UvA-DARE is a service provided by the library of the University of Amsterdam (https://dare.uva.nl)
Affine Markov processes on a general state space
Veerman, E.
Publication date
2011
Link to publication
Citation for published version (APA):
Veerman, E. (2011). Affine Markov processes on a general state space. Uitgeverij BOXPress.
General rights
It is not permitted to download or to forward/distribute the text or part of it without the consent of the author(s) and/or copyright holder(s), other than for strictly personal, individual use, unless the work is under an open content license (like Creative Commons).
Disclaimer/Complaints regulations
If you believe that digital publication of certain material infringes any of your rights or (privacy) interests, please let the Library know, stating your reasons. In case of a legitimate complaint, the Library will make the material inaccessible and/or remove it from the website. Please Ask the Library: https://uba.uva.nl/en/contact, or a letter to: Library of the University of Amsterdam, Secretariat, Singel 425, 1012 WP Amsterdam, The Netherlands. You will be contacted as soon as possible.
Chapter
4
Affine processes
After having developed a general framework for jump-diffusions and Markov pro-cesses in the first three chapters, we proceed with affine jump-diffusions and affine processes in the remaining chapters. Affine processes are Markov processes where the Fourier transform of the transition functions depend in an “exponential affine” manner on the initial state. It turns out that the coefficients in this affine expres-sion solve a system of ODEs, so-called generalized Riccati equations. A complete theory has been developed for affine processes on the state space Rm+× Rp−mand
S+p, see [17, 10]. Here we will consider general state spaces of the form X × Rp−m.
Similar as in [17, 10], we show that an affine process on an arbitrary state space is a Feller process and characterize it as the solution of a martingale problem with affine parameters, called affine jump-diffusion. The main theorem is Theorem 4.4, which extends [17, Theorem 2.7] and [10, Theorem 2.4] from affine processes on the particular state spaces Rm+× R
p−m and Sp
+ to affine processes on an arbitrary
state space of the form X × Rp−m. The required parameter restrictions, called
admissibility conditions, are given in a general form, equivalent with the positive maximum principle as discussed in Section 2.4. These conditions are explicitly worked out for a polyhedral and quadratic state space in the next chapter.
After having given the formal definition and having presented the main result in Section 4.1, we prove the main theorem in Sections 4.2, 4.3 and 4.4. For a large part we follow the techniques and ideas from [17, 10], but contrary to these papers, we do not analyze the solutions to the generalized Riccati equations directly with the use of the admissibility conditions. Rather we apply probabilistic methods, using the general results from the previous chapters, in particular Propositions 2.7 and 2.9.
In Section 4.5 we provide conditions under which the Fourier-Laplace transform of an affine process with a general state space does not vanish. Results for this in the literature are only known for affine process on the canonical state space, as treated in [17]. We note that the theory for affine matrix-valued processes as developed in [10], uses the Laplace transform instead of the Fourier-Laplace transform, that is, only real-valued parameters are considered. From Proposition 4.33 together with Theorem 6.5 in Chapter 6, we infer that many affine processes, including continuous diffusions, have a non-vanishing Fourier-Laplace transform, irrespective of the underlying state space. In particular our results apply to matrix-valued affine diffusions.
4.1
Definition and characterization
For the definition of an affine process we follow [38] rather than [17, 10].
Definition 4.1. A stochastically continuous Markov process (X, (Px)x∈E) is called
an affine process with state space E if for all t ≥ 0, u ∈ iRpthere exist Φ(t, u) ∈ C, ψ(t, u) ∈ Cpsuch that
Ptfu(x) = Φ(t, u) exp(ψ(t, u)>x), for all x ∈ E. (4.1)
Throughout we let χ : Rp → Rp be a continuous truncation function with the
property that the coordinates χi(z) only depends on zi, for z ∈ Rp. For example,
χi(z) = 0 if zi= 0 (1 ∧ |zi|) zi |zi| if zi6= 0. (4.2)
Definition 4.2. The martingale problem for A given by (2.2) with Ω given by (2.8) is called an affine martingale problem if b, c, γ and K are affine in the sense that b(x) = a0+ p X i=1 aixi c(x) = A0+ p X i=1 Aixi γ(x) = γ0+ p X i=1 γixi (|z|2∧ 1)K(x, dz) = K0(dz) + p X i=1 Ki(dz)xi, (4.3)
4.1. Definition and characterization 71
for some ai
∈ Rp, Ai ∈ Sp, γi
∈ R and finite signed measures Ki
on Rp\{0}. If
for all x ∈ E, Px is the unique solution for (A, δx) in Ω, then (X, (Px)x∈E) (with
X the coordinate process) is called an affine jump-diffusion with state space E.
The equivalence between affine processes and affine jump-diffusions has been proved in [17] resp. [10] for the state space Rm× Rp−m resp. Sp
+, the cone of
positive semi-definite matrices. Our aim is to extend these results to general state spaces of the form X × Rp−m.
We point out that restrictions should be imposed on the parameters as well as on the state space for the existence of an affine jump-diffusion. First, the right-hand side of the last three equalities in (4.3) has to be positive for all x ∈ E. Second, the positive maximum principle as discussed in Section 2.4 yields necessary conditions for (b, c, γ, K), given in Proposition 2.13. This motivates the following definition.
Definition 4.3. For i = 0, . . . , p let us be given ai∈ Rp, Ai∈ Sp, γi∈ R and finite
signed measures Ki on Rp\{0}. We call (ai, Ai, γi, Ki) an admissible parameter
set if b, c, γ, K given by (4.3) satisfy the following properties, called admissibility conditions.
(i) For x ∈ E we have c(x) ∈ S+p, γ(x) ∈ R+ and K(x, dz) is a positive measure
with supp K(x, dz) ⊂ E − x.
(ii) For f ∈ Cc2(E), x ∈ E with f (x) = supy∈Ef (y) ≥ 0 it holds that
(a) ∇f (x)>c(x) = 0,
(b) R ∇f (x)>χ(z)K(x, dz) is well-defined and finite,
(c) ∇f (x)>b(x) −R ∇f (x)>χ(z)K(x, dz) + 12tr (∇2f (x)c(x)) ≤ 0.
As in Chapter 3, we assume throughout that the state space E is of the form X × Rp−m, with X a closed convex set as in Section 3.2. In addition, we assume
that U as given by (3.3) has a smooth boundary, in the sense that there exists an analytic function η : Rm→ R such that for all u ∈ Re U it holds that
η(u) = 0 if and only if u 6∈ Re U◦. (4.4)
Examples are given below. It is worth pointing out that convexity of the state space is needed for applying the results from the appendix in Proposition 4.13 to deduce the exponential affine expression for the Fourier-Laplace transform.
We now state the main result, which establishes equivalence between regular affine processes and affine jump-diffusions. The theorem below extends [17, The-orem 2.7] and [10, TheThe-orem 2.4] from particular state spaces Rm
+ × Rp−m and S p +
Theorem 4.4. Let E be given as above. Then the following holds:
(i) Let (X, (Px)x∈E) be a regular affine process with state space E. Then X
is a Feller process. The domain of the infinitesimal generator A satisfies Cc2(E) ⊂ D(A) and for f ∈ Cc2(E), Af takes the form (2.2), where b, c, γ
and K are of the affine form given by (4.3) for some admissible parameters (ai, Ai, γi, Ki). In particular, Px solves the affine martingale problem for
(A, δx), when (X, (Px)x∈E) is in canonical form.
(ii) Conversely, let (ai, Ai, γi, Ki) be an admissible parameter set and let A be
the linear operator given by (2.2) with b, c, γ, K of the form (4.3). Then the corresponding affine martingale problem is well-posed and its solution (X, (Px)x∈E) is the unique regular affine Feller process whose generator
ex-tends A.
In either case, for u ∈ U × iRp−m there exists t0(u) ∈ (0, ∞] such that
Ptfu(x) =
exp(ψ0(t, u) + ψ(t, u)>x), for t ∈ [0, t0(u)),
0, for t ∈ [t0(u), ∞),
(4.5)
where (ψ0(·, u), ψ(·, u)) : [0, t0(u)) → C×Cpsolves the system of generalized Riccati
equations given by ∂tψi(t, u) = Ri(ψ(t, u)), ψi(0, u) = ui, i = 0, . . . , p, (4.6) with u0:= 0 and Ri(y) = y>ai+12y>Aiy − γi+ Z ey>z− 1 − y>χ(z) |z|2∧ 1 K i(dz), (4.7)
which is well-defined for y ∈ U × iRp−m.
The proof of this theorem is spread over the next sections. In Proposition 4.19 we prove (i), while Proposition 4.31 proves (ii). The last assertion is shown in Proposition 4.18. It is worth noting that if (ψ0(·, u), ψ(·, u)) : [0, T ] → C × Cp
solves the above system of generalized Riccati equations, then the function v ∈ C1,2([0, T ] × E) defined by
v(t, x) = exp(ψ0(T − t, u) + ψ(T − t, u)>x),
solves Kolmogorov’s backward equation
∂tv + Av = 0, v(T, ·) = fu,
as one easily verifies. Hence, if in addition v ∈ C0([0, T ] × E), then Proposition 2.9
4.2. Preliminary results 73
Remark 4.5. There are no examples known of affine processes where the Fourier-Laplace transform Ptfuvanishes for some t ≥ 0, u ∈ U ×iRp−m, equivalently, where
the solutions to the generalized Riccati equations explode in finite time. In fact, in Section 4.5 we show that the explosion-time is infinite in most cases. It is however still an open question whether this is the case in general.
Remark 4.6. The regularity condition in Theorem 4.4 can probably be dispensed with. For the state space Rm
+ × Rp−m it is shown in [38] that regularity is a
consequence of stochastic continuity together with the exponential affine form of the Fourier transform. Probably their methods also work for the more general state spaces we consider.
Examples of sets X ⊂ Rmfor which we can apply Theorem 4.4 are the following:
• X = Rm
+, the cone of non-negative numbers, with Re U = −X and η(u) =
u1· u2· . . . · um,
• X ∼= S+q (with m = q(q + 1)/2), the cone of positive semi-definite matrices,
with Re U = −X and η(u) = det(vech−1(u)). Here, we identify symmetric matrices with vectors by the half-vectorization operator vech, which is the linear operator that stacks the elements of the upper triangle of a symmetric matrix into a column vector.
