Introduction to Malliavin Calculus

Vlad Bally

26.03.2007 0.1. Introduction

We give a short introduction to Malliavin calculus which …nishes with the proof of Hörmandr’stheorem (which was the outstanding result obtained by P. Malliavin in his initial papers). We will not give all the complete proofs because this turns out to be rather heavy. In particular we shall skip the proofs of the convergence theorems which are used in order to check that a functional is smooth in Mal- lliavin sense We will also skip proofs for the evaluation of the Sobolev norms on the Wiener space. But we will discuss rather thoroughly the non degeneracy conditions. The paper is organized as follows. In the …rst section we introduce the simple functionals and the simple processes and we de…ne the Malliaivin derivative and the Skorohod integral for these …nite dimensional objects. Then we derive the duality formula. This permits to prove that the operators are closable and to extend them to the in…nite dimensional space. Then we prove the integration by parts formula and the basic calculus rules. We give the application of this formula to the analysis of the density of the law of a random variable. And we also prove the Clark Ocone formula. So the line of this presentation si to consider …rst a …nite dimensional setting and to de…ne there the di¤erential operators coming on in the Malliavin calculus and then to extend the calculus to functionals on the Wiener space. In the second section we give the expression of the di¤erential operators in terms of the Wiener chaos decomposition and we precise the domains of the operators. In fact we keep to a rather basic level and we give just a minimum of results in order to understand the connection with the decomposition in Wiener chaos. But in order to get a real idea about what analysis on the Wiener space is one has to look in some other texts as Nualart or Shigekawa (see also M. Santz for a short and simple presentation of the Malliavin calculus in which the Wiener chaos decomposition represents the starting point). Finally in the third section we discuss the case of di¤usion processes. We present …rst the elliptic case and then we give the result under Hörmander condition. 0.2. Basic notions

0.2.1. The Malliavin derivative and the Skorohod integral in the …nite dimensional framework On a probability space ( ;F;P ) we consider a d dimensional Brownian motion 1 d W = (W ; :::; W ) and we denote by Ft; t 0 the …ltration associated to W that is Ft = (Ws; s t): We work with t [0; 1]: We will use the following notation. k n 2 For each n N we denote t = k2 ; k N and 2 n 2 k;i = W i(tk+1) W i(tk ); i = 1; :::; d; k = 0; :::; 2n 1: n n n k k;1 k;d 0 2n 1 We also denote n = (n ; :::n ) and n = (n; :::; n ): Simple Functionals. A simple functional of order n is a random variable of the form d2n F = f(n); f C1(R ;R) 2 p d2n where Cp1(R ;R) designs the space of the in…nitely di¤erentiable functions which have polynomial growth and all their derivatives have also polynomial growth. We denote by Sn the space of the simple functionals of order n and by S = n N Sn the space of the "simple functionals". Notice that Sn Sn+1 and [ 2 2 2 S is a linear subspace of L ( ;F1;P ) which is dense in L ( ;F1;P ) (here comes on the fact that F1 = (Ws; s 1)): Simple Processes. A process U : [0; 1] R is called a simple process of order n if ! 2n 1 Ut = Uk1 k k+1 (t);Uk Sn: [tn;tn ) 2 k=0 X We denote by Pn the space of the simple processes of order n and P = nPn: [ Notice that Pn Pn+1:: d 2 Let U P : For each …xed ! ; t Ut is an element of L ([0; 1];B([0; 1]); dt : d 2 2 ! R ) =: Hd: The scalar product on this space is given by

1 d i i U; V = Us Vs ds h i 0 i=1 Z X for U = (U 1; :::; U d);V = (V 1; :::; V d): We will also denote

1 d p p i 2 p=2 L (Hd) = U : Hd : E U = E( U ds) < : f ! k kHd s 1g Z0 i=1 X

d p Notice that for each p N;P is a linear subspace of L (Hd) which is dense p 2 in L (Hd) Notice also that a simple process is generally not adapted. d The Malliavin derivatives. D : Sn P : Let F = f(n) Sn: We de…ne ! n 2 2n 1 i @f 1 d DsF = (n)1 k k+1 (s);DF = (D F; :::; D F ): @xk;i [tn;tn ) k=0 X A more intuitive notation: we denote

