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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.
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Chapter
6
The affine transform formula
In this chapter we revisit the exponential affine expression for the Fourier-Laplace transform of affine processes, in the literature known as the affine transform for-mula. We aim to extend the domain of validity beyond U × iRp−m, that is, we are looking for conditions such that
Exfu(Xt) = exp(ψ0(t, u) + ψ(t, u)>x) (6.1)
not only holds for u ∈ U × iRp−m, but also for u outside that domain, with (ψ 0, ψ) again solutions to the generalized Riccati equations. Results in the literature concerning this are only proved for affine processes living on the state space Rm
+× Rp−m, see [23, 24, 36, 40] amongst others. In this chapter we generalize these to arbitrary state spaces of the form E = X × Rp−m.
Section 6.1 is similar to [36]. We provide conditions for the validity of the affine transform formula (6.1) under existence of the right -hand side, for affine processes on a general state space. For the proof of the main result, Theorem 6.4, we use methods from stochastic calculus as derived in Chapter 2. In addition, we return to the question of non-vanishing Fourier-Laplace transforms as discussed in Section 4.5 and derive tractable conditions for this in Theorem 6.5.
In Section 6.2 we aim to establish the validity of the affine transform formula (6.1) under existence of the left -hand side. Under a strong moment condition on the jump-measures, we establish full range of validity for (6.1) in Theorem 6.7, again without restrictions on the state space. The same result is deduced for continuous affine diffusions on the state space Rm
+× Rp−min [23]. Contrary to the proof in that paper, we do not use the specific form of the admissibility conditions.
Instead, we heavily rely on Theorem 6.4. Whether the moment conditions on the jump-measures can be relaxed, is still an open problem.
Throughout this chapter we restrict ourselves to affine processes that are con-servative (i.e. Px(Xt ∈ E) = 1 for all t ≥ 0, x ∈ E), for which we provide a sufficient condition in the following proposition, cf. [17, Section 9] and [10, Re-mark 2.5]. As in Chapter 4, we assume throughout that the state space E is of the form E = X × Rp−m, with X as in Section 3.2.
Proposition 6.1. Let (ai, Ai, γi, Ki) be an admissible parameter set such that for all i = 0, . . . , m we have γi= 0 and
Z
|zI||Ki|(dz) < ∞. (6.2) Then the corresponding affine process X is conservative.
Proof. Let the Fourier-Laplace transform of X be given by (4.5). We have to show that Ptf0(x) = 1 for all t ≥ 0, x ∈ E, i.e. ψ0(t, 0) = 0 and ψ(t, 0) = 0 for all t ≥ 0. We already have ψJ(t, 0) = 0, so it remains to show that ψI(t, 0) = 0 for all t ≥ 0. Note that both ψI(t, 0) and the zero-function are Re U -valued solutions to the ODE
∂tg(t) = RI(g(t), 0), g(0) = 0.
It holds that y 7→ RI(y, 0) is Lipschitz continuous on Re U in view of (6.2). Hence the solution to the ODE in the above display is unique, so that ψI(t, 0) = 0 for all t ≥ 0.
6.1
Exponential martingales
In this section we establish validity of the affine transform formula for real -valued u, under existence of a solution (ψ0, ψ) of the generalized Riccati equations and a mild integrability condition on the jump measures. The results are of the same spirit as those of [36]. In our method we use tools from stochastic calculus, in particular equivalent change of measures and stochastic exponentials. Similar as in [36], we will see that the process
exp(ψ0(T − t, u) + ψ(T − t, u)>Xt)
is a stochastic exponential. In order to establish the affine transform formula, we need to show that this stochastic exponential is a proper martingale, for which the
6.1. Exponential martingales 139
results from Section 2.5 will prove much helpful. We start with translating Propo-sition 2.23 to the affine setting, see PropoPropo-sition 6.2 below. This yields sufficient conditions for the martingale property of a stochastic exponential
E(h(X−) · Xc+ w(X−, z) ∗ (µX− νX))
of an affine jump-diffusion X, with general parameters h and w. Next we choose the appropriate h and w to obtain our main result in Theorem 6.4.
Proposition 6.2. Let X be a (conservative) affine jump-diffusion on the proba-bility space (DE[0, ∞), (Ft+X), P) with generator A given by (2.2), where (b, c, γ, K) is of the affine form (4.3), for some admissible parameter set (ai, Ai, 0, Ki). Let us be given a bounded continuous function h : E → Rp and a measurable function w : E × Rp→ (−1, ∞) such that for all i = 0, . . . , m we have
(I) x 7→ (w(x, z) + 1)Ki(dz) is weakly continuous (and well-defined)
(II) x 7→R χ(z)w(x, z)Ki(dz) is continuous (and well-defined).
