Weak reflection principle for Lévy processes - Semantic Scholar
Weak reflection principle for L´evy processes Erhan Bayraktar∗ and Sergey Nadtochiy† August 15, 2013
Abstract In this paper, we develop a new mathematical technique which can be used to express the joint distribution of a Markov process and its running maximum (or minimum) through the distribution of the process itself. This technique is an extension of the classical reflection principle for Brownian motion, and it is obtained by weakening the assumptions of symmetry required for the standard reflection principle to work. We call this method a weak reflection principle and show that it provides solutions to many problems for which the classical reflection principle is typically used. In addition, unlike the standard reflection principle, the new method works for a much larger class of stochastic processes which, in particular, do not possess any strong symmetries. Here, we review the existing results which establish the weak reflection principle for a large class of time-homogeneous diffusions on a real line and, then, proceed to develop this method for all L´evy processes with one-sided jumps (subject to some admissibility conditions). Finally, we demonstrate the applications of the weak reflection principle in Financial Mathematics, Computational Methods, and Inverse Problems.
1 1.1
Introduction Standard reflection principle and its applications
We start with a brief review of the classical reflection principle for Brownian motion. Denote by B x the Brownian motion on a real line, started form x. Given arbitrary levels U > 0 and K < U , we can compute the joint distribution of Bt = Bt0 and its running maximum Mt = supu∈[0,t] Bu as follows: U P (Bt ≤ K, Mt > U ) = P (Bt−TU +TU ≤ K, TU < t) = P Bt−T ≤ K, TU < t U U = P 2U − Bt−T ≤ K, TU < t = P (Bt ≥ 2U − K, Mt > U ) = P (Bt ≥ 2U − K) , (1) U where TU is the first hitting time of U by B, and BsU = BTU +s . The above formula first appeared in the work of L. Bachelier [1], followed by a more rigorous treatment, for example, by P. L´evy [10]. Notice that the above derivations are based on the following well-known properties of Brownian motion: • strong Markov property: BTU +t −BTU is a standard Brownian motion (started from zero), independent of FTU (where the filtration is generated by B); • continuity: since the paths of B are continuous, BTU = U and, in view of the above, BTU +t is a Brownian motion started from U , and independent of FTU ; ∗ [email protected]; Department of Mathematics, University of Michigan. This author is supported by National Science Foundation under grant DMS-0955463. † [email protected]; Department of Mathematics, University of Michigan.
1
• symmetry: the distribution of BtU is symmetric with respect to the initial level U , i.e Law BtU Law 2U − BtU .
=
We will come back to the above observations in the next subsection, but, first, let us outline several applications of the standard reflection principle. One obvious application is the computation of the joint distribution. Since the marginal distribution of a Brownian motion is available in closed form, the above formula gives us a closed form expression for the joint distribution of the process and its running maximum, at any given time. A more subtle application, which requires the use of reflection principle itself (rather than the resulting formula for the joint distribution) comes from Financial Mathematics. Namely, the reflection principle turns out to be very useful in the problem of hedging barrier options. To simplify the notation, in the remainder of this section, we will consider U = 0. Let us assume that the risk-neutral evolution of the underlying is described by a Brownian motion B x started from x ≤ 0 (we assume no discounting). Consider a up-and-out option, written on this underlying, with a terminal payoff function h, such that supp(h) ⊂ (−∞, 0). The payoff of such option, at the time of maturity T , is given by h(BTx )1{sup
u∈[0,T ]
x 0. Then, the target equation E h (Xt ) = E g (Xt ) ,
for all t > 0,
has the solution g(x) = h (S(x)), where the functions g and h have supports on the opposite sides of zero. Thus, the reflection principle can be extended easily to any strong Markov process that possesses a strong symmetry and does not jump across the barrier (in this case, zero). This was observed by Carr and Lee [4], whose transformed put-call symmetry condition is equivalent to the existence of a strong symmetry for the underlying stochastic process (see also [12] for equivalent characterization of symmetric martingales). In particular, the reflection principle can be extended to any diffusion martingale whose coefficient is an even function, since any such process possesses a strong symmetry, with S : x 7→ −x. It is also shown in [4] that, if a given strong Markov process does not jump across the barrier and possesses a strong symmetry S, then, the reflection principle holds for this process run on an independent continuous 1 In fact, Carr and Chou [3] allow the Brownian motion to have a constant drift, which can be eliminated by a Girsanov change of measure.
3
stochastic clock (the filtration, in this case, is generated by both the original process and the subordinator). Even though the resulting process may not be Markov, it is easy to see that, conditional on any A ⊂ FT0 , the future values of the time-changed process have a strong symmetry with the same mapping S. However, the condition of strong symmetry excludes many processes important for applications: some of such examples are discussed in [5]. In the present paper, we develop a weak formulation of the reflection principle, which can be applied to a large class of stochastic processes that do not posses any strong symmetries. This new formulation, albeit weaker than the standard one, is sufficient to solve the problems outlined in Subsection 1.1. Herein, we restrict our analysis to strong Markov processes which do not jump across the given upper (lower) barrier from below (above). Then, in view of the above discussion, it suffices to consider the unconditional, as opposed to conditional, distributions of the process. Consider a stochastic process X, defined on a real line and started from zero. Consider two spaces, B − and B + , consisting of Lebesgue measurable test functions h, such that h(Xt ) has finite expectation, for all t ≥ 0, and all h ∈ B − have support in (−∞, 0) while all h ∈ B + have support in (0, ∞). We say that X possesses an upper weak symmetry (with respect to zero)2 if there exists a mapping W+ from a space of test functions B0− ⊂ B − into B + , such that E h (Xt ) = E W+ [h] (Xt ) , for all t > 0, for any h ∈ B0− . Analogously, one can define the lower weak symmetry, along with the mapping W− : B0+ → B − . We do not insist on a particular choice of the spaces of test functions B0± , but, in what follows, we choose the spaces that contain, at least, all smooth functions with compact support. We will refer to the mappings W± as the mirror image transformations, although one should remember that, in the present setting, the mirror may not be straight: if the distribution of Xt is not symmetric with respect to zero, the image of a straight line may be a curve with nonzero curvature. Thus, the weak reflection principle consists of the application of a strong Markov property and (semi) continuity, as well as the mirror image transformation W± . We will discuss the questions of existence, uniqueness, and numerical approximation of the mappings W± in the remainder of this paper. But, first, let us provide some motivation for the above constructions. The weak reflection principle, in particular, allows us to solve the problems outlined in Subsection 1.1, for a class of processes that may not posses any strong symmetries. Without loss of generality, let us focus on the weak reflection principle with an upper barrier. Namely, assuming that process X is strongly Markov, does not jump across 0(> X0 ) from below, and possesses upper weak symmetry, we obtain the following results. 1. We can express the joint distribution of the process and its running maximum via the distribution of the process itself: for any K ∈ R, there exists a measurable function p(K; ·), such that ! P Xt ≤ K, sup Xu > 0
= E p (K; Xt )
u∈[0,t]
2. We can solve the static hedging problem in a model where X is the underlying: for any h ∈ B0− , we have + E h(XT )1{sup t ∈ [0, T ] Xu 0, the functions µ and σ belong to C 3 (R), the functions themselves and their first three derivatives have finite limits at −∞, and, for any k = 1, 2, 3, the functions e(3−k)x |µ(k) (x)| and e(3−k)x σ (k) (x) are bounded over all x > 0. Then, Theorem 2.8 in [5] provides an explicit integral expression for the mirror image transformation associated with X. For the sake of completeness, we present a simplified corollary of this theorem here. Theorem 1. (Carr-Nadtochiy 2011) Let X be as above, and let h be a once weakly differentiable function, with support in (−∞, 0), such that its derivative has a modification with finite variation over (−∞, 0). Then, there exists a continuous and exponentially bounded function g, with support in (0, ∞), such that E h(Xt ) = E g(Xt ),
for all t > 0
(4)
Moreover, for any large enough γ > 0, the function g can be computed as follows: Z z Z γ+i∞ Z 0 wψ1 (x, w) ψ1 (z, w) µ(y) 2 exp −2 dy h(z)dzdw, g(x) = 2 πi γ−i∞ ∂x ψ1 (0, w) − ∂x ψ2 (0, w) −∞ σ 2 (z) 0 σ (y) where ψ1 and ψ2 are the fundamental solutions of the associated Sturm-Liouville equation ∂2 ∂ 1 2 σ (x) 2 ψ(x, w) + µ(x) ψ(x, w) − w2 ψ(x, w) = 0, 2 ∂x ∂x
(5)
determined uniquely, for all complex w with large enough Re(w) > 0, by the following conditions: ψ1 (·, w) is square integrable on (−∞, 0), ψ2 (·, w) is square integrable on (0, ∞), and ψ1 (0, w) = ψ2 (0, w) = 1. The above result shows that all regular enough diffusions possess weak symmetry, with the mirror image transformation given by an explicit integral transform. Of course, in order to implement this transform numerically, one needs to know the fundamental solutions of the associated Sturm-Liouville equation. These functions 5
can be approximated efficiently by expanding them into power series of w (cf. [13]). Alternatively, one can notice that, if µ ≡ 0 and σ is piecewise constant, then, ψi ’s are piecewise linear-exponential (linear combinations of exponentials). Thus, we can approximate any function σ with piecewise constant ones, then, compute ψi ’s in a closed form, and, finally, obtain g via numerical integration. Examples of functions g corresponding to a piecewise linear function h can be found in [5].
2 2.1
Mirror Image Transformation for Spectrally Negative L´evy Processes Problem formulation
Consider a L´evy process (Xt )t≥0 , given by its initial condition X0 = 0 and the Laplace exponent ψ: Z 0 σ2 2 eλx − 1 − λx Π(dx), λ + ψ(λ) = µλ + 2 −∞
(6)
where Π is the L´evy measure of X, and EeλXt = etψ(λ) , for all complex λ for which both sides of the above equation are well defined. To make sure that the above expressions are well defined, at least, for all λ with positive real part, as well as to simplify some of the derivations that follow, we make the following assumption on the L´evy triplet, (µ, σ, Π). Assumption 1. We assume that µ ∈ R, σ > 0, and Π is a σ-finite Borel measure on (−∞, 0), such that Z 0 x2 Π(dx) < ∞. −∞
The process X is called spectrally negative because it is only allowed to have negative (i.e. downward) jumps (cf. [9]). The reason for such a restriction is that the process must not jump across the upper barrier, in order for the weak reflection principle to hold. Of course, in the case of a lower barrier, one needs to consider X with only positive jumps. Our goal is to construct a mirror image transformation for the process X. Namely, for any given admissible function h : R → R, with supp(h) ⊂ (−∞, 0), we would like to find a measurable function g : R → R, with supp(g) ⊂ (0, ∞), such that Eh(Xt ) = Eg(Xt ), ∀ t > 0 To ensure that the expectation of h(Xt ) is well define, we need to make some additional assumptions on h and Π. ˆ ∈ L1 (R), such that the Assumption 2. We assume that supp(h) ⊂ (−∞, 0) and that there exist ζ ≥ 0 and h ζy ˆ function y 7→ e h(y), defined for all y ∈ R, is a Fourier transform of h. In addition, we assume that Z −1 |x|e−ζx Π(dx) < ∞ −∞
Due to uniqueness of the Laplace inverse, our problem is equivalent to: find a measurable function g : R → R, with supp(g) ⊂ (0, ∞), such that Z ∞ Z ∞ E e−λt h(Xt )dt = E e−λt g(Xt )dt, (7) 0
0
for all large enough λ > 0. 6
2.2
Matching expectations in Laplace space
Denote by G(x, λ, y) the Green’s function of X. For the definition and properties of this function we refer the reader to [2], [11], as well as the survey article [9], and the references therein. Using the definition of Green’s function, we can rewrite equation (7) as follows: Z 0 Z ∞ h(y)G(0, λ, y)dy = g(y)G(0, λ, y)dy, (8) −∞
0
for all large enough λ > 0. The right hand side of the above is a result of the integral transform with kernel G(0, λ, y). Thus, our problem amounts to finding an inverse transformation and showing that the left hand side of the above lies in its domain. Note that, so far, we d onto know for which λ the left hand side of the above is well defined. For all real λ ≥ 0, the Green’s function G(0, λ, y) has the following representation: G(0, λ, y) = Φ0 (λ)e−Φ(λ)y − W λ (−y),
(9)
where W λ is the λ-scale function of X (cf. [9]), with W λ (y) = 0, for y ≤ 0, and Φ(λ) := sup{q ≥ 0 : ψ(q) = λ},
λ ≥ 0,
with ψ being the Laplace exponent of X. Notice that, for all real λ > ψ(−ζ) ∨ 0, we have: Z 0 Z ∞ −ζy G(0, λ, y)e dy = e−λt E e−ζXt 1{Xt ≤0} dt −∞
(10)
0
Z
∞
≤
e−λt Ee−ζXt dt =
Z
0
∞
e−(λ−ψ(−ζ))t dt < ∞
0
Note also that ψ(−ζ) is well defined due to Assumption 2. Thus, Assumption 2 and estimate (10) imply that Z ∞ Z 0 Z ∞ Υ(λ) := E e−λt h(Xt )dt = G(0, λ, y)h(y)dy = Φ0 (λ)eΦ(λ)y − W λ (y) h(−y)dy, (11) −∞
0
0
is well defined for all real λ > ψ(−ζ) ∨ 0. In addition, (8) is equivalent to Z ∞ Φ0 (λ)e−Φ(λ)y g(y)dy = Υ(λ) 0
Notice that ψ is continuous and strictly increasing on |µ|/σ 2 , ∞ , and it explodes at infinity. Hence, we can change the variables and reformulate the problem as: find a measurable function g, such that, for all large enough real λ > 0, the following holds Z ∞ e−λy g(y)dy = ψ 0 (λ)Υ(ψ(λ)), (12) 0
with 0
0
Z
0
ψ (λ)Υ(ψ(λ)) = ψ (λ)
Z G(0, ψ(λ), y)h(y)dy =
−∞
∞
eλy − ψ 0 (λ)W ψ(λ) (y) h(−y)dy
0
The right hand side of the above is well defined for all real λ > Φ (ψ(−ζ) ∨ 0). 7
(13)
Notice that the mirror image problem (12), now, looks exactly like the Laplace transform inversion. However, there is a major difference. In the classical problem of inverting a Laplace transform, we, typically, know that the right hand side is a Laplace transform of some function, and we need to find a transformation that would recover this function. In the present case, we need to prove that ψ 0 (λ)Υ(ψ(λ)) is, indeed, a Laplace transform of some function, which is not obvious a priori. In addition, we need to propose a method to recover this function from the right hand side. Of course, there exist several methods for inverting the Laplace transform (cf. [14], [6]). However, the conditions, which are required for some of these methods to succeed, are expressed through the original function (in our case, g), rather than the transformed one (in the present case, ψ 0 (λ)Υ(ψ(λ))). Since, a priori, we know very little about function g (e.g. we do not even know if it exists), we cannot apply any of the existing results on the Laplace transform inversion to solve the mirror image problem. Instead, we will show, by hand, that the desired function g exists and can be recovered via the classical Bromwich integral (cf. [6]).
