Derivative operator for functionals of unit jump processes

In § 5.1 we restricted the path perturbationθ^ε_u(ω) = ω + εhui_. to directions u ∈ E_d with hui=R

Iu_tdt= 0. These perturbations do not influence the final state of the path X1◦θ^ε_u= X1, an essential factor in the derivation of the duality formula (5.33) that characterized the reciprocal class of the Wiener measure.

In the unit jump context the perturbation ¯θ^ε_¯v(ω⁰, ω) = ω + Y^{ε ¯v}(ω⁰) as defined in (2.32) is a random addition of jumps of size one. This addition of jumps will influence the final state X₁◦θ¯^ε_¯v = X1+ Y₁^{ε ¯v}, regardless of the direction of perturbation ¯v ∈ ¯E. For the study of the reciprocal class of a Poisson process, we therefore have to introduce a different perturbation.

6.2.1. Definition and properties of the jump-time derivative.

We follow an idea of perturbation of the jump-times that was introduced independently by Carlen, Pardoux [CP90] and Elliott, Tsoi [ET93].

Definition 6.9. Given a bounded, measurable u: I → R and small ε ≥ 0 we define the perturbation π^ε_u: J1(I) → J1(I), (x, {t1, . . . , tm}) 7→ (x, {t1+ εhuit1, . . . , tm+ εhuitm} ∩ I).

Let Q be any probability on J1(I). A functional F ∈ L²(Q) is called differentiable in direction u if the limit

(6.10) D_uF := − lim

ε→0

1

ε F ◦π^εu−F

exists in L²(Q). If F is differentiable in all directions u ∈ E, we denote by DF = (DtF)_t∈I ∈ L²(dt ⊗ Q) the unique process such that

D_uF= Z

I

D_tFu_tdt= hDF, ui holds Q-a.s. for every u ∈ E.

Note that the perturbation is well defined forε small enough, since −1 < εutfor all t ∈ I implies that the mapping t 7→ t+ εhuitis strictly monotone. We defined a true derivative operator in the following sense.

Lemma 6.11. The standard rules of differential calculus apply:

• Let F, G and FG be differentiable in direction u, then the product rule holds:

D_u(FG)= GDuF+ FDuG.

• Let F be differentiable in direction u, φ ∈ C^∞_b (R), then φ(F) is differentiable in direction u and the chain rule holds:

D_u(φ(F)) = φ⁰(F)D_uF.

Proof. Let us first proof the product rule: For F and G given in the statement we get

−D_u(FG) = lim

ε→0

1

ε (F ◦π^ε_u)(G ◦π^ε_u) − FG

= G lim

ε→0

1

ε F ◦π^ε_u−F
+ F lim

ε→0

1

ε G ◦π^ε_u−G
+ lim

ε→0

1

ε F ◦π^ε_u−F

G ◦π^ε_u−G
. The last term converges to zero by the Cauchy-Schwarz inequality:

Q

1

ε F ◦π^ε_u−F

G ◦π^ε_u−G
²!

≤ Q

1

ε(F ◦π^ε_u−F)

²!¹₂ Q

 G ◦π^ε_u−G
2¹₂ . As for the proof of the chain rule, we just have to use the Taylor expansion ofφ

1

εφ(F ◦ π^ε_u) −φ(F) = φ⁰(F)1

ε(F ◦π^ε_u−F)+1 ε

(F ◦π^ε−F)²O(1)

∧K,

for some K> 0. Then the second term converges to zero.  6.2.2. Fundamental examples and counterexamples of derivability.

By Lemma 6.2 the knowledge of the jump-times and initial condition is equivalent to the knowledge of the canonical process X. Instead of cylindric functionals F(X)= f (Xt1, . . . , Xtn) of the canonical process, it is natural in the context of unit jump processes to consider functionals of the jump-times:

SJ1 :=n

F : J1(I) → R, F((x, {t1, . . . , tm}))= f (x, t1, . . . , tn∧m, 1, . . . ), f ∈ C^∞_b (R × Iⁿ), n ∈ No . For F ∈ S_J1 we have F(X) = f (X0, T1∧ 1, . . . , Tn∧ 1) under the convention ∞ ∧ 1 = 1. We use the shorthand F(X)= f∧1(X₀, T1, . . . , Tn) := f (X0, T1∧ 1, . . . , Tn∧ 1) ∈ S_J1.