i;s = k;i for s [tk ; tk+1): n n 2 n n Then i @F DsF = i;s (n): @n

The operator D does not depend on n : for F Sn Sn+1 we have 2 @F @F i;s (n) = i;s (n+1): @n @n+1

d d This permits to de…ne (well) D : S = nSn P = nP by [ ! [ n

i @F DsF = i;s (n) for F Sn: @n 2

d The Skorohod integral (divergence operator). i : Pn Sn; i = 1; :::; d : 2n 1 ! let U = uk(n)1 k k+1 ( ) Pn: We de…ne k=0 [tn;tn ) 2 P 2n 1 k;i @uk 1 i(U) = (uk(n) (n) ): n @xk;i 2n k=0 X If U = (U 1; :::; U d) P d then we de…ne 2 n d (U) = i(U ) i=1 X d so : P S: Finally we check that the de…nition of i does not depend on n and we extend! this operator to P: Now : P d S: ! Remark 1. Suppose that U Pn is adapted. It follows that uk(n) does not de- k 2 2n 1 k;i pend on n and consequently @xk;i uk = 0: It follows that i(U) = k=0 uk(n)n = 1 U dW i: So for adapted processes the Skorohod integral coincides with the Itô 0 s s P integral. For this reason we will use the notation R 1 i i(U) = UsdWs : Z0 1 i But we keep in mind that generally 0 UsdWs is not an Itô integral but the Sko- rohod integral (an anticipative integral). R Proposition 1. (The Duality relation) Let F S and U P d: We have 2 2 E( DF;U ) = E(F (U)): h i d Proof. We assume (without loss of generality) that F Sn and U Pn with the same n: We write 2 2 2n 1 d k;i k;i i k;i k;i 1 E( DF;U ) = E(@ k;i f( ; )u ( ; ) h i x n n k n n 2n k=0 i=1 X X k;i k;i where n designs all the random variables involved in n; except n : In par- k;i k;i k;i n ticular ; are independent and N(0; h) with h = 2 so we obtain n n n k;i k;i i k;i k;i E(@xk;i f(n ; n )uk(n ; n )) 1 k;i i k;i y2=2h = E( @xk;i f( ; y)u ( ; y)e dy) p2h n k n ZR 1 k;i i k;i y i k;i y2=2h = E( f(n ; y)(@xk;i uk(n ; y) uk(n ; y))e dy) p2h R h Z k;i k;i k;i i k;i k;i n i k;i k;i = E(@ k;i f( ; )(u ( ; ) @ k;i u ( ; ))): x n n k n n h x k n n We come back and we obtain the formula.

0.2.2. Extension to the in…nite dimensional framework. We have

2 d 2 D : S L (R) P L (Hd); d 2 ! 2 : P L (Hd) S L (R) ! and we want to "extend" these operators. These are unbounded operators. This means that one may not …nd a constant C such that

1 d 2 i 2 2 2 DF 2 = E( D F ds) C F = CE F ; F S: k kL (Hd) s k k2 j j 8 2 Z0 i=1 X

EX. Give an example of sequence Fn S; n N such that sup Fn < and 2 2 n k k2 1 sup DF 2 = : n n L (Hd) Thenk ink order to be1 able to obtain a "well de…ned" extension of these operators we have to check that they are "closable", which means the following. Consider 2 a sequence Fn S; n N such that limn Fn = 0 in L (R) and limn DFn = G in 2 2 2 L (Hd): Then G = 0:

Lemma 2. D and are closable.

Proof. We consider the sequence Fn; n N as above and we want to prove that G = 0: We take U P d and we use the2 duality relation in order to obtain 2

E( G; U ) = lim E( DFn;U ) = lim E(Fn(U)) = 0: h i n h i n

d 2 Since P is dense in L (Hd) we conclude that G = 0: EX. Prove the is closable.