2.3
A priori estimates
In this subsection, we establish the analytic continuation of ψ 0 (λ)Υ(ψ(λ)) to a complex half plane of the form HR = {w : Re(w) > R}, and provide some useful estimates of its absolute value. Remark 1. Notice that Υ(λ) can be easily extended to a half plane, via its probabilistic representation given by the first identity in (11). However, under the change of variables λ 7→ ψ(λ), the half plane transforms into a smaller domain which is not sufficient for our purposes. To obtain an analytic extension of Υ(ψ(λ)), one would need the probabilistic representation in (11) to hold in a domain where the real part of λ is unbounded from below, which is typically impossible. More precisely, the conditions that function h has to satisfy, in order for the probabilistic representation to hold in the desired domain, are extremely restrictive and rather implicit. For example, if X is a martingale, none of the convex functions h, except zero, are admissible. In addition, even for those functions h for which the probabilistic representation in (11) is well defined (although we do not know how to characterize this set explicitly), the standard estimates of the integrals involved in this representation do not provide sufficient information about the asymptotic behavior of Υ(ψ(λ)), as |λ| → ∞, which is needed to solve the mirror image problem (12). Alternatively, one might be tempted to use the analytic continuation of W ψ(λ) to extend Υ(ψ(λ)) to a complex half plane via (13). Indeed, it is well known (cf. [9]) that W λ can be extended analytically to the entire complex plane. However, to the best of our knowledge, there exist no estimates of this extension (more precisely, we need an estimate of eλy −ψ 0 (λ)W ψ(λ) (y)), which, in particular, would guarantee that the integral in the right hand side of (13) is well defined for all λ in a half plane HR . In fact, it is easy to construct an example of a L´evy process whose infinitesimal generator has a nontrivial spectrum, and, hence, the integral in the right hand side of (13) is not well defined for some λ. The difficulties described above explain why we are forced to construct the analytic continuation of Υ(ψ(λ)), and investigate its asymptotic behavior, by hand. It turns out that Fourier transform offers a natural way to obtain the desired analytic continuation. Recall that, due to Assumption 2, Z −ζy ˆ h(y) = e e−iyz h(z)dz R
Due to estimate (10), we can interchange the limit and integration to obtain Z ∞ eλy − ψ 0 (λ)W ψ(λ) (y) h(−y)dy ψ 0 (λ)Υ(ψ(λ)) = 0
8
(14)
Z Z
∞
= R
ˆ eλy − ψ 0 (λ)W ψ(λ) (y) e(ζ+iz)y dy h(z)dz,
0
for all large enough real λ > 0. Our next goal is to extend the above representation to a complex half plane HR , and estimate its absolute value from above. Let us analyze the inner integral in the right hand side of (14). Notice that it can be viewed as Z ∞ eλy − ψ 0 (λ)W ψ(λ) (y) e−wy dy, 0
evaluated at w = −ζ − iz. For λ > |µ|/σ 2 and w ∈ (λ, ∞), the above integral can be computed explicitly, using the definition of the scale function W ψ(λ) (cf. [9]): Z ∞ Z ∞ Z ∞ λy 0 ψ(λ) −wy λy−wy 0 e − ψ (λ)W (y) e dy = e dy − ψ (λ) e−wy W ψ(λ) (y)dy (15) 0
0 0
=
0 0
ψ (λ) ψ(w) − ψ(λ) − ψ (λ)(w − λ) w−λ 1 − = 2 w − λ ψ(w) − ψ(λ) (w − λ) ψ(w) − ψ(λ)
The right hand side of the above is analytic everywhere in w ∈ H−ζ = {w : Re(w) > −ζ}, except, possibly, the zeros of ψ(w)−ψ(λ), where it has poles (on every compact, there is at most a finite number of such points). The left hand side of the above is analytic in H−ζ (we can differentiate with respect to w inside the integral, and the resulting expression is bounded, uniformly over w, by an absolutely integrable function of y). Therefore, the right hand side, in fact, does not have any poles in H−ζ , and the above equality holds for all w ∈ H−ζ . By continuity, the equality holds at w = −ζ − iz: Z ∞ ψ 0 (λ) 1 eλy − ψ 0 (λ)W ψ(λ) (y) e(ζ+iz)y dy = − , (16) ψ(λ) − ψ(−ζ − iz) λ + ζ + iz 0 for all large enough real λ > 0. Next, we need to show that, for every z ∈ R, the right hand side of the above is analytic in λ ∈ HR = {w : Re(w) > R}, for a large enough R > 0, and that, for every fixed λ ∈ HR , it is absolutely bounded over z ∈ R. To do this, we will need the following lemmas, which, in fact, are also crucial for the proofs of the main theorems. These lemmas describe the asymptotic behavior of 1/(ψ(λ) − ψ(−ζ − iz)) and they can be viewed as the core technical result of this paper. Their proofs are given in Appendix A. Lemma 1. For all u, z ∈ R and v > 0, we have Im (ψ(v + iu) − ψ(−ζ − iz)) = µ(u + z) + (vu − zζ)σ 2 + o(uv) + sign(u) o(u2 ) + sign(z) o(z 2 ) + o(z), where we denote by o (f (u, v, z)) any function of (u, v, z), such that o (f (u, v, z)) /f (u, v, z) is absolutely bounded over all u, z ∈ R and v > 0, and it converges to zero, as |f (u, v, z)| → ∞. Lemma 2. For all u, z ∈ R and v > 0, we have Im (ψ(v + iu) − ψ(−ζ − iz)) = µ(u+z)+(vu−zζ)σ 2 +o(uv)+o(z 2 −u2 )+sign(z+u) o (z + u)2 +o(z), where o (f (u, v, z)) has the same meaning as in Lemma 1. Lemma 3. For all u, z ∈ R and v > 0, we have Re (ψ(v + iu) − ψ(−ζ − iz)) = µ(v+ζ)+ v 2 + z 2 − u2 σ 2 /2+o(z 2 −u2 )+ o(uv) + o(v 2 ) +o(z)+o(1), where o (f (u, v, z)) has the same meaning as in Lemma 1. 9
Lemma 4. For any ε ∈ (0, 1), there exist R1 > 0, R2 > 0, c1 > 0, and c2 > 0, such that the following inequality |ψ(v + iu) − ψ(−ζ − iz)| ≥ c1 v(|z| + v)1{||u|−√z2 +v2 |≤εv} + c2 u2 − (v 2 + z 2 ) 1{||u|−√z2 +v2 |>εv} , holds for all u, v, z ∈ R, with |z| > R1 and v > R2 . Lemma 5. There exists a constant R1 > 0, such that, for any v > R1 , there exist R2 = R2 (v) > 0, R3 = R3 (v) > 0, and c = c(v) > 0, such that the following inequality 2 |ψ(v + iu) − ψ(−ζ − iz)| ≥ c (z 2 − u2 )2 + z 2 , holds for all u, z ∈ R, with |z| > R2 and |u| > R3 . Lemma 4 yields that, for any ε ∈ (0, 1), there exist N > 0, R > 0 and ci ’s, such that the following estimates hold, for all u, all |z| ≥ N , and all v ≥ R: 2 c4 1 c3 √ √ ψ(v + iu) − ψ(−ζ − iz) ≤ v 2 (|z| + v)2 1{||u|− z2 +v2 |≤εv} + (u2 − (v 2 + z 2 ))2 1{||u|− z2 +v2 |>εv} c5 √ 1 2 2 v 2 ((|u| − εv)2 + v) {||u|− z +v |≤εv} c4 √ + 4 1 2 2 z + (2v 2 − 1)z 2 + v 2 (v 2 − 1) + z 2 + v 2 − u2 {||u|− z +v |>εv} c5 √ ≤ 2 1 2 2 v ((|u| − εv)2 + v) {||u|− z +v |≤εv} c4 √ √ 1 + 2 2 z 4 + 2v 2 z 2 + v 4 − z 2 − v 2 + εv z 2 + v 2 + |u| {||u|− z +v |>εv} c5 c6 ≤ 2 + v ((|u| − εv)2 + v) (z 2 + v 2 )2 + εv|u| ≤
(17)
Since the right hand side of (16) is continuous, as a function of z, the above estimate implies that there exists R0 ≥ R > 0, such that the following inequalities hold for all λ = v + iu, with v ≥ R0 : Z ψ 0 (λ) 1 ˆ ψ(λ) − ψ(−ζ − iz) − λ + ζ + iz |h(z)|dz R Z Z c6 |λ| 1 0 ˆ ˆ |h(z)|dz | h(z)|dz + |ψ (λ)| ≤ 2 |λ| − c7 |z|≤N |z|>N ψ(λ) − ψ(−ζ − iz) Z 1 ˆ |h(z)|dz + R λ + ζ + iz ! Z Z c8 1 1 1 ˆ ˆ √ + p p ≤ + c9 |λ| |h(z)|dz + c10 |h(z)|dz 2 2 |λ| v (||u| − εv| + v) v + v|u| v + (u + z)2 |z|>N R ! 1 |λ| |λ| 1 √ + p ≤ c11 + + |λ| v (||u| − εv| + v) v 2 + v|u| v
10
In the above, we also used the fact that, for all large enough |λ|, with positive real part, we have: |ψ 0 (λ)| ≤ c12 |λ|,
|ψ(λ)| ≥ c13 |λ|2
Therefore, due to (14), the following representation Z ψ 0 (λ) 1 0 ˆ ψ (λ)Υ(ψ(λ)) = − h(z)dz, ψ(λ) − ψ(−ζ − iz) λ + ζ + iz R
(18)
is well defined and holds for all large enough real λ > 0. In fact, the right hand side of the above is analytic, as a function of λ ∈ HR0 = {λ : Re(λ) > R0 }. To see this, we differentiate, formally, inside the integral and apply the same estimates as above, to show that the integral of the derivative is absolutely convergent, for any λ ∈ HR0 . Hence, the above expression is analytic as a function of λ ∈ HR0 . We have proved the following proposition. Proposition 1. For any ε ∈ (0, 1), there exist constants R > 0 and c ≥ 0, such that the function λ → 7 ψ 0 (λ)Υ(ψ(λ)) (defined in (13), for all large enough real λ > 0) can be extended analytically to HR = {λ : Re(λ) > R} via (18), and its extension satisfies: |λ| |λ| 1 1 + p + + , |ψ 0 (λ)Υ(ψ(λ))| ≤ c p 2 |λ| Re(λ) ||Im(λ)| − εRe(λ)| + Re(λ) Re(λ) Re(λ) + Re(λ)|Im(λ)| for all λ ∈ HR .
2.4
Main results
Now, we have everything we need to solve the mirror image problem (12). First, for any r > 0, we introduce the following function of y ∈ R: 1 R αy 0 2πi Gr e ψ (α)Υ(ψ(α))dα, y ≥ 0, gr (y) = (19) 0, y < 0, where ψ 0 (α)Υ(ψ(α)) is given by (18), and Gr is a vertical interval on a complex plane: Gr = [γ − ir, γ + ir], with a large (but fixed) constant γ > 0. To ensure that gr is well defined, we assume that γ > R, where R is the constant appearing in Proposition 1 (the constant ε ∈ (0, 1) is fixed arbitrarily). Our next result shows that, if we use gr , in lieu of g, in the left hand side of (12), then, it will converge to the right hand side, as r → ∞. Proposition 2. For all large enough real λ > 0, we have: Z ∞ lim e−λy gr (y)dy = ψ 0 (λ)Υ(ψ(λ)) r→∞
0
Proof: We fix arbitrary λ > γ and apply the Fubini’s theorem, to obtain: Z ∞ Z Z ∞ Z 1 1 ψ 0 (α)Υ(ψ(α)) e−λy gr (y)dy = e−λy eαy dyψ 0 (α)Υ(ψ(α))dα = dα 2πi Gr 0 2πi Gr λ−α 0 11
Next, we complete the contour of integration Gr on the right, by a rectangle, Cr , with the horizontal sides Hr1 and Hr2 , of length r, and the vertical side, Vr , of length 2r. When r is large enough, the real number λ > γ is inside this closed contour. Since ψ 0 (α)Υ(ψ(α)) is analytic inside the contour, the integral is equal to the residue: Z ψ 0 (α)Υ(ψ(α)) dα = 2πiψ 0 (λ)Υ(ψ(λ)), λ−α Gr ∪Cr Z Z ψ 0 (α)Υ(ψ(α)) ψ 0 (α)Υ(ψ(α)) 0 dα = 2πiψ (λ)Υ(ψ(λ)) − dα λ−α λ−α Gr Cr It only remains to show that the integral over Cr vanishes for large r. Applying Proposition 1, we obtain: Z
Cr
0 Z 0 2 Z ψ (α)Υ(ψ(α)) ψ (α)Υ(ψ(α)) ψ 0 (α)Υ(ψ(α)) X |dα| dα ≤ |dα| + j λ−α λ−α λ−α Vr j=1 Hr ≤ c1
γ+r
1 1 r r 1 √ + + + dv r r vr(1 − ε) v 2 + vr v γ Z r r r 1 1 1 + √ + 2+ du +c2 r r r r r −r r
Z
Applying the straightforward estimates, as well as the dominated convergence theorem, we easily conclude that the right hand side of the above converges to zero, as r → ∞. Finally, we present the main result of this paper. Theorem 2. Let h and X satisfy Assumptions 1 and 2, and let gr be given by (19), with γ > R1 , where R1 is the constant appearing in Lemma 5. Then, there exists a continuous and exponentially bounded function g : (0, ∞) → R, such that lim gr (y) = g(y), r→∞
for all y > 0, and Eg(Xt ) = Eh(Xt ), for all t > 0. The proof of the above theorem is given in Appendix B. Theorem 2 shows that any spectrally negative L´evy process X, satisfying Assumptions 1 and 2, possesses an upper weak symmetry, with the space of test functions B0− consisting of all function that satisfy Assumption 2. The associated mirror image transformation is given by Z γ+i∞ Z 1 ψ 0 (α) 1 + αx ˆ W [h](x) = e − h(z)dzdα (20) 2πi γ−i∞ ψ(α) − ψ(−ζ − iz) α + ζ + iz R In view of (12), the uniqueness of the mirror image transformation follows from the uniqueness of the Laplace inversion (cf. [14]). To implement this transformation, one needs to find a Fourier inverse of y 7→ eζy h(y) and evaluate the above integral numerically. It is shown in the next section that B0− s is large enough to include many functions of interest, for which the aforementioned Fourier inverse can be computed explicitly.