• X = {x ∈ Rm : x
1 ≥ |(x2, . . . , xm)|}, the Lorentz cone, with Re U = −X
and η(u) = u2 1− Pm i=1u 2 i, • X = {x ∈ Rm : x 1 ≥ P m i=1x 2
i}, the parabolic state space, with Re U =
R−−× Rm−1∪ {0} and η(u) = u1.
Note that the first three examples are all self-dual cones, that is,
X = {x ∈ Rm: y>x ≥ 0, for all y ∈ X },
contrary to the parabolic state space, which is not even a cone. In the next chapter we provide explicit admissibility conditions for the polyhedral state space (which includes Rm
+× Rp−m), the Lorentz cone and the parabolic state space. We refer to
[10] for the form of the admissible parameters for matrix-valued affine processes.
4.2
Preliminary results
In this section we obtain some preliminary results which we apply both in Sec-tion 4.3 as well as in SecSec-tion 4.4 to prove Theorem 4.4. First we characterize the support of an affine transition kernel K(x, dz).
Lemma 4.7. For x ∈ E, let K(x, dz) be a finite positive measure on Rp given
by K(x, dz) = K0(dz) +Pp
i=1Ki(dz)xi, for some finite signed measures Ki on
Rp. Then it holds that supp K(x, ·) ⊂ E − x for all x ∈ E if and only if there exists a closed convex cone C ⊂ E such that supp K(x, ·) ⊂ C for all x ∈ E and E + C ⊂ E.
Proof. The “if”-part is trivial. To prove the “only if”-part, we may assume without loss of generality that 0 ∈ E◦. Then it holds that K0(dz) = K(0, dz), whence supp K0⊂ E by assumption. For x ∈ E we have supp K(x, ·) ⊂ supp K0. Indeed,
let O ⊂ Rp be an open set such that K0(O) = 0. Since 0 ∈ E◦, there exists ε > 0 such that −εx ∈ E. This gives K(−εx, O) = −εK(x, O) ≥ 0, so necessarily K(x, O) = 0. For x ∈ E◦ we have also supp K(x, ·) ⊃ supp K0. To see this, let O ⊂ Rp be an open set such that K(x, O) = 0 for some x ∈ E◦. Then we have
K(x, O) = K0(O) +Pp
i=1K i(O)x
i = 0. If K0(O) > 0, then there is an i such
that Ki(O)x
i < 0. Since x ∈ E◦, there exists ε 6= 0 such that y := x + εei ∈ E
and Ki(O)ε < 0. This yields K(y, O) = K(x, O) + Ki(O)ε < 0, a contradiction. It follows that for x ∈ E◦ we have x + supp K0 = x + supp K(x, ·) ⊂ E, i.e.
E◦+ supp K0⊂ E. Since x + E◦ is open for all x ∈ supp K0, we even have
E◦+ supp K0⊂ E◦.
Iterating the above relation yields E◦+ n supp K0 ⊂ E◦
for all n ∈ N, where n supp K0denotes the n-fold iterated sumset. Taking closures gives E+n supp K0⊂
E. The assertion follows if we let C be the closure of
( k X i=1 riyi: ri≥ 0, yi∈ supp K0, k ∈ N ) .
The previous lemma enables us to derive that A(Cc2(E)) ⊂ C0(E), in case the
parameters of A given by (2.2) are of the affine form.
Proposition 4.8. Let A be the linear operator given by (2.2) with b, c, γ, K of the affine form (4.3). Then it holds that A(C2
c(E)) ⊂ C0(E).
Proof. Continuity follows from Lemma 2.12, since the continuity assumptions in (2.17) are fulfilled in view of the affine form of the parameters. It remains to show that Af vanishes at infinity for f ∈ C2
c(E), which comes down to showing that
x 7→ Z
4.2. Preliminary results 75
vanishes at infinity for f ∈ C2
c(E), since this is obvious for the other terms. By
Lemma 4.7 it holds that supp K(x, dz) ⊂ C for all x ∈ E, for some closed convex cone C that satisfies E + C ⊂ E. Recalling E = X × Rp−m, we may assume
C = CI × Rp−m, for some closed convex cone CI ⊂ Rmsuch that X + CI ⊂ CI.
We first show that for all M > 0 there exists N > 0 such that |x + z| ≥ M for all |x| ≥ N with x ∈ X and z ∈ CI. Arguing by contradiction, suppose there
exists M > 0 such that for all n ∈ N there is xn∈ X , zn ∈ CI, with |xn| ≥ n and
|xn+ zn| < M . Since X + CI ⊂ X , we have xn+ 2zn ∈ X . Now take u ∈ Re U◦.
Then u>x
n→ −∞, in view of Lemma 3.5. But this gives
u>(xn+ 2zn) = −u>xn+ 2u>(xn+ zn) → ∞,
as |xn+ zn| < M . This contradicts the fact that supx∈Xu>x < ∞.
We are now ready to show that the expression g(x) in (4.8) vanishes at infinity for f ∈ C2
c(E). Let f ∈ Cc2(E) be arbitrary and suppose f (x) = 0 for all |x| ≥ M ,
for some M > 0. By the previous there exists N > 0 such that |xI+zI| ≥ M for all
|xI| ≥ N with x ∈ E and z ∈ C. Hence g(x) = 0 for all x ∈ E with |xI| ≥ M ∨ N .
Let |x| → ∞. Then clearly g(x) → 0 in case |xI| → ∞. It remains to show that
g(x) → 0 in case |xJ| → ∞.
Since K(x, dz) is a positive measure for all x ∈ E and xJ is unrestricted, we
have Ki= 0 for all i ∈ J . Hence we can write
g(x) = m X i=0 Z f (x + z) − f (x) − ∇f (x)>χ(z) |z|2∧ 1 K i(dz)x i,
where we write x0= 1 for convenience. Let |xJ| ≥ M + 1. If |x + z| < M , then
|z| ≥ |x| − |x + z| ≥ |xJ| − M ≥ 1. Since g(x) = 0 for |xI| ≥ N , we infer that
|g(x)| ≤ N kf k∞ m
X
i=0
|Ki|(z : |z| ≥ |xJ| − M ).
This tends to zero if |xJ| → ∞, as |Ki| is a finite measure.
We recall [17, Lemma A.2] on analyticity of the Fourier-Laplace transform and give two important corollaries.
Lemma 4.9. Let µ be a finite measure on Rp and suppose there exists an open set V ⊂ Rp such that R f
u(x)µ(dx) < ∞ for all u ∈ V . Then
u 7→ Z
fu(x)µ(dx)
Corollary 4.10. Let µ be a finite measure on E. Fix z ∈ Rp−m. Then
y 7→ Z
fy,iz(x)µ(dx)
is analytic on U◦.
Corollary 4.11. Let K be a finite signed measure on E. Fix z ∈ Rp−m. Then
y 7→
Z e(y,iz)>ζ − 1 − (y, iz)>χ(ζ)
|ζ|2∧ 1 K(dζ)
is analytic on U◦.
A set V ⊂ Cpis called a set of uniqueness if every analytic function on a domain (i.e. open and connected set) W ⊃ V is uniquely determined by its restriction to V , in other words, if the zero-function is the only analytic function that vanishes on V . Examples of sets of uniqueness are the open discs in Rp, see for instance [8,
Theorem IV.3.7]. We employ this frequently in what follows.
The next propositions are crucial in our analysis of affine processes and the affine martingale problem. We use analyticity together with the exponential form of solutions to Cauchy’s functional equation (A.1), as discussed in the appendix, to extend the exponential affine expression of a Fourier-Laplace transform.
Proposition 4.12. Suppose 0 ∈ X◦. Let (µx)x∈X be a collection of finite
mea-sures on X with µ0(X ) > 0 and define the Fourier-Laplace transforms µbx(u) =
R fu(z)µx(dz) for u ∈ U . Suppose there exists an open set B ⊂ Re U◦ on which
the µbx take the form
b
µx(u) = Φ(u) exp(ψ(u)>x), u ∈ B.
for some Φ(u) > 0 and ψ(u) ∈ Re U . Then Φ(u) and ψ(u) can be extended to u ∈ Re U such that the equality in the above display holds for all u ∈ Re U . In addition, if ψ(u) ∈ Re U◦ for some u ∈ B, then ψ(u) ∈ Re U◦ for all u ∈ Re U◦.
Proof. For u ∈ B we have the equality
b
µx(u)µby(u) =µb0(u)µbx+y(u),
for all x, y ∈ X such that x + y ∈ X . Since both sides are analytic on U◦ by Corol-lary 4.10, the equality extends to U◦ as B is a set of uniqueness. By Lemma 3.10 we have equality on the whole of U . Note that µb0(u) 6= 0 for all u ∈ Re U , as
µ0(X ) > 0. Fix u ∈ Re U and define the function g : X → R+ by
4.2. Preliminary results 77
Then g satisfies Cauchy’s functional equation (A.1) on X . Since it is bounded and non-negative, g is of the form g(x) = eλ>x
for some λ ∈ Rm, in view of Lemma A.2.
From this we infer that Φ and ψ can be extended to the whole of Re U such that
b
µx(u) = Φ(u) exp(ψ(u)>x),
with Φ(u) > 0, for u ∈ Re U .
It remains to show that either ψ(u) ∈ Re U◦for all u ∈ Re U◦ or ψ(u) 6∈ Re U◦ for all u ∈ Re U◦. Suppose ψ(u0) 6∈ Re U◦for some u0∈ Re U◦. By Lemma 3.5 this
is equivalent with fψ(u0)6∈ C0(X ). Take ε > 0 such that B(u0, ε) ⊂ Re U . It holds
that B(0, ε) ∩ Re U◦ is open and non-empty, by Corollary 3.6. Let v ∈ B(0, ε) ∩ Re U◦. Then it holds that u0− v ∈ B(u0, ε) ⊂ Re U , whence C := kfu0−vk∞< ∞.
It follows that
fu0(x) = fu0−v(x)fv(x) ≤ Cfv(x), for all x ∈ X .