(III) R (|z|2∨ 1)(w(x, z) + 1)|Ki|(dz) ≤ C(1 + |x|), for all x ∈ E, some C > 0. (IV) R (|z|2∧1)−1|χ(z)||w(x, z)||Ki|(dz)|x
i| ≤ C(1+|x|), for all (x0, x) ∈ {1}×E, some C > 0.
(V) x 7→R (w(x, z)−log(w(x, z)+1))(|z|2∧1)−1|Ki|(dz) is bounded on compacta. Then
E(h(X−) · Xc+ w(X−, z) ∗ (µX− νX))
is a P-martingale and the martingale problem for eA given by (2.23) is well-posed in DE[0, ∞).
Proof. We have to check the conditions of Proposition 2.23. Condition (i) follows from the affine structure of the parameters (b, c, K) together with the continuity of h and assumptions (I) and (II). To show condition (ii), let f ∈ C2
c(E) be arbitrary. Note that continuity of eAf is a consequence of Lemma 2.12. It remains to show that eAf vanishes at infinity. We follow the proof of Proposition 4.8, with Ki(dz) replaced by (w(x, z) + 1)Ki(dz). Let g be given by (4.8) and let M > 0 be such that f (x) = 0 for x > M . Similar as in the proof of Proposition 4.8 there exists N > 0 such that g(x) = 0 for |xI| ≥ N and
|g(x)| ≤ N kf k∞ m X i=0 Z {|z|≥|xJ|−M } (w(x, z) + 1)|Ki|(dz),
for |xJ| ≥ M + 1. By assumption (III) we have Z {|z|≥|xJ|−M } (w(x, z) + 1)|Ki|(dz) ≤ Z |z|2 (|xJ| − M )2 (w(x, z) + 1)|Ki|(dz) ≤ 1 + |x| (|xJ| − M )2 ,
for all i = 0, . . . , m. This yields that g(x) → 0 for |x| → ∞, so that eAf ∈ C0(E). Condition (iii) follows from the continuity of c and h and from assumption (V). Finally, condition (iv) is a consequence of the affine structure of (b, c, K) together with the assumption that h is bounded and assumptions (III) and (IV).
Remark 6.3. In case x 7→ w(x, z) is continuous for all z ∈ Rp, then assumption (I) can be replaced by
(I’) x 7→R (w(x, z) + 1)Ki(dz) is continuous (and well-defined),
in view of an extended version of the dominated convergence theorem (see [34, Theorem 1.21]).
The following theorem is the main result of this section and extends more or less [36, Theorem 5.1] from the canonical state space Rm
+ × Rp−m to arbitrary state spaces. We note that the conditions for [36, Theorem 5.1] are slightly weaker, as part of its proof relies on a conservativeness result for time-inhomogeneous affine processes with canonical state space, based on [22]. Relaxing our condi-tions involves developing a theory for time-inhomogeneous affine processes with an arbitrary state space. This is left for further research.
Theorem 6.4. Let (X, (Px)x∈E) be a conservative affine process with generator A given by (2.2), where (b, c, γ, K) is of the affine form (4.3), for some admissible parameter set (ai, Ai, 0, Ki). Fix u ∈ Rp, T > 0 and let (ψ0(·, u), ψ(·, u)) : [0, T ] → R×Rpbe a solution to the system of generalized Riccati equations as given by (4.6) and (4.7). Assume that
(i) supt≤TR |z|2eψ(t,u)
>z
|Ki|(dz) < ∞ for i = 0, . . . , m, (ii) supt≤TR |ψ(t, u)>z||Ki|(dz) < ∞ for i = 0, . . . , m, (iii) t 7→R
{|z|>1}e
ψ(t,u)>zKi(dz) is continuous for i = 0, . . . , m.
Then it holds that
Exexp(u>Xt) = exp(ψ0(t, u) + ψ(t, u)>x),
6.1. Exponential martingales 141
Proof. It is enough to show the assertion for t = T . Extend (ψ0(·, u), ψ(·, u)) to a C1
-function on (−∞, T ] and define f : R+× E → R by f(t, x) = exp(ψ0(T − t, u) + ψ(T − t, u)>x). Since f (T, x) = exp(u>x), it suffices to show that f (t, Xt) is a Px-martingale for all x. Indeed, in that case we have
Exexp(u>XT) = Exf (T, XT) = Exf (0, X0) = exp(ψ0(T, u) + ψ(T, u)>x). Let x0 ∈ E be arbitrary and write P = Px0. First note that Yt := (t, Xt) is a
P-semimartingale with differential characteristics
(eb(t, Xt),ec(t, Xt), eK(Xt, ds, dz)), with eb(t, x) = 1 b(x) ! , ec(t, x) = 0 0 0 c(x) ! , K(x, ds, dz) = δe 0(ds) × K(x, dz).