12
2.5
Example
Consider a L´evy process X which is a sum of a scaled Brownian motion and a negative Gamma process. In other words, Xt = σBt − Γt , where B is a standard Brownian motion and Γ is a Gamma process with parameters α > 0 and β > 0 (cf. [2], [11]). In this case, the characteristic triplet of X, as defined in (6), is given by µ = −β/α, σ = σ, Π(dx) = β|x|−1 e−α|x| 1(−∞,0) (x)dx , and, in particular, 1 2 2 σ λ − β log (1 + λ/α) 2 As a test function h, we choose the hockey-stick function h(x) = (K − x)+ , as it represents the payoff of a put option and, therefore, is relevant for the static hedging application. Note that Assumption 1 is always satisfied for the process X, and Assumption 2 holds with any ζ ∈ (0, α). In addition, it is easy to see that the inverse Fourier transform of the function x 7→ eζx h(x) is given by ψ(λ) =
exp(K(ζ + ix)) ˆ h(x) = 2π(ζ + ix)2 With the above expressions at hand, one can evaluate the integral in (20) numerically, to obtain the mirror image function g. The results of these computations are presented in Figure 1.
3
Summary and extensions
We have presented a new mathematical technique, which we christened weak reflection principle and which is an extension of the well known reflection principle for Brownian motion. This new form of reflection principle is obtained by weakening the notion of symmetry that is required for the classical reflection principle to hold. Namely, it is based on the notions of upper and lower weak symmetries, defined via the associated mirror image transformations. We started by reviewing the existing results which provide an explicit integral expression for the mirror image transformation for any time-homogeneous diffusion process on a real line (subject to some regularity conditions). Finally, for the most of the paper we focused on constructing the mirror image transformation for spectrally negative L´evy processes, thus, extending the weak reflection principle to this new class of stochastic processes. The weak reflection principle provides solutions to various problems for which the classical reflection principle can be used, even when the underlying process is not a Brownian motion and does not possess any strong symmetries. Examples of such problems are discussed at the end of Subsection 1.2. In particular, the weak reflection principle is a perfect tool for constructing the exact static hedging strategies of barrier options (in fact, this problem motivated the development of the method in the first place). Another application of this method is the computation of the joint distribution of a process and its running maximum (minimum). Of course, while this problem is quite relevant for diffusions, in the case of L´evy processes, there exist several alternative computational methods, based on the Wiener-Hopf factorization (cf. [8]). Note that the weak reflection principle may not always be the most efficient way to compute the joint distribution of a L´evy process and its running maximum: after all, we still need to know the marginal distribution of the process, and, in addition, the numerical implementation of (20) requires some computational effort. However, in the cases when the marginal distribution is known (or is easy to compute), and provided that we need to find the joint 13
Figure 1: Mirror image of a put payoff h(x) = (K − x)+ , with K = −0.2, α = β = σ = 1, and ζ = 0.9.
distribution for multiple time horizons or multiple initial values, it is advantageous to reduce the problem of computing the joint distribution to the computation of an expectation with respect to the marginal distribution of the L´evy process. The last application suggested in Subsection 1.2 is related to the parabolic partial integrodifferential equation (PIDE) associated with a L´evy process. Namely, the mirror image transformation allows us to modify the initial condition of this PIDE, on one half line only, so that its solution remains constant at x = 0, for all times (see the discussion at the end of Subsection 1.2 for a more detailed explanation of the analogous problem in the case of diffusions). It is also worth mentioning that the technical Lemmas 1-5, and the resulting Proposition 1, describe a domain on which the resolvent function of the L´evy process is well defined. This domain is rather large, and, in particular, the real parts of its elements are unbounded from below. Thus, our results provide a non-trivial estimate of the spectrum of the integro-differential operator associated with any admissible spectrally-negative L´evy process. Recall that this operator is non-local and non-symmetric, which makes it very hard to describe its spectrum using the general theory (see also Remark 1 for a description of the associated difficulties). So far, the weak symmetry and the weak reflection principle have only been established for diffusion processes and L´evy processes with one-sided jumps. However, we conjecture that these results can be extended to a larger class of time-homogeneous Markov processes – possibly, all jump-diffusions satisfying some regularity conditions. Such an extension would allow us to solve the aforementioned problems for a larger class of stochastic processes. Another important extension is related to the domain with respect to which the weak symmetry is defined. Notice that, in the present case, we split the real line into two half lines and study the weak symmetry of the process with respect to the (unique) boundary point. It is interesting to extend these results to the case of a compact interval, whose boundary consists of two points (provided the underlying process does 14
not jump across the boundary points). In financial applications, this problem would correspond to the static hedging of double barrier options. More generally, one can investigate domains in higher dimension and try to establish the weak symmetry with respect to their boundaries.
4
Appendix A
Proof of Lemma 1 Im (ψ(v + iu) − ψ(−ζ − iz)) Z
= µ(u + z) + (vu − zζ) σ 2 +
0
(evx sin(ux) − ux) Π(dx) +
−∞
0
−∞
0
e−ζx sin(zx) − zx Π(dx) = zζ
Z
−∞
0
e−ζx sin(zx) − zx Π(dx),
−∞
−∞
Notice that Z 0 Z (evx sin(ux) − ux) Π(dx) = vu Z
Z
0
−∞
evx − 1 sin(ux) 2 x Π(dx) + u2 vx ux
Z
0
−∞
1 − eζx sin(zx) 2 −ζx x e Π(dx) + z 2 ζx zx
Z
sin(ux) − ux 2 x Π(dx), u2 x2 0
−∞
sin(zx) − zx 2 x Π(dx) z 2 x2
Since x2 and x2 e−ζx are integrable with respect to Π(dx), due to the dominated convergence theorem, the above integrals are absolutely bounded and vanish as the corresponding functions of (u, v, z) go to infinity. Proof of Lemma 2 We follow the proof of Lemma 1, except that, at the end, we apply the following additional estimate: Z 0 Z 0 sin(ux) − ux 2 sin(zx) − zx 2 2 2 u x Π(dx) + z x Π(dx) 2 x2 u z 2 x2 −∞ −∞ Z 0 = (2 sin((z + u)x/2) cos((z − u)x/2) − (z + u)x) Π(dx) −∞ 2
2
Z
0
= (z − u )
2 −∞
+(z + u)2
Z
sin((z + u)x/2) cos((z − u)x/2) − 1 2 x Π(dx) (z + u)x (z − u)x
0
−∞
2 sin((z + u)x/2) − (z + u)x 2 x Π(dx) (z + u)2 x2
Applying the dominated convergence theorem we complete the proof of the lemma. Proof of Lemma 3 Re (ψ(v + iu) − ψ(−ζ − iz)) = µ(v + ζ) + v 2 − ζ 2 + z 2 − u2 σ 2 /2 Z
0
+
(e
Z
e−ζx cos(zx) − 1 + ζx Π(dx),
cos(ux) − 1 − vx) Π(dx) −
−∞
Notice that
0
Z
vx
−∞
0
(evx cos(ux) − 1 − vx) Π(dx) −
Z
−∞
0
e−ζx cos(zx) − 1 + ζx Π(dx)
−∞
Z
0
=
evx (cos(ux) − 1)Π(dx) +
−∞
Z
0
−∞
15
(evx − 1 − vx) Π(dx)
Z
0
−
e
−ζx
Z
e−ζx − 1 + ζx Π(dx)
−∞
−∞ 0
0
(cos(zx) − 1)Π(dx) − 0
Z 0 vx − 1 cos(ux) − 1 2 e − 1 − vx 2 2 (cos(ux) − 1)Π(dx) + uv = x Π(dx) x Π(dx) + v vx ux v 2 x2 −∞ −∞ −∞ Z 0 Z 0 Z 0 1 − eζx cos(zx) − 1 2 −ζx 1 − eζx + ζxeζx 2 −ζx 2 − (cos(zx)−1)Π(dx)−zζ x e x e Π(dx)−ζ Π(dx) ζx zx ζ 2 x2 −∞ −∞ −∞ Z 0 = (cos(ux) − cos(zx))Π(dx) + o(uv) + o(v 2 ) + o(z) + o(1) Z
Z
e
vx
−∞
= 2(z 2 − u2 )
Z
0
−∞
sin((u + z)x/2) sin((z − u)x/2)) 2 x Π(dx) + o(uv) + o(v 2 ) + o(z) + o(1) (z + u)x (z − u)x 2 2 = o(z − u ) + o(uv) + o(v 2 ) + o(z) + o(1),
where we applied the dominated convergence theorem. Proof of Lemma 4 Using Lemmas 2 and 3, we obtain: |ψ(v + iu) − ψ(−ζ − iz)| ≥ |Im (ψ(v + iu) − ψ(−ζ − iz))| 1{||u|−√z2 +v2 |≤εv} (21) + |Re (ψ(v + iu) − ψ(−ζ − iz))| 1{||u|−√z2 +v2 |>εv} = µ(u + z) + (vu − zζ)σ 2 + o(vu) + o(z) + o(z 2 − u2 ) + sign(z + u) o (z + u)2 1{||u|−√z2 +v2 |≤εv} + µ(v + ζ) + v 2 + z 2 − u2 σ 2 /2 + o(z 2 − u2 ) + o(uv) + o(v 2 ) + o(z) + o(1) 1{||u|−√z2 +v2 |>εv} Let us estimate the first term in the above: µ(u + z) + vuσ 2 − zζσ 2 + o(vu) + o(z) + o(z 2 − u2 ) + sign(z + u) o (z + u)2 1 √ {||u|− z2 +v2 |≤εv} ≥ v|u|σ 2 + µu + z(µ − ζσ 2 ) + o(v(|z| + v)) + o(z) + o(v|z| + v 2 ) + sign(z + u) o (z + u)2 · 1{||u|−√z2 +v2 |≤εv,u≥0} + v|u|σ 2 − µu − z(µ − ζσ 2 ) + o(v(|z| + v)) + o(z) + o(v|z| + v 2 ) − sign(z + u) o (z + u)2 · 1{||u|−√z2 +v2 |≤εv,uεv} ! 2 2 o(uv) + o(v 2 ) + o(z) 2 µ(v + z) + o(u − z ) + 2 2 2 ≥ u − (v + z ) σ /2 + 1{||u|−√z2 +v2 |>εv} |u2 − (v 2 + z 2 )| ≥ c2 u2 − (v 2 + z 2 ) 1{||u|−√z2 +v2 |>εv} , which holds for all large √ enough |z| and v > 0, with some positive constants ci . In the above we made use of the fact that, if |u| − z 2 + v 2 > εv, then, p 2 u − (v 2 + z 2 ) ≥ εv |u| + z 2 + v 2 , |u| ≥ (1 − ε)v, v v2 |u2 − z 2 | √ 2 −z 2 |≤v} + ≤ 1 1 + {|u |u2 − (v 2 + z 2 )| v2 − v εv |u| + v 2 + z 2 1 1 ≤ + 1+ 1{|u2 −z2 |>v} , v−1 ε and, hence,
! 1{|u2 −z2 |>v}
µ(v + z) + o(u2 − z 2 ) + o(uv) + o(v 2 ) + o(z) |u2 − (v 2 + z 2 )|
can be made arbitrarily small by choosing large enough v. Proof of Lemma 5 First, we notice that, since ψ is analytic, we have |ψ(v − iu)| = |ψ(v + iu)|, and, hence, it suffices to consider only u > 0. Using Lemma 3, we obtain the following inequalities: 2
(Re (ψ(v + iu) − ψ(−ζ − iz)))
2 = µ(v + ζ) + v 2 + z 2 − u2 σ 2 /2 + o z 2 − u2 + o(uv) + o v 2 + o(z) + o(1) 2 ≥ v 2 + (v + ζ)2µ/σ 2 + z 2 − u2 σ 4 /4 + v 2 + z 2 − u2 + (v + ζ)2µ/σ 2 o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||≤ε} + v 2 + z 2 − u2 + (v + ζ)2µ/σ 2 o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} 2 2 ≥ v 2 + (v + ζ)2µ/σ 2 σ 4 /4 + v 2 + (v + ζ)2µ/σ 2 (z 2 − u2 )σ 4 /2 + z 2 − u2 σ 4 /4 + v 2 + u o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||≤ε} + v 2 + z 2 − u2 + (v + ζ)2µ/σ 2 o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} 2 2 = v 2 + (v + ζ)2µ/σ 2 σ 4 /4 + z 2 − u2 σ 4 /4 + v 2 + (v + ζ)2µ/σ 2 z 2 σ 4 /2 17
− v 2 + (v + ζ)2µ/σ 2 u2 σ 4 /2 + v 2 + u uo (v) + o v 2 + o(u) 1{|u−|z||≤ε} + v 2 + |z 2 − u2 | o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} , 2 ≥ v 4 (1 − ε0 )σ 4 /4 + z 2 − u2 σ 4 /4 + v 2 z 2 (1 − ε0 )σ 4 /2 − u2 v 2 (1 + ε0 )σ 4 /2 +u2 o (v) + o u2 + uo v 3 + v 2 o (u) + o v 4 + v 2 + |z 2 − u2 | o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} , which hold for any (fixed) ε, ε0 > 0 and all large enough u, v and |z|. Next, using Lemmas 1, 2, we obtain the following inequalities: 2