Hence µbx(u0) ≤ Cµbx(v) for all x ∈ X . Since µbx(u0) = Φ(u0)fψ(u0)(x) 6∈ C0(X )
by assumption, we also have µbx(v) = Φ(v)fψ(v)(x) 6∈ C0(X ). Lemma 3.5 yields
that ψ(v) 6∈ Re U◦. Recall the assumption that there exists an analytic function η : Rm
→ R such that for all u ∈ Re U we have (4.4). This yields that η(ψ(v)) = 0 for all v ∈ B(0, ε) ∩ Re U◦. Note that u 7→ ψ(u) is analytic on Re U◦, in view of Corollary 4.10, whence u 7→ η(ψ(u)) is analytic on Re U◦. Since B(0, ε) ∩ Re U◦is
a set of uniqueness, it follows that η(ψ(u)) = 0 for all u ∈ Re U◦. In other words, ψ(u) 6∈ Re U◦ for all u ∈ Re U◦, as we needed to show.
The next proposition enables us to extend the exponential affine form of the Fourier-Laplace transform from U◦× iRp−m
to U × iRp−m.
Proposition 4.13. Let (X, (Px)x∈E) be a Markov process that is stochastically
right-continuous at 0 (i.e. pt(x, dz) w
→ δx(dz) as t ↓ 0). Assume that for all
u ∈ U◦× iRp−m it holds that x 7→ P
tfu(x) is continuous and
Ptfu(x + z)Ptfu(y + z) = Ptfu(x + y + z)Ptfu(z), (4.9)
for all t ≥ 0, for all z ∈ E and all x, y ∈ E − z such that x + y ∈ E − z. Then Ptfu is of the form (4.1) for all u ∈ U × iRp−m. In addition, Ptfu(x) > 0 for all
u ∈ Re U × {0}, x ∈ E, t ≥ 0.
Proof. Note that by Lemma 3.10, we can extend the equality in (4.9) to the whole of U × iRp−m. We first prove that P
tfuis of the form (4.1) for all u ∈ Re U × {0}.
Ptfu(x) = 0 for real-valued u if and only if Xt= ∆, Px-a.s. Therefore, if Ptfu(x) =
0 for some u ∈ Re U × {0}, then it is zero for all u ∈ Re U × {0}. Fix z ∈ E◦. Since X is stochastically right-continuous at 0, it holds that Ptfu(z) > 0 for all
u ∈ Re U × {0}, for t small enough. Define
T = inf{t ≥ 0 : Ptfu(z) = 0, for some u ∈ Re U × {0}}.
For t < T and u ∈ Re U × {0} it holds that gt,u : E − z → R : x 7→ Ptfu(x +
z)/Ptfu(z) is a non-negative, bounded function that satisfies Cauchy’s functional
equation (A.1) on E − z, in view of (4.9). Since 0 ∈ (E − z)◦, we can apply
Lemma A.2 to deduce that gt,u(x) is of the form gt,u = fψ(t,u) for some ψ(t, u) ∈
Rp. This gives that Ptfu = Φ(t, u)fψ(t,u), for some Φ(t, u) > 0, in particular
Ptfu(x) > 0 for all x ∈ E, t < T . Note that also ψ(t, u) ∈ Re U × {0}, as
x 7→ Ptfu(x) is bounded. By the Markov property we deduce that
Pt+sfu(z) = PsPtfu(z) = Φ(t, u)Psfψ(t,u)(z) > 0,
for s, t < T . This implies that T = ∞. Hence Ptfu is of the form (4.1) for all
t ≥ 0, u ∈ Re U × {0}, with Φ(t, u) > 0. This also proves the second assertion of the proposition that Ptfu(x) > 0.
Next we show that x 7→ Ptfu(x) is continuous for all u ∈ U × iRp−m, t ≥ 0.
Let xn → x, with xn, x ∈ E and let t ≥ 0 be arbitrary. Define measures µn(dz) =
pt(xn, dz), µ(dz) = pt(x, dz). We have to show that µn w → µ. By assumption we have Z fu(z)µn(dz) = Ptfu(xn) → Ptfu(x) = Z fu(z)µ(dz), for all u ∈ U◦× iRp−m.
It holds that u 7→ Ptfu(x) is continuous on U◦× iRp−m by Lemma 3.7 and
µn(E) = Ptf0(xn) = Φ(t, 0)fψ(t,0)(xn) → Φ(t, 0)fψ(t,0)(x) = Ptf0(x) = µ(E),
in view of the previous paragraph. From Lemma 3.13 we infer that µn w
→ µ. Now let t ≥ 0 and u ∈ U × iRp−m be arbitrary. If for all x ∈ E◦ it holds that Ptfu(x) = 0, then by continuity in x and the fact that E = E◦ we have
Ptfu(x) = 0 for all x ∈ E. Hence Ptfu(x) is of the form (4.1) with Φ(t, u) = 0
and ψ(t, u) arbitrary. Now suppose there exists z ∈ E◦ such that Ptfu(z) 6= 0.
Then it holds that f : E − z → C : x 7→ Ptfu(x + z)/Ptfu(z) is a continuous
function that satisfies Cauchy’s functional equation (A.1) on E − z, in view of (4.9). Since 0 ∈ (E − z)◦, we can apply Lemma A.2 to deduce that f (x) is of the form f (x) = eλ>x
for some λ ∈ Cp. It follows that P
tfu(x) is of the form (4.1),
4.3. The infinitesimal generator 79
4.3
The infinitesimal generator
In this section we show that a regular affine process is a Feller process and charac-terize its infinitesimal generator by applying Theorem 3.20. As a first step we show that the symbol of a regular affine process has a L´evy-Khintchine representation (3.11) with affine coefficients (b, c, γ, K).
Proposition 4.14. Let (X, (Px)x∈E) be a regular affine process with state space
E. Then the symbol Ψ is of the form (3.11) for affine b, c, γ and K.
Proof. Since (X, (Px)x∈E) is a regular affine process, we have
Ψ(u, x) = ∂t+ t=0Φ(t, u) + ∂t+
t=0ψ(t, u)>x, u ∈ U◦× iRp−m, x ∈ E, so in particular x 7→ Ψ(u, x) is affine for all x ∈ E. Proposition 3.14 yields
Ψ(u, x) = u>b(x) +12u>c(x)u − γ(x) + Z
(eu>z− 1 − u>χ(z))K(x, dz), for x ∈ E, u ∈ U◦× iRp−m
, for certain b(x) ∈ Rp, c(x) ∈ Sp+, γ(x) ∈ R+, K(x, dz)
a transition kernel from E to Rp\{0} with supp K(x, dz) ⊂ E − x and satisfying
(3.10). By the proof of Lemma 3.10 we can take limits to deduce that
iy>b(x) −12y>c(x)y − γ(x) + Z
(eiy>z− 1 − iy>χ(z))K(x, dz)
is affine for all y ∈ Rp. Putting y = 0 gives that γ(x) is affine. Dividing the
left- and right-hand side by yi, putting yl = 0 for l 6= i and letting yi tend to
zero gives that bi(x) is affine for all i. Dividing the left- and right-hand side by
yiyj for i, j ≤ p, putting yl = 0 for l 6= i, j, yi = yj = m and letting m → ∞,
we deduce that cij(x) is affine. Hence γ(x), b(x) and c(x) are affine, so also
R (eiy>z− 1 − iy>
χ(z))K(x, dz) is affine in x for all y ∈ Rp. To show that K(x, dz)
is affine in x, we fix k ∈ E◦arbitrary and take ε > 0 such that
{x ∈ Rp: k
i≤ xi≤ ki+ ε for all i} ⊂ E.
Write eK(x, dz) = (|z|2∧ 1)K(x, dz) (which is a finite measure for all x) and define
finite signed measures Ki on Rp\{0} by
K0(dz) = eK(k, dz) − p X i=1 ( eK(k + εei, dz) − eK(k, dz))ki/ε Ki(dz) = ( eK(k + εei, dz) − eK(k, dz))/ε, for i = 1, . . . , p.
Then it follows that Z eu>z− 1 − u>χ(z) |z|2∧ 1 K(x, dz) =e Z eu>z− 1 − u>χ(z) |z|2∧ 1 (K 0(dz) + p X i=1 Ki(dz)xi),
for all u ∈ iRp, x ∈ E, since the left-hand side is affine and is uniquely determined
by the values at x = k and x = k + εei, i = 1, . . . , p. Equality of the left- and
right-hand side at these points follows from the identity
K0(dz) + p X i=1 Ki(dz)xi= eK(k, dz)(1 + p X i=1 (ki− xi)/ε) + p X i=1 e K(k + εei, dz)(xi− ki)/ε.
Note that the right-hand side is a non-negative measure for x ∈ Bk, where Bk is
given by
Bk:= {x ∈ Rp: ki≤ xi≤ ki+ ε/p for all i}.
By uniqueness of the L´evy triplet (see [33, II.2.44]), this yields that eK(x, dz) = K0(dz) +Pp
i=1K i(dz)x
i for x ∈ Bk. Since k ∈ E◦ is chosen arbitrarily, we have
an affine expression for eK(x, dz) on a neighborhood of each x ∈ E◦. From this
it follows that eK(x, dz) is affine on the whole of E = E◦. Hence K(x, dz) =
(|z|2∧ 1)−1
e
K(x, dz) is affine.
Next we establish the exponential affine form of Ptfu(x) on the whole of U ×
iRp−m.
Proposition 4.15. Let (X, (Px)x∈E) be an affine process with state space E. Then
Ptfu(x) is of the form (4.1), for all u ∈ U × iRp−m, t ≥ 0.
Proof. By assumption, Ptfu(x) is of the form (4.1) for all u ∈ iRp. Hence for all
u ∈ iRp, (4.9) holds for all z ∈ E and all x, y ∈ E − z such that x + y ∈ E − z. First we extend this equality to U × iRp−m. Fix z ∈ E, x, y ∈ E − z such that
x + y ∈ E − z and let w ∈ Rp−m be arbitrary. Define
g(v) = Ptfv,iw(x + z)Ptfv,iw(y + z) − Ptfv,iw(x + y + z)Ptfv,iw(z), v ∈ U.
