Lectures on the parametrix method II Sensitivities and ...

Lectures on the parametrix method IISensitivities and approximations

Alex Kulik

Wroc law University of Science and Technology

NOMP II, 23.03.2021

Alex Kulik Lectures on the parametrix method II 1/23 1 / 23

Differentiability of the solution to the parametrix integral equation

Recall that we have started from the Cauchy problem{(∂s + Ls,x)ps,t(x, y) = 0, s ∈ (−∞, t),ps,t(x, y)→ δy(x), s↗ t,

(1)

with the 2nd order PDO

Ltf(x) = a(t, x) · ∇f(x) +1

2b(t, x) · ∇2f(x),

which we have replaced by the integral equation

ps,t(x, y) = p0s,t(x, y) +

∫ t

s

∫Rdps,r(x, z)Υr,t(z, y) dzdr, (2)

Υs,t(x, y) = (∂s + Ls,x)p0s,t(x, y).

The solution to (2) is given by

ps,t(x, y) = p0s,t(x, y) +

∫ t

s

∫Rdp0s,r(x, z)Ξr,t(z, y) dzdr, (3)

Ξt(x, y) =∑k≥1

Υ~ks,t (x, y).


The further aim is to show that the solution to the integral equation (2) is the uniquesolution to (1)

Naive explanation: p0s,t(x, y) is C1 in s and C2 in x. Differentiating in (3) formally, weget

∇xps,t(x, y) = ∇xp0s,t(x, y) +

∫ t

s

∫Rd∇xp0s,r(x, z)Ξr,t(z, y) dzdr (4)

∇2xxps,t(x, y) = ∇2

xxp0s,t(x, y) +

∫ t

s

∫Rd∇2xxp

0s,r(x, z)Ξr,t(z, y) dzdr (5)

∂sps,t(x, y) = ∂sp0s,t(x, y)− Ξs,t(x, y) +

∫ t

s

∫Rd∂sp

0s,r(x, z)Ξr,t(z, y) dzdr (6)

Combining ∇x,∇2xx with coefficients we get then

(∂sps,t(x, y) + Ls,xps,t(x, y))

= Υs,t(x, y)− Ξs,t(x, y) +

∫ t

s

∫Rd

Υs,r(x, z)Ξr,t(z, y) dzdr

=

(Υ−

∞∑k=1

Υ~k +

∞∑k=1

Υ ~ Υ~k

)s,t

(x, y) = 0.


Warm up: parametrix-based expansions for sensitivities w.r.t. externalparameters

Let coefficients of the diffusion depend on external parameter θ = (α, β):

a = a(α; t, x), b = b(β; t, x).

Then

ps,t(θ;x, y) = ps,t(θ;x, y) +

∞∑k=1

(p0 ~ Υ~k)s,t(θ;x, y)

and, provided that the corresponding series converge,

∇θps,t(θ;x, y) = ∇θps,t(θ;x, y) +

∞∑k=1

∇θ(p0 ~ Υ~k)s,t(θ;x, y)

with

∇θ(p0 ~ Υ~k

)=(∇θp0

)~ Υ~k +

k−1∑j=0

p0 ~ Υ~j ~ (∇θΥ) ~ Υ~(k−j−1).


Using Fact 3 we get that

∂αp0s,t(θ;x, y) = (t− s)

∑i

∂αai(α; t, y)Φit−s(a(θ, t, y), b(θ, t, y);x, y)

∂βp0s,t(θ;x, y) =

t− s2

∑i,j

∂βbij(β; t, y)Φi,jt−s(a(θ, t, y), b(θ, t, y);x, y)

have orders (t− s)1/2 and (t− s)0, respectively. Similar but more cumbersomecalculations give that ∂αΥs,t(θ;x, y), ∂βΥs,t(θ;x, y) have orders (t− s)−1/2+δ and(t− s)−1+δ, and the parametrix series for ∇θps,t(θ;x, y) converges.

∂αps,t(θ;x, y) ≈O(t−s)1/2+δ (t− s)∑i

∂αai(α; s, x)Φit−s(a(s, x), b(s, x);x, y)

∂βps,t(θ;x, y) ≈O(t−s)δt− s

2

∑i,j

∂βbij(β; s, x)Φi,jt−s(a(θ, s, x), b(θ, s, x);x, y)

The higher order approximations are also available.


