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BSDEs with terminal conditions that have bounded Malliavin derivative ...and driver that have arbitrary growth Patrick Cheridito 1 Kihun Nam 2 1 ORFE, Princeton University 2 PACM, Princeton University BSDEs, Numerics, and Finance July 2012, Oxford University

BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

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Page 1: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

BSDEs with terminal conditions that havebounded Malliavin derivative...and driver that have arbitrary growth

Patrick Cheridito1 Kihun Nam2

1ORFE, Princeton University

2PACM, Princeton University

BSDEs, Numerics, and FinanceJuly 2012, Oxford University

Page 2: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Outline

1 MotivationThe Problem We StudiedPrevious Work

2 Our ResultsMain ResultsSketch of ProofSemilinear parabolic PDEsLipschitzness and Bounded Malliavin derivativeSemilinear Parabolic PDEs with Neumann BC or withDirichlet BC

Page 3: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Outline

1 MotivationThe Problem We StudiedPrevious Work

2 Our ResultsMain ResultsSketch of ProofSemilinear parabolic PDEsLipschitzness and Bounded Malliavin derivativeSemilinear Parabolic PDEs with Neumann BC or withDirichlet BC

Page 4: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Question

Yt = ξ +

∫ T

tf (s,Ys,Zs)ds −

∫ T

tZsdWs

Let W be n dimensional Brownian motion on (Ω, (Ft )0≤t≤T ,P)where (Ft ) be the filtration generated by W . ξ is FT measurableL2(dP) random variable. f is progressively measurable andf (·,0,0) ∈ H2.

(Pardoux and Peng) f is globally Lipschitz in(y , z)⇒ ∃!(Y ,Z ) ∈ S2 ×H2(Rn).(Kobylanski) ξ is bounded andf (t , y , z) = a0(t , y , z)y + f0(t , y , z) where a0 is bounded andf0has growth like b + ρ(|y |)|z|2 for an increasing functionρ⇒ ∃(Y ,Z ) ∈ H∞ ×H2

Can we assume superquadratic growth in z with/withoutsuperlinear growth in y?⇒ Under “regularity” condition on ξ, yes.

Page 5: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Outline

1 MotivationThe Problem We StudiedPrevious Work

2 Our ResultsMain ResultsSketch of ProofSemilinear parabolic PDEsLipschitzness and Bounded Malliavin derivativeSemilinear Parabolic PDEs with Neumann BC or withDirichlet BC

Page 6: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Previous Works

Lipschitz Driver: Pardoux & Peng(1990)Quadratic growth in z

Kobylanski(2000): bounded ξBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ξ

Superquadratic growth in zDelbaen et al.(2010): convex driver under Markovianframework and bdd ξ.Cheridito & Stadje(2011): convex driver and Lipschitz ξ.Richou(2011, preprint): Markovian framework.

Superlinear growth in yLepeltier & San Martin(1997): |y |

√log|y | like growth.

Briand & Carmona(2000): monotonicity conditionQuadratic in z and Polynomial in y

Kobylanski(2000): bounded ξFrei & Dos Reis(2012): bounded ξ and|f (s, y , z)| ≤ C(1 + |y |+ (1 + |y |k )|z|2)

Page 7: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Outline

1 MotivationThe Problem We StudiedPrevious Work

2 Our ResultsMain ResultsSketch of ProofSemilinear parabolic PDEsLipschitzness and Bounded Malliavin derivativeSemilinear Parabolic PDEs with Neumann BC or withDirichlet BC

Page 8: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Main Theorem

We assume the following conditions.(A1) ξ ∈ D1,2 with |Di

rξ| ≤ Ci dr ⊗ dP a.e.(A2) f (·,0,0) ∈ H4

(A3) For a nondecreasing function ρ,|f (s, y , z)− f (s, y ′, z)| ≤ B|y − y ′||f (s, y , z)− f (s, y , z ′)| ≤ ρ(|z| ∨ |z ′|)|z − z ′|

(A4) f (·, y , z) ∈ L1,2a and |Dr f (s, y , z)| ≤ φi(s),dr ⊗ dP-a.e. where

φi are Borel measurable with∫ T

0 φ2i (t)dt <∞

(A5) |Dr f (s, y , z)− Dr f (s, y ′, z ′)| ≤ K Rr ,s (|y − y ′|+ |z − z ′|) for all

|z|, |z ′| ≤ R where for a.a r , K Rr ,. ∈ H4 such that∫ T

0

∥∥K Rr ,.∥∥4H4 dr <∞

Then, our BSDE has a unique solution (Y ,Z ) ∈ S4 ×H∞(Rn).Moreover, the bound of Z i

t is(

Ci +∫ T

t φi(s)ds)

eB(T−t).