Proposition 6.12. Let Q be an arbitrary probability on the space of unit jump paths J1(I). Then all functionals F(X)= f∧1(X₀, T1, . . . , Tn) ∈ S_J1are differentiable in direction of any bounded function u : I → R. The derivative is given by DuF(X)=R

ID_tF(X) utdt with D_tF(ω) = DtF((x, {t1, . . . , tm})) := −

m∧nX

i=1

∂_1+if (x, t1, . . . , tn∧m, 1, . . . )1_[0,ti](t).

Proof. Let F(X) = f∧1(X₀, T1, . . . , Tn) ∈ S_J1 and a bounded u : I → R be arbitrary. By definition of the perturbation

F(X) ◦π^ε_u= f∧1(X₀, T1+ εhuiT1, . . . , Tn+ εhuiTn),

where T_i+ εhuiTi := ∞ if Ti= ∞ already. We use the Taylor expansion to get

−1

ε F(X) ◦π^ε_u−F(X)
= −

η∧n

X

i=1

∂_1+if∧1(X0, T1, . . . , Tn)huiT_i+ O(ε).

Then dominated convergence gives the L²(Q)-limit. 

Unfortunately it is fairly easy to find functionals that are not differentiable. The following example explains why functionals of the type F(X)= f (Xt1, . . . , Xtn) ∈ S are in general not differentiable.

Example 6.13. Let P be the law of a Poisson process. Take F = Xt ∈ L²(P) for some fixed t ∈ I and chose u ∈ E such that u ≥ 0 on I and hui_t > 0. We define the time-inversion related to the perturbationπ^ε_uby

(6.14)

Z

[0,τ^εu(t)]

(1+ εus)ds= t.

Thenτ^ε_u : I → [0, τ^ε_u(1)] is a deterministic function and (τ^ε_u)⁻¹(t) = t + εhuit. The perturbed functional is

F ◦π^ε_u= X_τ^ε_u(t),

since by u ≥ 0 and huit > 0 we have τ^ε_u(t) < t and τ^ε_u(t) → t ifε → 0. Since X is stochastically continuous under P we get

− lim

ε→0

1

ε(X_τ^ε_u(t)−Xt)= 0 holds P-a.s.

Thus if a derivative of X_t in direction u in the sense of Definition 6.9 exists, it has to be zero: The L²(P)-convergence of the perturbed functional X_τ^ε_u(t)would imply the almost sure convergence of a subsequence to the L²(P)-limit. But since the almost sure limit exists and is zero, the L²(P)-limit needs to be zero too. But since t −τ^ε_u(t)= εhui_τ^ε_u(t)we get

E

1

ε(Xt−X_τ^ε_u(t)) − 0

2!

= 1 ε²

∞

X

k=0

k²e^{−εhuiτεu(t)}(εhui_τ^ε_u(t))^k

k! = 1

ε² ε²hui²_τε

u(t)+ εhui_τ^ε_u(t) , which diverges forε → 0. Thus Xtis not differentiable in direction u if huit> 0.

In the next example we present a class of jump-time functionals that are not generally differentiable.

Example 6.15. Given a sequence( fj)j≥0of functions f_j ∈ C^∞

b(R × I^j) we define (6.16) F(ω) = F((x, {t1, . . . , tm})) :=

∞

X

j=0

fj(x, t1, . . . , tj)1^{_j=m}.