(Im (ψ(v + iu) − ψ(−ζ − iz)))
2 = µ(u + z) + uvσ 2 − zζσ 2 + o(uv) + o(z) + o(z 2 − u2 ) + o (z + u)2 1{z≤0} 2 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) + o(u2 ) + o(z 2 ) 1{z>0} 2 ≥ uvσ 2 + µ(u + z) − zζσ 2 + o(vu) + o(z) + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤0} + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z>0} ≥ u2 v 2 σ 4 + 2uvσ 2 µ(u + z) − zζσ 2 + o(vu) + o(z) + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤−u−ε00 } +vo u2 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 )1{−u+ε00 ≤z≤0} + uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z|ζ−µ/σ2 |>u(v−ε00 )} , which hold for any (fixed) ε00 > 0 and all large enough u, v and |z|. In the above, we also made use of the fact that µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{−u−ε00 ≤z≤−u+ε00 } = vo u2 and uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{0≤z|ζ−µ/σ2 |≤u(v−ε00 )} ≥ 0 hold for all large enough u, v and |z|. Finally, choosing ε = ε0 = ε00 ∈ (0, 1), we collect the above, to obtain: 2
2
2
|ψ(v + iu) − ψ(−ζ − iz)| = (Re (ψ(v + iu) − ψ(−ζ − iz))) + (Im (ψ(v + iu) − ψ(−ζ − iz))) ≥ v 4 (1 − ε)σ 4 /4 + z 2 − u2
2
σ 4 /4 + v 2 z 2 (1 − ε)σ 4 /2 − u2 v 2 (1 + ε)σ 4 /2 +u2 v 2 σ 4 + 2uvσ 2 µ(u + z) − zζσ 2 + o(vu) + o(z) + vo u2 +u2 o (v) + o u2 + uo v 3 + v 2 o (u) + o v 4 18
+ v 2 + |z 2 − u2 |
o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤−u−ε} + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 )1{−u+ε≤z≤0} + uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z|ζ−µ/σ2 |>u(v−ε)} 2 ≥ v 4 (1 − ε)σ 4 /4 + z 2 − u2 σ 4 /20 + v 2 z 2 (1 − ε)σ 4 /2 + u2 v 2 (1 − ε)σ 4 /2 +2u2 vµσ 2 + 2uvzσ 2 (µ − ζσ 2 ) + o (vu)2 + uvo(z) + vo u2 +u2 o (v) + uo v 3 + v 2 o (u) + o v 4
2 σ 4 + v 2 + |z 2 − u2 | o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} 20 2 σ 4 + z 2 − u2 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤−u−ε} 20 2 σ 4 + z 2 − u2 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 )1{−u+ε≤z≤0} 20 4 σ 2 + z 2 − u2 + uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z|ζ−µ/σ2 |>u(v−ε)} 20 Let us estimate the above terms separately: + z 2 − u2
v 4 (1 − ε)σ 4 /4 + z 2 − u2
2
σ 4 /20 + v 2 z 2 (1 − ε)σ 4 /2 + u2 v 2 (1 − ε)σ 4 /2
+2u2 vµσ 2 + 2uvzσ 2 (µ − ζσ 2 ) + o (vu)2 + uvo(z) + vo u2 +u2 o (v) + uo v 3 + v 2 o (u) + o v 4 2 ≥ v 4 (1 − ε − δ)σ 4 /4 + z 2 − u2 σ 4 /20 + v 2 z 2 (1 − ε)σ 4 /2 + u2 v 2 (1 − ε − δ)σ 4 /2 +uo v 3 − c1 uv|z|, 2 σ 4 + v 2 + |z 2 − u2 | o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} 20 ! o (|z| + u) + o(uv) + o v 2 + o(z) σ4 v2 2 2 2 ≥ z −u + +1 , 20 |z| + u |z| + u
z 2 − u2
(22)
2 σ 4 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤−u−ε} (23) 20 ! σ4 µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (|z| − u)2 2 2 2 ≥ z −u + , 20 u + |z| (|z| − u)|z 2 − u2 |
z 2 − u2
2 σ 4 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 )1{−u+ε≤z≤0} 20 ! 2 2 4 µ(u + z) + uvσ − zζσ + o(vu) + o(z) o(|z| + u) σ 2 + , ≥ z 2 − u2 20 |z| + u (u − |z|)2
z 2 − u2
19
(24)
z 2 − u2
2 σ 4 + uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z|ζ−µ/σ2 |>u(v−ε)} 20 (25) σ4 uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(z 2 ) 2 2 2 , ≥ z −u + 20 |z| + u z2
where we fixed arbitrary δ ∈ (0, 1 − ε) and assumed that u, v and |z| are large enough. It only remans to notice that, for any v > 0, there exist R1 > 0 and R2 > 0, such that, for all |z| > R1 and u > R2 , the right hand sides of (22)–(25) are nonnegative, and, in addition, u2 v 2 (1 − ε − δ)σ 4 /2 + uo v 3 ≥ u2 v 2 (1 − ε − δ)σ 4 /4 Thus, we conclude that, for any large enough v > 0, there exist R1 > 0, R2 > 0 and {ci > 0}, such that, for all |z| > R1 and u > R2 , the following holds 2
|ψ(v + iu) − ψ(−ζ − iz)| ≥ c2 v 4 + c3 (z 2 − u2 )2 + 2c4 v 2 z 2 + c5 u2 v 2 − c1 uv|z| √ √ √ = c2 v 4 + c3 (z 2 − u2 )2 + c4 v 2 z 2 + ( c4 v|z| − c5 uv)2 + uv|z|(2 c4 c5 v − c1 ) ≥ c6 (z 2 − u2 )2 + z 2
5
Appendix B
Proof of Theorem 2 Using the representation (18), and applying Fubini’s theorem and integration by parts, we obtain: Z 2πigr (y) = eαy ψ 0 (α)Υ(ψ(α))dα Gr
Z
Z = Gr
eαy
R
ψ (α) ˆ h(z)dzdα − ψ(α) − ψ(−ζ − iz) 0
Z
Z Gr
R
Gr
0
(eαy )
ψ (α) 1 ˆ dαh(z)dz − ψ(α) − ψ(−ζ − iz) y
eαy
R
Z Z
1 ˆ h(z)dzdα α + ζ + iz
1 ˆ dαh(z)dz α + ζ + iz Z exp((γ + ir)y)ψ 0 (γ + ir) exp((γ − ir)y)ψ 0 (γ − ir) ˆ 1 = − h(z)dz y R ψ(γ + ir) − ψ(−ζ − iz) ψ(γ − ir) − ψ(−ζ − iz) Z 1 exp((γ − ir)y) exp((γ + ir)y) ˆ + − h(z)dz y R γ + ζ − ir + iz γ + ζ + ir + iz Z Z Z Z 2 1 exp(αy)ψ 00 (α) 1 exp(αy) (ψ 0 (α)) ˆ ˆ − dαh(z)dz + dαh(z)dz y R Gr ψ(α) − ψ(−ζ − iz) y R Gr (ψ(α) − ψ(−ζ − iz))2 Z Z 1 exp(αy) ˆ − dαh(z)dz y R Gr (α + ζ + iz)2
=
1 y
Z Z
0
R
0
(eαy )
Gr
(26)
To show that the first integral in the right hand side of (26) converges to zero, we notice that, for any N > 0, there exist r0 > 0, c1 > 0 and c2 > 0, such that, for all r > r0 , we have: Z exp((γ + ir)y)ψ 0 (γ + ir) exp((γ − ir)y)ψ 0 (γ − ir) ˆ − h(z)dz (27) ψ(γ + ir) − ψ(−ζ − iz) ψ(γ − ir) − ψ(−ζ − iz) R 20
≤ c1 eγy
Z |z|≤N
r ˆ |h(z)|dz+ r2 − c2
exp((γ + ir)y)ψ 0 (γ + ir) exp((γ − ir)y)ψ 0 (γ − ir) ˆ |h(z)|dz − ψ(γ − ir) − ψ(−ζ − iz) |z|>N ψ(γ + ir) − ψ(−ζ − iz)
Z
We choose N to be large enough, so that the estimate in Lemma 5 can be applied for all |z| > N : exp((γ + ir)y)ψ 0 (γ + ir) exp((γ − ir)y)ψ 0 (γ − ir) r γy ψ(γ + ir) − ψ(−ζ − iz) − ψ(γ − ir) − ψ(−ζ − iz) ≤ c3 e p(z 2 − r2 )2 + z 2 It is easy to see that the first term in the right hand side of (27) converges to zero, as r → ∞. In addition, (z 2 − r2 )2 + z 2 = (z 2 − r2 + 1/2)2 + r2 − 1/4 ≥ r2 − 1/4 hence, we can apply the dominated convergence theorem to show that the second term in the right hand side of (27) vanishes, as r → ∞. Thus, we conclude that the first integral in the right hand side of (26) converges to zero, as r → ∞. Similarly, the second integral in the right hand side of (26) vanishes, as r → ∞, due to the dominated convergence theorem. As r → ∞, the fourth, fifth and sixth integrals in the right hand side of (26) converge, uniformly over y changing on any compact in (0, ∞). Let us prove convergence of the fourth integral. Consider arbitrary (large) r0 > r and proceed as follows: Z Z Z Z exp(αy)ψ 00 (α) |ψ 00 (γ + iu)| ˆ ˆ dαh(z)dz du|h(z)|dz ≤ 2eγy R Gr0 \Gr ψ(α) − ψ(−ζ − iz) R |u|∈[r,r 0 ] |ψ(γ + iu) − ψ(−ζ − iz)| Z
N
Z
r0
1 du 2−c u 5 −N r Z Z r0 1 γy ˆ +c4 e ψ(γ + iu) − ψ(−ζ − iz) du|h(z)|dz, |z|>N r ≤ c4 eγy
ˆ |h(z)|dz
(28)
where, again, we choose N > 0 to be such that the estimate in Lemma 5 can be applied for all |z| > N , and r, r0 are assumed to be large enough. It is easy to see that the first term in the right hand side of (28) vanishes, as r, r0 → ∞. To show that the second term in the right hand side of (28) vanishes, as r, r0 → ∞, we make use of Lemma 5: Z Z r0 Z Z r0 1 1 ˆ ˆ du|h(z)|dz p ≤ c du|h(z)|dz, 5 ψ(γ + iu) − ψ(−ζ − iz) 2 (u − z 2 )2 + z 2 |z|>N r |z|>N r and apply the dominated convergence theorem: Z 0 Z r 1 1 p √ du ≤ du < ∞, 4 2 2 2 2 r u +1 (u − z ) + z R where we assumed that N > 1. Next, we use Lemma 5 to prove convergence of the fifth integral in the right hand side of (26): Z |z|>N
Z r
r0
2
| exp((γ + iu)y)| |ψ 0 (γ + iu)| |ψ(γ + iu) − ψ(−ζ − iz)|
21
2
ˆ du|h(z)|dz
≤ c12 e
γy
Z
Z
z>N
r0
u2 ˆ du|h(z)|dz (u2 − z 2 )2 + z 2
r
The right hand side of the above converges to zero, as r, r0 → ∞, due to the dominated convergence theorem: Z 0 Z Z r z+1 u2 u2 u2 ≤ du du + du 2 2 2 2 2 2 2 2 2 2 2 2 r (u − z ) + z z−1 (u − z ) + z u∈[0,z−1]∪[z+1,∞) (u − z ) + z Z
1+1/z
≤ 1−1/z
zu2 du + 2 2 z (u − 1)2 + 1
Z u∈[0,z−1]∪[z+1,∞)
Z
2
≤ 2 (1 + 1/N ) du + |u|>1
u2 du (u − z)2 (u + z)2
1 du < ∞ u2
To show that the last integral in the right hand side of (26) converges, as r → ∞, we notice that Z Z Z Z r0 exp(αy) 1 γy ˆ ˆ dαh(z)dz ≤ e du|h(z)|dz, 2 + (z + u)2 R Gr0 \Gr (α + ζ + iz)2 (γ + ζ) R r and apply the dominated convergence theorem once more: Z r
r0
1 du ≤ (γ + ζ)2 + (z + u)2
Z R
1 du < ∞. γ 2 + u2
Thus, we have shown that, for y > 0, gr (y) converges to g(y), as r → ∞. To see that the limiting function, g, is continuous, we notice that the estimates, derived above, yield that the convergence of the fourth, fifth and sixth terms in the right hand side of (26) is uniform over y changing on a compact in (0, ∞). The above estimates also imply that |gr (y)| is bounded by a constant times exp(γy), uniformly over r. Then, Proposition 2, along with the dominated convergence, yield: Z ∞ Z ∞ e−λy g(y)dy = lim e−λy gr (y)dy = ψ 0 (λ)Υ(ψ(λ)), 0
r→∞
0
for all large enough real λ > 0. As shown in Subsection 2.2, a change of variables turns the left and right hand sides of the above into the Laplace transforms of Eg(Xt ) and Eh(Xt ), respectively. Due to the uniqueness of Laplace transform, the expectations have to coincide for all t > 0. This completes the proof of the theorem.
References ´ [1] L. Bachelier. Th´eorie math´ematique de jeu. Annales Scientifiques de l’E.N.S., 18:143–201, 1901. [2] J. Bertoin. L´evy processes. Cambridge University Press, 1996. [3] P. Carr and A. Chou. Breaking barriers. Risk, 7:45 – 49, 1997. [4] P. Carr and R. Lee. Put-call symmetry: extensions and applications. Mathematical Finance, 19(4):523– 560, 2009. [5] P. Carr and S. Nadtochiy. Static hedging under time-homogeneous diffusions. SIAM Journal on Financial Mathematics, 2(1):794–838, 2011. 22
[6] B. Davies. Integral transforms and their applications, 3rd ed. Springer-Verlag, 2002. [7] E. Dynkin. Theory of Markov processes. Oxford-London-New York-Paris: Pergamon Press, 1961. [8] A. Kuznetsov, A. Kyprianou, J. Pardo, and K. van Schaik. A Wiener-Hopf monte carlo simulation technique for L´evy processes. The Annals of Applied Probability, 21(6):2171–2190, 2011. [9] A. Kuznetsov, A. Kyprianou, and V. Rivero. The theory of scale functions for spectrally negative L´evy processes. In L´evy Matters II, pages 97 – 186. Springer Lecture Notes in Mathematics, 2013. [10] P. L´evy. Sur certains processus stochastiques homog´enes. Compositio Mathematica., 7:283–339, 1940. [11] K. Sato. L´evy processes and infinitely divisible distributions. Cambridge University Press, 1999. [12] M. Tehranchi. Symmetric martingales and symmetric smiles. Stochastic Processes and Their Applications, 119(10):3785–3797, 2009. [13] E. Titchmarsh. Eigenfunction expansions associated with second-order differential equations, Part 1. Oxford, Clarendon Press, 1946. [14] D. Widder. The Laplace transform. Princeton University Press, 1946.