Second, f (Yt) is a special semimartingale by Proposition 1.3. Indeed, let νY be the compensator of the random measure µY associated to the jumps of Yt. Then we have |f (s, z)|1{|(s,z)|>1}∗ νYt = Z t 0 Z |f (s, z)|1{|(s,z)|>1}K(Xe r, ds, dz)dr = Z t 0 Z |f (0, z)|1{|z|>1}K(Xr, dz)dr = Z t 0 Z {|z|>1} eψ0(T ,u)+ψ(T ,u)>zK(X r, dz)dr ≤ Z t 0 C(1 + |Xr|)dr,
for some C > 0, which follows from the affine structure of K(x, dz) and the fact that R
{|z|>1}e
ψ(T ,u)>z|Ki|(dz) < ∞ by assumption. Since Rt
0C(1 + |Xr|)dr is locally integrable, Proposition 1.3 yields that f (Yt) is special.
We now apply Theorem 1.2 to decompose f (Yt) as the sum of a continuous local martingale Mc
t, a purely discontinuous local martingale Mtdand a continuous process of bounded variation At. Let Yc be the continuous martingale part of Y and note that Yc= (0, Xc). It holds that
Mtc= ∇f (Yt−) · Yc= f (t, Xt−) · (ψ(T − t, u) · Xtc),
and
Mtd= (f (Yt−+(0, z))−f (Yt−))∗(µY−νY) = f (t, Xt−)·((eψ(T −t,u)
>z
Moreover, v(t, x) = exp(ψ0(T − t, u) + ψ(T − t, u)>x) solves Kolmogorov’s back-ward equation ∂sv + Av = 0, so that At = 0. Hence Theorem 1.2 yields the decomposition
f (t, Xt) = f (0, X0) + f (t, Xt−) · (ψ(T − t, u) · Xtc+ (e
ψ(T −t,u)>z− 1) ∗ (µY − νY)), whence
f (t, Xt) = f (0, X0)E (h(Yt−) · Ytc+ w(Yt−, z) ∗ (µY − νY)),
with h(t, x) = (0, ψ(T − t, u)) and w(t, x, z) = eψ(T −t,u)>z− 1. One verifies that the assumptions of Proposition 6.2 are met, which yields that f (t, Xt) is a P-martingale, as we needed to show.
Although Theorem 6.4 concerns validity of the affine transform formula for real -valued u, it enables us to establish validity for complex-valued u as well. In conjunction with Proposition 4.32 it provides us in particular tractable conditions for validity of the affine transform formula on the whole of U × iRp−m, as stated in the next theorem. Note that these conditions are automatically fulfilled for continuous affine diffusions.
Theorem 6.5. Let (X, (Px)x∈E) be a conservative affine process with admissible parameter set (ai, Ai, 0, Ki) and with Fourier-Laplace transform P
tfu(x) given by (4.5). Suppose there exists an open set V ⊂ Rp containing Re U × {0} such that
Z
ev>z|Ki|(dz) < ∞, for all v ∈ V, i = 0, . . . , m. (6.3) Then t0(u) := inf{t > 0 : Ptfu= 0} = ∞ for all u ∈ U × iRp−m.
Proof. In view of Proposition 4.32 it suffices to show that u 7→ Ptfu(x) is analytic on U ×iRp−mfor some x ∈ E and for all t ∈ [0, ε), some ε > 0, and that t 7→ ψ(t, u) is analytic on [0, t0(u)) for all u ∈ U × iRp−m. The latter follows from standard ODE results (e.g. [16, Theorem 10.4.5]), since y 7→ Ri(y) is analytic on U × iRp−m, due to (6.3) and Lemma 4.9.