(cf)If K Rr ,s is bounded, we can replace (A2) to f (·,0,0) ∈ H2 and

Y ∈ S2 instead of S4.

Page 9: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Main Theorem

We assume the following conditions.(A1) ξ ∈ D1,2 with |Di

rξ| ≤ Ci dr ⊗ dP a.e.(A2) f (·,0,0) ∈ H4

(A3) For a nondecreasing function ρ,|f (s, y , z)− f (s, y ′, z)| ≤ B|y − y ′||f (s, y , z)− f (s, y , z ′)| ≤ ρ(|z| ∨ |z ′|)|z − z ′|

(A4) f (·, y , z) ∈ L1,2a and |Dr f (s, y , z)| ≤ φi(s),dr ⊗ dP-a.e. where

φi are Borel measurable with∫ T

0 φ2i (t)dt <∞

(A5) |Dr f (s, y , z)− Dr f (s, y ′, z ′)| ≤ K Rr ,s (|y − y ′|+ |z − z ′|) for all

|z|, |z ′| ≤ R where for a.a r , K Rr ,. ∈ H4 such that∫ T

0

∥∥K Rr ,.∥∥4H4 dr <∞

Then, our BSDE has a unique solution (Y ,Z ) ∈ S4 ×H∞(Rn).Moreover, the bound of Z i

t is(

Ci +∫ T

t φi(s)ds)

eB(T−t).

(cf)If K Rr ,s is bounded, we can replace (A2) to f (·,0,0) ∈ H2 and

Y ∈ S2 instead of S4.

Page 10: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Main Corollary

If we assume stronger solution than (A1) and (A2), we cangeneralize (A3).

(A1’) (A1) and |ξ| ≤ E(A2’) |f (s, y , z)| ≤ G(1 + |y |) + ψ(|y |, |z|)|z| for nondecreasing

function ψ(A3’) For nondecreasing functions ρ1 and ρ2,

|f (s, y , z)− f (s, y ′, z)| ≤ ρ1(|y | ∨ |y ′|)|y − y ′||f (s, y , z)− f (s, y , z ′)| ≤ ρ2(|y |, |z| ∨ |z ′|)|z − z ′|

(A5’) |Dr f (s, y , z)− Dr f (s, y ′, z ′)| ≤ K Rr ,s (|y − y ′|+ |z − z ′|) for all

|y |, |y ′|, |z|, |z ′| ≤ R where for a.a r , K Rr ,. ∈ H4

In addition, we assume (A4). Then, our BSDE has a uniquesolution (Y ,Z ) ∈ S∞ ×H∞(Rn). Moreover,|Yt | ≤ (E + 1)eG(T−t) − 1 := d(t) for all t a.s.|Z i

t | ≤(

Ci +∫ T

t φi(s)ds)

eρ1(d(t))(T−t) =: ei(t) for almost all t a.s.

ex)f (y , z) = |z|3 + |z|1+|z| |y |

2

Page 11: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Main Corollary

If we assume stronger solution than (A1) and (A2), we cangeneralize (A3).

(A1’) (A1) and |ξ| ≤ E(A2’) |f (s, y , z)| ≤ G(1 + |y |) + ψ(|y |, |z|)|z| for nondecreasing

function ψ(A3’) For nondecreasing functions ρ1 and ρ2,

|f (s, y , z)− f (s, y ′, z)| ≤ ρ1(|y | ∨ |y ′|)|y − y ′||f (s, y , z)− f (s, y , z ′)| ≤ ρ2(|y |, |z| ∨ |z ′|)|z − z ′|