23
Abstract In this paper, we develop a new mathematical technique which can be used to express the joint distribution of a Markov process and its running maximum (or minimum) through the distribution of the process itself. This technique is an extension of the classical reflection principle for Brownian motion, and it is obtained by weakening the assumptions of symmetry required for the standard reflection principle to work. We call this method a weak reflection principle and show that it provides solutions to many problems for which the classical reflection principle is typically used. In addition, unlike the standard reflection principle, the new method works for a much larger class of stochastic processes which, in particular, do not possess any strong symmetries. Here, we review the existing results which establish the weak reflection principle for a large class of time-homogeneous diffusions on a real line and, then, proceed to develop this method for all L´evy processes with one-sided jumps (subject to some admissibility conditions). Finally, we demonstrate the applications of the weak reflection principle in Financial Mathematics, Computational Methods, and Inverse Problems.
1 1.1
Introduction Standard reflection principle and its applications
We start with a brief review of the classical reflection principle for Brownian motion. Denote by B x the Brownian motion on a real line, started form x. Given arbitrary levels U > 0 and K < U , we can compute the joint distribution of Bt = Bt0 and its running maximum Mt = supu∈[0,t] Bu as follows: U P (Bt ≤ K, Mt > U ) = P (Bt−TU +TU ≤ K, TU < t) = P Bt−T ≤ K, TU < t U U = P 2U − Bt−T ≤ K, TU < t = P (Bt ≥ 2U − K, Mt > U ) = P (Bt ≥ 2U − K) , (1) U where TU is the first hitting time of U by B, and BsU = BTU +s . The above formula first appeared in the work of L. Bachelier [1], followed by a more rigorous treatment, for example, by P. L´evy [10]. Notice that the above derivations are based on the following well-known properties of Brownian motion: • strong Markov property: BTU +t −BTU is a standard Brownian motion (started from zero), independent of FTU (where the filtration is generated by B); • continuity: since the paths of B are continuous, BTU = U and, in view of the above, BTU +t is a Brownian motion started from U , and independent of FTU ; ∗ [email protected]; Department of Mathematics, University of Michigan. This author is supported by National Science Foundation under grant DMS-0955463. † [email protected]; Department of Mathematics, University of Michigan.
1
• symmetry: the distribution of BtU is symmetric with respect to the initial level U , i.e Law BtU Law 2U − BtU .
=
We will come back to the above observations in the next subsection, but, first, let us outline several applications of the standard reflection principle. One obvious application is the computation of the joint distribution. Since the marginal distribution of a Brownian motion is available in closed form, the above formula gives us a closed form expression for the joint distribution of the process and its running maximum, at any given time. A more subtle application, which requires the use of reflection principle itself (rather than the resulting formula for the joint distribution) comes from Financial Mathematics. Namely, the reflection principle turns out to be very useful in the problem of hedging barrier options. To simplify the notation, in the remainder of this section, we will consider U = 0. Let us assume that the risk-neutral evolution of the underlying is described by a Brownian motion B x started from x ≤ 0 (we assume no discounting). Consider a up-and-out option, written on this underlying, with a terminal payoff function h, such that supp(h) ⊂ (−∞, 0). The payoff of such option, at the time of maturity T , is given by h(BTx )1{sup
u∈[0,T ]
x 0. Then, the target equation E h (Xt ) = E g (Xt ) ,
for all t > 0,
has the solution g(x) = h (S(x)), where the functions g and h have supports on the opposite sides of zero. Thus, the reflection principle can be extended easily to any strong Markov process that possesses a strong symmetry and does not jump across the barrier (in this case, zero). This was observed by Carr and Lee [4], whose transformed put-call symmetry condition is equivalent to the existence of a strong symmetry for the underlying stochastic process (see also [12] for equivalent characterization of symmetric martingales). In particular, the reflection principle can be extended to any diffusion martingale whose coefficient is an even function, since any such process possesses a strong symmetry, with S : x 7→ −x. It is also shown in [4] that, if a given strong Markov process does not jump across the barrier and possesses a strong symmetry S, then, the reflection principle holds for this process run on an independent continuous 1 In fact, Carr and Chou [3] allow the Brownian motion to have a constant drift, which can be eliminated by a Girsanov change of measure.
3
stochastic clock (the filtration, in this case, is generated by both the original process and the subordinator). Even though the resulting process may not be Markov, it is easy to see that, conditional on any A ⊂ FT0 , the future values of the time-changed process have a strong symmetry with the same mapping S. However, the condition of strong symmetry excludes many processes important for applications: some of such examples are discussed in [5]. In the present paper, we develop a weak formulation of the reflection principle, which can be applied to a large class of stochastic processes that do not posses any strong symmetries. This new formulation, albeit weaker than the standard one, is sufficient to solve the problems outlined in Subsection 1.1. Herein, we restrict our analysis to strong Markov processes which do not jump across the given upper (lower) barrier from below (above). Then, in view of the above discussion, it suffices to consider the unconditional, as opposed to conditional, distributions of the process. Consider a stochastic process X, defined on a real line and started from zero. Consider two spaces, B − and B + , consisting of Lebesgue measurable test functions h, such that h(Xt ) has finite expectation, for all t ≥ 0, and all h ∈ B − have support in (−∞, 0) while all h ∈ B + have support in (0, ∞). We say that X possesses an upper weak symmetry (with respect to zero)2 if there exists a mapping W+ from a space of test functions B0− ⊂ B − into B + , such that E h (Xt ) = E W+ [h] (Xt ) , for all t > 0, for any h ∈ B0− . Analogously, one can define the lower weak symmetry, along with the mapping W− : B0+ → B − . We do not insist on a particular choice of the spaces of test functions B0± , but, in what follows, we choose the spaces that contain, at least, all smooth functions with compact support. We will refer to the mappings W± as the mirror image transformations, although one should remember that, in the present setting, the mirror may not be straight: if the distribution of Xt is not symmetric with respect to zero, the image of a straight line may be a curve with nonzero curvature. Thus, the weak reflection principle consists of the application of a strong Markov property and (semi) continuity, as well as the mirror image transformation W± . We will discuss the questions of existence, uniqueness, and numerical approximation of the mappings W± in the remainder of this paper. But, first, let us provide some motivation for the above constructions. The weak reflection principle, in particular, allows us to solve the problems outlined in Subsection 1.1, for a class of processes that may not posses any strong symmetries. Without loss of generality, let us focus on the weak reflection principle with an upper barrier. Namely, assuming that process X is strongly Markov, does not jump across 0(> X0 ) from below, and possesses upper weak symmetry, we obtain the following results. 1. We can express the joint distribution of the process and its running maximum via the distribution of the process itself: for any K ∈ R, there exists a measurable function p(K; ·), such that ! P Xt ≤ K, sup Xu > 0
= E p (K; Xt )
u∈[0,t]
2. We can solve the static hedging problem in a model where X is the underlying: for any h ∈ B0− , we have + E h(XT )1{sup t ∈ [0, T ] Xu 0, the functions µ and σ belong to C 3 (R), the functions themselves and their first three derivatives have finite limits at −∞, and, for any k = 1, 2, 3, the functions e(3−k)x |µ(k) (x)| and e(3−k)x σ (k) (x) are bounded over all x > 0. Then, Theorem 2.8 in [5] provides an explicit integral expression for the mirror image transformation associated with X. For the sake of completeness, we present a simplified corollary of this theorem here. Theorem 1. (Carr-Nadtochiy 2011) Let X be as above, and let h be a once weakly differentiable function, with support in (−∞, 0), such that its derivative has a modification with finite variation over (−∞, 0). Then, there exists a continuous and exponentially bounded function g, with support in (0, ∞), such that E h(Xt ) = E g(Xt ),
for all t > 0
(4)
Moreover, for any large enough γ > 0, the function g can be computed as follows: Z z Z γ+i∞ Z 0 wψ1 (x, w) ψ1 (z, w) µ(y) 2 exp −2 dy h(z)dzdw, g(x) = 2 πi γ−i∞ ∂x ψ1 (0, w) − ∂x ψ2 (0, w) −∞ σ 2 (z) 0 σ (y) where ψ1 and ψ2 are the fundamental solutions of the associated Sturm-Liouville equation ∂2 ∂ 1 2 σ (x) 2 ψ(x, w) + µ(x) ψ(x, w) − w2 ψ(x, w) = 0, 2 ∂x ∂x
(5)
determined uniquely, for all complex w with large enough Re(w) > 0, by the following conditions: ψ1 (·, w) is square integrable on (−∞, 0), ψ2 (·, w) is square integrable on (0, ∞), and ψ1 (0, w) = ψ2 (0, w) = 1. The above result shows that all regular enough diffusions possess weak symmetry, with the mirror image transformation given by an explicit integral transform. Of course, in order to implement this transform numerically, one needs to know the fundamental solutions of the associated Sturm-Liouville equation. These functions 5
can be approximated efficiently by expanding them into power series of w (cf. [13]). Alternatively, one can notice that, if µ ≡ 0 and σ is piecewise constant, then, ψi ’s are piecewise linear-exponential (linear combinations of exponentials). Thus, we can approximate any function σ with piecewise constant ones, then, compute ψi ’s in a closed form, and, finally, obtain g via numerical integration. Examples of functions g corresponding to a piecewise linear function h can be found in [5].
2 2.1
Mirror Image Transformation for Spectrally Negative L´evy Processes Problem formulation
Consider a L´evy process (Xt )t≥0 , given by its initial condition X0 = 0 and the Laplace exponent ψ: Z 0 σ2 2 eλx − 1 − λx Π(dx), λ + ψ(λ) = µλ + 2 −∞
(6)
where Π is the L´evy measure of X, and EeλXt = etψ(λ) , for all complex λ for which both sides of the above equation are well defined. To make sure that the above expressions are well defined, at least, for all λ with positive real part, as well as to simplify some of the derivations that follow, we make the following assumption on the L´evy triplet, (µ, σ, Π). Assumption 1. We assume that µ ∈ R, σ > 0, and Π is a σ-finite Borel measure on (−∞, 0), such that Z 0 x2 Π(dx) < ∞. −∞
The process X is called spectrally negative because it is only allowed to have negative (i.e. downward) jumps (cf. [9]). The reason for such a restriction is that the process must not jump across the upper barrier, in order for the weak reflection principle to hold. Of course, in the case of a lower barrier, one needs to consider X with only positive jumps. Our goal is to construct a mirror image transformation for the process X. Namely, for any given admissible function h : R → R, with supp(h) ⊂ (−∞, 0), we would like to find a measurable function g : R → R, with supp(g) ⊂ (0, ∞), such that Eh(Xt ) = Eg(Xt ), ∀ t > 0 To ensure that the expectation of h(Xt ) is well define, we need to make some additional assumptions on h and Π. ˆ ∈ L1 (R), such that the Assumption 2. We assume that supp(h) ⊂ (−∞, 0) and that there exist ζ ≥ 0 and h ζy ˆ function y 7→ e h(y), defined for all y ∈ R, is a Fourier transform of h. In addition, we assume that Z −1 |x|e−ζx Π(dx) < ∞ −∞
Due to uniqueness of the Laplace inverse, our problem is equivalent to: find a measurable function g : R → R, with supp(g) ⊂ (0, ∞), such that Z ∞ Z ∞ E e−λt h(Xt )dt = E e−λt g(Xt )dt, (7) 0
0
for all large enough λ > 0. 6
2.2
Matching expectations in Laplace space
Denote by G(x, λ, y) the Green’s function of X. For the definition and properties of this function we refer the reader to [2], [11], as well as the survey article [9], and the references therein. Using the definition of Green’s function, we can rewrite equation (7) as follows: Z 0 Z ∞ h(y)G(0, λ, y)dy = g(y)G(0, λ, y)dy, (8) −∞
0
for all large enough λ > 0. The right hand side of the above is a result of the integral transform with kernel G(0, λ, y). Thus, our problem amounts to finding an inverse transformation and showing that the left hand side of the above lies in its domain. Note that, so far, we d onto know for which λ the left hand side of the above is well defined. For all real λ ≥ 0, the Green’s function G(0, λ, y) has the following representation: G(0, λ, y) = Φ0 (λ)e−Φ(λ)y − W λ (−y),
(9)
where W λ is the λ-scale function of X (cf. [9]), with W λ (y) = 0, for y ≤ 0, and Φ(λ) := sup{q ≥ 0 : ψ(q) = λ},
λ ≥ 0,
with ψ being the Laplace exponent of X. Notice that, for all real λ > ψ(−ζ) ∨ 0, we have: Z 0 Z ∞ −ζy G(0, λ, y)e dy = e−λt E e−ζXt 1{Xt ≤0} dt −∞
(10)
0
Z
∞
≤
e−λt Ee−ζXt dt =
Z
0
∞
e−(λ−ψ(−ζ))t dt < ∞
0
Note also that ψ(−ζ) is well defined due to Assumption 2. Thus, Assumption 2 and estimate (10) imply that Z ∞ Z 0 Z ∞ Υ(λ) := E e−λt h(Xt )dt = G(0, λ, y)h(y)dy = Φ0 (λ)eΦ(λ)y − W λ (y) h(−y)dy, (11) −∞
0
0
is well defined for all real λ > ψ(−ζ) ∨ 0. In addition, (8) is equivalent to Z ∞ Φ0 (λ)e−Φ(λ)y g(y)dy = Υ(λ) 0
Notice that ψ is continuous and strictly increasing on |µ|/σ 2 , ∞ , and it explodes at infinity. Hence, we can change the variables and reformulate the problem as: find a measurable function g, such that, for all large enough real λ > 0, the following holds Z ∞ e−λy g(y)dy = ψ 0 (λ)Υ(ψ(λ)), (12) 0
with 0
0
Z
0
ψ (λ)Υ(ψ(λ)) = ψ (λ)
Z G(0, ψ(λ), y)h(y)dy =
−∞
∞
eλy − ψ 0 (λ)W ψ(λ) (y) h(−y)dy
0
The right hand side of the above is well defined for all real λ > Φ (ψ(−ζ) ∨ 0). 7
(13)
Notice that the mirror image problem (12), now, looks exactly like the Laplace transform inversion. However, there is a major difference. In the classical problem of inverting a Laplace transform, we, typically, know that the right hand side is a Laplace transform of some function, and we need to find a transformation that would recover this function. In the present case, we need to prove that ψ 0 (λ)Υ(ψ(λ)) is, indeed, a Laplace transform of some function, which is not obvious a priori. In addition, we need to propose a method to recover this function from the right hand side. Of course, there exist several methods for inverting the Laplace transform (cf. [14], [6]). However, the conditions, which are required for some of these methods to succeed, are expressed through the original function (in our case, g), rather than the transformed one (in the present case, ψ 0 (λ)Υ(ψ(λ))). Since, a priori, we know very little about function g (e.g. we do not even know if it exists), we cannot apply any of the existing results on the Laplace transform inversion to solve the mirror image problem. Instead, we will show, by hand, that the desired function g exists and can be recovered via the classical Bromwich integral (cf. [6]).