Derivatives in x

Lemma

The identity (4) for ∇xps,t(x, y) holds true and

|∇xps,t(x, y)| ≤ C(t− s)−1/2ϕc(t−s)(x, y), ϕt(x, y) = ϕt(y − x).

The key estimate:∂xip

0s,t(x, y) = Φit−s(a(s, x), b(s, x);x, y)

has order (t− s)−1/2, hence the integral∫ t

s

∫Rd|∇xp0s,r(x, z)||Ξr,t(z, y)| dzdr

is finite.


Lemma

The identity (5) for ∇2xxps,t(x, y) holds true and

|∇2xxps,t(x, y)| ≤ C(t− s)−1ϕc(t−s)(x, y)

Here the situation is more subtle because

∂2xixjp

0s,t(x, y) = Φi,jt−s(a(t, y), b(t, y);x, y)

has order (t− s)−1, and the integral∫ t

s

∫Rd|∂2xixjp

0s,r(x, z)||Ξr,t(z, y)| dzdr

has no means to be finite. Before proving the lemma, let us first establish certain Holdercontinuity properties of the kernels involved into the parametrix construction.


Proposition

For any δ < γ,

|p0t (x1, y)− p0t (x2, y)| ≤ C |x1 − x2|2δ

tδ

(ϕct(x1, y) + ϕct(x2, y)

),

|Υ(x1, y)−Υt(x2, y)| ≤ C|x1 − x2|2δt−1+γ−δ(ϕct(x1, y) + ϕct(x2, y)

)|Ξ(x1, y)− Ξt(x2, y)| ≤ C|x1 − x2|2δt−1+γ−δ


)|pt(x1, y)− pt(x2, y)| ≤ C |x1 − x2|

2δ

tδ


).

Similar statement holds true for the increments w.r.t. y.

The proof follows the same ‘parametrix’ strategy we have already discussed.


Proof of the Lemma

We have∂2xixjp

0s,t(x, y) = Φi,jt−s(a(t, y), b(t, y);x, y)

and the following two facts:

Φi,jt−s(a(t, y), b(t, y);x, y)− Φi,jt−s(a(t, x), b(t, x);x, y)

has order (t− s)−1+γ ; ∫Rd

Φi,jt−s(a(t, x), b(t, x);x, z) dz = 0.

These facts combined give∣∣∣∣∫Rd

Φi,jt−s(a(t, z), b(t, z);x, z) dz

∣∣∣∣ ≤ C(t− s)−1+γ .


Then∣∣∣∣∫Rd∇2xxp

0s,r(x, z)Ξr,t(z, y) dz

∣∣∣∣ ≤ ∣∣∣∣∫Rd∇2xxp

0s,r(x, z)Ξr,t(x, y) dz

∣∣∣∣+

∣∣∣∣∫Rd∇2xxp

0s,r(x, z)(Ξr,t(x, y)− Ξr,t(z, y)) dz

∣∣∣∣≤ C(r − s)−1+δ(t− r)−1+γ−δϕc(t−r)(x, y)

≤ C(r − s)−1+δ(t− r)−1+γ−δϕc(t−s)/2(x, y)

for r < t+s2

, and∣∣∣∣∫Rd∇2xxp


∣∣∣∣ ≤ C(r − s)−1(t− r)−1+γϕc(t−s)(x, y)

for r > t+s2. These estimates provide∫ t

s

∣∣∣∣∫Rd∇2xxp


∣∣∣∣ dr ≤ C(t− s)−1+γϕc(t−s)(x, y).


PMP and the properties of the evolutionary family of operators

We have proved the following

Proposition

For any f ∈ C∞(Rd) and t ∈ R, function u(s, x) = Ps,tf(x) satisfies{(∂s + Ls,x)u(s, x) = 0, s ∈ (−∞, t),u(s, x)→ f(x) s↗ t.

In other words, u(s, x) is harmonic for (∂s + Ls) on (−∞, t)× Rd and u(t, x) = f(x).

Each operator Ls satisfies the Positive Maximum Principle (PMP) on C2∞:

f(x∗) ≥ f(x), x ∈ Rd, f(x∗) ≥ 0 =⇒ Lsf(x∗) ≥ 0.


