmussen et al. (2004) for the calculations of (3.9) and (3.10). Acknowledgments
W (q) (ε)
Eb e−qτε 1{U 1 1
=
=b+ε} 1
τε
ω(q) (b
+ ε; b − ε)
,
(A.5)
and
Eb e−qτε 1{U 1
1
τε1
= Z (q) (ε) −
=b−ε}
z (q) (b + ε; b − ε)
ω(q) (b + ε; b − ε)
W (q) (ε). (A.6)
Here ω(q) (b + ε; b − ε) and z (q) (b + ε; b − ε) are given as follows.
ω(q) (b + ε; b − ε) = W (q) (2ε) + α
b+ε
˜ (q) (b + ε − y)W (q)′ (y − b + ε)dy, W
b
(A.7)
z (q) (b + ε; b − ε)
= Z (q) (2ε) + α q
b+ε
˜ (q) (b + ε − y)W (q) (y − b + ε)dy, W
b
The authors are grateful to an anonymous reviewer for various useful comments and suggestions on an earlier version, which help to improve the structure and text of the paper. They also thank the editor Elias S. Shiu for helpful comments and suggestions and kindly sending an article about writing.
where W (q) (x) =
Recalling the process X given by (2.1), we denote the first jump time of the Poisson process Nt by T1 which is an exponential random variable with parameter λ. We denote the continuous part of X by X˜ = (X˜ t )t ≥0 which is given by X˜ t = X0 + µt + σ Wt ,
t ≥ 0.
(A.1)
The process U˜ related to X˜ is defined by dU˜ t = dX˜ t − α 1{U˜ t
t > 0,
σ 2 (βq − γq )
,
x > 0,
2(eβq x − eγ˜q x ) ˜
˜ (q) (x) = W
Appendix
2(eβq x − eγq x )
, x > 0, σ 2 (β˜ q − γ˜q ) ˜ (q) (x) = 0, x ≤ 0, W (q) (x) = W x (q) Z (x) = 1 + q W (q) (y)dy, x ∈ R.
(A.8)
0
In (A.8), βq > 0 and γq < 0 are the two roots of the equation:
µx +
σ2 2
x2 = q,
(A.9)
and β˜ q > 0 and γ˜q < 0 are the two roots of the equation: (A.2)
(µ − α)x +
σ2 2
x2 = q.
(A.10)
J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
Another function Z˜ q (x) defined as Z˜ (q) (x) = 1 + q
x
˜ (q) (y)dy, W
133
Remark A.4. Note that the limitations: limε↓0
x ∈ R,
1/2−Eb e−qτ˜ε 1{U˜
(A.11) and limε↓0
0
ε
1/2−Eb e−qτ˜ε 1{U˜
ε
τ˜ε =b−ε}
τ˜ε =b+ε}
do not depend on q.
will be used in the following.
are known as the q-scale functions of X˜ Y˜ , one can refer to
Proof. {The proof of V ′ (b−; b) = V ′ (b+; b)} From (3.18), it is obvious that both V ′ (b−; b) and V ′ (b+; b) exist. Therefore, to prove they are equal, we only need to show that
Chapter 8 of Kyprianou (2006) for more details about these functions.
lim
˜ (q) (x) and Z (q) (x) Z˜ (q) (x) Remark A.2. The functions W (q) (x) W
From Renaud (2013) and Kyprianou and Loeffen (2010), it is easy to show that Lemma A.1 holds for α < 0 as well. Actually, as Renaud stated, Theorem 2 in his paper is a restatement of Theorem 4 in Kyprianou and Loeffen (2010) which holds for α ∈ R. Therefore, recalling (A.2), we have U˜ t = X˜ t − α
t
1{U˜ s
0
ε↓0
(V (b + ε; b) + V (b − ε; b)) /2 − V (b; b) = 0. ε
From the definition of V (x; b), we have V (b; b) = Eb
0
1{U˜ s
(A.12)
Eb e
(A.13)
(A.14)
where the stopping time τ˜ε is defined as τ˜ε := inf{t ≥ 0, U˜ t > b + ε or U˜ t < b − ε} and
˜ =W
(2ε) + α
dt
t
0 1{Us
dt
ν e−ν t e−θ
t
ν e−ν t e−θ
t
0 1{Us
+∞ τ˜ε
0 1{Us
dt1{τ˜ε ≥T1 }
dt1{τ˜ε
(A.18)
τ˜ε
Eb
νe
−ν t −θ
e
t
0 1{Us
0
W
(q)
˜ (b + ε − y)W
(q)′
b
(y − b + ε)dy,
≤ Eb τ˜ε
Eb
νe
= 1 − Eb e−ν τ˜ε
ν e−ν t e−θ
t
0 1{Us
dt
≥
0
˜ (q) (y − b + ε)dy. W (q) (b + ε − y)W
b
−ν t
(A.19)
0
(A.15) b+ε
τ˜ε
b+ε
z˜ (q) (b + ε; b − ε)
= Z˜ (q) (2ε) + α q
0 1{Us
Because θ > 0, thus
ω˜ (q) (b + ε; b − ε) (q)
τ˜ε
+ Eb ˜ (q) (ε), W
e
t
+∞
ω˜ (q) (b + ε; b − ε)
−ν t −θ
ν e−ν t e−θ
+ Eb
τ˜ε =b−ε}
= Z˜ (q) (ε) −
dt
0
z˜ (q) (b + ε; b − ε)
0 1{Us
+∞ t ν e−ν t e−θ 0 1{Us
and
Eb e−qτ˜ε 1{U˜
νe
τ˜ε
1{U˜ =b+ε} = (q) , τ˜ε ω˜ (b + ε; b − ε)
t
0
= Eb
˜ (q) (ε) W
τ˜ε
= Eb
Corollary A.1. For q > 0, −qτ˜ε
ν e−ν t e−θ 0
Then, comparing (A.4) and (A.12), we obtain the following corollary from Lemma A.1.
