Recall that {Xt}t≥0is a jump-diffusion process of the form in (3.2) and −ieρ1, · · ·, −ieρµ2, −iρ1, · ·
·, −iρµ1 are the roots of r − ψ(z) = 0 with Re (eρµ2) ≤ · · · ≤ Re (eρ1) ≤ 0 < Re (ρ1) ≤ Re (ρ2) ≤
· · · ≤ Re (ρµ1). We assume further that Zr is separable (i.e., m1= m2 = · · · = mµ1 = em1 =
· · · = emµ2 = 1). Under these assumptions, our goal is to solve the optimal stopping problem (1.1) for a given continuous function g and r > 0. To do this, we first give an explicit formula for solutions of the averaging problems in (2.4) and (2.5). Then we find sufficient conditions on g that guarantees the existence of the optimal stopping boundary for the problem (1.1).
We observe, by Theorem 3.3, that the distribution of inf0≤s≤erXsis given by
P
Also the distribution of sup0≤s≤erXsis given by
P
Throughout this thesis, we follow the convention thatQµ1
i=1,i6=k(. . .) = 1 (resp.,Qµ2 We also need the following facts.
Lemma 4.1 Suppose µ1≥ 1 and µ2≥ 1. Then
(a) For i = 1, .., µ1, we have
(d) For any complex numbers Ap,m and ω, we have
µ2 (e) For any complex numbers As, we obtain
µ1
(f) The following identities hold:
Proof. (a) The identities in part (a) follows from the facts that −ieρ1, ···, −ieρµ2and−iρ1, ···, −iρµ1 fractional decomposition to the right hand side of (4.15) gives
v1
(c) Note that we have
ψr−(u) = fractional decomposition to the right hand side of (4.17) gives
v2 Given any complex numbers Ap,m and w, we have
µ2
where the last equality follows from (4.11). The proof is complete. This together with (4.12) yields (4.13).
(f) From Theorem 3.3, we have the following observation.
By applying the Wiener-Hopf factorization formula and combining with (4.16) and (4.18), we see that for b = 0,
For the case b = 0, our results follow by multiplying both sides of (4.21) by u, letting u → ∞ and using (4.20). For the case b 6= 0, we obtain our result by multiplying both sides of (4.22) by u2, letting u → ∞ and using (4.20).
Observe that if v1 ≥ 1 (i.e., there are upside jumps for the process X), then the function ψ(iz) is a real analytic function in (−β1, 0) with ψ(0) = 0 and limz↓−β1ψ(iz) = ∞. Hence, we have that 0 < ρ1< β1.
Definition 4.2 We write g ∈ π0 if the function g : R → R is absolutely continuous on every
We shows below that Qg is a solution of the average problem (2.4).
Theorem 4.3 For any g ∈ π0 and r > 0,
Q(4)g = de0
Taking account of g ∈ π0 and using integration by parts, we obtain eρsx
(For details, see the Appendix). For simplicity, we write Is,k,j−`(1) = eρsx
By this and the identity Again, by using integration by parts together with g ∈ π0, we get
eρsx
E
Consequently, using (4.20), (4.14) and (4.25), we see that for the case b 6= 0,
Eh (4.32), (4.24) and (4.29) that
Z ∞
The last equality comes from (b) of Lemma 4.1. Using (4.35), (4.24), and (4.30), we have
It follows from (4.25) (4.38), (4.39), (4.40), and (4.41) that E
= g(x). The proof is complete.
We write g ∈ R if g : R → R is a L1-integrable function such that the Fourier transform bg,
satisfies the integrability conditionR∞
−∞(1 + |ω|3)|bg(ω)|dω < ∞. As noted in Surya [27], the set R belongs to the class of Cb3. This implies that every element in R also is in π0. In [27], Surya showed that if g ∈ R, the function 2π1 R∞
−∞eiωx bψg(ω)−
r(ω)dω solves the American put-type averaging problem. In the following, we show that the function 2π1 R∞
−∞eiωx bψg(ω)+
r(ω)dω coincides with the Qg(x) given by (4.23).
