Qu adratic B SDEs w ith convex genera to rs › procstoc › doc › jena06-slides.pdf · Qu...
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Quadratic BSDEs with convex generators
Philippe Briandjoint work with Ying Hu
IRMAR, Université Rennes 1, FRANCEhttp://perso.univ-rennes1.fr/philippe.briand/
Workshop on Stochastic Equations and Related TopicsJena, July 24–28, 2006
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 1/28
Backward Stochastic Di!erential Equation
Yt = ! +! T
tf (s, Ys, Zs) ds !
! T
tZs dBs (E!,f )
• ! is the terminal value : FT–measurable• f is the generator
• (Y , Z ) is the unknown• (Y , Z ) has to be adapted to F
Pardoux–Peng, ’90If f is Lipschitz w.r.t. (y, z) and
E"|!|2 +
! T
0|f (s, 0, 0)|2ds
#< +"
(E!,f ) has a unique square integrable solution.
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 2/28
Nonlinear Feynman-Kac’s Formula
Semilinear PDE (P)
"tu(t, x) + Lu(t, x) + f (t, x, u(t, x),#xu#(t, x)) = 0, u(T , .) = g,
Lu(t, x) =12trace(##!#2
xu(t, x)) + b(t, x) ·#xu(t, x).
Linear part =! SDE
X t0,x0t = x0 +
! t
t0
b$s, X t0,x0
s%
ds +! t
t0
#$s, X t0,x0
s%
dBs
Nonlinear part =! BSDE (B)
Y t0,x0t = g
&X t0,x0
T
'+
! T
tf$s, X t0,x0
s , Y t0,x0s , Z t0,x0
s%
ds !! T
tZ t0,x0
s dBs
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 3/28
Nonlinear Feynman-Kac’s Formula
If u is smooth solution to (P)&
u&
t, X t0,x0t
',#xu#
&t, X t0,x0
t
''solves the BSDE (B)
Feynman-Kac’s Formulau(t, x) := Y t,x
t is a (viscosity) solution to (P).
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 4/28
Quadratic BSDEs
A real valued BSDE
Yt = ! +! T
tf (s, Ys, Zs) ds +
! T
tZs dBs (E!,f )
• B is a Brownian motion in Rd ;• ! is FT–measurable;• the generator f : [0, T ]$ !$R $Rd !% R is measurable and
• (y, z) !"# f (t, y, z) is continuous• f is quadratic with respect to z:
|f (t, y, z)| $ !(t) + "|y| + #2 |z|2
where " % 0, # > 0 and ! is a nonnegative process.
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 5/28
The bounded case
If ! and $ – or more generally |$|1 :=! T
0$(s) ds – are bounded
• Existence• Uniqueness, Comparison Theorem• Stability
References:
• M. Kobylanski (1997 & 2000);• J.-P. Lepeltier and J. San Martin (superlinear framework, 1998);• M.-A. Morlais (non brownian setting, preprint)
These results yieldThe Nonlinear Feynman-Kac Formula
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 6/28
The unbounded case
• Boundedness of ! and $ is not necessary to construct a solution;• Exponential moment is enough !
Theorem (Ph. B. & Ying HU)
Let % := |!| +! T
0$(s) ds and let us assume that E
(exp
$&e"T %
%)< +".
Then, (E!,f ) has at least a solution such that
|Yt | &1&
log E$exp
$&e"T%
% ** Ft%.
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 7/28
Construction : f " 0, ! " 0
(Y n, Zn) minimal solution
Y nt = ! ' n +
! T
t1s"#n f (s, Y n
s , Zns ) ds !
! T
tZn
s dBs
#n = inf+
t ( 0 :! t
0$(s) ds ( n
,
Step 1: a priori estimate
0 & Y nt &
1&
E-
exp"&e"T
-! +
! T
0$(s) ds
.# **** Ft
.