2.3
A priori estimates
In this subsection, we establish the analytic continuation of ψ 0 (λ)Υ(ψ(λ)) to a complex half plane of the form HR = {w : Re(w) > R}, and provide some useful estimates of its absolute value. Remark 1. Notice that Υ(λ) can be easily extended to a half plane, via its probabilistic representation given by the first identity in (11). However, under the change of variables λ 7→ ψ(λ), the half plane transforms into a smaller domain which is not sufficient for our purposes. To obtain an analytic extension of Υ(ψ(λ)), one would need the probabilistic representation in (11) to hold in a domain where the real part of λ is unbounded from below, which is typically impossible. More precisely, the conditions that function h has to satisfy, in order for the probabilistic representation to hold in the desired domain, are extremely restrictive and rather implicit. For example, if X is a martingale, none of the convex functions h, except zero, are admissible. In addition, even for those functions h for which the probabilistic representation in (11) is well defined (although we do not know how to characterize this set explicitly), the standard estimates of the integrals involved in this representation do not provide sufficient information about the asymptotic behavior of Υ(ψ(λ)), as |λ| → ∞, which is needed to solve the mirror image problem (12). Alternatively, one might be tempted to use the analytic continuation of W ψ(λ) to extend Υ(ψ(λ)) to a complex half plane via (13). Indeed, it is well known (cf. [9]) that W λ can be extended analytically to the entire complex plane. However, to the best of our knowledge, there exist no estimates of this extension (more precisely, we need an estimate of eλy −ψ 0 (λ)W ψ(λ) (y)), which, in particular, would guarantee that the integral in the right hand side of (13) is well defined for all λ in a half plane HR . In fact, it is easy to construct an example of a L´evy process whose infinitesimal generator has a nontrivial spectrum, and, hence, the integral in the right hand side of (13) is not well defined for some λ. The difficulties described above explain why we are forced to construct the analytic continuation of Υ(ψ(λ)), and investigate its asymptotic behavior, by hand. It turns out that Fourier transform offers a natural way to obtain the desired analytic continuation. Recall that, due to Assumption 2, Z −ζy ˆ h(y) = e e−iyz h(z)dz R
Due to estimate (10), we can interchange the limit and integration to obtain Z ∞ eλy − ψ 0 (λ)W ψ(λ) (y) h(−y)dy ψ 0 (λ)Υ(ψ(λ)) = 0
8
(14)
Z Z
∞
= R
ˆ eλy − ψ 0 (λ)W ψ(λ) (y) e(ζ+iz)y dy h(z)dz,
0
for all large enough real λ > 0. Our next goal is to extend the above representation to a complex half plane HR , and estimate its absolute value from above. Let us analyze the inner integral in the right hand side of (14). Notice that it can be viewed as Z ∞ eλy − ψ 0 (λ)W ψ(λ) (y) e−wy dy, 0
evaluated at w = −ζ − iz. For λ > |µ|/σ 2 and w ∈ (λ, ∞), the above integral can be computed explicitly, using the definition of the scale function W ψ(λ) (cf. [9]): Z ∞ Z ∞ Z ∞ λy 0 ψ(λ) −wy λy−wy 0 e − ψ (λ)W (y) e dy = e dy − ψ (λ) e−wy W ψ(λ) (y)dy (15) 0
0 0
=
0 0
ψ (λ) ψ(w) − ψ(λ) − ψ (λ)(w − λ) w−λ 1 − = 2 w − λ ψ(w) − ψ(λ) (w − λ) ψ(w) − ψ(λ)
The right hand side of the above is analytic everywhere in w ∈ H−ζ = {w : Re(w) > −ζ}, except, possibly, the zeros of ψ(w)−ψ(λ), where it has poles (on every compact, there is at most a finite number of such points). The left hand side of the above is analytic in H−ζ (we can differentiate with respect to w inside the integral, and the resulting expression is bounded, uniformly over w, by an absolutely integrable function of y). Therefore, the right hand side, in fact, does not have any poles in H−ζ , and the above equality holds for all w ∈ H−ζ . By continuity, the equality holds at w = −ζ − iz: Z ∞ ψ 0 (λ) 1 eλy − ψ 0 (λ)W ψ(λ) (y) e(ζ+iz)y dy = − , (16) ψ(λ) − ψ(−ζ − iz) λ + ζ + iz 0 for all large enough real λ > 0. Next, we need to show that, for every z ∈ R, the right hand side of the above is analytic in λ ∈ HR = {w : Re(w) > R}, for a large enough R > 0, and that, for every fixed λ ∈ HR , it is absolutely bounded over z ∈ R. To do this, we will need the following lemmas, which, in fact, are also crucial for the proofs of the main theorems. These lemmas describe the asymptotic behavior of 1/(ψ(λ) − ψ(−ζ − iz)) and they can be viewed as the core technical result of this paper. Their proofs are given in Appendix A. Lemma 1. For all u, z ∈ R and v > 0, we have Im (ψ(v + iu) − ψ(−ζ − iz)) = µ(u + z) + (vu − zζ)σ 2 + o(uv) + sign(u) o(u2 ) + sign(z) o(z 2 ) + o(z), where we denote by o (f (u, v, z)) any function of (u, v, z), such that o (f (u, v, z)) /f (u, v, z) is absolutely bounded over all u, z ∈ R and v > 0, and it converges to zero, as |f (u, v, z)| → ∞. Lemma 2. For all u, z ∈ R and v > 0, we have Im (ψ(v + iu) − ψ(−ζ − iz)) = µ(u+z)+(vu−zζ)σ 2 +o(uv)+o(z 2 −u2 )+sign(z+u) o (z + u)2 +o(z), where o (f (u, v, z)) has the same meaning as in Lemma 1. Lemma 3. For all u, z ∈ R and v > 0, we have Re (ψ(v + iu) − ψ(−ζ − iz)) = µ(v+ζ)+ v 2 + z 2 − u2 σ 2 /2+o(z 2 −u2 )+ o(uv) + o(v 2 ) +o(z)+o(1), where o (f (u, v, z)) has the same meaning as in Lemma 1. 9
Lemma 4. For any ε ∈ (0, 1), there exist R1 > 0, R2 > 0, c1 > 0, and c2 > 0, such that the following inequality |ψ(v + iu) − ψ(−ζ − iz)| ≥ c1 v(|z| + v)1{||u|−√z2 +v2 |≤εv} + c2 u2 − (v 2 + z 2 ) 1{||u|−√z2 +v2 |>εv} , holds for all u, v, z ∈ R, with |z| > R1 and v > R2 . Lemma 5. There exists a constant R1 > 0, such that, for any v > R1 , there exist R2 = R2 (v) > 0, R3 = R3 (v) > 0, and c = c(v) > 0, such that the following inequality 2 |ψ(v + iu) − ψ(−ζ − iz)| ≥ c (z 2 − u2 )2 + z 2 , holds for all u, z ∈ R, with |z| > R2 and |u| > R3 . Lemma 4 yields that, for any ε ∈ (0, 1), there exist N > 0, R > 0 and ci ’s, such that the following estimates hold, for all u, all |z| ≥ N , and all v ≥ R: 2 c4 1 c3 √ √ ψ(v + iu) − ψ(−ζ − iz) ≤ v 2 (|z| + v)2 1{||u|− z2 +v2 |≤εv} + (u2 − (v 2 + z 2 ))2 1{||u|− z2 +v2 |>εv} c5 √ 1 2 2 v 2 ((|u| − εv)2 + v) {||u|− z +v |≤εv} c4 √ + 4 1 2 2 z + (2v 2 − 1)z 2 + v 2 (v 2 − 1) + z 2 + v 2 − u2 {||u|− z +v |>εv} c5 √ ≤ 2 1 2 2 v ((|u| − εv)2 + v) {||u|− z +v |≤εv} c4 √ √ 1 + 2 2 z 4 + 2v 2 z 2 + v 4 − z 2 − v 2 + εv z 2 + v 2 + |u| {||u|− z +v |>εv} c5 c6 ≤ 2 + v ((|u| − εv)2 + v) (z 2 + v 2 )2 + εv|u| ≤
(17)
Since the right hand side of (16) is continuous, as a function of z, the above estimate implies that there exists R0 ≥ R > 0, such that the following inequalities hold for all λ = v + iu, with v ≥ R0 : Z ψ 0 (λ) 1 ˆ ψ(λ) − ψ(−ζ − iz) − λ + ζ + iz |h(z)|dz R Z Z c6 |λ| 1 0 ˆ ˆ |h(z)|dz | h(z)|dz + |ψ (λ)| ≤ 2 |λ| − c7 |z|≤N |z|>N ψ(λ) − ψ(−ζ − iz) Z 1 ˆ |h(z)|dz + R λ + ζ + iz ! Z Z c8 1 1 1 ˆ ˆ √ + p p ≤ + c9 |λ| |h(z)|dz + c10 |h(z)|dz 2 2 |λ| v (||u| − εv| + v) v + v|u| v + (u + z)2 |z|>N R ! 1 |λ| |λ| 1 √ + p ≤ c11 + + |λ| v (||u| − εv| + v) v 2 + v|u| v
10
In the above, we also used the fact that, for all large enough |λ|, with positive real part, we have: |ψ 0 (λ)| ≤ c12 |λ|,
|ψ(λ)| ≥ c13 |λ|2
Therefore, due to (14), the following representation Z ψ 0 (λ) 1 0 ˆ ψ (λ)Υ(ψ(λ)) = − h(z)dz, ψ(λ) − ψ(−ζ − iz) λ + ζ + iz R
(18)
is well defined and holds for all large enough real λ > 0. In fact, the right hand side of the above is analytic, as a function of λ ∈ HR0 = {λ : Re(λ) > R0 }. To see this, we differentiate, formally, inside the integral and apply the same estimates as above, to show that the integral of the derivative is absolutely convergent, for any λ ∈ HR0 . Hence, the above expression is analytic as a function of λ ∈ HR0 . We have proved the following proposition. Proposition 1. For any ε ∈ (0, 1), there exist constants R > 0 and c ≥ 0, such that the function λ → 7 ψ 0 (λ)Υ(ψ(λ)) (defined in (13), for all large enough real λ > 0) can be extended analytically to HR = {λ : Re(λ) > R} via (18), and its extension satisfies: |λ| |λ| 1 1 + p + + , |ψ 0 (λ)Υ(ψ(λ))| ≤ c p 2 |λ| Re(λ) ||Im(λ)| − εRe(λ)| + Re(λ) Re(λ) Re(λ) + Re(λ)|Im(λ)| for all λ ∈ HR .
2.4
Main results
Now, we have everything we need to solve the mirror image problem (12). First, for any r > 0, we introduce the following function of y ∈ R: 1 R αy 0 2πi Gr e ψ (α)Υ(ψ(α))dα, y ≥ 0, gr (y) = (19) 0, y < 0, where ψ 0 (α)Υ(ψ(α)) is given by (18), and Gr is a vertical interval on a complex plane: Gr = [γ − ir, γ + ir], with a large (but fixed) constant γ > 0. To ensure that gr is well defined, we assume that γ > R, where R is the constant appearing in Proposition 1 (the constant ε ∈ (0, 1) is fixed arbitrarily). Our next result shows that, if we use gr , in lieu of g, in the left hand side of (12), then, it will converge to the right hand side, as r → ∞. Proposition 2. For all large enough real λ > 0, we have: Z ∞ lim e−λy gr (y)dy = ψ 0 (λ)Υ(ψ(λ)) r→∞
0
Proof: We fix arbitrary λ > γ and apply the Fubini’s theorem, to obtain: Z ∞ Z Z ∞ Z 1 1 ψ 0 (α)Υ(ψ(α)) e−λy gr (y)dy = e−λy eαy dyψ 0 (α)Υ(ψ(α))dα = dα 2πi Gr 0 2πi Gr λ−α 0 11
Next, we complete the contour of integration Gr on the right, by a rectangle, Cr , with the horizontal sides Hr1 and Hr2 , of length r, and the vertical side, Vr , of length 2r. When r is large enough, the real number λ > γ is inside this closed contour. Since ψ 0 (α)Υ(ψ(α)) is analytic inside the contour, the integral is equal to the residue: Z ψ 0 (α)Υ(ψ(α)) dα = 2πiψ 0 (λ)Υ(ψ(λ)), λ−α Gr ∪Cr Z Z ψ 0 (α)Υ(ψ(α)) ψ 0 (α)Υ(ψ(α)) 0 dα = 2πiψ (λ)Υ(ψ(λ)) − dα λ−α λ−α Gr Cr It only remains to show that the integral over Cr vanishes for large r. Applying Proposition 1, we obtain: Z
Cr
0 Z 0 2 Z ψ (α)Υ(ψ(α)) ψ (α)Υ(ψ(α)) ψ 0 (α)Υ(ψ(α)) X |dα| dα ≤ |dα| + j λ−α λ−α λ−α Vr j=1 Hr ≤ c1
γ+r
1 1 r r 1 √ + + + dv r r vr(1 − ε) v 2 + vr v γ Z r r r 1 1 1 + √ + 2+ du +c2 r r r r r −r r
Z
Applying the straightforward estimates, as well as the dominated convergence theorem, we easily conclude that the right hand side of the above converges to zero, as r → ∞. Finally, we present the main result of this paper. Theorem 2. Let h and X satisfy Assumptions 1 and 2, and let gr be given by (19), with γ > R1 , where R1 is the constant appearing in Lemma 5. Then, there exists a continuous and exponentially bounded function g : (0, ∞) → R, such that lim gr (y) = g(y), r→∞
for all y > 0, and Eg(Xt ) = Eh(Xt ), for all t > 0. The proof of the above theorem is given in Appendix B. Theorem 2 shows that any spectrally negative L´evy process X, satisfying Assumptions 1 and 2, possesses an upper weak symmetry, with the space of test functions B0− consisting of all function that satisfy Assumption 2. The associated mirror image transformation is given by Z γ+i∞ Z 1 ψ 0 (α) 1 + αx ˆ W [h](x) = e − h(z)dzdα (20) 2πi γ−i∞ ψ(α) − ψ(−ζ − iz) α + ζ + iz R In view of (12), the uniqueness of the mirror image transformation follows from the uniqueness of the Laplace inversion (cf. [14]). To implement this transformation, one needs to find a Fourier inverse of y 7→ eζy h(y) and evaluate the above integral numerically. It is shown in the next section that B0− s is large enough to include many functions of interest, for which the aforementioned Fourier inverse can be computed explicitly.