Next let v0∈ Re U◦ be arbitrary. Note that for q in the horizontal strip R + [0, i]
we have −iqv0 ∈ U , while for q ∈ R + (0, i] we have −iqv0 ∈ U◦. Indeed, write
q = a + ib, with a ∈ R, b ∈ [0, 1]. Then −iqv0= bv0− iav0, which is clearly in U .
Since v0∈ U◦, it follows from Corollary 3.6 that −iqv0 ∈ U◦ if b ∈ (0, 1]. Hence,
4.3. The infinitesimal generator 81
R + (0, i) and continuous on R + [0, i] in view of Lemma 4.9. In addition, it is real-valued on R (indeed, it is zero on R). By the Schwarz reflection principle [8, IX 1.1] we therefore deduce that h(q) = 0 for all q ∈ R + [0, i], in particular for q = i, so that g(v0) = 0. Since v0∈ Re U◦was chosen arbitrarily, we have g(v) = 0
for all v ∈ Re U◦. Lemma 4.9 yields that g(v) = 0 for all v ∈ U , as Re U◦ is a set
of uniqueness. Hence (4.9) holds for all u ∈ U × iRp−m.
Next we show that x 7→ Ptfu(x) is continuous for all t ≥ 0, u ∈ U × iRp−m.
Fix t ≥ 0 and let xn → x0, with xn, x0 ∈ E. Define µn = Pxn ◦ X
−1
t and
µ = Px0 ◦ X
−1
t . Since Ptfu(x) is of the form (4.1) for u ∈ iRp, we have in
particular that x 7→ Ptfu(x) is continuous for u ∈ iRp. This yields that µn w
→ µ. Hence R f (z)µn(dz) → R f (z)µ(dz) for all f ∈ Cb(E). Choosing f = fu with
u ∈ U × iRp−m, we infer that x 7→ P
tfu(x) is continuous for all u ∈ U × iRp−m.
Now the result follows from Proposition 4.13.
Let (X, (Px)x∈E) be an affine process. In view of the previous proposition,
we can extend Φ(t, u) and ψ(t, u) to u ∈ U × iRp−m such that (4.1) holds for all
u ∈ U × iRp−m. We define the set O by
O = {(t, u) ∈ R+× U × iRp−m: Φ(t, u) 6= 0},
and we write O◦= O ∩ ([0, ∞) × U◦× iRp−m) and
t0(u) = inf{t ≥ 0 : (t, u) 6∈ O},
for u ∈ U × iRp−m. Note that t
0(u) > 0 for all u by stochastic continuity and
the fact that P0fu = fu, while t0(u) = ∞ for all u ∈ Re U × {0} in view of
Proposition 4.13. In addition, ψ(t, u) ∈ U × iRp−m for all (t, u) ∈ O, since x 7→
Ptfu(x) is bounded. The semigroup property of Ptyields the following result, cf.
[38, Proposition 3.4] and [10, Lemma 3.3]
Proposition 4.16. Let (X, (Px)x∈E) be an affine process and let Φ(t, u) and
ψ(t, u) be such that (4.1) holds for all u ∈ U × iRp−m. The following
proper-ties hold:
(i) If t, s ≥ 0 and (t + s, u) ∈ O for some u ∈ U × iRp−m, then also (t, u) ∈ O and (s, ψ(t, u)) ∈ O, and it holds that
Φ(t + s, u) = Φ(t, u)Φ(s, ψ(t, u)), ψ(t + s, u) = ψ(s, ψ(t, u)).
Proof. The first assertion follows from the semigroup property of Pt, as it yields
Φ(t + s, u)fψ(t,u)= Pt+sfu= PsPtfu= Φ(t, u)Psfψ(t,u)
= Φ(t, u)Φ(s, ψ(t, u))fψ(s,ψ(t,u)).
It remains to show (ii). Arguing by contraction, assume ψ(t, u0) 6∈ U◦× iRp−m
for some (t, u0) ∈ O◦. Let v0 = Re u0,I. We first show that ψI(t, v0, 0) 6∈ U◦. It
holds that xI 7→ Ptfu0(xI, 0) 6∈ C0(X ) in view of Lemma 3.5. Since
|Ptfu0(x)| ≤ Ptfv0,0(x) = Φ(t, v0, 0)fψ(t,v0,0)(x),
and since Φ(t, v0, 0) 6= 0 by (i), we also have xI 7→ fψ(t,v0,0)(xI, 0) 6∈ C0(X ). Hence
ψI(t, v0, 0) 6∈ Re U◦, again by Lemma 3.5.
Define a collection of measures on X by µt,xI(dzI) = pt(xI, 0, dzI, R
p−m), where
pt(x, dz) denote the Markov transition functions of (X, (Px)x∈E). Write bµt,xI for
the Fourier-Laplace transform and let φ(t, v) = ψI(t, v, 0) for v ∈ U . Then it holds
that
b
µt,xI(v) = Ptfφ(t,v),0(xI, 0) = Φ(t, v, 0)fφ(t,v)(xI),
for xI ∈ X , v ∈ U . Since φ(t, v0) 6∈ Re U◦and v0∈ Re U◦, Proposition 4.12 yields
that φ(t, v) 6∈ Re U◦for all v ∈ Re U◦. Note that the semigroup property (i) carries over to φ for real parameters, as ψJ(t, u) = 0 for all u ∈ Re U × {0}. Hence
φ(t, v0) = φ(t/2, φ(t/2, v0)).
From this we infer that v1 := φ(t/2, v0) 6∈ Re U◦. Indeed, if v1 ∈ Re U◦, then,
since φ(t/2, v1) = φ(t, v0) 6∈ Re U◦, Proposition 4.12 gives φ(t/2, v) 6∈ Re U◦ for
all v ∈ Re U◦, in particular for v = v0. This contradicts the fact that v1∈ Re U◦.
Hence φ(t/2, v0) 6∈ Re U◦. Repeating this argument yields φ(t/2k, v0) 6∈ Re U◦
for all k ∈ N. Stochastic continuity gives that t 7→ φ(t, v0) is continuous, whence
taking the limit we obtain that v0= limk→∞φ(t/2k, v0) 6∈ Re U◦. This contradicts
the assumption that v0∈ Re U◦.
The above proposition enables us to show differentiability of t 7→ Ptfu(x) for
(t, u) ∈ O and to determine the derivative.
Proposition 4.17. Let (X, (Px)x∈E) be a regular affine process with symbol Ψ
given by (3.6). Then Ψ(u, x) can be continuously extended to u ∈ U × iRp−m,
∂tPtfu(x) exists for (t, u) ∈ O and it holds that
∂tPtfu(x) = Ptfu(x)Ψ(ψ(t, u), x). (4.10)
4.3. The infinitesimal generator 83
Proof. By Proposition 4.14, Ψ(u, x) is of the form (3.11) for affine b, c, γ and K. The support of K(x, dz) is contained in a closed convex cone C satisfying E + C ⊂ E, in view of Lemma 4.7. Note that for all u ∈ U × iRp−m, z ∈ C, we have
Re (u>z) ≤ 0. Indeed, iterating E + C ⊂ E yields E + nC ⊂ E for all n ∈ N, i.e. x + nz ∈ E for all x ∈ E, z ∈ C, n ∈ N. For v ∈ Re U we have supxI∈Xv
>x I < ∞,
so necessarily Re (u>z) = Re u>IzI ≤ 0 for u ∈ U × iRp−m, z ∈ C. Therefore,
the right-hand side of (3.11) is continuous in u on U × iRp−m, by dominated
convergence. Hence we can extend Ψ(u, x) continuously to u ∈ U × iRp−m.
Fix x ∈ E. We first show that ∂tPtfu(x) exists and is continuous on [0, t0(u))
for u ∈ U◦× iRp−m. By Proposition 4.16 it holds for (t, u) ∈ O◦ that P tfu =
Φ(t, u)fψ(t,u) with ψ(t, u) ∈ U◦× iRp−m. It follows by the semigroup property of
Ptand the definition of the symbol Ψ that
∂t+Ptfu(x) = ∂h+
h=0PhPtfu(x)
= Φ(t, u) ∂h+
h=0Phfψ(t,u)(x)
= Φ(t, u)fψ(t,u)(x)Ψ(ψ(t, u), x)
= Ptfu(x)Ψ(ψ(t, u), x),
(4.11)
for (t, u) ∈ O◦. Since t 7→ Ptfu(x) is continuous by stochastic continuity, also
t 7→ ψ(t, u) is continuous for t < t0(u). In addition, u 7→ Ψ(u, x) is continuous.
Hence t 7→ Ptfu(x) has a continuous right-hand side derivative for (t, u) ∈ O◦
in view of (4.11). This yields that t 7→ Ptfu(x) is continuously differentiable on
[0, t0(u)) for all u ∈ U◦× iRp−m, see [56, Section IX.3].
Now we show that ∂tPtfu(x) exists and is continuous on [0, t0(u)) for all u ∈
U × iRp−m, using a limit argument. Fix (t, u0) ∈ O, x ∈ E. Let v ∈ U◦ be
arbitrary, define the line
C = {λv + (1 − λ)u0,I: λ ∈ [0, 1]} ⊂ U,
and let vn = n1v + (1 −n1)u0,I ∈ C. Define the sequence (un) in U◦× iRp−m by
un,I= vn, un,J = u0,J. Then un → u0 and
sup n∈N kfunk∞< ∞. Therefore, sup s≥0,n∈N |Psfun| < ∞, and Psfun(x) → Psfu0(x),
as n → ∞, for all s ≥ 0. Since Ptfu0(x) 6= 0, we have Ptfun(x) 6= 0 for all n large
enough, whence (t, un) ∈ O◦ for n large enough. By Proposition 4.16(i) we have
(s, un) ∈ O◦ for all s ≤ t, if n is large enough. By Lemma 3.11, (s, u) 7→ Psfu(x)
is jointly continuous on R+× C × iRp−m. It follows that (s, u) 7→ Φ(s, u) and
(s, ) 7→ ψ(s, u) are jointly continuous on [0, t] × C × iRp−m. Hence it holds that
inf
s≤t,n∈N|Φ(s, un)| > 0 and s≤t,n∈Nsup |ψ(s, un)| < ∞,
as well as
ψ(s, un) → ψ(s, u0)
for s ≤ t. Since u 7→ Ψ(u, x) is continuous on U × iRp−m, it follows that Psfun(x)Ψ(ψ(s, un), x)
is uniformly bounded in n and in s ≤ t, whence the same holds for ∂sPsfun(x) in
view of (4.11). Thus we can interchange limit and integration to obtain
Ptfu0(x) − fu0(x) = limn→∞(Ptfun(x) − fun(x)) = lim n→∞ Z t 0 ∂sPsfun(x)ds = Z t 0 lim n→∞∂sPsfun(x)ds = Z t 0 lim n→∞Psfun(x)Ψ(ψ(s, un), x)ds = Z t 0 Psfu0(x)Ψ(ψ(s, u0), x)ds.