It remains to show analyticity of u 7→ Ptfu(x). It holds that y 7→ Ri(y) is locally Lipschitz continuous on V , whence by Proposition 4.20 for all u ∈ V there exists t0(u) ∈ (0, ∞] and a unique solution (ψ0(·, u), ψ(·, u)) : [0, t0(u)) → R × V to the generalized Riccati equations as given by (4.6) and (4.7), and such that
6.2. Full range of validity 143
is open. Now take T > 0 arbitrary. Then D(T ) := {u ∈ V : T < t0(u)} is open and contains Re U × {0} (since t0(u) = ∞ for u ∈ Re U × {0}). The conditions of Theorem 6.4 are fulfilled for (ψ0(·, u), ψ(·, u)) : [0, T ] → R × V with u ∈ D(T ). Hence
Exexp(u>Xt) < ∞
for all x ∈ E, u ∈ D(T ), t ≤ T . From Lemma 4.9 we derive that u 7→ Exexp(u>Xt) is analytic on D(T ) + iRp, for all x ∈ E, t ≤ T , whence u 7→ P
tfu(x) is analytic on U × iRp−m⊂ D(T ) + iRp. This yields the result.
6.2
Full range of validity
Throughout this section we assume that all exponential moments of the jump-measures Ki of an affine jump-diffusion are finite, that is,
Z
ev>z|Ki
|(dz) < ∞, for all v ∈ Rp, i = 0, . . . , m. (6.4) Under this condition, we establish the full-range of validity of the affine transform formula for affine jump-diffusion with a general state space. From Theorem 6.4 we immediately derive that if a solution (ψ0(·, u), ψ(·, u)) of the Riccati equations exist on some time interval [0, T ] for u ∈ Rp, then it holds that Exexp(u>Xt) < ∞ for all x ∈ E, t ≤ T and the affine transform formula (6.1) is valid. In this section we prove that the converse also holds, i.e. if Exexp(u>Xt) < ∞ for all x ∈ E, t ≤ T , then there exists a solution (ψ0(·, u), ψ(·, u)) on [0, T ] and (6.1) holds. This result, as stated in Theorem 6.7 below, extends [23, Theorem 3.3], which is restricted to affine diffusions with a canonical state space Rm
+ × Rp−m. We use the notation from [23, Lemma 2.3], which we state as a proposition for ease of reference. In addition, as X is a special semimartingale, we take as a “truncation function” χ(z) = z.
Proposition 6.6. Assume (6.4). Let K be a placeholder for either R or C. It holds that
(i) For all u ∈ Kpthere exists t
0(u) > 0 and a unique solution (ψ0(·, u), ψ(·, u)) : [0, t0(u)) → K×Kp to the system of generalized Riccati equations (4.6), where either t0(u) = ∞ or limt↑t0(u)kψ(t, u)k = ∞. In particular t0(0) = ∞.
(ii) The set
is open in [0, ∞) × Kp and the ψ
i are analytic on DK. In addition, for all t ≥ 0
DK(t) := {u ∈ Kp: (t, u) ∈ DK}
is an open neighborhood of 0 and DK(t2) ⊂ DK(t1) for 0 ≤ t1≤ t2.
Proof. This follows from Proposition 4.20 and [16, Theorem 10.4.5]), see also [23, Lemma 2.3].
In the following, S(DR(t)) denotes the strip DR(t) + iRp.
Theorem 6.7. Let (X, (Px)x∈E) be a conservative affine process with generator A given by (2.2), where (b, c, γ, K) is of the affine form (4.3), for some admissible parameter set (ai, Ai, 0, Ki) relative to the truncation function χ(z) = z. Assume (6.4) and let the notation of Proposition 6.6 be in force. Then for t > 0 it holds that
(i) DR(t) = M (t), where
M (t) = {u ∈ Rp: Ex(exp(u>Xt)) < ∞ for all x ∈ E}. (ii) S(DR(t)) ⊂ DC(t).
(iii) The affine transform formula (6.1) holds for all u ∈ S(DR(t)).
(iv) DR(t) and DR are convex sets.
(v) M (t) ⊂ M (s) for 0 ≤ s ≤ t.
Proof. Theorem 6.4 yields DR(t) ⊂ M (t). The proof of DR(t) ⊃ M (t) is the content of the first subsection below, while the second subsection is devoted to the proof of (ii) and (iii). Assertions (iv) and (v) follow from (i).
6.2.1
Real-valued parameters
Let T > 0. The idea to prove M (T ) ⊂ DR(T ), is to show that ψ(T, u) explodes when u ∈ DR(T ) approaches the boundary ∂DR(T ). This is not immediate as the following example demonstrates.
Example 6.8. Consider the Riccati equation ˙x = x2. Its solution x with initial condition u ∈ C is given by x(t, u) = u/(1 − ut) and we have t0(u) = u−1 for u ∈ R>0 and t0(u) = ∞ otherwise. Hence DC(T ) = {u ∈ C : u 6∈ [T−1, ∞)} and ∂DC(T ) = [T−1, ∞). Obviously x(T, u) does not explode if u ∈ DC(T ) tends to
6.2. Full range of validity 145
u0∈ (T−1, ∞). If we take real and imaginary part, then we obtain a 2-dimensional system of Riccati equations given by
˙ x1= x21− x 2 2 ˙ x2= 2x1x2.