(A5’) |Dr f (s, y , z)− Dr f (s, y ′, z ′)| ≤ K Rr ,s (|y − y ′|+ |z − z ′|) for all

|y |, |y ′|, |z|, |z ′| ≤ R where for a.a r , K Rr ,. ∈ H4

In addition, we assume (A4). Then, our BSDE has a uniquesolution (Y ,Z ) ∈ S∞ ×H∞(Rn). Moreover,|Yt | ≤ (E + 1)eG(T−t) − 1 := d(t) for all t a.s.|Z i

t | ≤(

Ci +∫ T

t φi(s)ds)

eρ1(d(t))(T−t) =: ei(t) for almost all t a.s.

ex)f (y , z) = |z|3 + |z|1+|z| |y |

2

Page 12: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Outline

1 MotivationThe Problem We StudiedPrevious Work

2 Our ResultsMain ResultsSketch of ProofSemilinear parabolic PDEsLipschitzness and Bounded Malliavin derivativeSemilinear Parabolic PDEs with Neumann BC or withDirichlet BC

Page 13: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Sketch of Proof

1 Assume that (A1-(A4) hold and (A5) holds with K independentof R, f is differentiable in y and z,and ρ being constant. Then,

Theorem (El Karoui et al(1997))

We have a unique solution (Y ,Z ) in S4(R)×H4(Rn). Moreover,(Y ,Z ) ∈ L1,2

a (Rn+1) and for all i = 1, . . . ,n,(Di

r Yt ,Dir Zt ) = (U r

t ,Vrt ) dr ⊗ dt ⊗ dP-a.e. and Z i

t = U tt dt ⊗ dP-a.e.

where U rt = 0 and V r

t = 0 for 0 ≤ t < r ≤ T , and, for each fixed r ,(U r

t ,Vrt )r≤t≤T is the unique pair in S2(R)×H2(Rn) solving the

BSDE

U rt = Di

rξ +

∫ T

t[∂y f (s,Ys,Zs)U r

s + ∂z f (s,Ys,Zs)V rs + Di

r f (s,Ys,Zs)]ds

−∫ T

tV r

s dWs.

Page 14: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Sketch of Proof

2 Using Ito formula on eBt |U rt |2, we can bound Dr Yt and Zt .

3 Show the solution exists uniquely for small time with Zbounded by big enough S. <= cutoff argument.

4 By comparison theorem, we get better bound for Z which isnot blowing up.

5 Iterate the procedure to time 0.6 Using mollification, we can show the claim for Lipschitz driver.7 Generalize to K R using the explicit bound of Z .8 Under (A1’)-(A3’), (A4) and (A5’), by comparison theorem, we

can bound Y and prove the corollary.

Page 15: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Outline

1 MotivationThe Problem We StudiedPrevious Work

2 Our ResultsMain ResultsSketch of ProofSemilinear parabolic PDEsLipschitzness and Bounded Malliavin derivativeSemilinear Parabolic PDEs with Neumann BC or withDirichlet BC

Page 16: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Markovian BSDEs and semilinear parabolic PDEs

For some B > 0, consider

Y t ,xs = h(X t ,x

T ) +

∫ T

sg(r ,X t ,x

r ,Y t ,xr ,Z t ,x

r )dr −∫ T

sZ t ,x

r dWr (1)

where X t ,xs = x +

∫ s

tb(r ,X t ,x

r )dr +

∫ s

tσ(r)dWr

where |b(s, x)− b(s, y)| ≤ B|x − y |, |b(s, x)| ≤ B(1 + |x |), and|σ| ≤ B.

(B1) h : Rm → R is Lipschitz with coefficient B.(B2)

∫ T0 |g(s,0,0,0)|2ds <∞.

(B3) For a nondecreasing function ρ,

|g(s, x , y , z)− g(s, x , y ′, z)| ≤ B|y − y ′||g(s, x , y , z)− g(s, x , y , z ′)| ≤ ρ

(|z| ∨ |z ′|

)|z − z ′|

(B4) |g(s, x , y , z)− g(s, x ′, y , z)| ≤ B|x − x ′|(B5) |g(s, x , y , z)− g(s, x ′, y , z)− g(s, x , y ′, z ′) + g(s, x ′, y ′, z ′)| ≤

BR|x − x ′| (|y − y ′|+ |z − z ′|) for all |z|, |z ′| ≤ R

Page 17: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

(B1’) (B1) with |h| ≤ B(B2’) |g(s, x , y , z)| ≤ B(1 + |y |) + ψ(|y |, |z|)|z|.(B3’) For nondecreasing functions ρ1 and ρ2,

|g(s, x , y , z)− g(s, x , y ′, z)| ≤ ρ1(|y | ∨ |y ′|)|y − y ′||g(s, x , y , z)− g(s, x , y , z ′)| ≤ ρ2