De…nition 3. We say that F DomD if there exists a sequence Fn S; n N such that 2 2 2 2 2 lim Fn = F in L (R); lim DFn = G in L (Hd) n n 2 for some G L (Hd): In this case we de…ne DF = G = limn DFn: 2

The de…nition does not depend on the sequence Fn; n N because D is closable. 2

2 Remark 2. Notice that we may just assume that limn DFn = G in L (Hd) weakly 2 (that is limn E( DFn;U ) = E( G; U ) for every U L (Hd): In order to prove it we use Mazur’stheorem:h i we passh to ai subsequence, and2 then, for each n N there n kn n kn n 2 exist k 0; k = n; :::; kn such that k=n k = 1 and Gn := k=n k DFk G kn n ! strongly. Notice that Gn = DF n with F n = k=n k Fk; which is still a simple kn Pn P functional. And F F n 2 k=n k F Fk 2 supk n F Fk 2 0: k Pk k k ! P

2 2 Notice that F;F 0 L (R) ; F F 0 L (R) and consequently DomD is not closed to multiplication,2 and then it is not2 an algebra. This is unpleasant and so we de…ne:

De…nition 4. Let p N: We say that F DompD if there exists a sequence 2 2 Fn S; n N such that 2 2 p p lim Fn = F in L (R); lim DFn = G in L (Hd) n n p for some G L (Hd): In this case we de…ne DF = G = limn DFn: 2

So DomD = Dom2D: And since and p p for kkp0 kkp kkL 0 (Hd) kkL (Hd) p p0; we have DompD Domp D: It is easy to see that the de…nition of DF 0 does not depend on p: We put Dom D = p N DompD and we then Dom D is an algebra. 1 \ 2 1 We would like to see DompD as a normed space so we de…ne

F = F + DF p : k k1;p k kp k kL (Hd) Then it is clear that 1;p DompD = Sk k :

1;p Notice that the signi…cance of "F Sk k " is the following: there exist Fn p 2 2 S; n N such that Fn F in L (R) and (Fn)n N is a Cauchy sequence in 2 ! 2 p : Then the above equality is obvious because L (Hd) is a complete space. k k1;p It is also obvious that DompD is complete. Indeed consider a Cauchy sequence

(Fn)n N with respect to 1;p : In particular this sequence is Cauchy with respect 2 k k p to and so there exists F L (R) such that Fn F in : Moreover, k kp 2 ! k kp since Fn DompD we may …nd a sequence of simple functionals Fn0 such that 2 1 Fn Fn0 1;p so that (Fn0 )n N is Cauchy with respect to 1;p and Fn0 F k k n 2 k k ! in p : So F DompD: kInk particular,2 for p = 2 the above norm is given by the scalar product

1 d i i F;G 1;2 = E(FG) + E( DsF DsGds) h i 0 i=1 Z X so that Dom2D is a Hilbert space. We denote

1;p 1;p 1; 1;p D = DompD = Sk k ;D 1 = p N D : \ 2 Sometimes it is unpleasant to compute limn DFn = G. So we give the following criterion. 2 Proposition 5. Let F L (R): Suppose that there exists a sequence Fn 2 2 2 DomD (in particular Fn S) such that limn Fn = F in L (R) and sup Fn 2 n k k1;2 C < : Then F DomD and F C: Moreover, if sup Fn Cp then 1 2 k k1;2 n k k1;p F Cp: k k1;p Proof. Any bounded set in a Hilbert space is relatively compact so we may 1;2 …nd F 0 D such that Fn F 0 weakly (we pass to a subsequence). We use now Mazur’stheorem:2 one may! construct convex combinations of the functionals Fn in order to obtain "strong convergence" instead of week convergence. Let n us be more precise: for each n N there exist k 0; k = n; :::; kn such that kn n kn 2n k=n k = 1 and F n := k=n k Fk F 0 strongly (with respect to 1;2): In 2 ! k k particular F n F 0 in L (R): Notice that P ! P

kn kn n n F F n 2 = k (F Fk) k F Fk 2 sup F Fk 2 0: k k k n k k ! k=n 2 k=n X X 1;2 It follows that F 0 = F and so F D : We also have 2

kn n F = lim F n lim k Fn C: k k1;2 n 1;2 n k k1;2 k=n X

Let us now prove the assertion concerning the p norm. Passing to a subse- 2 quence we may assume that ( F n F ; DF n DF ) (0; 0) almost surely. Hd ! Since sup F C we may use unifromly-integrability in order to derive n n 1;p p F n F in for p0 < p: Then F sup F n sup F n Cp: 1 ;p0 1;p0 n 1;p n 1;p ! k k k k 0 And …nally F 1;p supp

De…nition 6. Let p N: We say that U Domp if there exists a sequence d 2 2 Un P ; n N such that 2 2 p p lim Un = U in L (Hd); lim (Un) = F in L (R) n n

p for some F L (R): In this case we de…ne (U) = F = limn (Un): 2 On P d we consider the norm

U = U p + (U) k k;p k kL (Hd) k kp and we have dkk;p Domp = P :

Remark 3. Let p = 2: We may replace the strong convergence limn (Un) = F in L2(R) by week convergence (the same reasoning based on Mazur’stheorem as for the derivatives).