In this case DR(T ) = {u ∈ R2: u 6∈ [T−1, ∞)} and again x(t, u) does not explode if u ∈ DR(T ) tends to u0∈ (T−1, ∞). Note that the Riccati equations are of the form (4.6) (excluding the equation for ψ0) with
A1= 1 0 0 −1 ! , A2= 0 1 1 0 ! , a = 0.
However, they are not related to an affine diffusion where the state space has non-empty interior. Indeed, the corresponding diffusion matrix would be
c(x) = x1 x2 x2 −x1
! ,
which is positive semi-definite if and only if x = 0.
In Proposition 6.9 below we derive a formula that relates solutions to Riccati equations to the expectation of the corresponding affine jump-diffusion. This will turn out to be most useful in Proposition 6.12 to derive that M (T ) ⊂ DR(T ), which proves Theorem 6.7 (i). Note that convexity of the state space is explicitly used in the following proof.
Proposition 6.9. Consider the situation of Theorem 6.7. Define the non-negative function k : E × Rp → R by
k(x, y) =12y>c(x)y + Z
(ey>z− 1 − y>z)K(x, dz), (6.5) for x ∈ E and y ∈ Rp. Then for all x ∈ E, u ∈ Rp, t < t0(u) it holds that
ψ0(t, u) + ψ(t, u)>x = u>ExXt+ Z t
0
k(ExXt−s, ψ(s, u))ds, (6.6) and ExXtsolves the linear ODE
Proof. Fix u ∈ Rp and write ψ(·) instead of ψ(·, u). We can write the ODE for (ψ0, ψ) as an inhomogeneous linear ODE, namely
∂t ψ0(t) ψ(t) ! = A ψ0(t) ψ(t) ! + g(t), with A = 0 a 0> 0 a> ! ,
where we write a for the (p × p)-matrix with columns ai, i = 1, . . . , p and g = (g0, g1, . . . , gp) is the function given by
gi= 12ψ>Aiψ +
Z eψ>z− 1 − ψ>z |z|2∧ 1 K
i(dz), i = 0, . . . , p.
By an application of a variation of constants, the solution can be written as ψ0(t) ψ(t) ! = eAt 0 u ! + Z t 0 eA(t−s)g(s)ds, which yields ψ0(t) + ψ(t)>x = ψ0(t) ψ(t) !> 1 x ! =0 u>eA>t 1 x ! + Z t 0 g(s)>eA>(t−s) 1 x ! ds. (6.8)
Write f (t, x) for the solution to the linear ODE (6.7) with f (0, x) = x. Then we have y(t) z(t) ! := eA>t 1 x ! = 1 f (t, x) ! . Indeed, since ∂t y(t) z(t) ! = A> y(t) z(t) ! = 0 a0y(t) + az(t) ! ,
it holds that y(t) = 1 and ∂tz(t) = az(t) + a0 = b(z(t)) with z(0) = x, whence z(t) = f (t, x). Noting that g> 1 x ! =12ψ>c(x)ψ + Z (eψ>z− 1 − ψ>z)K(x, dz), for all x ∈ E, and ExXt∈ E for all x ∈ E, t ≥ 0, by convexity of E, we obtain (6.6) from (6.8) after we have shown that ExXt= f (t, x). The latter follows from Proposition 1.6, as it yields ExXt= x + Ex Z t 0 (a0+ aXs)ds = Z t 0 (a0+ aExXs)ds.
6.2. Full range of validity 147
To prove Theorem 6.7 (i) we first prove that Exexp(u>XT) = ∞ for T = t0(u) < ∞, for some x ∈ E, see Lemma 6.11. Then in Proposition 6.12 we extend this property to T > t0(u). The following lemma will be needed.
Lemma 6.10. for cn∈ Rp it holds that lim n→∞kcnk = ∞ ⇒ ∃x ∈ {−1, 1} p, ε > 0 : lim sup n→∞ inf y∈B(x,ε)c > ny = ∞. (6.9)
Proof. If limn→∞kcnk = ∞, then there exists a subsequence cnk such that all
components cnk,i are convergent in [−∞, ∞]. In addition, one of them converges
to either +∞ or −∞. Define x ∈ Rp by taking x
i= −1 if cnk,i→ −∞ and xi= 1
otherwise. Then obviously for y ∈ B(x, ε) with 0 < ε < 1 we have inf
y∈B(x,ε)c >
nky → ∞, as k → ∞.