Proposition

Let a function u(s, x) be harmonic for (∂s + Ls) on (−∞, t)× Rd with Ls satisfying thePMP.Then u(t, x) ≥ 0 implies u(s, x) ≥ 0, s < t.

Corollary

I. For any f ≥ 0 and s ≤ t, Ps,tf(x) ≥ 0.

II. For any s ≤ r ≤ t, Ps,t = Ps,rPr,t.

III. For any s ≤ t and f ∈ C2∞,

Ps,tf(x) = f(x) +

∫ t

s

Ps,rLrf(x) dr.

To prove, apply Proposition to

I. u(s, x) = Ps,tf(x);

II. u(s, x) = ±Ps,tf(x)∓ Ps,rPr,tf(x), s ≤ r;

III. u(s, x) = ±Ps,tf(x)∓ f(x)∓∫ tsPs,rLrf(x) dr.


Summary

Assume that

A.1 Coefficients a(t, x), b(t, x) are bounded.

A.2 Coefficient b(t, x) is uniformly elliptic: for some β > 0,

v>b(t, x)v ≥ β|v|2.

A.3 Coefficient b(t, x) is Holder continuous: for some γ ∈ (0, 12],

|b(t, x)− b(t′, x)| ≤ C|t− t′|γ , |b(t, x)− b(t, x′)| ≤ C|x− x′|2γ .

Then the kernel ps,t(x, y) is the unique Fundamental Solution to the Cauchy problem for∂s + Ls in the class of kernels fs,t(x, y) such that for any ε > 0, s < t

|fs,t(x, y)| ≤ Ceε|y−x|2

.

The corresponding operator family {Ps,t} is positivity preserving, evolutionary, andsatisfies Forward and Backward Kolmogorov’s equations


Back to Processes: SDEs and Martingale Problems

The kernel ps,t(x, y) is the transition density for a Markov process X = {Xt} with thegenerating family

Ltf(x) = a(t, x) · ∇f(x) +1

2b(t, x) · ∇2f(x).

The process X is the unique weak solution to the SDE

dXt = a(t,Xt) dt+ σ(t,Xt) dWt, σ(t, x)σ(t, x)∗ = b(t, x).

Existence of some weak solution to SDE follows by compactness argument

By the Ito formula, for any weak solution X and u ∈ C1,2,

u(r,Xr)−∫ r

s

(∂v + Lv)u(v,Xv) dv

is a martingale.

Taking u(r, x) = Pr,tf(x), we get that

u(r,Xr), r ∈ [s, t]

is a martingale, which yields

P(Xt ∈ A|Fs) = Ps,t(Xs, A).


In the time homogeneous setting, the above argument is closely related to the notion ofthe martingale problem.D. W. Stroock, S. R. S. Varadhan (1979), Multidimensional Diffusion Processes,Springer, BerlinEthier, S.N., Kurtz, T.G. (1986) Markov Processes: Characterization and Convergence.Wiley, New YorkRecall that a process X is said to be a solution to the martingale problem (L,C2

∞(Rd)),if for every f ∈ C2

∞(Rd) the process

f(Xt)−∫ t

0

Lf(Xs) ds, t ≥ 0

is a martingale w.r.t. the natural filtration of X. The martingale problem is said to bewell-posed, if

1 for any probability measure π on Rd there exists a solution X to the martingaleproblem with X0 ∼ π, and

2 any two solutions with same π have the same distribution.

The above argument shows that the martingale problem for

Lf(x) = a(x) · ∇f(x) +1

2b(x) · ∇2f(x), f ∈ C2

∞(Rd)

is well posed.


Some other possibilities

We have|(∇x)kps,t(x, y)| ≤ C(t− s)−k/2ϕc(t−s)(x, y), k = 1, 2,

which yields|∂sps,t(x, y)| ≤ C(t− s)−1ϕc(t−s)(x, y).

I. In the time-homogeneous setting, this can be extended to

|(∂t)mpt(x, y)| ≤ C(t− s)−mϕc(t−s)(x, y), m ≥ 1.

Knopova, V., Kulik, A. (2018) Parametrix construction of the transition probabilitydensity of the solution to an SDE driven by α-stable noise. Annales de l’Institut HenriPoincare: Probabilites et Statistiques 54

pt(x, y) = p0t (x, y) +

∫ t/2

0

∫Rdp0t−s(x, z)Ξs(z, y) dzds

+

∫ t/2

0

∫Rdp0s(x, z)Ξt−s(z, y) dz ds.