12
2.5
Example
Consider a L´evy process X which is a sum of a scaled Brownian motion and a negative Gamma process. In other words, Xt = σBt − Γt , where B is a standard Brownian motion and Γ is a Gamma process with parameters α > 0 and β > 0 (cf. [2], [11]). In this case, the characteristic triplet of X, as defined in (6), is given by µ = −β/α, σ = σ, Π(dx) = β|x|−1 e−α|x| 1(−∞,0) (x)dx , and, in particular, 1 2 2 σ λ − β log (1 + λ/α) 2 As a test function h, we choose the hockey-stick function h(x) = (K − x)+ , as it represents the payoff of a put option and, therefore, is relevant for the static hedging application. Note that Assumption 1 is always satisfied for the process X, and Assumption 2 holds with any ζ ∈ (0, α). In addition, it is easy to see that the inverse Fourier transform of the function x 7→ eζx h(x) is given by ψ(λ) =
exp(K(ζ + ix)) ˆ h(x) = 2π(ζ + ix)2 With the above expressions at hand, one can evaluate the integral in (20) numerically, to obtain the mirror image function g. The results of these computations are presented in Figure 1.
3
Summary and extensions
We have presented a new mathematical technique, which we christened weak reflection principle and which is an extension of the well known reflection principle for Brownian motion. This new form of reflection principle is obtained by weakening the notion of symmetry that is required for the classical reflection principle to hold. Namely, it is based on the notions of upper and lower weak symmetries, defined via the associated mirror image transformations. We started by reviewing the existing results which provide an explicit integral expression for the mirror image transformation for any time-homogeneous diffusion process on a real line (subject to some regularity conditions). Finally, for the most of the paper we focused on constructing the mirror image transformation for spectrally negative L´evy processes, thus, extending the weak reflection principle to this new class of stochastic processes. The weak reflection principle provides solutions to various problems for which the classical reflection principle can be used, even when the underlying process is not a Brownian motion and does not possess any strong symmetries. Examples of such problems are discussed at the end of Subsection 1.2. In particular, the weak reflection principle is a perfect tool for constructing the exact static hedging strategies of barrier options (in fact, this problem motivated the development of the method in the first place). Another application of this method is the computation of the joint distribution of a process and its running maximum (minimum). Of course, while this problem is quite relevant for diffusions, in the case of L´evy processes, there exist several alternative computational methods, based on the Wiener-Hopf factorization (cf. [8]). Note that the weak reflection principle may not always be the most efficient way to compute the joint distribution of a L´evy process and its running maximum: after all, we still need to know the marginal distribution of the process, and, in addition, the numerical implementation of (20) requires some computational effort. However, in the cases when the marginal distribution is known (or is easy to compute), and provided that we need to find the joint 13
Figure 1: Mirror image of a put payoff h(x) = (K − x)+ , with K = −0.2, α = β = σ = 1, and ζ = 0.9.
distribution for multiple time horizons or multiple initial values, it is advantageous to reduce the problem of computing the joint distribution to the computation of an expectation with respect to the marginal distribution of the L´evy process. The last application suggested in Subsection 1.2 is related to the parabolic partial integrodifferential equation (PIDE) associated with a L´evy process. Namely, the mirror image transformation allows us to modify the initial condition of this PIDE, on one half line only, so that its solution remains constant at x = 0, for all times (see the discussion at the end of Subsection 1.2 for a more detailed explanation of the analogous problem in the case of diffusions). It is also worth mentioning that the technical Lemmas 1-5, and the resulting Proposition 1, describe a domain on which the resolvent function of the L´evy process is well defined. This domain is rather large, and, in particular, the real parts of its elements are unbounded from below. Thus, our results provide a non-trivial estimate of the spectrum of the integro-differential operator associated with any admissible spectrally-negative L´evy process. Recall that this operator is non-local and non-symmetric, which makes it very hard to describe its spectrum using the general theory (see also Remark 1 for a description of the associated difficulties). So far, the weak symmetry and the weak reflection principle have only been established for diffusion processes and L´evy processes with one-sided jumps. However, we conjecture that these results can be extended to a larger class of time-homogeneous Markov processes – possibly, all jump-diffusions satisfying some regularity conditions. Such an extension would allow us to solve the aforementioned problems for a larger class of stochastic processes. Another important extension is related to the domain with respect to which the weak symmetry is defined. Notice that, in the present case, we split the real line into two half lines and study the weak symmetry of the process with respect to the (unique) boundary point. It is interesting to extend these results to the case of a compact interval, whose boundary consists of two points (provided the underlying process does 14
not jump across the boundary points). In financial applications, this problem would correspond to the static hedging of double barrier options. More generally, one can investigate domains in higher dimension and try to establish the weak symmetry with respect to their boundaries.
4
Appendix A
Proof of Lemma 1 Im (ψ(v + iu) − ψ(−ζ − iz)) Z
= µ(u + z) + (vu − zζ) σ 2 +
0
(evx sin(ux) − ux) Π(dx) +
−∞
0
−∞
0
e−ζx sin(zx) − zx Π(dx) = zζ
Z
−∞
0
e−ζx sin(zx) − zx Π(dx),
−∞
−∞
Notice that Z 0 Z (evx sin(ux) − ux) Π(dx) = vu Z
Z
0
−∞
evx − 1 sin(ux) 2 x Π(dx) + u2 vx ux
Z
0
−∞
1 − eζx sin(zx) 2 −ζx x e Π(dx) + z 2 ζx zx
Z
sin(ux) − ux 2 x Π(dx), u2 x2 0
−∞
sin(zx) − zx 2 x Π(dx) z 2 x2
Since x2 and x2 e−ζx are integrable with respect to Π(dx), due to the dominated convergence theorem, the above integrals are absolutely bounded and vanish as the corresponding functions of (u, v, z) go to infinity. Proof of Lemma 2 We follow the proof of Lemma 1, except that, at the end, we apply the following additional estimate: Z 0 Z 0 sin(ux) − ux 2 sin(zx) − zx 2 2 2 u x Π(dx) + z x Π(dx) 2 x2 u z 2 x2 −∞ −∞ Z 0 = (2 sin((z + u)x/2) cos((z − u)x/2) − (z + u)x) Π(dx) −∞ 2
2
Z
0
= (z − u )
2 −∞
+(z + u)2
Z
sin((z + u)x/2) cos((z − u)x/2) − 1 2 x Π(dx) (z + u)x (z − u)x
0
−∞
2 sin((z + u)x/2) − (z + u)x 2 x Π(dx) (z + u)2 x2
Applying the dominated convergence theorem we complete the proof of the lemma. Proof of Lemma 3 Re (ψ(v + iu) − ψ(−ζ − iz)) = µ(v + ζ) + v 2 − ζ 2 + z 2 − u2 σ 2 /2 Z
0
+
(e
Z
e−ζx cos(zx) − 1 + ζx Π(dx),
cos(ux) − 1 − vx) Π(dx) −
−∞
Notice that
0
Z
vx
−∞
0
(evx cos(ux) − 1 − vx) Π(dx) −
Z
−∞
0
e−ζx cos(zx) − 1 + ζx Π(dx)
−∞
Z
0
=
evx (cos(ux) − 1)Π(dx) +
−∞
Z
0
−∞
15
(evx − 1 − vx) Π(dx)
Z
0
−
e
−ζx
Z
e−ζx − 1 + ζx Π(dx)
−∞
−∞ 0
0
(cos(zx) − 1)Π(dx) − 0
Z 0 vx − 1 cos(ux) − 1 2 e − 1 − vx 2 2 (cos(ux) − 1)Π(dx) + uv = x Π(dx) x Π(dx) + v vx ux v 2 x2 −∞ −∞ −∞ Z 0 Z 0 Z 0 1 − eζx cos(zx) − 1 2 −ζx 1 − eζx + ζxeζx 2 −ζx 2 − (cos(zx)−1)Π(dx)−zζ x e x e Π(dx)−ζ Π(dx) ζx zx ζ 2 x2 −∞ −∞ −∞ Z 0 = (cos(ux) − cos(zx))Π(dx) + o(uv) + o(v 2 ) + o(z) + o(1) Z
Z
e
vx
−∞
= 2(z 2 − u2 )
Z
0
−∞
sin((u + z)x/2) sin((z − u)x/2)) 2 x Π(dx) + o(uv) + o(v 2 ) + o(z) + o(1) (z + u)x (z − u)x 2 2 = o(z − u ) + o(uv) + o(v 2 ) + o(z) + o(1),
where we applied the dominated convergence theorem. Proof of Lemma 4 Using Lemmas 2 and 3, we obtain: |ψ(v + iu) − ψ(−ζ − iz)| ≥ |Im (ψ(v + iu) − ψ(−ζ − iz))| 1{||u|−√z2 +v2 |≤εv} (21) + |Re (ψ(v + iu) − ψ(−ζ − iz))| 1{||u|−√z2 +v2 |>εv} = µ(u + z) + (vu − zζ)σ 2 + o(vu) + o(z) + o(z 2 − u2 ) + sign(z + u) o (z + u)2 1{||u|−√z2 +v2 |≤εv} + µ(v + ζ) + v 2 + z 2 − u2 σ 2 /2 + o(z 2 − u2 ) + o(uv) + o(v 2 ) + o(z) + o(1) 1{||u|−√z2 +v2 |>εv} Let us estimate the first term in the above: µ(u + z) + vuσ 2 − zζσ 2 + o(vu) + o(z) + o(z 2 − u2 ) + sign(z + u) o (z + u)2 1 √ {||u|− z2 +v2 |≤εv} ≥ v|u|σ 2 + µu + z(µ − ζσ 2 ) + o(v(|z| + v)) + o(z) + o(v|z| + v 2 ) + sign(z + u) o (z + u)2 · 1{||u|−√z2 +v2 |≤εv,u≥0} + v|u|σ 2 − µu − z(µ − ζσ 2 ) + o(v(|z| + v)) + o(z) + o(v|z| + v 2 ) − sign(z + u) o (z + u)2 · 1{||u|−√z2 +v2 |≤εv,uεv} ! 2 2 o(uv) + o(v 2 ) + o(z) 2 µ(v + z) + o(u − z ) + 2 2 2 ≥ u − (v + z ) σ /2 + 1{||u|−√z2 +v2 |>εv} |u2 − (v 2 + z 2 )| ≥ c2 u2 − (v 2 + z 2 ) 1{||u|−√z2 +v2 |>εv} , which holds for all large √ enough |z| and v > 0, with some positive constants ci . In the above we made use of the fact that, if |u| − z 2 + v 2 > εv, then, p 2 u − (v 2 + z 2 ) ≥ εv |u| + z 2 + v 2 , |u| ≥ (1 − ε)v, v v2 |u2 − z 2 | √ 2 −z 2 |≤v} + ≤ 1 1 + {|u |u2 − (v 2 + z 2 )| v2 − v εv |u| + v 2 + z 2 1 1 ≤ + 1+ 1{|u2 −z2 |>v} , v−1 ε and, hence,
! 1{|u2 −z2 |>v}
µ(v + z) + o(u2 − z 2 ) + o(uv) + o(v 2 ) + o(z) |u2 − (v 2 + z 2 )|
can be made arbitrarily small by choosing large enough v. Proof of Lemma 5 First, we notice that, since ψ is analytic, we have |ψ(v − iu)| = |ψ(v + iu)|, and, hence, it suffices to consider only u > 0. Using Lemma 3, we obtain the following inequalities: 2
(Re (ψ(v + iu) − ψ(−ζ − iz)))