Step 2: taking the limit in nDi!cult step: Localization procedure
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 8/28
The localization procedure
Main IdeaWork on the interval [0, 'k ] where
'k = inf/
t ( 0 :1&
E-
exp"&e"T
-! +
! T
0$(s) ds
.# **** Ft
.( k
0' T
Set Y nk (t) = Y n
t!!k, Zn
k (t) = 1t"!k Znt
Y nk (t) = Y n
$k +! $k
t#$k
1s"#n f (s, Y nk (s), Zn
k (s)) ds !! $k
t#$k
Znk (t) dBs
For fixed k
(Y nk )n$N is nondecreasing, 0 & Y n
k (t) & k
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 9/28
The localization procedurek fixed, limn#+$
Yk(t) = !k +! $k
t#$k
f (s, Yk(s), Zk(s)) ds !! $k
t#$k
Zk(s) dBs, !k = supn%1
Y n$k
• By construction
Yk(t) = Yk+1(t ' 'k), Zk(t) = 1t"$k Zk+1(t)
• Define (Y , Z ) by
Yt = Yk(t), Zt = Zk(t) if t & 'k
Yt = !k +! $k
t#$k
f (s, Ys, Zs) ds !! $k
t#$k
Zs dBs
• k !% +" gives a solution
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 10/28
Remarks
• superlinear framework and ! ) L1: Ph. B. & Y. Hu• Monotone continuous generators: Ph. B., J.-P. Lepeltier & J. San Martin• Reflected Quadratic BSDEs: J.-P. Lepeltier & M. Xu• Stochastic control : M. Fuhrman, Y. Hu & G. Tessitore
QuestionsUniqueness ? Stability ? Feynman-Kac’s formula ?
AnswersWhen f is convex (or concave) w.r.t. z
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 11/28
Motivation: Stochastic Control Problem
Controlled di!usion process
Xt = x +! t
0b(s, Xs) ds +
! t
0#(s, Xs)[dWs + r(us) ds]
where u takes its values in a nonempty closed set C .
Minimize the cost functional
J (u) = E"
g(XT) +! T
0G(t, Xt , ut) dt
#
over all the admissible controls u.
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 12/28
Motivation
Associated BSDE
Yt = g(XT) +! T
tf (s, Xs, Zs) ds !
! T
tZs dBs
Xt = x +! t
0b(s, Xs) ds +
! t
0#(s, Xs) dBs
f (t, x, z) = inf {G(t, x, u) + r(u)z : u ) C}
Important feature of the generator
z *!% f (t, x, z) is concave
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 13/28
Assumptions (H)
There exist ( ( 0, & ( 0 and a nonnegative process $ s.t. P–a.s.
• f is Lipschitz w.r.t. y: for any t, z,
|f (t, y, z)! f (t, y&, z)| & ( |y ! y&|
• quadratic growth in z:
|f (t, y, z)| & $(t) + (|y| + &
2 |z|2
• for any t, y, z *!% f (t, y, z) is a convex function;• ! is FT–measurable and
+) > 0, E"
exp-
)
"|!| +
! T
0$(s) ds
#.#< +".
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 14/28
Some estimatesProposition(E!,f ) has a solution (Y , Z ) s.t.
+p ( 1, E"
supt$[0,T] ep|Yt | +& ! T
0|Zs|2 ds
'p/2#& C
where C depends only on p, T and the exponential moments of |!| + |$|1.
• The estimate for Y comes directly from
|Yt | &1&
log E$exp
$&e"T(|!| + |$|1)
% ** Ft%
• For Z , standard computation starting from Itô’s formula to
1&2
&e%|Yt | ! 1! &|Yt |
'
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 15/28
Comparison theorem
TheoremLet (Y , Z ) and (Y &, Z &) be solution to (E!,f ) and (E!!,f !) where (!, f ) satisfies(H) and Y , Y & belongs to E (E := exponential moment of all order).If ! & !& and f & f & then
+t ) [0, T ], Yt & Y &t
If moreover, Yt = Y &t then
P-
! = !&,
! T
tf (s, Y &
s , Z &s) ds =
! T
tf &(s, Y &
s , Z &s) ds
.> 0.