The right-hand side is continuously differentiable in t, whence the same holds for the left-hand side.
It remains to show strong regularity of X. Let K be a compact subset of Rp−m and fix y ∈ U◦, x ∈ E. Lemma 3.11 yields that (t, z) → Ptfy,iz(x) is
jointly continuous. Hence there exists t > 0 such that Psfy,iz 6= 0 for all s ≤ t,
z ∈ K. This gives that (s, z) 7→ ψ(s, y, iz) is jointly continuous on [0, t] × K, whence ∂sPsfy,iz(x) is uniformly bounded on [0, t] × K, in view of (4.11). This
implies strong regularity of X.
By the affine structure of the symbol Ψ, the coefficients in the exponential affine expression for Ptfu(x) necessarily satisfy a system of ODEs, as shown in the
4.3. The infinitesimal generator 85
Proposition 4.18. Let (X, (Px)x∈E) be a regular affine process with state space
E and symbol Ψ(u, x) of the form (3.11), for affine b, c, γ and K given by (4.3). Then (4.5) holds for u ∈ U × iRp−m, where (ψ
0(·, u), ψ(·, u)) : [0, t0(u)) → C × Cp
solves the system of generalized Riccati equations given by (4.6), where Ri is given
by (4.7) and well-defined on U .
Proof. By Lemma 4.7 we have supp Ki
(·) ⊂ C for some C ⊂ Rpsatisfying E +C ⊂
E. Iterating this relation gives E + nC ⊂ E for all n ∈ N. This yields Re y>z ≤ 0 for y ∈ U , z ∈ C, whence Ri(y) given by (4.7) is well-defined for y ∈ U .
By Proposition 4.17, ∂tPtf (x) exists for (t, u) ∈ O, whence we can define
ψ0(t, u) =
Z t
0
∂sΦ(s, u)
Φ(s, u) ds,
for t < t0(u). Then ψ0(t, u) is continuously differentiable in t, ψ0(0, u) = 0 and
Φ(t, u) = exp(ψ0(t, u)). To see the latter, note that the quotient of the left- and
right-hand side has derivative 0 and equality holds for t = 0, whence it holds for all t ∈ [0, t0(u)). It follows that
∂tPtfu(x) = ∂texp(ψ0(t, u) + ψ(t, u)>x) = (∂tψ0(t, u) + ∂tψ(t, u)>x)Ptfu(x),
for (t, u) ∈ O, x ∈ E. Combining this with (4.10) and (3.11) yields
∂tψ0(t, u) + ∂tψ(t, u)>x = Ψ(ψ(t, u), x)
= ψ(t, u)>b(x) +12ψ(t, u)>c(x)ψ(t, u) − γ(x)
+ Z
(eψ(t,u)>z− 1 − ψ(t, u)>χ(z))K(x, dz). Since b(x), c(x), γ(x), K(x, dz) are affine and given by (4.3), separating first order terms in x gives (4.6) and (4.7).
We are now ready to prove the Feller property and determine the infinitesimal generator.
Proposition 4.19. Every regular affine process X is a Feller process. The domain of its infinitesimal generator A satisfies C2
c(E) ⊂ D(A) and for f ∈ Cc2(E), Af
takes the form (2.2), where b, c, γ and K are of the affine form given by (4.3) for some admissible parameters (ai, Ai, γi, Ki).
Proof. We first prove the Feller property. Let H be given by (3.14). It suffices to show that Pth ∈ C0(E) for all h ∈ H, t ≥ 0, in view of Proposition 3.17. Let
h : x 7→ R fy,iz(x)g(z)dz ∈ H for some y ∈ U◦, g ∈ Cc(Rp−m) and let t ≥ 0 be
arbitrary. By Fubini we can write
Pth(x) =
Z
Ptfy,iz(x)g(z)dz =
Z
Φ(t, y, iz)fψI(t,y,iz)(xI)fψJ(t,y,iz)(xJ)g(z)dz.
(4.12)
The integral can be restricted to those z for which (t, y, iz) ∈ O◦. By Proposi-tion 4.16 we then have ψI(t, y, iz) ∈ U◦, whence xI 7→ fψI(t,y,iz)(xI) ∈ C0(X ) in
view of Lemma 3.5. Dominated convergence yields Pth(x) → 0 if |xI| → ∞.
To show that Pth(x) → 0 if |xJ| → ∞, it suffices to show that ψJ(t, y, iz) is
of the form iM z for some invertible matrix M , in view of the Riemann-Lebesgue lemma applied to (4.12). We shall see M = exp(a>J Jt).
By Proposition 4.14, the symbol Ψ(u, x) is of the form (3.11), for affine b, c, γ and K given by (4.3). Since c(x), γ(x), K(x, dz) are positive for all x ∈ E and since xJ is unrestricted, it holds that Ai= 0, γi= 0 and Ki(dz) = 0 for all i ∈ J .
Therefore, by Proposition 4.18, we have that
∂sψJ(s, y, iz) = a>IJψI(s, y, iz) + a>J JψI(s, y, iz),
for all s < t0(y, iz). Since ψJ(s, y, iz) ∈ iRp−m, it follows that aIJ = 0. Indeed,
necessarily we have a>IJψI(s, y, iz) ∈ iRp−m for all s < t0(y, iz), y ∈ U◦. Taking
s = 0 we see that in particular a>
IJy = 0 for all y ∈ Re U◦, whence aIJ = 0. Hence
the above display yields
ψJ(t, y, iz) = i exp(a>J Jt)z,
as we needed to show.
Thus X is a Feller process. Next we verify the conditions of Theorem 3.20. Proposition 4.17 yields strong regularity, while Proposition 4.8 gives A(C2
c(E)) ⊂
C0(E), where A is the linear operator given by (2.2). It remains to show A(H) ⊂
C0(E), where H is given by (3.14).
Let h : x 7→ R fy,iz(x)g(z)dz ∈ H for some y ∈ U◦, g ∈ Cc∞(Rp−m). By
Lemmas 3.18 and 3.19 we can write
Ah(x) = Z
fy,iz(x)
(y, iz)>b(x) +12(y, iz)>c(x)(y, iz) − γ(x)
+ Z
(e(y,iz)>ζ − 1 − (y, iz)>χ(ζ))K(x, dζ)
g(z)dz.
Note that the expression in parentheses, without the first term (y, iz)>b(x), is uniformly bounded in z on compacta by an affine expression in xI, say k(xI) :=
4.4. Existence under admissibility 87
k0+ k>xI, some k0 ∈ R, k ∈ Rm. Since y ∈ U◦, it holds that fy(xI)k(xI) → 0
if |xI| → ∞. If |xJ| → ∞, then the above display tends to zero by the
Riemann-Lebesgue lemma. To see the latter, use the equality Z fy,iz(x)xJg(z)dz = Z (∂zfy,iz(x))g(z)dz = − Z fy,iz(x)∂zg(z)dz.
Hence Ah ∈ C0(E), as we needed to show.
Thus Theorem 3.20 is applicable. This yields the asserted properties of the infinitesimal generator of X. The admissibility of the parameters is a consequence of Proposition 2.13.
4.4
Existence under admissibility
Throughout this section we let Ω be given by (2.8), (ai, Ai, γi, Ki) be an admissible
parameter set and we let A be defined by (2.2) with b, c, γ, K given by (4.3). Our aim is to show that the affine martingale problem for A is well-posed in Ω and that (X, (Px)x∈E) is a stochastically continuous affine process, where Px
denotes the unique solution of the martingale problem for (A, δx). Our approach
is as follows. Existence of a solution of the martingale problem is an immediate consequence of Proposition 2.15. To establish uniqueness, we show that that the Fourier-Laplace transform is of the exponential affine form as given in (4.5), for u ∈ U◦× iRp−m, where (ψ
0, ψ) is a unique solution to the system of generalized
Riccati equations given by (4.6) and (4.7). For the existence and uniqueness of the latter we apply general ODE results, as summarized in the proposition below. In the remainder of our analysis, we extensively make use of the results obtained in Chapter 2. In addition we use that X is not only right-continuous, but also quasi-left continuous under Px, which implies that Px(Xt= Xt−) = 1 for all t > 0,
see [21, Theorem 4.3.12].
Proposition 4.20. Let H be a Banach space, O ⊂ H an open set and F : O → H a locally Lipschitz continuous function. Then for all u ∈ O there exists t0(u) ∈
(0, ∞] and a unique solution x(·, u) : [0, t0(u)) → O of the ODE
∂tx(t, u) = F (x(t, u)), x(t, 0) = u,
with either t0(u) = ∞ or t0(u) < ∞ and
lim t↑t0(u) x(t, u) ∈ ∂O or lim t↑t0(u) |x(t, u)| = ∞. In addition, D = {(t, u) ∈ R+× O : t < t0(u)}
is an open set in R+× H and x : D → O : (t, u) 7→ x(t, u) is continuous.
Proof. See for instance [2, Theorems 7.6 and 8.3] or [16].
To apply the above proposition, we need the following lemma. As a corollary we show that admissibility remains preserved under restriction of the parameters to the first m components.
Lemma 4.21. Let (ai, Ai, γi, Ki) be an admissible parameter set. Then necessar-ily it holds that
∀j ∈ J : Aj = 0, γj= 0, Kj = 0, (4.13)
∀i ∈ I, ∀j ∈ J : aij = 0. (4.14)
As a consequence, the system of generalized Riccati equations given by (4.6) and (4.7) (excluding i = 0) can be decomposed into
∂tψI(t, u) = RI(ψ(t, u)), ψI(0, u) = uI, (4.15)
∂tψJ(t, u) = a>J JψJ(t, u), ψJ(0, u) = uJ, (4.16)
so that a solution ψ necessarily satisfies ψJ(t, u) = exp(a>J Jt)uJ.