(|y |, |z| ∨ |z ′|

)|z − z ′|

(B5’) |g(s, x , y , z)− g(s, x ′, y , z)− g(s, x , y ′, z ′) + g(s, x ′, y ′, z ′)| ≤BR|x − x ′| (|y − y ′|+ |z − z ′|) for all |y |, |y ′|, |z|, |z ′| ≤ R

TheoremIf (B1)-(B5) hold, then BSDE (1) has a unique solution such thatX t ,x is a square integrable strong solution and(Y t ,x ,Z t ,x ) ∈ S4 ×H∞(Rn) .

TheoremIf (B1’)-(B3’),(B4), and (B5’) hold, then BSDE (1) has a uniquesolution such that X t ,x is a square integrable strong solution,(Y t ,x ,Z t ,x ) ∈ S∞ ×H∞(Rn).

Page 18: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Sketch of Proof

It is easy to check conditions of main theorem and corollary usingthe following two lemmas.

LemmaX t ,x ∈ D1,2 with |Dr X t ,x | ≤ BeBT .

Proof.Differentiability is proved in the reference and we can useGronwall’s inequality to Malliavin derivative of above SDE ofX t ,x .

Lemma (Proposition 1.1.4 of Nualart(2006))

If φ is L-Lipschitz and X ∈ D1,2, then φ(X ) ∈ D1,2 and there existsrandom vector R bounded by L such that Dφ(X ) = R(DX ).

Page 19: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Viscosity Solution

TheoremAssume either (B1)-(B5) or (B1’)-(B3’), (B4), (B5’). Then,u(t , x) := Y t ,x

t is a unique viscosity solution of

ut (t , x) + Ltu(t , x) + g(t , x ,u(t , x), (∇uσ)(t , x)) = 0 (2)u(T , x) = g(x)

whereLt =

12

∑i,j

(σσT )i,j(t)∂xi∂xj +∑

i

bi(t , x)∂xi .

Proof.By standard result, it is known u is a viscosity solution. By Ishii andLions(1990), under our conditions, viscosity solution is unique.

Page 20: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Classical Solution

Assume either (i) or (ii) holds.(i) Assume (B1)–(B5). Also, suppose that b(s, x),h(x), and

g(s, x , y , z) are C3 in (x , y , z) and they have bounded firstorder, second order, and third order derivatives in

(s, x , y , z) ∈ [0,T ]× Rm × R× Rn : |z| ≤ Mz

where Mz is the bound of Z t ,x in our previous theorem.(ii) Assume (B1’)–(B3’), (B4), (B5’). Also, suppose that

b(s, x),h(x), and g(s, x , y , z) are C3 in (x , y , z) and they havebounded first order, second order, and third order derivativesin

(s, x , y , z) ∈ [0,T ]× Rm × R× Rn : |y | ≤ M ′y , |z| ≤ M ′z

where M ′y and M ′z are the bounds of Y t ,x and Z t ,x in ourprevious theorem, respectively.

Page 21: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

TheoremThen, u(t , x) in previous theorem is actually of classC1,2([0,T ]× Rm;R) which solves (2). Moreover, if (i) holds,∇u(t , x)σ(t) is bounded for almost every t. If (ii) holds, then u(t , x)is bounded for all t and ∇u(t , x)σ(t) is bounded for almost every t.

Proof.Apply the standard result in El Karoui et al.(1997) with appropriatebound.

cf) When g(z) = a|z|d with d > 1, these results closely relatedwith Amour and Ben-Artzi(1998), Gilding et al.(2003) which dealswith classical solutions. For classical solution, Amour andBen-Artzi assumed bounded terminal condition with C2

b . Gilding etal. assumed the boundedness of terminal condition and fixeda = 1.

Page 22: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Outline

1 MotivationThe Problem We StudiedPrevious Work

2 Our ResultsMain ResultsSketch of ProofSemilinear parabolic PDEsLipschitzness and Bounded Malliavin derivativeSemilinear Parabolic PDEs with Neumann BC or withDirichlet BC

Page 23: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Lipschitz⇒Bdd Malliavin derivative

We provide a sufficient condition for (A1).