And a similar argument as above gives

2 Proposition 7. Let U L (Hd): Suppose that there exists a sequence Un 2 d 2 2 Dom2 (in particular Un P ) such that limn Un = U in L (Hd) and sup Un 2 n k k;2 C < : Then U Dom2 and U C: Moreover, if sup Un Cp then 1 2 k k;2 n k k;p U Cp: k k;p Proof: EX. By passage to the limit we readily obtain the extension of the duality relation:

Lemma 8. Let F DomD and U Dom2: Then 2 2 E( DF;U ) = E(F (U)): h i We give now an alternative characterization of the domain of :

2 Proposition 9. Let U L (Hd): Then U Dom2() if and only if 2 2 a) E( DF;U ) C F ; F D1;2: j h i j k k2 8 2 In this case

b) (U) 2 = sup E( DF;U ) : k k F =1;F S j h i j k k2 2 1;2 Proof. Suppose that U Dom2() and F D : Then using the duality 2 2 relation we obtain E( DF;U ) = E(F (U)) (U) 2 F 2 so we have a) with C = (U) : jSinceh S isi densej inj L2(R) wej obtain k k k k k k2

sup E( DF;U ) = sup E(F (U)) = (U) 2 : F =1;F S j h i j F =1;F S j j k k k k2 2 k k2 2 Assume now that a) holds true. Then using Riesz theorem we produce G L2(R) d 2 such that E( DF;U ) = E(FG): We take now Un P such that Un U in 2 h i 2 ! L (Hd) and consequently E( DF;Un ) E( DF;U ): It follows that E(F (Un)) h i ! h i ! E(FG): And this is su¢ cient in order to obtain U Dom2 (see the remark after 2 the de…nition of Dom2()): Finally we give the following computational rules:

1;2 Proposition 10. (Chain rule) Let F = (F1; :::; Fm) with Fi D and let C1(Rm; R): Then (F ) D1;2 and 2 2 b 2 m

D(F ) = @k(F )DFk: k=1 X 1; 1 m If Fi D 1 then the conclusion is true for C (R ; R): 2 2 p

Proof. If Fk S; k = 1; :::; m then (F ) S and the standard chain rule 2 1;2 2 n gives the formula. And if Fk D then we take Fk S; n N such that n 2 2 2n Fk Fk 1;2 0: Since has bounded derivatives we obtain (F ) (F ) 2 0k: Next wek write! k k !

m m m n n n D(F ) @k(F )DFk = @k(F )DF @k(F )DFk 0: k ! k=1 2 k=1 k=1 2 L (Hd) L (Hd) X X X m This yields (F ) Dom2D and D(F ) = k=1 @k(F )DFk: In order to prove the 2 n 2 1 2 last convergence we have to treat terms of the form an = E( @k(F ) @(F ) 0 DsFk ds) n 2 1 n 2 P j j j j and bn = E( @k(F ) 0 Ds(Fk Fk ) ds): Since @k is bounded it is clear that j j j j n R bn 0: In order to treat an we pass to a subsequence such that F F almost surely! and then we useR Lebesgue’sTheorem. ! In the above argument the fact that @k is bounded comes on crucially because 1; we are not able to use Holder inequalities. But if Fi D 1 then we may use 2 Holder’sinequality in order to get the same conclusion in the case when @k has polynomial growth.