Proof. For x ∈ E we have that xJ is unrestricted, whence (4.13) follows from the
positiveness of c(x), γ(x) and K(x, dz). To show (4.14) we proceed as follows. Recall E = X × Rp−m. Since X is closed and convex, we can write
X = ∞ \ i=1 {x ∈ Rm: ζ i+ ηi>x ≥ 0},
for some ζi∈ R, ηi∈ Rm. Without loss of generality we may assume 0 ∈ X , so that
ζi≥ 0 for all i. We first show that the linear span of {ηi : i ∈ N} is equal to Rm.
Arguing by contradiction, suppose it is not, then there exists y ∈ Rm\{0} such
that ηi>y = 0 for all i. It holds that λy ∈ X for all λ ∈ R, since ζi+ ηi>λy = ζi≥ 0.
Let u ∈ U◦ 6= ∅. Since |λy| → ∞ for |λ| → ∞, it follows from Lemma 3.5 that u>λy → −∞ for |λ| → ∞, which is absurd.
Note that E can be written as
E = ∞ \ i=1 {x ∈ Rp : ζ i+ ηi>xI ≥ 0}.
We apply Proposition 2.16 to derive the assertion. Let y ∈ X be fixed. Since Kj = 0 for j ∈ J , it follows from property (iii) of Proposition 2.16 that there
exists C ∈ R such that for all i and all z ∈ Rp−m it holds that
η>i a0I + X k∈I ηi>akIyk+ X j∈J ηi>a j Izj= ηi>bI(y, z) ≥ C.
4.4. Existence under admissibility 89
Hence η>i ajI = 0 for all i ∈ N, j ∈ J. Since the linear span of {ηi: i ∈ N} is equal
to Rm, it follows that aj
I = 0 for all j ∈ J , i.e. aij = 0 for all i ∈ I, j ∈ J , as we
needed to show.
Corollary 4.22. Let (ai, Ai, γi, Ki) be an admissible parameter set with respect
to the state space E = X × Rp−m. Define parameters on X by e ai= aiII, Aei= AIIi , eγi= γi, Kei(G) = Z 1G(zI) |zI|2∧ 1 |z|2∧ 1K i(dz), for i = 0, . . . , m, G ∈ B(Rm\{0}). Then ( e ai, eAi, e
γi, eKi) are admissible parameters
with respect to the state space X . In addition, (ψ0(t, uI, 0), ψ(t, uI, 0)) is a solution
to the system of generalized Riccati equations corresponding with (ai, Ai, γi, Ki) if
and only if ψJ(t, uI, 0) = 0 and (ψ0(t, uI, 0), ψI(t, uI, 0)) is a solution to the system
of generalized Riccati equations corresponding with (eai, eAi,eγi, eKi).
Proof. Let eb,ec,eγ, eK be given by (4.3) with (eai, eAi,eγi, eKi) instead of (ai, Ai, γi, Ki) and m instead of p. By Lemma 4.21 we have (4.13), so it holds that
eb(xI) = bI(x), ec(xI) = cII(x), eγ(xI) = γ(x), K(xe I, dzI) = K(x, dzI, R
p−m),
for all x ∈ E × Rp−m. Recall that the coordinates χ
i(z) of the truncation function
χ only depend on zi. Define the continuous truncation function χ on Re m by
e
χ(zI) = χI(zI, 0). The properties of Definition 4.3 can now easily be checked to
see that (eai, eAi,eγi, eKi) are admissible. The second assertion of the proposition follows from Lemma 4.21, as it implies that
Ri(y) = yI>ea i+1 2y > I AeiyI+eγi+ Z eyI>zI− 1 − y> I χ(ze I) |zI|2∧ 1 e Ki(dzI), for i = 0, . . . , m and y ∈ Rm× {0}.
Proposition 4.20 takes the following form when applied to the system of gen-eralized Riccati equations we consider.
Proposition 4.23. Let (ai, Ai, γi, Ki) be an admissible parameter set. For all
u ∈ U◦× iRp−m there exist t
0(u) ∈ (0, ∞] and a unique solution (ψ0(·, u), ψ(·, u)) :
[0, t0(u)) → C × U◦× iRp−m to the system of generalized Riccati equations given
by (4.6) and (4.7) such that either t0(u) = ∞ or t0(u) < ∞ and
lim
t↑t0(u)
ψI(t, u) ∈ ∂U or lim t↑t0(u)
In addition,
D = {(t, u) ∈ R+× U◦× iRp−m: t < t0(u)}
is an open set in R+× Cm× iRp−m and ψ : D → U◦× iRp−m: (t, u) 7→ ψ(t, u) is
continuous.
Proof. Let R : U × iRp−m→ Cpbe the function with elements R
i, for i = 1, . . . , p.
Then it holds that R(U ×iRp−m
) ⊂ Cm
×iRp−min view of Lemma 4.21. Moreover,
one easily verifies that y 7→ R(y) is locally Lipschitz continuous on U◦× iRp−m.
Now the existence and uniqueness of ψ, the asserted properties of t0(u) and D
and the continuity of (t, u) 7→ ψ(t, u), follow from Proposition 4.20 with H = Cm× iRp−m, O = U◦× iRp−mand F = R. The existence and uniqueness of ψ0is
an easy consequence, as it satisfies
ψ0(t, u) =
Z t
0
R0(ψ(s, u))ds.
Having showed the existence and uniqueness of a solution (ψ0, ψ) to the system
of generalized Riccati equations, we now establish the exponential affine expression with coefficients ψi for the Fourier-Laplace transform Exexp(u>Xt), with (t, u) ∈
D.
Proposition 4.24. Let (ai, Ai, γi, Ki) be an admissible parameter set and for all
x ∈ E, let Pxbe a solution of the affine martingale problem for (A, δx) in Ω. With
the notation of Proposition 4.23, it holds that
Exexp(u>Xt) = exp(ψ0(t, u) + ψ(t, u)>x), for all x ∈ E, (t, u) ∈ D. (4.18)
Proof. Fix (T, y, iz0) ∈ D and let g ∈ Cc∞(Rp−m) with supp g ⊂ {z : (T, y, iz) ∈ D}
be arbitrary. Then ψ0(t, y, iz) and ψ(t, y, iz) exist for all t ∈ [0, T ], z ∈ supp g and
are continuous in z by Proposition 4.23, whence v : [0, T ] × E → C given by v(t, x) :=
Z
exp(ψ0(T − t, y, iz) + ψ(T − t, y, iz)>x)g(z)dz,
is well-defined. As noted in the paragraph succeeding Theorem 4.4, the function
(t, x) 7→ exp(ψ0(T − t, y, iz) + ψ(T − t, y, iz)>x),
solves Kolmogorov’s backward equation. However, Proposition 2.9 is not imme-diately applicable, as the above function is not in C0 in general. Therefore, we
4.4. Existence under admissibility 91
consider the integrated function v(t, x) instead. Using a limit argument, we obtain the desired result, as we will see below.
We claim that v ∈ C01,2([0, T ] × E) and that v solves Kolmogorov’s backward equation (2.16) with f (x) =R fy,iz(x)g(z)dz. To see the latter, it suffices to show
that the operator ∂t+ A commutes with integration, since then we have
(∂t+ A)v(t, x) =
Z
(∂t+ A) exp(ψ0(T − t, y, iz) + ψ(T − t, y, iz)>x)g(z)dz = 0.
Write
f (t, x, z) = exp(ψ0(T − t, y, iz) + ψ(T − t, y, iz)>x)g(z).
Note that f ∈ C1,2,0
([0, T ] × E × Rp−m). Indeed, (t, u) 7→ ψ(t, u) is jointly
con-tinuous by Proposition 4.23, whence also the derivatives ∂tψi(t, u) = Ri(ψ(t, u))
are jointly continuous in (t, u) for i = 0, . . . , p. Note also that f is bounded in view of Corollary 3.8, since {ψI(t, y, iz) : t ∈ [0, T ], z ∈ supp g} is contained in a
compact set of U◦ by continuity of (t, u) 7→ ψ(t, u) and the assumption that g has compact support. Hence the second part of the claim follows from Lemma 3.18 and Lemma 3.19.
For the first part of the claim it remains to show that v vanishes at infinity. Let (xn) be a sequence in E with |xn| → ∞ for n → ∞. Then |xI,n| → ∞ or
|xJ,n| → ∞. If |xI,n| → ∞, then v(t, xn) → 0 for all t, since fu ∈ C0(E) for
all u ∈ U◦ by Lemma 3.5 and ψI(t, u) ∈ U◦ for all (t, u) ∈ D. If |xJ,n| → ∞,
then v(t, xn) → 0 follows from the Riemann-Lebesgue lemma, as ψJ(t, y, iz) =
exp(a>
J Jt)iz for (t, y, iz) ∈ D by Lemma 4.21.
Thus we can apply Proposition 2.9 to v(t, x). Together with Fubini this yields that
Z
Exfy,iz(Xt)g(z)dz =
Z
exp(ψ0(t, y, iz) + ψ(t, y, iz)>x)g(z)dz, (4.19)
for all t ∈ [0, T ] and g ∈ Cc∞(Rp−m) with supp g ⊂ {z : (T, y, iz) ∈ D}.
Take a sequence (gn) of non-negative functions in Cc∞(Rp−m) with support
contained in the unit ball B(z0, 1/n), withR gn(z)dz = 1 for all n ∈ N and such
that gn(z)dz converges weakly to δz0(dz). Since D is open in R+× C
m
× iRp−m, it
holds that supp gn ⊂ {z : (T, y, iz) ∈ D} for n large enough. By weak convergence
we have
Z
f (z)gn(z)dz → f (z0), for all f ∈ Cb(Rp−m).
Choosing f (z) = Exfy,iz(Xt) resp. f (z) = exp(ψ0(t, y, iz) + ψ(t, y, iz)>x) yields
Exfy,iz0(Xt) = exp(ψ0(t, u0) + ψ(t, u0)
>x), for all x ∈ E, t ∈ [0, T ]).