DefinitionWe call a random variable ξ Lipschitz continuous in the Brownianmotion W with constants C1, . . . ,Cn ∈ R+ if ξ = ϕ(W ) for afunction ϕ : C([0,T ];Rn)→ R satisfying

|ϕ(v)− ϕ(w)| ≤ sup0≤t≤T

n∑i=1

Ci |v i(t)− w i(t)|.

TheoremLet ξ be Lipschitz continuous in W with constantsC1, . . . ,Cn ∈ R+. Then ξ ∈ D1,2 and |Di

tξ| ≤ Ci dt ⊗ dP-a.e. for alli = 1, . . . ,n. Converse is not TRUE.

Page 24: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

ExampleAssume T = n = 1. Define

g(t) :=∞∑

k=1

(−1)k−12k11−21−k<t≤1−2−k, h(t) :=

∫ t

0g(s)ds,

and set

ξ :=

∫ 1

0h(t)dWt .

Then Dξ = h is bounded by 1.On the other hand, it follows from integration by parts that

ξ = − limk→∞

∫ 1−2−2k

0g(t)Wtdt .

Page 25: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Outline

1 MotivationThe Problem We StudiedPrevious Work

2 Our ResultsMain ResultsSketch of ProofSemilinear parabolic PDEsLipschitzness and Bounded Malliavin derivativeSemilinear Parabolic PDEs with Neumann BC or withDirichlet BC

Page 26: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Semilinear Parabolic PDEs with Neumann BC

Let O be a bounded open subset in Rm with smooth boundary. Letthe forward process in FBSDE (1) is changed to

X t ,xs = x +

∫ s

tb(X t ,x

r )dr +

∫ s

tσ(X t ,x

r )dWr +

∫ s

tn(X t ,x

r )dκt ,xr

κt ,xs =

∫ s

t1X t,x

r ∈∂Odκt ,x

r , κt ,x· is increasing

ExampleAssume O is an open bounded box, b is Lipschitz, σ is a constant.Then, X t ,x is Lipschitz in underlying Brownian motion. Therefore,X t ,x ∈ D1,2 and DX t ,x is bounded.

This gives a viscosity solution of semilinear parabolic PDE withNeumann boundary condition by Pardoux and Zhang(1998).

Page 27: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Semilinear Parabolic PDEs with Dirichlet BC

Consider the following Markovian BSDEs which is indexed by(t , x) ∈ [0,T ]× Rm with a random terminal timeτ := infs ≥ t : X t ,x

s /∈ O ∧ T for a bounded open set O ⊂ Rp.

Y t ,xs = h(X t ,x

τ ) +

∫ τ

s∧τg(r ,X t ,x

r ,Y t ,xr ,Z t ,x

r )dr −∫ τ

s∧τZ t ,x

r dWr

where X t ,xs = x +

∫ s

tb(r ,X t ,x

r )dr +

∫ s

tσ(r ,X t ,x

r )dWr

ExampleAssume O = (a,b) ⊂ R. Let

X t ,xs := x + σ(Ws −Wt )

τ := inf

s ≥ t : X t ,xs /∈ O

∧ T

where σ is a constant. Then, X t ,xτ is Lipschitz in underlying BM.

Page 28: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

This gives a probabilistic representation of a classical solution ofsemilinear parabolic PDE.(Ladyzenskaja at al. 1968). Peng(1992)used strong version of Ladyzenskaja’s theorem to prove this forLipschitz driver. We used our result to recover part of originalLadyzenskaja’s theorem.

Page 29: BSDEs with terminal conditions that have bounded Malliavin ...people.maths.ox.ac.uk/cohens/BNF/Nam.pdfBriand & Hu(2006,2008), Delbaen et al.(2011): unbounded ˘ Superquadratic growth

Summary

The regularity of the terminal condition gives the bound of Z(and Y ) in our BSDE. Then we can apply most of standardBSDE results.The Lipschitzness is a sufficient, but not necessary conditionfor bounded Malliavin derivative and this is often easier tocheck than Malliavin boundedness.Comparison theorem is crucial in the global existence ofsolution. In multidimensional superquadratic BSDE, we havesolution only for small time if we assume the terminalcondition has bounded Malliavin derivative.

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