Time discretization of quadratic and superquadratic Markovian BSDEs with unbounded terminal conditions

Transcription

1 Time discreizaion of quadraic and superquadraic Markovian BSDEs wih unbounded erminal condiions Adrien Richou Universié Bordeaux 1, INRIA équipe ALEA Oxford

2 framework Le (Ω, F, P) be a probabiliy space, (W ) R + be a Brownian moion in R d, (F ) R + be his augmened naural filraion, T be a nonnegaive real number. We consider an SDE X = x + b(s, X s )ds + σ(s, X s )dw s, 0 0 wih sandard assumpions on b and σ, and a Markovian BSDE Y = g(x T ) + f (s, X s, Y s, Z s )ds Z s dw s.

3 Time discreizaion We consider a ime discreizaion of he BSDE. We denoe he ime sep by h = T /n and ( k = kh) 0 k n sands for he discreizaion imes. For X we ake he Euler scheme : X0 n = x X n k+1 = X n k + hb( k, X n k ) + σ( k, X n k )(W k+1 W k ), 0 k n. For (Y, Z ) we use he classical dynamic programming equaion Y n n = g(x n n ) Z n k = 1 h E k [Y n k+1 (W k+1 W k )], 0 k n 1, Y n k = E k [Y n k+1 ] + he k [f ( k, X n k, Y n k+1, Z n k )], 0 k n 1, where E k sands for he condiional expecaion given F k.

4 Remarks on simulaion We need o compue condiional expecaions. We have a speed of convergence. Theorem (J. Zhang [2004], B. Bouchard, N. Touzi [2004]) Le us assume ha g and f are Lipschiz funcions wih respec o x, y, z and. We define he approximaion error by e(n) = sup E Y n n 1 k Y k 2 + E 0 k n Then e(n) = O(1/n). k=0 k+1 k Z n k Z 2 d.

5 Quesion Wha happens when f has a quadraic or a superquadraic growh wih respec o z?

6 Simplificaions and resricion Simplificaions : Resricion : X = x + b(x s )ds + σ(x s )dw s, 0 0 Y = g(x T ) + X = x + 0 Y = g(x T ) + f (Z s )ds b(x s )ds + 0 f (Z s )ds σdw s, Z s dw s. Z s dw s.

7 A simple lemma Lemma Le us assume ha g is a Lipschiz funcion (no necessarily bounded) and f is locally Lipschiz : f (z) f (z ) C(1 + h( z ) + h( z )) z z, wih h : R R a nondecreasing funcion. Then here exiss a unique soluion (Y, Z ) S 2 M 2 such ha Z is bounded.

8 Skech of he proof Le us inroduce he BSDE Y M = g(x T ) + f (ρ M (Zs M ))ds Zs M dw s, wih ρ M he projecion on he cenered euclidean ball of radius M. We have Zs M = Ys M ( X s ) 1 σ and Y M = g(x T ) X T + = g(x T ) X T We can use Girsanov : and Z M s Y M [ E QM s (f ρ M )(Z M s ) Z M s ds Z M s (dw s (f ρ M )(Z M s )ds). = E QM [ g(x T ) X T ] ] g(x T ) X T ( X s ) 1 σ C. Z M s dw s

9 A remark for non markovian framework Remark Le us assume here ha he erminal condiion is g(x) wih g Lipschiz : g(x 1 ) g(x 2 ) C sup xs 1 xs 2. 0 s T Then he resul says rue, Z is bounded. Proof : we jus have o approximae g(x) by g n (X 1, X 2,..., X n ) (see J. Ma, J. Zhang [2002]). Resul already proved in P. Cheridio, M. Sadje [2012].

10 A generalizaion of he lemma We have g C Z C, ha is o say : he growh of Z is linked wih he growh of he derivaive of he erminal condiion. Generalizaion : Could we have g C(1 + x r ) Z C(1 + X r )?

11 Theorem Le us assume g(x) g(x ) C(1 + x r + x r ) x x f (z) f (z ) C(1 + z l + z l ) z z, wih l 1 (l = 1 : quadraic case).

12 Theorem Le us assume g(x) g(x ) C(1 + x r + x r ) x x f (z) f (z ) C(1 + z l + z l ) z z, wih l 1 (l = 1 : quadraic case). When rl < 1, here exiss a soluion (Y, Z ) in S 2 M 2 s.. Z C(1 + X r ). This soluion is unique amongs soluions s.. Y S 2 and E [e C ] T 0 Zs 2l ds < +. When rl = 1 he resul says rue only when T is small enough.

13 Remarks In he quadraic case, he uniqueness resul is new (f is no assume o be convex or concave). In he superquadraic case, exisence and uniqueness resuls are new (in he paper of X. Bao, F. Delbaen and Y. Hu (2010), g is bounded). Uniqueness resul allows us o obain a Feynman-Kac formula. In he pah-dependen framework, he resul says rue (?) wih he esimae Z C(1 + sup X s r ). 0 s

14 Remark on he uniqueness lemma Y 1 Y 2 = = f (Zs 1 ) f (Zs 2 )ds Z 1 s Z 2 s (dw s β s ds). Z 1 s Z 2 s dw s We have always a uniqueness resul in he class of processes such ha we are allowed o apply Girsanov. In our case, his class is no empy because we can apply Novikov s condiion : E [e C ] [ ] T 0 Zs 2l ds CE e C sup 0 s T Xs 2rl < + when rl < 1 or rl = 1 and C small enough.

15 Time approximaion Le us consider an iniial approximaion of (Y, Z ) : Le us denoe Y M = g(ρ M (X T )) + f (Zs M )ds Zs M dw s. (Y, Z ) e 1(M) (Y M, Z M ) e 2(n,M) (Y M,n, Z M,n ) and (Y, Z ) e(n,m) (Y M,n, Z M,n ).

16 Time approximaion Theorem We have e(m, n) e 1 (M) + e 2 (M, n) C e C 1M 2 + CeC2M2rl n. When rl < 1, we can choose M such ha e(m, n) = o((1/n) 1 ε ) for all ε > 0. When rl = 1, we can choose M such ha e(m, n) = 0((1/n) C 1 C 1 +C 2 ).

17 A parial resul in he general case When l = 1 and he erminal condiion is bounded, we can hrea he case X = x + b(x s )ds + σ(x s )dw s, 0 0 wih he growh assumpion σ(x) C(1 + x κ ). Lemma Z C(1 + X r+κ ).

18 Time approximaion Theorem When 2κ 1 r, we have e(m, n) C e C 1M 2 + CeC2M2. n When 2κ < 1 r, we can ake C 1 as small as we wan and we can choose M such ha e(m, n) = o((1/n) 1 ε ) for all ε > 0. When 2κ = 1 r, we can choose M such ha e(m, n) = 0((1/n) C 1 C 1 +C 2 ).

19 Two ineresing (?) open quesions Is i possible o obain a good speed of convergence for he ime discreizaion scheme when g is locally Lipschiz, unbounded and σ(x) is bounded? g is Lipschiz and σ(x) is Lipschiz wih linear growh? For he second poin i is already shown ha we can choose M such ha C e(m, n) (log n) k, for all k N. See P. Imkeller, G. dos Reis [2010], A. R. [2012].