Our next aim is to show that Exexp(u>Xt) = 0 for all t ∈ [t0(u), ∞), x ∈ E,
for u ∈ U◦× iRp−m. In order to establish this, we show that |ψ
I(t, u)| → ∞ if
t ↑ t0(u) < ∞ and take limits in (4.18). The following lemma on continuity of
t 7→ Exexp(u>Xt) is therefore required. Note that quasi-left continuity of X does
not automatically yield continuity of the expectation, in view of Remark 3.4. As an immediate corollary we obtain that Exexp(u>Xt0(u)) = 0 for x ∈ E
◦, in case
|ψI(t, u)| → ∞ as t ↑ t0(u) < ∞.
Lemma 4.25. Let (ai, Ai, γi, Ki) be an admissible parameter set and for all x ∈ E, let Pxbe a solution of the affine martingale problem for (A, δx) in Ω. In addition,
let the notation of Proposition 4.23 be in force. Then for all u ∈ U◦× iRp−m,
x ∈ E, it holds that t 7→ Exexp(u>Xt) is continuous.
Proof. Fix u ∈ U◦× iRp−m and x ∈ E. By Proposition 4.24 we have
Exexp(u>Xt) = exp(ψ0(t, u) + ψ(t, u)>x), for all t < t0(u). (4.20)
Hence t 7→ Exexp(u>Xt) is continuous for t < t0(u). Now let T ≥ t0(u).
Proposi-tion 2.7 together with ProposiProposi-tion 4.24 gives
Ex(exp(u>XT)|FεX) = Ex(exp(u>(X ◦ θε)T −ε)|FεX)
= exp(ψ0(T − ε, u) + ψ(T − ε, u)>Xε), Px-a.s.,
for all ε ∈ (T − t0(u), T ], whence
Exexp(u>XT) = exp(ψ0(T − ε, u))Exexp(ψ(T − ε, u)>Xε), (4.21)
for all ε ∈ (T − t0(u), T ]. Fix such ε and note that t 7→ ψ0(t, u) and t 7→ ψ(t, u) are
continuous for t ∈ [0, t0(u)) (as they are differentiable), so in particular they are
continuous at T −ε. Since Exexp(u>Xε) is continuous in u for u ∈ U◦×iRp−mand
since ψI(t, u) ∈ U◦for all t < t0(u), it follows from (4.21) that t 7→ Exexp(u>Xt)
is continuous at t = T , which yields the result.
Corollary 4.26. Let (ai, Ai, γi, Ki) be an admissible parameter set and for all x ∈ E, let Px be a solution of the affine martingale problem for (A, δx) in Ω. In
addition, let the notation of Proposition 4.23 be in force. Suppose T := t0(u) < ∞
for some u ∈ U◦× iRp−m and |ψ
I(t, u)| → ∞ for t ↑ T . Then it holds that
Exexp(u>XT) = 0, for all x ∈ E◦.
Proof. Proposition 4.24 together with Lemma 4.25 gives
Exexp(u>XT) = lim
t↑Texp(ψ0(t, u) + ψ(t, u) >x),
4.4. Existence under admissibility 93
for all x ∈ E. Note that if limt↑TRe ψI(t, u) exists and is finite, then necessarily
limt↑TIm ψI(t, u) exists and is finite, since the limit in the above display exists.
Reversing this yields that limt↑T|Re ψI(t, u)| = ∞, since limt↑T|ψI(t, u)| = ∞ by
assumption. Now suppose Ex0exp(u
>X
T) 6= 0 for some x0∈ E◦. Then we have
lim
t↑T(Re ψ0(t, u) + Re ψI(t, u) >x
0) > −∞.
Since limt↑T|Re ψI(t, u)| = ∞, there exists i ∈ I, a sequence tn ↑ T and ε ∈ R,
such that
lim
n→∞εRe ψi(tn, u) = ∞.
We can take ε such that B(x0, |ε|) ⊂ E. Then y = x0+ εei∈ E and it holds that
Re ψ0(tn, u) + Re ψI(tn, u)>y = (Re ψ0(tn, u) + Re ψI(tn, u)>x0) exp(εRe ψi(tn, u))
→ ∞.
Hence Eyexp(u>XT) = ∞, which is a contradiction.
In view of Proposition 4.23, there are two possibilities for the limit-behavior of ψ(t, u) in case t0(u) < ∞, as stated in (4.17). In the next lemmas we deduce from
Proposition 4.12 that
lim
t↑t0(u)
ψI(t, u) ∈ ∂U
is impossible.
Lemma 4.27. Suppose E = X (i.e. m = p). Let (ai, Ai, γi, Ki) be an admissible
parameter set and for all x ∈ E, let Px be a solution of the affine martingale
problem for (A, δx) in Ω. In addition, let the notation of Proposition 4.23 be in
force. Then it holds that t0(u) = ∞ for all u ∈ Re U◦.
Proof. Fix u0∈ Re U◦ and suppose T := t0(u0) < ∞. Proposition 4.24 gives
Exexp(u>0Xt) = exp(ψ0(t, u0) + ψ(t, u0)>x), for all x ∈ E, t < T ,
so by Lemma 4.25 we have
Exexp(u>0XT) = lim
t↑Texp(ψ0(t, u0) + ψ(t, u0)
>x), for all x ∈ E.
From Proposition 4.12 we infer that x 7→ Exfu(Xt) ∈ C0(E) for all u ∈ Re U◦,
t < T . Proposition 2.7 together with Proposition 4.24 gives
Ex(exp(u>0XT)|FεX) = Ex(exp(u>0(X ◦ θε)T −ε)|FεX)
for all ε ∈ (0, T ]. Thus for all x ∈ E, ε ∈ (0, T ], we have
Exexp(u>0XT) = exp(ψ0(T − ε, u0))Exexp(ψ(T − ε, u0)>Xε), (4.22)
which yields that x 7→ Exexp(u>0XT) ∈ C0(E). This implies that limt↑Tψ(t, u0) ∈
∂U is impossible, as fu ∈ C0(E) if and only if u ∈ U◦ by Lemma 3.5. Hence
limt↑T|ψ(t, u0)| = ∞, which yields that
Exexp(u>0XT) = 0, for all x ∈ E◦,
in view of Corollary 4.26. However, (4.22) then implies Exexp(ψ(T − ε, u0)>Xε) =
0 for all x ∈ E◦, ε ∈ (0, T ]. Since ψ(T − ε, u0) ∈ Rp, this yields that Xε = ∆,
Px-a.s., for all ε ∈ (0, T ], x ∈ E◦, so by right-continuity of Xt we have X0 = ∆,
Px-a.s. for all x ∈ E◦, which is absurd.
Lemma 4.28. Let (ai, Ai, γi, Ki) be an admissible parameter set and for all x ∈ E,
let Pxbe a solution of the affine martingale problem for (A, δx) in Ω. In addition,
let the notation of Proposition 4.23 be in force. Then for all u ∈ U◦× iRp−m it
holds that either t0(u) = ∞ or t0(u) < ∞ and
lim
t↑t0(u)
|ψI(t, u)| = ∞.
Proof. Fix u0= (w, iy) ∈ U◦× iRp−mand suppose T := t0(u0) < ∞. It holds that
|Exexp(u>0Xt)| ≤ Exexp(Re u>0Xt)
= exp(ψ0(t, w, 0) + ψI(t, w, 0)>xI,
for all x ∈ E, t ∈ [0, t0(Re u0)), in view of Lemma 4.21. Hence for all t ∈
[0, t0(Re u0)) we have
Exexp(u>0Xt) → 0, as |xI| → ∞, (4.23)
by Lemma 3.5. Corollary 4.22 together with Lemma 4.27 gives t0(Re u0) = ∞
(existence of a solution of the affine martingale problem with state space X is a consequence of Proposition 2.15 in view of the admissibility). Thus (4.23) holds for all t ≥ 0, in particular it holds for t = T . Suppose that limt↑TψI(t, u)
ex-ists and is finite, then (ψ0(T, u), ψ(T, u)) := limt↑T(ψ0(t, u), ψ(t, u)) exists and
Proposition 4.24 together with Lemma 4.25 yields that
Exexp(u>XT) = exp(ψ0(T, u) + ψ(T, u)>x).
From (4.23) we infer that ψI(T, u) ∈ U◦, in view of Lemma 3.5. Proposition 4.23
4.4. Existence under admissibility 95
Combining Lemma 4.28 with Corollary 4.26 we thus obtain
Exexp(u>Xt) = 0, if t = t0(u) < ∞, u ∈ U◦× iRp−m, for all x ∈ E◦. (4.24)
In case Exexp(u>Xt) is continuous in x, then the expectation vanishes for all
x ∈ E and hence also for all t ≥ t0(u) by a a version of the Markov-property,
see the following lemma. The continuity of x 7→ Exexp(u>Xt) is proved in the
succeeding proposition.
Lemma 4.29. Let (ai, Ai, γi, Ki) be an admissible parameter set and for all x ∈ E, let Px be a solution of the affine martingale problem for (A, δx) in Ω. In
addition, let the notation of Proposition 4.23 be in force. Suppose for some u ∈ U◦× iRp−m we have T := t
0(u) < ∞ and Exexp(u>XT) = 0 for all x ∈ E. Then
Exexp(u>Xt) = 0 for all x ∈ E and t ≥ T .
Proof. Note that for all x ∈ E and all solutions Qxof the affine martingale problem
for (A, δx) it holds that
EQxexp(u >X T) = lim t↑Texp(ψ0(t, u) + ψ(t, u) >x) = lim t↑TEPxexp(u >X t) = 0,
by Lemma 4.25. Hence by Proposition 2.7 it follows that
Ex(exp(u>XT +t)|FtX) = Ex(exp(u>(X ◦ θt)T)|FtX) = 0, Px-a.s.,
for all x ∈ E, t ≥ 0. Taking expectations yields the result.
Proposition 4.30. Let (ai, Ai, γi, Ki) be an admissible parameter set and for all
x ∈ E, let Px be a solution of the affine martingale problem for (A, δx) in Ω. In
addition, let the notation of Proposition 4.23 be in force. It holds that
x 7→ Exexp(u>Xt)
is continuous for all u ∈ U◦× iRp−m, t ≥ 0.