Proof. By the Fourier inversion formula, it is sufficient to prove that Qbg(ω) :=
To computeR∞ To prove these, observe that
¯¯
Notice that g ∈ R implies g ∈ Cb3. Therefore g is uniformly continuous on R. Combining this with g ∈ L1 yields
|x|→∞lim g(x) = 0. (4.50)
This together with the Dominated Convergence Theorem, implies that the right hand side of (4.49) converges to zero, as t → −∞ or t → ∞. Therefore, we justify (4.47) and (4.48). From By using integration by parts along with (4.50), we have for µ2≥ 1
Z ∞
and Z ∞
Combining (4.45), and (4.51)-(4.56), we see that
Qbg(ω) =
This together with (3.4) and (a) of Lemma 4.1 implies that
Qbg(ω) = bg(ω)
Remark 4.5 It follows from (4.57) and (4.58) that
Proposition 4.6 Assume {Xt}t≥0 is a jump-diffusion process of the form (3.2) with ckj > 0, βk > 0, and αp > 0 for 1 ≤ k ≤ v1, 1 ≤ j ≤ nk, 1 ≤ p ≤ v2. Consider the reward function
Also, we will show that for 1 ≤ k ≤ v1 and 1 ≤ ` ≤ nk, From identities (4.17) and (4.18), we acquire
E[euIr] = We obtain (4.61) by multiplying both sides by u, letting u → ∞ in (4.64) and using the fact that µ2=Pv2 This yields (4.63). Using (4.61)-(4.63) and (4.60), we obtain
t→blima+Qg(t) < 0. (4.65) To prove (a), notice that by the assumption in (a), we have limx→+∞Qg(x) > 0 and Qg(x) ∈ C(ba, +∞). These together with (4.65) and the intermediate-value theorem, imply that there exists at least one x∗ in (ba, ∞).
has an unique solution on (ba, ∞), we have Qg(x) < 0, for x ∈ (ba, x∗). For x > x∗, we have Qg(x) = g(x)h(x) and hence Q0g(x) = g0(x)h(x) + g(x)h0(x). Since each term of the right hand side is nonnegative, and g(x) and h0(x) are positive, we obtain Q0g(x) > 0 for x > x∗and hence Qg(x) is increasing on (x∗, ∞).
Combining Theorem 2.7, Theorem 4.3, and Proposition 4.6 gives the following main result.
Theorem 4.7 Assume {Xt}t≥0 is a jump-diffusion process of the form (3.2) with ckj > 0, βk > 0, and αp > 0 for 1 ≤ k ≤ v1, 1 ≤ j ≤ nk, 1 ≤ p ≤ v2. Consider a payoff function g(x) ∈ bπ0 with Qg(x) given by (4.23). Assume that the following conditions hold:
(a) There exists α > 0 such that limx→∞Qg(x) ≥ α.
(b) (g(u+x)g(x) )0< 0 and (gg(x)0(x))0 < 0 for any x > ba and u > 0.
Then the optimal stopping time for the optimal stopping problem (1.1) is given by τ∗= inf{t >
0 : Xt> x∗} and the value function is given by V (x) = Ex(e−rτ∗g(Xτ∗)) =
Z ∞
x∗−x
Qg(x + m)fMr(m)dm.
Here x∗ is the unique solution of the equation Qg(x) = 0 in (ba, +∞) and fMr is given by (4.4).
Remark 4.8 The inspiration of giving the explicit formula for solutions of the averaging prob-lem (2.4) comes from the well-known result. If g ∈ C02(Rn) then
Ex
· Z ∞
0
e−rt(r − A)g(Xt)dt
¸
= g(x).
Furthermore, observe that Ex
· Z ∞
0
e−rt(r − A)g(Xt)dt
¸
=E
· (r − A
r )g(x + Xer)
¸
=E
· E
µ (r − A
r )g(x + Mr+ Ir)|Mr
¶¸
=E
·
Qg(x + Mr)
¸
where
Qg(z) = E
· (r − A
r )g(z + Ir)
¸ .
In fact, the identity (4.23) follows from expanding the above expression of Qg(x). In addition, using similar argument, we also obtain
g(x) = Ex
· Z ∞
0
e−rt(r − A)g(Xt)dt
¸
= E
·
Pg(x + Ir)
¸
where
Pg(z) = E
· (r − A
r )g(z + Mr)
¸ .
Observe that if v2≥ 1 (i.e., there are downside jumps for the process X), then the function ψ(iz) is a real analytic function in (0, α) with ψ(0) = 0 and limz↑α1ψ(iz) = ∞. Hence, we have that 0 < −eρ1< α.
Definition 4.9 We write g ∈ π1 if the function g : R → R is absolutely continuous on every compact interval and moreover, for v2 ≥ 1, there exist A1 > 0, A2 > 0 and θ ∈ (0, −eρ1) such that |g(x)| ≤ A1+ A2e−θx, ∀x ∈ R.