In the canonical form such functionals are conveniently given by F(X)= f_η(X₀, T1, . . . , T_η). Using an alternating sequence of ( f_j)_j≥0 we present a specific example of such functionals that is not differentiable with respect to the bridge of a Poisson process. Take f2 j ≡ 1 and f_{2 j+1} ≡ 0 for j ∈ N and u ≡ 1 ∈ E, then

1 ε



F(X◦π^ε_u) − F(X)

 = 1

ε(1_η◦π^ε_u=η−1+ 1_η◦π^ε_u=η−3+ . . . ),

and with respect to a Poisson bridge P^0,1(. ) = P( . |X0 = 0, X1 = 1) the a.s. limit of the above is zero since P^0,1(T₁ = 1) = 0. Remark that τ^ε_u(t) = t/(1 + ε) for u ≡ 1. We use P^0,1(η = 1) = 1 to compute

E^0,1

1

ε(F(X ◦π^ε_u) − F(X)) − 0

²!¹₂

= 1

εP^0,1(η ◦ π^ε_u= 0)¹²

= 1

εP^0,1(X_τ^ε_u(1)= 0)¹²

= 1 ε



P(X_τ^ε_u(1)= 0, X1= 1|X0= 0)/P(X1= 1|X0= 0)¹₂

= 1 ε

√ε

√ 1+ ε,

which diverges for ε → 0. We conclude that functionals of the form (6.16) are not in general differentiable in the sense of Definition 6.9.

In what follows, we are in particular interested in the differentiability of functionals of the stochastic integral type. Let us remark that the stochastic integral over the canonical unit jump process is always well defined as the finite sum

Z

I

u_sdX_s:=

η

X

i=1

u(T_i) for any measurable u : I × J1(I) → R.

We define the compensated integral (6.17) δ(u) :=

Z

I

us(dXs−ds), u : I × J1(I) → R predictable, u(., ω) ∈ L¹(dt), ∀ω ∈ J1(I).

It will be the dual operator of the derivative operator D with respect to a Poisson process.

In Example 6.13 we have seen thatδ(1[0,t]) = Xt−X₀−t is not differentiable with respect to the law P of a Poisson process. In the last two examples we develop conditions that guarantee differentiability for the compensated integrals δ(u).

Example 6.18. Let Q be an arbitrary probability on J1(I), v ∈ C¹_b(I) be deterministic and the total number of jumpsη ∈ L²(Q) be square integrable. Then the functional δ(v) is differentiable in direction of every u ∈ E with hui= 0.

Proof. We know that δ(v) =

η

X

i=1

v(Ti) − Z

I

vsds, and δ(v) ◦ π^ε_u=

η◦π^εu

X

i=1

v(Ti+ εhuiTi) − Z

I

vsds.

Under the assumption hui= 0 we have t+ εhuit= t − ε

Z

[t,1]u_sds ≤ t+ εK(1 − t) ≤ 1, ∀t ∈ I,

ifεK < 1, where K > 0 is a global bound for the absolute value of u. Therefore η ◦ π^ε_u= η identically on J1(I) for smallε > 0, and we can apply the differentiability of v ∈ C¹_b(I) to see that

(6.19) lim

ε→0

1

ε δ(v) ◦ π^ε_u−δ(v)
= Xη

i=1

v⁰(T_i)hui_Ti holds Q-a.s.

Since the increment of X is square-integrable and v⁰_.hui_.is bounded we can apply dominated

convergence to get the L²(Q)-convergence. 

In particular we have seen that the derivative ofδ(v) in direction u ∈ E with hui = 0 is given by (6.19). We define the process

(6.20) Dδ(v) = (Dtδ(v))_t∈I, with Dtδ(v) = − Xη

i=1

v⁰(T_i)1[0,Ti](t),

even if δ(v) is not differentiable in directions of bounded u with hui , 0. An additional assumption on v will allow differentiation in every direction.

Example 6.21. Let Q be a probability on J1(I) such thatη ∈ L²(Q) and v ∈ C¹_c(I) in the sense that v= 0 in a neighborhood of t = 0 and t = 1. Then the functional δ(v) is differentiable in every direction u ∈ E with respect to Q.

Proof. By assumption there exists a constantδ > 0 such that vt = 0 for all t ∈ [2δ, 1 − 2δ]^c.

which in combination with the above implies the result. 

We define the derivative of Dδ(v) of δ(v) as given in Example 6.21 by (6.20).

In document Reciprocal classes of Markov processes (an approach with duality formulae) (Page 94-98)