Proof. Fix y ∈ Rp−m. For convenience, we write t0(v) instead of t0(v, iy), with
v ∈ U◦. For t ≥ 0 we write
Gt= {v ∈ U◦: x 7→ Exfv,iy(Xt) is not continuous},
At= {v ∈ Gt: t0(v) < t}.
We have to show Gt = ∅ for all t ≥ 0. We divide the proof in a couple of steps.
Step 0. For all t ≥ 0, v ∈ Gt, it holds that t0(v) ≤ t and v ∈ Gt0(v). Indeed, if
t < t0(v), then Proposition 4.23 yields (4.18), so in particular x 7→ Exfv,iy(Xt) is
continuous. If v 6∈ Gt0(v), then Exfv,iy(Xt0(v)) = 0 for all x ∈ E, in view of (4.24)
and the continuity in x. Lemma 4.29 implies that Exfv,iy(Xt) = 0 for all x ∈ E,
t ≥ t0(v). In particular, the expectation is continuous in x, so that v 6∈ Gt.
Step 1. For all t ≥ 0 it holds that Gt is open. Indeed, let vn → v, with
vn ∈ U◦\Gt, v ∈ U◦, for some t ≥ 0. Then it holds that
|Exfvn,iy(Xt) − Exfv,iy(Xt)| = Z (fvn(zI) − fv(zI))fiy(zJ)Px(Xt∈ dz) ≤ kfvn− fvk∞→ 0,
by Lemma 3.7. Thus x 7→ Exfv,iy(Xt)) is continuous, as it is the uniform limit of
a sequence of continuous functions. Hence v ∈ U◦\Gt, so that Gtis open.
Step 2. For all t ≥ 0 it holds that Atis dense in Gt. Arguing by contradiction,
suppose there is an open ball B ⊂ Gtwith At∩ B = ∅. Then we have t0(v) = t for
all v ∈ B in view of Step 0. By (4.24) it holds that
E0fv,iy(Xt) = 0, for all v ∈ B.
Since the expectation is analytic in v ∈ U◦ by Corollary 4.10 and B is a set of uniqueness, the above equality extends to U◦. From Lemma 3.10 we infer that
E0fv,iy(Xt) = 0, for all v ∈ U,
in particular for v = 0, i.e.
E0f0,iy(Xt) = 0. (4.25)
Let v ∈ B and 0 < ε < t be arbitrary and write u(ε) = ψ(ε, v, iy) ∈ U◦. By the semigroup property of the flow we have
ψ(s, u(ε)) = ψ(s + ε, v, iy),
for s < t − ε. It follows that t0(u(ε)) = t − ε, so (4.24) yields
E0fψ(ε,v,iy)(Xt−ε) = E0fu(ε)(Xt−ε) = 0.
Define C(ε) := {ψI(ε, v, iy) : v ∈ B}. Since ψJ(ε, v, iy) = i exp(a>J Jε)y =: iy(ε),
we infer that
4.4. Existence under admissibility 97
It remains to show that C(ε) is open, since in that case, by the same arguments as for (4.25), it holds that
E0f0,iy(ε)(Xt−ε) = 0, for all ε < t.
Letting ε ↑ t, we deduce from the right-continuity of X that 1 = E0f0,iy(t)(X0) = 0,
which is absurd.
To show that C(ε) is open, it suffices to show that the continuous function v 7→ ψI(ε, v, iy) has a continuous inverse, for fixed ε < t. By the semigroup
property of the flow we have
(v, iy) = ψ(−ε, ψ(ε, v, iy)).
Since ψJ(ε, v, iy) = y(ε) does not depend on v, we can write
v = ψ(−ε, ψI(ε, v, iy), y(ε)),
whence v 7→ ψI(ε, v, iy) has inverse w 7→ ψ(−ε, w, y(ε)). The assertion follows
from [47, 4.17].
Step 3. We are now ready to prove that Gt = ∅ for all t ≥ 0. Arguing by
contradiction, let v ∈ Gt. Then also v ∈ Gt0(v). Since Gt0(v) is open by Step 1,
there exists a compact set K ⊂ Gt0(v) with v ∈ K. We construct a sequence
vn ∈ K recursively as follows.
v1= v (4.26)
tn = t0(vn) (4.27)
vn+1∈ Atn∩ K. (4.28)
The choice of vn+1 is possible in view of Step 2. Note that t0(vn) is strictly
decreasing. Write t0 for the limit. Since vn∈ K for all n, it has a limit point v0.
Denote the subsequence of vn that converges to v0 also by vn. By right-continuity
of X, Lemma 3.7 and dominated convergence, we have
E0fvn,iy(Xtn) → E0fv0,iy(Xt0).
The left-hand side is equal to 0 for all n in view of (4.24), whence the right-hand side equals 0, wherefore t0 ≥ t0(v0). It follows that t0(vn) > t0(v0), so that
vn 6∈ Gt0(v0) for all n ∈ N by Step 0. However, v0∈ K ⊂ Gt, so that v0∈ Gt0(v0)
We are now able to deduce that the Fourier-Laplace transform is exponential affine for all t ≥ 0, u ∈ U◦× iRp−m. Indeed, (4.24) together with Proposition 4.30
and Lemma 4.29 gives
Exexp(u>Xt) = 0, for all t ∈ [t0(u), ∞), u ∈ U◦× iRp−m, for all x ∈ E.
Hence by Proposition 4.24 we have
Exexp(u>Xt) = Φ(t, u) exp(ψ(t, u)>x), (4.29)
for all u ∈ U◦× iRp−m, x ∈ E, t ≥ 0, where we write
Φ(t, u) =
exp(ψ0(t, u)), for t ∈ [0, t0(u)),
0, for t ∈ [t0(u), ∞),
and we take ψ(t, u) ∈ U◦× iRp−m arbitrary for t ∈ [t
0(u), ∞). The uniqueness of
the ψi(t, u) for t < t0(u) enable us to prove the uniqueness of the affine martingale
problem. The following proposition concludes the proof of Theorem 4.4 (ii).
Proposition 4.31. Let Ω be given by (2.8), (ai, Ai, γi, Ki) be an admissible pa-rameter set and let A be defined by (2.2) with b, c, γ, K given by (4.3). Then the affine martingale problem for A is well-posed in Ω and (X, (Px)x∈E) is a regular
affine process, where Px denotes the unique solution of the martingale problem for
(A, δx).
Proof. By Proposition 2.15 there exists for all x ∈ E∆ a solution Px of the
mar-tingale problem for (A, δx) in Ω. To show uniqueness of Px, assume Qxis another
solution of the martingale problem for (A, δx) in Ω, with x ∈ E. We show by
induction that for all n ∈ N and all 0 ≤ t1≤ . . . ≤ tn, u1, . . . , un∈ U◦× iRp−mit
holds that EPx n Y k=1 fuk(Xtk) = EQx n Y k=1 fuk(Xtk). (4.30)
The assertion holds for n = 1 in view of the paragraph preceding the proposition. Proceding by induction, assume (4.30) holds for some n ∈ N. Let tn+1 ≥ tn and
un+1∈ U◦× iRp−mbe arbitrary and write ∆tn= tn+1− tn. From Proposition 2.7
and the exponential affine expression for the characteristic function, we infer that
EPx(fun+1(Xtn+1)|Ftn) = EPx(fun+1(X ◦ θtn)∆tn)|Ftn)
4.4. Existence under admissibility 99
with ψ(∆tn, un+1) ∈ U◦× iRp−m. The same holds for Qx in lieu of Px. By the
induction hypothesis (4.30) we obtain
EPx n+1 Y k=1 fuk(Xtk) = EPx n Y k=1 fuk(Xtk)EPx(fun+1(Xtn+1)|Ftn) ! = Φ(∆tn, un+1)EPx n Y k=1 fuk(Xtk)fψ(∆tn,un+1)(Xtn) = Φ(∆tn, un+1)EQx n Y k=1 fuk(Xtk)fψ(∆tn,un+1)(Xtn) = EQx n+1 Y k=1 fuk(Xtk).
Thus by induction (4.30) holds for all n ∈ N, 0 ≤ t1 ≤ . . . ≤ tn, u1, . . . , un ∈
U◦×iRp−m. By Lemma 4.9 we can take limits and obtain (4.30) for all u
1, . . . , un∈
U × iRp−m. In particular we have equality for the characteristic functions. It follows that all finite dimensional distributions of X are unique, which yields well-posedness of the martingale problem, see [34, Proposition 3.2].
By Proposition 2.8 it holds that (X, (Px)x∈E) is a Markov process. Write
pt(x, dz) for the Markov transition functions and let Ptbe the corresponding
semi-group on B(E) as given by Ptf (x) =R f (z)pt(x, dz). Then for u ∈ U◦× iRp−m,
(4.29) yields that x 7→ Ptfu(x) is continuous and (4.9) holds for all z ∈ E and all
x, y ∈ E − z such that x + y ∈ E − z. Since X is right-continuous, it follows from Proposition 4.13 that Ptfu(x) is of the form (4.1) for all u ∈ U × iRp−m. Clearly
X is regular.
It remains to show stochastic continuity, which comes down to continuity of t 7→ Ptfu(x) for u ∈ iRp. By right-continuity of X and the special form of Ω as
given by (2.8), we have right-continuity of t 7→ fu(Xt) for u ∈ iRp. This yields
right-continuity of t 7→ Ptfu(x) for u ∈ iRp, by dominated convergence. To show
left-continuity, let x ∈ E and t > 0 be arbitrary and let the stopping times T∆,
Texpl and Tjump be given by (2.4), (2.5) and (2.6). Take 0 < s < t such that
Px(Texpl= s) = 0, which is possible as Texpl has at most countably many atoms.
Quasi-left continuity of X yields
lim
r↑sfu(Xr)1{s<T∆}= fu(Xs)1{s<T∆}, for all u ∈ U × iR p−m
, Px-a.s.,
and since Ω is given by (2.8) we also have
lim
r↑sfu(Xr)1{s>T∆}= 0 = fu(Xs)1{s>T∆}, for all u ∈ U × iR p−m