Lemma 6.11. Consider the situation of Theorem 6.7. Let u ∈ Rp and suppose T := t0(u) < ∞. Then there exists x ∈ E such that Exexp(u>XT) = ∞.
Proof. Without loss of generality we may assume that {−1, 1}p ⊂ E◦ (thus by convexity also 0 ∈ E◦). Since kψ(t, u)k → ∞ for t ↑ T and in view of (6.9), there exists a ball B := B(x0, ε) ⊂ E (with x0∈ {−1, 1}p ⊂ E◦, ε > 0) and a sequence tn ↑ T such that
inf
y∈Bψ(tn, u)
>y → ∞ as n → ∞.
Moreover, it holds that ψ0(t, u) ≥ u>E0(Xt) for t < T by Proposition 6.9. In particular we have lim inft↑Tψ0(t, u) > −∞. Hence
lim
n→∞y∈Binf(ψ0(tn, u) + ψ(tn, u)
>y) = ∞.
By right-continuity of X, it follows that lim
n→∞(ψ0(tn, u) + ψ(tn, u) >X
T −tn) = ∞, Px0-a.s.
The Markov property and Theorem 6.4 give Exexp(u>XT) = Ex EXT −texp(u
>X
t) = Exexp(ψ0(t, u) + ψ(t, u)>XT −t), for 0 ≤ t < T , x ∈ E. Applying the previous together with Fatou’s Lemma we get
Ex0exp(u >X T) = lim inf n→∞ Ex0exp(u >X T) = lim inf n→∞ Ex0exp(ψ0(tn, u) + ψ(tn, u) >X T −tn)
≥ Ex0lim infn→∞ exp(ψ0(tn, u) + ψ(tn, u)
>X
Proposition 6.12. Consider the situation of Theorem 6.7. Let T ≥ 0. Then M (T ) = DR(T ) and (6.1) holds for u ∈ M (T ), t ≤ T .
Proof. In view of Theorem 6.4 it is sufficient to prove M (T ) ⊂ DR(T ). Without loss of generality we may assume that {−1, 1}p ⊂ E◦
. Let u ∈ Rp and suppose t0(u) < ∞. We need to show that for all T ≥ t0(u) there exists x ∈ E such that Exexp(u>XT) = ∞. Lemma 6.11 gives the result for T = t0(u). Therefore, let T > t0(u). Arguing by contradiction, assume Exexp(u>XT) < ∞ for all x ∈ E. Then by Jensen’s inequality we have
Exexp(λu>XT) ≤ (Exexp(u>XT))λ≤ 1 + Exexp(u>XT) < ∞, (6.10) for all 0 ≤ λ ≤ 1, x ∈ E. Let λ∗ = inf{λ ≥ 0 : λu 6∈ DR(T )}. Note that 0 < λ∗≤ 1 and λ∗u 6∈ D
R(T ), since u 6∈ DR(T ) and DR(T ) is an open neighborhood of 0. Considering λ∗u instead of u, we may assume without loss of generality that λ∗ = 1. In the following, we let un = λnu, for arbitrary λn ∈ [0, 1) such that λn ↑ 1 as n → ∞, so that un ∈ DR(T ) and un → u. We divide the proof into a couple of steps.
Step 1. If for some t ≤ T and x ∈ E we have lim
n→∞(ψ0(t, un) + ψ(t, un)
>x) = ∞, (6.11)
then lim supn→∞kψ(t, un)k = ∞. To prove this, suppose (6.11) holds for some t ≤ T , but lim supn→∞kψ(t, un)k < ∞. Then limn→∞ψ0(t, un) = ∞ and (6.11) holds for all x. Since un∈ DR(T ) ⊂ DR(t), the Markov property and Theorem 6.4 give Exexp(u>nXT) = Ex EXT −texp(u > nXt) = Exexp(ψ0(t, un) + ψ(t, un)>XT −t). Fatou’s Lemma yields
∞ = Exlim inf n→∞ exp(ψ0(t, un) + ψ(t, un) >X T −t) ≤ lim inf n→∞ Exexp(ψ0(t, un) + ψ(t, un) >X T −t) = lim inf n→∞ Exexp(u > nXT),
which contradicts (6.10) as un= λnu with 0 ≤ λn < 1. Step 2. It holds that
lim sup n→∞
6.2. Full range of validity 149
Indeed, since un∈ DR(T ) ⊂ DR(t0(u)), Fatou’s Lemma together with Theorem 6.4 gives
Exexp(u>Xt0(u)) ≤ lim infn→∞ Exexp(u
> nXt0(u))
= lim inf
n→∞ exp(ψ0(t0(u), un) + ψ(t0(u), un) >x),
for all x ∈ E. In view of Lemma 6.11 there exists an x0 ∈ E such that we have Ex0exp(u
>X
t0(u)) = ∞, whence
ψ0(t0(u), un) + ψ(t0(u), un)>x0→ ∞, as n → ∞. Step 1 yields (6.12).