For any g ∈ π1, we define Pg(x) by the formula Pg(x)
=
v2
X
p=1
`p
X
m=1
Xm k=1
−µ(αp)mecpm r(m − k)!
×
·µ 1{µ1≥1}
µ1
X
j=1
djρj
(αp+ ρj)k + 1{a<0andb=0}1{k=1}d0
¶ Z 0
−∞
(−t)m−kg(t + x)eαptdt
¸
+
·
1{µ1≥1}b2 2r
µXµ1
j=1
djρj
¶
− 1{a<0andb=0}
ad0
r
¸ g0(x)
+
· 1{µ1≥1}
µ1
X
j=1
djρj
r (a +b2ρj
2 ) + 1{a<0andb=0}
d0
r (λ + µ + r)
¸
g(x). (4.66)
Remark 4.10 Pg(−x) = bQbg(x) where bQbg is given in (4.23) for bg(x) = g(−x) and the process
−Xt.
In the following, we study some properties of Pg(x). We write g ∈ bπ1 if g ∈ π1 and g is non-increasing and g ∈ C1(−∞, ba), where {g > 0} = (−∞, ba) for some ba > −∞.
Proposition 4.11 Assume {Xt}t≥0is a jump-diffusion process of the form (3.2) with ecpm> 0, βk > 0, and αp > 0 for 1 ≤ p ≤ v2, 1 ≤ m ≤ `p, 1 ≤ k ≤ v1. Consider the reward function g ∈ bπ1 with {g > 0} = (−∞, ba) and assume Pg is given by the formula in (4.66). Then (a) If there exists β > 0 such that limx→−∞Pg(x) ≥ β, then there exists x∗ < ba such that
Pg(x∗) = 0.
(b) If (g(t+x)g(x) )0 > 0 and (gg(x)0(x))0 < 0 for any x < ba and t < 0, then there exists at most one x∗∈ (−∞, ba) such that Pg(x∗) = 0.
(c) If both conditions in (a) and (b) hold, then there exists a unique x∗ ∈ (−∞, ba) such that Pg(x∗) = 0. Moreover, Pg(x) is decreasing for x < x∗ and Pg(x) ≤ 0 for x∗≤ x < ba.
Proof. We first show that limt→ba−Pg(t) < 0. To do this, we first verify that for µ1 ≥ 1 and b 6= 0
µ1
X
j=1
djρj= Qµ1
η=1ρη
Qv1
k=1βk
> 0 (4.67)
and for a < 0 and b = 0 From identities (4.15) and (4.16), we have that
Ex[e−uMr] =
We obtain (4.67) by multiplying both sides by u, letting u → ∞ in (4.70) and using the fact that µ1=Pv1
This yields (4.69). Using (4.67)-(4.69) and (4.66), we obtain
t→blima−Pg(t) < 0. (4.71) To prove (a), notice that by the assumption in (a), we have limx→−∞Pg(x) > 0 and Pg(x) ∈ C(−∞, ba). These together with (4.71) and the intermediate-value theorem, imply that there exists at least one x∗ in (−∞, ba).
has an unique solution on (−∞, ba), we have Pg(x) < 0, for x ∈ (x∗, ba). For x < x∗, we have Pg(x) = g(x)h(x) and hence Pg0(x) = g0(x)h(x) + g(x)h0(x). By means of the facts that g0≤ 0, g > 0, h0< 0 and h > 0, we have that Pg0(x) < 0 for x < x∗ and hence Pg(x) is decreasing on (−∞, x∗).
Theorem 4.12 Assume {Xt}t≥0 is a jump-diffusion process of the form (3.2) with ecpm > 0, βk > 0, and αp > 0 for 1 ≤ p ≤ v2, 1 ≤ m ≤ `p, 1 ≤ k ≤ v1. Consider a payoff function g(x) ∈ bπ1 with Pg(x) given by (4.66). Assume that the following conditions hold:
(a) There exists β > 0 such that limx→−∞Pg(x) ≥ β.
(b) (g(t+x)g(x) )0> 0 and (gg(x)0(x))0< 0 for any x < ba and t < 0.
Then the optimal stopping time for the optimal stopping problem (1.1) is given by τ∗= inf{t >
0 : Xt< x∗} and the value function is given by
V (x) = Ex(e−rτ∗g(Xτ∗)) = Z x∗−x
−∞
Pg(x + z)fIr(z)dz.
Here x∗ is the unique solution of the equation Pg(x) = 0 in (−∞, ba) and fIr is given by (4.1).