In particular, (E!,f ) has a unique solution in the class E.
Main ideaEstimate of Yt ! µY &
t for µ ) (0, 1).
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 16/28
Proof: f independent of y
Set, for µ ) (0, 1), Ut = Yt ! µY &t , Vt = Zt ! µZ &
t .
Ut = UT +! T
tFs ds !
! T
tVs dBs, Fs = f (s, Zs)! µf & (s, Z &
s)
Ft = [f (t, Zt)! µf (t, Z &t )] + µ [f (t, Z &
t )! f & (t, Z &t )]
and *f (t) := f (t, Z &t )! f & (t, Z &
t ) & 0.
Zt = µZ &t + (1! µ)
Zt ! µZ &t
1! µ
f (t, Zt) = f1
t, µZ &t + (1! µ)
Zt ! µZ &t
1! µ
2
Convexity & µf (t, Z &t ) + (1! µ)f
1t, Zt ! µZ &
t1! µ
2
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 17/28
f (t, Zt)! µf (t, Z &t ) & (1! µ)f
1t, Vt
1! µ
2& (1! µ)$(t) +
&
2(1! µ)|Vt |2
Ft & µ*f (t) + (1! µ)$(t) +&
2(1! µ)|Vt |2
Second stepAn exponential change of variable to remove the quadratic term
Pt = ecUt , Qt = cPtVt , c ( 0
Pt = PT + c! T
tPs
&Fs !
c2 |Vs|2
'ds !
! T
tQs dBs
c = %1'µ yield
Pt & PT + &
! T
t
$$(s) + (1! µ)'1µ*f (s)
%Ps ds !
! T
tQs dBs
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 18/28
Pt & E-
exp"&
! T
t
$$(s) + (1! µ)'1µ*f (s)
%ds
#PT
*** Ft
.
PT = exp1
&
1! µ(! ! µ!&)
2= exp
1&
1! +
µ
1! µ*!
22
Pt & E-
exp"&
-! +
! T
0$(s) ds
.+ &
µ
1! µ
-*! +
! T
t*f (s) ds
.# *** Ft
.
In particular,
Yt ! µY &t &
1! µ
&log E
-exp
"&
-! +
! T
0$(s) ds
.# *** Ft
.
and sending µ to 1, we getYt !Y &
t & 0.
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 19/28
Strict Comparison
If Yt = Y &t , then Pt = e%Yt and
0 < E[Pt ] & E"
exp-
&
-! +
! T
0$(s) ds
.+ &
µ
1! µ
-*! +
! T
t*f (s) ds
.#.
Sending µ to 1,
0 < E"
exp-
&
"! +
! T
0$(s) ds
#.1&!+
R Tt &f (s) ds=0
#
Press if late
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 20/28
The general case
Ft = f (t, Yt , Zt)! µf &(t, Y &t , Z &
t ) = f (t, Yt , Zt)! µf (t, Y &t , Z &
t ) + µ*f (t)
f (t, Yt , Zt)! µf (t, Y &t , Z &
t )= f (t, Yt , Zt)! µf (t, Yt , Z &
t ) + µ (f (t, Yt , Z &t )! f (t, Y &
t , Z &t )) .
Convexity
f (t, Yt , Zt)! µf (t, Yt , Z &t ) & (1! µ)($(t) + (|Yt |) +
&
2(1! µ)|Vt |2
Linearization: a(t) = (Yt !Y &t )'1 (f (t, Yt , Z &
t )! f (t, Y &t , Z &
t )) 1Yt'Y !t (=0
µ (f (t, Yt , Z &t )! f (t, Y &
t , Z &t )) = µa(t) (Yt !Y &
t ) & a(t)Ut + (1! µ)(|Yt |
Ft & µ*f (t) + (1! µ)($(t) + 2(|Yt |) +&
2(1! µ)|Vt |2 + a(t)Ut .