+∞
t
(A.17)
ν 1 − Eb e−(ν+θ )τ˜ε . ν+θ
Then applying (A.16) yields
Remark A.3. Corollary A.1 can be extended to a Lévy process without positive jump, i.e., the process X˜ in (A.12) can be assumed to be a general Lévy process with only negative jump except that X˜ t − α t has monotone paths. In fact, in this case, we still have (A.4)–(A.6) and (A.12), from which this extension is derived. Alternatively, one can show this conclusion directly by using almost the same method as in Renaud (2013), for interested readers, see that paper for the details. Using L’Hospital’s rule twice, we can derive the following results from Corollary A.1 after some calculations.
Eb
τ˜ε 0
lim
ν e−ν t e−θ
t
0 1{Us
dt
= 0.
ε
ε↓0
(A.20)
Note that T1 is independent of X˜ and thus independent of τ˜ε . We have +∞
0 ≤ Eb
τ˜ε
ν e−ν t e−θ
t
0 1{Us
dt1{τ˜ε ≥T1 }
≤ Pb (τ˜ε ≥ T1 ) = 1 − Eb e−λτ˜ε ,
(A.21)
Corollary A.2.
lim
1/2 − Eb e−qτ˜ε 1{U˜ =b+ε} τ˜ε
1/2 − Eb e lim ε↓0
=
ε
ε↓0
−qτ˜ε
ε
1{U˜ =b−ε} τ˜ε
and using (A.16) again, we obtain
=
3α − 2µ 4σ
2
2µ − 3α 4σ 2
,
(A.16)
.
lim Eb ε↓0
+∞ τ˜ε
νe
−ν t −θ
e
t
0 1{Us
dt1{τ˜ε ≥T1 } /ε = 0.
(A.22)
Using the strong Markov property and that {Xt , t < T1 } and {X˜ t , t < T1 } have the same law under Pb , we have
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J. Zhou, L. Wu / Insurance: Mathematics and Economics 61 (2015) 125–134
Eb
+∞
ν e−ν t e−θ
τ˜ε
−ν τ˜ε −θ
τ˜ε
t
0 1{Us
1{U˜
References
dt1{τ˜ε
˜ {τ˜ε
= Eb e τ˜ −(ν+λ)τ˜ε −θ 0 ε 1{U˜
ε
−(ν+λ)τ˜ε −θ
+ Eb e
τ˜ε 0
1{U˜
{U˜ τ˜ε =b−ε} V (b − ε; b) ,
(A.23)
where the second equality comes from the fact that T1 is indepen˜ Similar to (A.19), we can show that dent of the process U.
1/2 − Eb e−(ν+λ)τ˜ε e lim
−θ
τ˜ε 0
1{U˜
1{U˜ =b+ε} τ˜ε
=
ε
ε↓0
1/2 − Eb e−(ν+λ)τ˜ε e lim ε↓0
−θ
τ˜ε 0
1{U˜
{U˜ τ˜ε =b−ε}
ε
3α − 2µ 4σ 2
=
,
(A.24)
2µ − 3α 4σ 2
.
Therefore, it follows from (A.18), (A.20), (A.22) and (A.24) that
(V (b + ε; b) + V (b − ε; b)) /2 − V (b; b) ε τ˜ −θ 0 ε 1{U˜
lim ε↓0
= V (b; b) × 0 = 0, thus we complete the proof.
(A.25)
Remark A.5. Note that the distribution of Z1 (see (2.1)) does not appear in the above proof. Thus the conclusion V ′ (b−; b) = V ′ (b+; b) holds for general jump diffusion processes.
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