Step 3. It holds that lim supn→∞kψ(T, un)k = ∞. To prove this, we show that there exists ε > 0 such that if lim supn→∞kψ(t0, un)k = ∞ for some t0∈ [t0(u), T ], then lim supn→∞kψ(t1, un)k = ∞ for t1= T ∧ (t0+ ε). By Step 2 and an iteration of the above implication, it follows that lim supn→∞kψ(T, un)k = ∞.
Write f (t, x) for the solution to the linear ODE (6.7) with f (0, x) = x. By continuity of f and the assumption {−1, 1}p ⊂ E◦, there exists ε > 0 such that f (−t, x) ∈ E for all x ∈ {−1, 1}p, 0 ≤ t ≤ ε. Let t
0 ∈ [t0(u), T ] and t1 = T ∧ (t0+ ε). Suppose lim supn→∞kψ(t0, un)k = ∞. Then in view of (6.9), there exist x ∈ {−1, 1}p and a subsequence of u
n (also denoted by un) such that
lim
n→∞ψ(t0, un)
>x = ∞.
As in the proof of Lemma 6.11 we have lim infn→∞ψ0(t0, un) > −∞. Hence
lim
n→∞(ψ0(t0, un) + ψ(t0, un)
>x) = ∞. (6.13)
Since t0− t1≥ −ε, we have y := f (t0− t1, x) ∈ E and by the semi-group property of the flow it holds that
EyXt1−s= f (t1− s, f (t0− t1, x)) = f (t0− s, x) = ExXt0−s, for s ≤ t0.
Let k be the non-negative function given by (6.5). It follows from Proposition 6.9 that ψ0(t1, un) + ψ(t1, un)>y = u>nEyXt1+ Z t1 0 k(EyXt1−s, ψ(s, un))ds ≥ u>nEyXt1+ Z t0 0 k(EyXt1−s, ψ(s, un))ds = u>nEyXt1+ Z t0 0 k(ExXt0−s, ψ(s, un))ds = u>n(EyXt1− ExXt0) + ψ0(t0, un) + ψ(t0, un) >x,
which tends to infinity as n → ∞. Step 1 yields lim supn→∞kψ(t1, un)k = ∞. Step 4. We are now able to conclude the proof. By Step 3 and (6.13) with t0 = T , there is an x ∈ {−1, 1}p and a subsequence of un (also denoted by un) such that
lim
n→∞(ψ0(T, un) + ψ(T, un)
>x) = ∞.
From (6.10) and Theorem 6.4 we obtain
1 + Exexp(u>XT) ≥ Exexp(u>nXT) = exp(ψ0(T, un) + ψ(T, un)>x), for all n. The right-hand side tends to infinity, whence Exexp(u>XT) = ∞, contrary to the assumption.
6.2.2
Complex-valued parameters
To prove Theorem 6.7 (ii) and (iii), we need continuity of x 7→ Exexp(u>Xt). Lemma 6.13. Consider the situation of Theorem 6.7. Then it holds that x 7→ Exexp(u>Xt) is continuous for all t ≥ 0, u ∈ S(DR(t)).
Proof. Let xn → x0, for some xn, x0∈ E. Define µn(dz) = pt(xn, dz) and µ(dz) = pt(x0, dz). Note that µn
w
→ µ as the characteristic functions converge in view of (4.5). By Skorohod’s Representation Theorem [34, Theorem 4.30] there exist random variables Yn, Y defined on a common probability space (Ω, F , P ) such that P ◦ Yn−1 = µn, P ◦ Y−1 = µ and Yn → Y , P -a.s. Now let u ∈ S(DR(t)) be arbitrary. It holds that | exp(u>Yn)| = exp(Re u>Yn) and
Z
exp(Re u>Yn)dP = exp(ψ0(t, Re u) + ψ(t, Re u)>xn)
→ exp(ψ0(t, Re u) + ψ(t, Re u)>x) = Z
exp(Re u>Y )dP,
for n → ∞, since Re u ∈ DR(t). An extended version of the Dominated Conver-gence Theorem [34, Theorem 1.21] yields
Exnexp(u >X t) = Z exp(u>Yn)dP → Z exp(u>Y )dP = Exexp(u>Xt), for n → ∞, whence x 7→ Exexp(u>Xt) is continuous.