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 21/28
The general case
Set Et = exp1! t
0a(s) ds
2, 3Ut = Et Ut and 3Vt = Et Vt . Then,
3Ut = 3UT +! T
t3Fs ds !
! T
t3Vs dBs
with, since |a(t)| & (,
3Ft & µEt *f (t) + (1! µ)Et($(t) + 2(|Yt |) +&e"T
2(1! µ)
*** 3Vt
***2
This is the same inequality as before.
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 22/28
Stability
Assume that (!n, fn) satisfies (H) with $n, (, & and
+) > 0, supn%1 E [exp {) (|!n| + |$n|1)}] < +".
TheoremIf !n !% ! P–p.s. and dt , dP–a.e., +(y, z), fn(t, y, z) !% f (t, y, z), then
+p ( 1, E"
exp&
p supt$[0,T] |Yt !Y nt |
'+
& ! T
0|Zs ! Zn
s |2 ds'p/2
#!% 0.
Proof.Same method as in the proof of comparison theorem to
Yt ! µY nt , Y n
t ! µYt
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 23/28
Application to PDEs
• Probabilistic representation for
"tu(t, x) + Lu(t, x) + f (t, x, u(t, x),#xu#(t, x)) = 0, u(T , .) = g,
Lu(t, x) =12trace(##!#2
xu(t, x)) + b(t, x) ·#xu(t, x).
• The SDE: X t0,x0 solution to
Xt = x0 +! t
t0
b(s, Xs) ds +! t
t0
#(s, Xs) dBs
• The BSDE: (Y t0,x0 , Z t0,x0) solution to
Yt = g&
X t0,x0T
'+
! T
tf$s, X t0,x0
s , Ys, Zs%
ds !! T
tZs dBs
• Nonlinear Feynman-Kac’s formula: u(t, x) := Y t,xt is a viscosity solution
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 24/28
Assumptions
• b, #, f and g are continuous;• b, # Lipschitz w.r.t. x
|b(t, x)! b(t, x &)| + |#(t, x)! #(t, x &)| & (|x ! x &|;
• restriction: # is bounded;• f is Lipschitz w.r.t. y
|f (t, x, y, z)! f (t, x, y&, z)| & (|y ! y&|;
• z *!% f (t, x, y, z) is convex;• -p < 2 s.t.
|g(x)| + |f (t, x, y, z)| & C$1 + |x|p + |y| + |z|2
%.
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 25/28
First properties
Propositionu(t, x) := Y t,x
t is continuous and
|u(t, x)| & C (1 + |x|p) .
Proof.• Since # is bounded
+) > 0, E4exp
&) supt$[0,T] |X t0,x0
t |p'5& eC(1+|x|p)
• a priori estimate Here
|u(t, x)| & C (1 + |x|p)
• Stability =. Continuity
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 26/28
u is a viscosity solution
DefinitionA continuous function u s.t. u(T , ·) = g is a viscosity subsolution(supersolution) if, whenever u ! + has a local maximum (minimum) at(t0, x0) where + is C1,2,
"t+(t0, x0) + L+(t0, x0) + f (t0, x0, u(t0, x0),#xu#(t0, x0)) ( 0, (& 0)
Solution = Subsolution + Supersolution
Propositionu(t, x) := Y t,x
x is a viscosity solution to the PDE.
Proof.• Markov property : u(t, X t0,x0
t ) = Y t0,x0t
• Comparison theorem
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 27/28
To do or at least try to do
• Replace the Lipschitz continuity in y by monotony• Weaken the integrability assumptions
|g(x)| + |f (t, x, y, z)| & C$1 + |x|2 + |y| + |z|2
%
• Prove uniqueness and stability without convexity
|f (t, y, z)! f (t, y, z &)| & C |z ! z &| (1 + |z| + |z &|)
• Conclude the talk #
Ph. Briand, Univ. Rennes 1 Quadratic BSDEs & convexity Jena, 28/07/2006 28/28