2 = µ(v + ζ) + v 2 + z 2 − u2 σ 2 /2 + o z 2 − u2 + o(uv) + o v 2 + o(z) + o(1) 2 ≥ v 2 + (v + ζ)2µ/σ 2 + z 2 − u2 σ 4 /4 + v 2 + z 2 − u2 + (v + ζ)2µ/σ 2 o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||≤ε} + v 2 + z 2 − u2 + (v + ζ)2µ/σ 2 o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} 2 2 ≥ v 2 + (v + ζ)2µ/σ 2 σ 4 /4 + v 2 + (v + ζ)2µ/σ 2 (z 2 − u2 )σ 4 /2 + z 2 − u2 σ 4 /4 + v 2 + u o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||≤ε} + v 2 + z 2 − u2 + (v + ζ)2µ/σ 2 o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} 2 2 = v 2 + (v + ζ)2µ/σ 2 σ 4 /4 + z 2 − u2 σ 4 /4 + v 2 + (v + ζ)2µ/σ 2 z 2 σ 4 /2 17
− v 2 + (v + ζ)2µ/σ 2 u2 σ 4 /2 + v 2 + u uo (v) + o v 2 + o(u) 1{|u−|z||≤ε} + v 2 + |z 2 − u2 | o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} , 2 ≥ v 4 (1 − ε0 )σ 4 /4 + z 2 − u2 σ 4 /4 + v 2 z 2 (1 − ε0 )σ 4 /2 − u2 v 2 (1 + ε0 )σ 4 /2 +u2 o (v) + o u2 + uo v 3 + v 2 o (u) + o v 4 + v 2 + |z 2 − u2 | o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} , which hold for any (fixed) ε, ε0 > 0 and all large enough u, v and |z|. Next, using Lemmas 1, 2, we obtain the following inequalities: 2
(Im (ψ(v + iu) − ψ(−ζ − iz)))
2 = µ(u + z) + uvσ 2 − zζσ 2 + o(uv) + o(z) + o(z 2 − u2 ) + o (z + u)2 1{z≤0} 2 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) + o(u2 ) + o(z 2 ) 1{z>0} 2 ≥ uvσ 2 + µ(u + z) − zζσ 2 + o(vu) + o(z) + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤0} + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z>0} ≥ u2 v 2 σ 4 + 2uvσ 2 µ(u + z) − zζσ 2 + o(vu) + o(z) + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤−u−ε00 } +vo u2 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 )1{−u+ε00 ≤z≤0} + uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z|ζ−µ/σ2 |>u(v−ε00 )} , which hold for any (fixed) ε00 > 0 and all large enough u, v and |z|. In the above, we also made use of the fact that µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{−u−ε00 ≤z≤−u+ε00 } = vo u2 and uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{0≤z|ζ−µ/σ2 |≤u(v−ε00 )} ≥ 0 hold for all large enough u, v and |z|. Finally, choosing ε = ε0 = ε00 ∈ (0, 1), we collect the above, to obtain: 2
2
2
|ψ(v + iu) − ψ(−ζ − iz)| = (Re (ψ(v + iu) − ψ(−ζ − iz))) + (Im (ψ(v + iu) − ψ(−ζ − iz))) ≥ v 4 (1 − ε)σ 4 /4 + z 2 − u2
2
σ 4 /4 + v 2 z 2 (1 − ε)σ 4 /2 − u2 v 2 (1 + ε)σ 4 /2 +u2 v 2 σ 4 + 2uvσ 2 µ(u + z) − zζσ 2 + o(vu) + o(z) + vo u2 +u2 o (v) + o u2 + uo v 3 + v 2 o (u) + o v 4 18
+ v 2 + |z 2 − u2 |
o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤−u−ε} + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 )1{−u+ε≤z≤0} + uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z|ζ−µ/σ2 |>u(v−ε)} 2 ≥ v 4 (1 − ε)σ 4 /4 + z 2 − u2 σ 4 /20 + v 2 z 2 (1 − ε)σ 4 /2 + u2 v 2 (1 − ε)σ 4 /2 +2u2 vµσ 2 + 2uvzσ 2 (µ − ζσ 2 ) + o (vu)2 + uvo(z) + vo u2 +u2 o (v) + uo v 3 + v 2 o (u) + o v 4
2 σ 4 + v 2 + |z 2 − u2 | o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} 20 2 σ 4 + z 2 − u2 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤−u−ε} 20 2 σ 4 + z 2 − u2 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 )1{−u+ε≤z≤0} 20 4 σ 2 + z 2 − u2 + uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z|ζ−µ/σ2 |>u(v−ε)} 20 Let us estimate the above terms separately: + z 2 − u2
v 4 (1 − ε)σ 4 /4 + z 2 − u2
2
σ 4 /20 + v 2 z 2 (1 − ε)σ 4 /2 + u2 v 2 (1 − ε)σ 4 /2
+2u2 vµσ 2 + 2uvzσ 2 (µ − ζσ 2 ) + o (vu)2 + uvo(z) + vo u2 +u2 o (v) + uo v 3 + v 2 o (u) + o v 4 2 ≥ v 4 (1 − ε − δ)σ 4 /4 + z 2 − u2 σ 4 /20 + v 2 z 2 (1 − ε)σ 4 /2 + u2 v 2 (1 − ε − δ)σ 4 /2 +uo v 3 − c1 uv|z|, 2 σ 4 + v 2 + |z 2 − u2 | o z 2 − u2 + o(uv) + o v 2 + o(z) 1{|u−|z||>ε} 20 ! o (|z| + u) + o(uv) + o v 2 + o(z) σ4 v2 2 2 2 ≥ z −u + +1 , 20 |z| + u |z| + u
z 2 − u2
(22)
2 σ 4 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (z + u)2 1{z≤−u−ε} (23) 20 ! σ4 µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 ) + o (|z| − u)2 2 2 2 ≥ z −u + , 20 u + |z| (|z| − u)|z 2 − u2 |
z 2 − u2
2 σ 4 + µ(u + z) + uvσ 2 − zζσ 2 + o(vu) + o(z) o(z 2 − u2 )1{−u+ε≤z≤0} 20 ! 2 2 4 µ(u + z) + uvσ − zζσ + o(vu) + o(z) o(|z| + u) σ 2 + , ≥ z 2 − u2 20 |z| + u (u − |z|)2
z 2 − u2
19
(24)
z 2 − u2
2 σ 4 + uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(u2 ) + o(z 2 ) 1{z|ζ−µ/σ2 |>u(v−ε)} 20 (25) σ4 uvσ 2 + µu − zσ 2 (ζ − µ/σ 2 ) + o(vu) + o(z) o(z 2 ) 2 2 2 , ≥ z −u + 20 |z| + u z2
where we fixed arbitrary δ ∈ (0, 1 − ε) and assumed that u, v and |z| are large enough. It only remans to notice that, for any v > 0, there exist R1 > 0 and R2 > 0, such that, for all |z| > R1 and u > R2 , the right hand sides of (22)–(25) are nonnegative, and, in addition, u2 v 2 (1 − ε − δ)σ 4 /2 + uo v 3 ≥ u2 v 2 (1 − ε − δ)σ 4 /4 Thus, we conclude that, for any large enough v > 0, there exist R1 > 0, R2 > 0 and {ci > 0}, such that, for all |z| > R1 and u > R2 , the following holds 2
|ψ(v + iu) − ψ(−ζ − iz)| ≥ c2 v 4 + c3 (z 2 − u2 )2 + 2c4 v 2 z 2 + c5 u2 v 2 − c1 uv|z| √ √ √ = c2 v 4 + c3 (z 2 − u2 )2 + c4 v 2 z 2 + ( c4 v|z| − c5 uv)2 + uv|z|(2 c4 c5 v − c1 ) ≥ c6 (z 2 − u2 )2 + z 2
5
Appendix B
Proof of Theorem 2 Using the representation (18), and applying Fubini’s theorem and integration by parts, we obtain: Z 2πigr (y) = eαy ψ 0 (α)Υ(ψ(α))dα Gr
Z
Z = Gr
eαy
R
ψ (α) ˆ h(z)dzdα − ψ(α) − ψ(−ζ − iz) 0
Z
Z Gr
R
Gr
0
(eαy )
ψ (α) 1 ˆ dαh(z)dz − ψ(α) − ψ(−ζ − iz) y
eαy
R
Z Z
1 ˆ h(z)dzdα α + ζ + iz
1 ˆ dαh(z)dz α + ζ + iz Z exp((γ + ir)y)ψ 0 (γ + ir) exp((γ − ir)y)ψ 0 (γ − ir) ˆ 1 = − h(z)dz y R ψ(γ + ir) − ψ(−ζ − iz) ψ(γ − ir) − ψ(−ζ − iz) Z 1 exp((γ − ir)y) exp((γ + ir)y) ˆ + − h(z)dz y R γ + ζ − ir + iz γ + ζ + ir + iz Z Z Z Z 2 1 exp(αy)ψ 00 (α) 1 exp(αy) (ψ 0 (α)) ˆ ˆ − dαh(z)dz + dαh(z)dz y R Gr ψ(α) − ψ(−ζ − iz) y R Gr (ψ(α) − ψ(−ζ − iz))2 Z Z 1 exp(αy) ˆ − dαh(z)dz y R Gr (α + ζ + iz)2
=
1 y
Z Z
0
R
0
(eαy )
Gr
(26)
To show that the first integral in the right hand side of (26) converges to zero, we notice that, for any N > 0, there exist r0 > 0, c1 > 0 and c2 > 0, such that, for all r > r0 , we have: Z exp((γ + ir)y)ψ 0 (γ + ir) exp((γ − ir)y)ψ 0 (γ − ir) ˆ − h(z)dz (27) ψ(γ + ir) − ψ(−ζ − iz) ψ(γ − ir) − ψ(−ζ − iz) R 20
≤ c1 eγy
Z |z|≤N
r ˆ |h(z)|dz+ r2 − c2
exp((γ + ir)y)ψ 0 (γ + ir) exp((γ − ir)y)ψ 0 (γ − ir) ˆ |h(z)|dz − ψ(γ − ir) − ψ(−ζ − iz) |z|>N ψ(γ + ir) − ψ(−ζ − iz)
Z
We choose N to be large enough, so that the estimate in Lemma 5 can be applied for all |z| > N : exp((γ + ir)y)ψ 0 (γ + ir) exp((γ − ir)y)ψ 0 (γ − ir) r γy ψ(γ + ir) − ψ(−ζ − iz) − ψ(γ − ir) − ψ(−ζ − iz) ≤ c3 e p(z 2 − r2 )2 + z 2 It is easy to see that the first term in the right hand side of (27) converges to zero, as r → ∞. In addition, (z 2 − r2 )2 + z 2 = (z 2 − r2 + 1/2)2 + r2 − 1/4 ≥ r2 − 1/4 hence, we can apply the dominated convergence theorem to show that the second term in the right hand side of (27) vanishes, as r → ∞. Thus, we conclude that the first integral in the right hand side of (26) converges to zero, as r → ∞. Similarly, the second integral in the right hand side of (26) vanishes, as r → ∞, due to the dominated convergence theorem. As r → ∞, the fourth, fifth and sixth integrals in the right hand side of (26) converge, uniformly over y changing on any compact in (0, ∞). Let us prove convergence of the fourth integral. Consider arbitrary (large) r0 > r and proceed as follows: Z Z Z Z exp(αy)ψ 00 (α) |ψ 00 (γ + iu)| ˆ ˆ dαh(z)dz du|h(z)|dz ≤ 2eγy R Gr0 \Gr ψ(α) − ψ(−ζ − iz) R |u|∈[r,r 0 ] |ψ(γ + iu) − ψ(−ζ − iz)| Z
N
Z
r0
1 du 2−c u 5 −N r Z Z r0 1 γy ˆ +c4 e ψ(γ + iu) − ψ(−ζ − iz) du|h(z)|dz, |z|>N r ≤ c4 eγy
ˆ |h(z)|dz
(28)
where, again, we choose N > 0 to be such that the estimate in Lemma 5 can be applied for all |z| > N , and r, r0 are assumed to be large enough. It is easy to see that the first term in the right hand side of (28) vanishes, as r, r0 → ∞. To show that the second term in the right hand side of (28) vanishes, as r, r0 → ∞, we make use of Lemma 5: Z Z r0 Z Z r0 1 1 ˆ ˆ du|h(z)|dz p ≤ c du|h(z)|dz, 5 ψ(γ + iu) − ψ(−ζ − iz) 2 (u − z 2 )2 + z 2 |z|>N r |z|>N r and apply the dominated convergence theorem: Z 0 Z r 1 1 p √ du ≤ du < ∞, 4 2 2 2 2 r u +1 (u − z ) + z R where we assumed that N > 1. Next, we use Lemma 5 to prove convergence of the fifth integral in the right hand side of (26): Z |z|>N
Z r
r0
2
| exp((γ + iu)y)| |ψ 0 (γ + iu)| |ψ(γ + iu) − ψ(−ζ − iz)|
21
2
ˆ du|h(z)|dz
≤ c12 e
γy
Z
Z
z>N
r0
u2 ˆ du|h(z)|dz (u2 − z 2 )2 + z 2
r
The right hand side of the above converges to zero, as r, r0 → ∞, due to the dominated convergence theorem: Z 0 Z Z r z+1 u2 u2 u2 ≤ du du + du 2 2 2 2 2 2 2 2 2 2 2 2 r (u − z ) + z z−1 (u − z ) + z u∈[0,z−1]∪[z+1,∞) (u − z ) + z Z
1+1/z
≤ 1−1/z
zu2 du + 2 2 z (u − 1)2 + 1
Z u∈[0,z−1]∪[z+1,∞)
Z
2
≤ 2 (1 + 1/N ) du + |u|>1
u2 du (u − z)2 (u + z)2
1 du < ∞ u2
To show that the last integral in the right hand side of (26) converges, as r → ∞, we notice that Z Z Z Z r0 exp(αy) 1 γy ˆ ˆ dαh(z)dz ≤ e du|h(z)|dz, 2 + (z + u)2 R Gr0 \Gr (α + ζ + iz)2 (γ + ζ) R r and apply the dominated convergence theorem once more: Z r
r0
1 du ≤ (γ + ζ)2 + (z + u)2
Z R
1 du < ∞. γ 2 + u2
Thus, we have shown that, for y > 0, gr (y) converges to g(y), as r → ∞. To see that the limiting function, g, is continuous, we notice that the estimates, derived above, yield that the convergence of the fourth, fifth and sixth terms in the right hand side of (26) is uniform over y changing on a compact in (0, ∞). The above estimates also imply that |gr (y)| is bounded by a constant times exp(γy), uniformly over r. Then, Proposition 2, along with the dominated convergence, yield: Z ∞ Z ∞ e−λy g(y)dy = lim e−λy gr (y)dy = ψ 0 (λ)Υ(ψ(λ)), 0
r→∞
0
for all large enough real λ > 0. As shown in Subsection 2.2, a change of variables turns the left and right hand sides of the above into the Laplace transforms of Eg(Xt ) and Eh(Xt ), respectively. Due to the uniqueness of Laplace transform, the expectations have to coincide for all t > 0. This completes the proof of the theorem.
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