Proposition 6.14. Consider the situation of Theorem 6.7 and let t > 0 be arbi-trary. Then S(DR(t)) ⊂ DC(t) and the affine transform formula (6.1) holds for all u ∈ S(DR(t)), x ∈ E.
6.2. Full range of validity 151
Proof. To show that S(DR(t)) ⊂ DC(t) we argue by contradiction. So assume there exists u ∈ S(DR(T ))\DC(T ) for some T > 0. Define λ0= inf{λ ≥ 0 : λu 6∈ DC(T )} and u0= λu. Then λ0> 0 and u06∈ DC(T ), since 0 ∈ DC(T ) and DC(T ) is open by Proposition 6.6 (ii). Define t0 = t0(u0) and let t < t0(u). Then we have u06∈ DC(t0), u0∈ DC(t) and λu0∈ DC(t0) for all λ ∈ [0, 1). By Proposition 6.12 it holds that
Exexp(u>Xt) = exp(ψ0(t, u) + ψ(t, u)>x),
for all u ∈ DR(t), x ∈ E. Both sides are analytic in u on DC(t) by Proposi-tion 6.6 (ii) and Lemma 4.9. Since DR(t) ∪ [0, u0] is a connected subset of DC(t), the equality in the above display extends to DR(t) ∪ [0, u0]. In particular it holds for u = u0. Hence
Exexp(u>0Xt) = exp(ψ0(t, u0) + ψ(t, u0)>x), for t < t0, x ∈ E. (6.14)
The left-hand side is continuous in t by quasi-left continuity of X. Following the proof of Corollary 4.26 one can derive that
Exexp(u>0Xt0) = 0,
for all x ∈ E◦, whence for all x ∈ E in view of Lemma 6.13.
We proceed with the argument that was used in the proof of Proposition 4.32. Fix δ ∈ (0, t0) and suppose there exists 0 < ε < t0− δ such that
ψ(t, u0) ∈ S(M (t0− δ)), for all t ∈ [0, δ + ε]. (6.15) Then Exexp(ψ(t, u0)>Xt0−δ) is well-defined for all t ∈ [0, δ + ε]. Let h < ε. Then
the Markov property gives
0 = Exexp(u>0Xt0+δ) = ExEXt0−δexp(u
> 0Xδ+h)
= exp(ψ0(δ + h, u0))Exexp(ψ(δ + h, u0)>Xt0−δ).
Hence Exexp(ψ(t, u0)>Xt0−δ) = 0 for all t ∈ [δ, δ + ε]. Since this expectation is
analytic in t on [0, δ + ε], we can extend this equality to the whole interval [0, δ + ε]. In particular Exexp(u>0Xt0−δ) = 0, which is absurd in view of (6.14).
It remains to show the existence of 0 < ε < t0− δ such that (6.15) holds. By (6.14), Jensen’s inequality and the Markov property, it holds for t < δ, x ∈ E that
Exexp(Re ψ0(t, u0) + Re ψ(t, u0)Xt0−δ) = Ex| exp(ψ0(t, u0) + ψ(t, u0) >X t0−δ)| = Ex|EXt0−δexp(u>0Xt)| ≤ ExEXt0−δexp(Re u>0Xt) = Exexp(Re u>0Xt0−δ+t).
Since Re u0 ∈ M (t0) ⊂ M (t0− δ), it follows from Proposition 6.12 together with the previous display that for t < δ, x ∈ E we have
Exexp(Re ψ(t, u0)Xt0−δ) ≤ exp(−Re ψ0(t, u0))Exexp(Re u
>
0Xt0−δ+t)
= exp(−Re ψ0(t, u0) + ψ0(t0− δ + t, Re u0)
+ ψ(t0− δ + t, Re u0)>x) < ∞. Fatou’s Lemma yields
Exexp(Re ψ(δ, u0)Xt0−δ) ≤ lim inf
t↑δ Exexp(Re ψ(t, u0)Xt0−δ)
≤ exp(−Re ψ0(δ, u0) + ψ0(t0, Re u0) + ψ(t0, Re u0)>x) < ∞,
for all x ∈ E. Hence ψ(t, u0) ⊂ S(M (t0− δ)) for all t ∈ [0, δ]. Since S(M (t0− δ)) is open and t 7→ ψ(t, u0) is continuous on [0, t0), we can replace t ∈ [0, δ] with t ∈ [0, δ + ε] for some small ε > 0 and the result follows.