∂tΥ~(k+1)t (x, y) =

∫ t/2

0

∫Rd

(∂tΥ~k)t−s(x, z)Υs(z, y) dzds

+

∫ t/2

0

∫Rd

Φ~ks (x, z)(∂tΥ)t−s(z, y) dzds

+

∫Rd

Υ~kt/2(x, z)Υt/2(z, y) dz.


II. Provided that a, b are Cm in x,

|(∇x +∇y)mps,t(x, y)| ≤ Cϕc(t−s)(x, y).

S.D. Eidel’man (1969) Parabolic Systems. North-Holland & Wolters-Noordhoff,Amsterdam.


Approximate fundamental solutions: the way to avoid (5)

Define for η > 0

ps,t,η(x, y) = p(0)s,t+η(x, y) +

∫ t

s

(p(0)s,r+η ∗ Ξr+η,t+η)(x, y) dr.

Denote

Ps,t,ηf(x) =

∫Rdps,t,η(x, y)f(y) dy,

∆s,t,ηf(x) = (∂s + Ls,x)Ps,t,ηf(x).

Lemma

Each Ps,t,ηf(x) is C1 in t and C2 in x, thus ∆s,t,ηf(x) is well defined. For eachf ∈ C∞(Rd),

Ps,t,ηf → Ps,tf, η → 0

uniformly on 0 ≤ t− s ≤ T and

∆s,t,ηf → 0, η → 0

uniformly on τ ≤ t− s ≤ T for any τ > 0.


(∂sps,t,η(x, y) + Ls,xps,t,η(x, y))

= Υs,t+η(x, y)− (p0s,s+η ∗ Ξs+η,t+η)(x, y)

+

∫ t

s

(Υs,r+η ∗ Ξr+η,t+η)(x, y) dr.

→

(Υ−

∞∑k=1

Υ~k +

∞∑k=1

Υ ~ Υ~k

)s,t

(x, y) = 0, η → 0.

We have proved the following

Proposition

For any f ∈ C∞(Rd) and t ∈ R, sequence uη(s, x) = Ps,t,ηf(x) satisfies{(∂s + Ls,x)uη(s, x)→ 0, s ∈ (−∞, t− τ),uη(s, x)→ u(s, x) := Ps,tf(x) .

In such a case, we call function u(s, x) approximate harmonic


Rebuilding the theory using approximate harmonic functions

A. Kulik (2019) Approximation in law of locally α-stable Levy-type processes bynon-linear regressions. Electronic Journal of Probability 24V. Knopova, A. Kulik, R. Schilling (2020) Construction and heat kernel estimates ofgeneral stable-like Markov processes, arXiv:2005.08491

Proposition

Let a function u(s, x) be approximate harmonic for (∂s + Ls) on (−∞, t)× Rd with Lssatisfying the PMP.Then u(t, x) ≥ 0 implies u(s, x) ≥ 0, s < t.

Corollary

I. For any f ≥ 0 and s ≤ t, Ps,tf(x) ≥ 0.

II. For any s ≤ r ≤ t, Ps,t = Ps,rPr,t.

III. For any s ≤ t and f ∈ C2∞,

Ps,tf(x) = f(x) +

∫ t

s

Ps,rLrf(x) dr.

We have a Markov process X with the transition probability Ps,t(x, dy) = ps,t(x, y)dy.


Proposition

Let a function u(s, x) be approximate harmonic for (∂s + Ls) and Y be a weak solutionto the SDE.Then u(r, Yr), r ∈ [s, t] is a martingale.

Mr,η := uη(r, Yr)−∫ r

s

(∂v + Lv)uη(v, Yv) dv → u(r, Yr), η → 0.

Taking u(r, x) = Ps,tf(x) we get that X, Y have the same laws.


Summary

I. ps,t(x, y) defines a Markov process X.

II. X is the unique weak solution to SDE.

III. ps,t(x, y) solves the Forward Equation.

IV. ps,t(x, y) solves the Backward Equation in the approximate sense.

These properties are based on the parametrix integral equation only, and allow easyextension to the Levy-type PDOs


Thank you!


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