1; Proposition 11. Let U Dom and F D 1: Then F U Dom and 2 1 2 2 1 (FU) = F (U) DF;U : h i Proof. EX. You test against G S and you use the duality relation and the 2 identity D(FG) = FDG + GDF: Finally we introduce the Ornstein Uhlenbeck operator L: We de…ne L : S S as L = D: For every F;G S we have the duality relation: ! 2 E(F LG) = E( DF;DG ) = E(GLF ): h i Then the same arguments as above ensure that L is closable and we may extend L :

De…nition 12. F DomL if there exists a sequence of simple functionals Fn; n 2 2 2 2 2 N such that Fn F in L (R) and LFn G in L (R) for some G L (R): Then ! ! 2 p we de…ne LF := G = limn LFn: If the above convergence holds in L (R); p 2 we say that F DompL: We put Dom L = p 2DompL: 2 1 \ p p p We may de…ne on S the norm F = F + LF and then DompL = k kL;p k kp k kp SkkL;p : And we may relax the convergence assumption on LFn; n N and replace 2 2 it by the assumption that (Fn)n and (LFn)n bounded in L (R) (respectively in Lp(R)):

Proposition 13. Let F = (F1; :::; Fm) with Fk Dom L; k = 1; :::; m: Then, 2 m 2 1 for every Cp (R ;R) we have (F ) Dom L and 2 2 1 m m

L(F ) = @k(F )LFk + @k@p(F ) DFk;DFp : h i k=1 k;p=1 X X Proof: EX.

0.2.3. Examples

i i i EX 1. F = Wt with t (0; 1): Then DsWt = 1[0;t)(s) 2 2 i 1 i 1;2 i i EX 2. L (0; 1): Then W () := 0 rdWr D and DsW () = (s) j i 2 2 and DsW () = 0 for j = i: If is a step function6 on dyadic intervalsR then W i() is a simple functional and we compute directly the derivative. If L2(0; 1) is a general function we 2 approximate it as follows: we denote by Gn the algebra generated by the dyadic intervals of order n: And we take n to be the conditional expectation of with respect to Gn (with respect to the Lebesgue measure on (0; 1)): Then n in L2(0; 1) and we pass to the limit in order to obtain the formula. ! 1 d 2 EX 3. We denote W () = (W (); :::; W ()): We take 1; :::; m L (0; 1) and F = (W ( ); :::; W ( )) with C1(Rdm;R) : Then F D1;2 and2 1 m 2 p 2 m i DsF = @xk;i (W (1); :::; W (m))k(s): k=1 X This is an immediate consequence of EX 2 and of the chain rule. In many text- books one considers directly F = (W (1); :::; W (m)) to be the simple functionals and de…nes the Malliaivn derivative by the above formula. In our case we have restricted ourselves to which is an indicator function of a dyadic interval. But …nally we obtain the same thing. 1 2 EX 4. Let U be an adapted process such that E 0 Ur dr < and Ii(U) = 1 i 1 j j 1 0 UrdWr ; i = 1; :::; d and I0(U) = 0 Urdr: We assume that for each r [0; 1] we 1;2 R 2 have Ur D and we assume that R 2 R

i) sup Ur 1;2 < : r 1 k k 1 n n n n Moreover we denote n(r) = k2 for k2 r < (k + 1)2 and Us := U n(s): We assume that 1 1 1 n 2 2 2 ii) Ur U dr = E ( Ur U (r) + DsUr DsU (r) ds)dr 0: k r k1;2 n n ! Z0 Z0 Z0 1;2 Then Ii(U) D ; i = 0; :::; d and 2 1 j j i ( ) D Ii(U) = ijUs + D UrdW ; j = 1; :::; d s s r Zs 0 with the convention that dWr = dr: n n In order to prove this we approximate Ii(U) by Ii(U ) with Ur = U n(r): By n 2 n ii) we have Ii(U ) Ii(U) in L (R): We check that ( ) holds for Ii(U ): And ! n using i) we obtain supn Ii(U ) 1;2 < : Then we use our criterion in order to 1;2 k k 1 get Ii(U) D : If we know ii) we are able to prove that ( ) holds for Ii(U) by passing to2 the limit. EX 5. We consider the di¤usion process

d t t i i i j i Xt = x + j(Xs)dWs + b (Xs)ds; i = 1; :::; m: j=1 0 0 X Z Z i i 1 m i 1;2 We assume that j; b Cb (R : R): Then we claim that Xt D and the derivative veri…es 2 2

d t m l i i i l k j DsXt = l(Xs) + @kj(Xr)DsXr dWr j=1 s k=1 X Z X t m i l k + @kb (Xr)DsXr dr: s k=1 Z X This may be proved using two types of approximations: 1. Euler scheme and 2. n;i Fix point argument. In both cases it is easy to check that supn Xt 1;2 < (here n;i 1 Xt designs the n0th approximation - Euler scheme or …x point approximation). This is an immediate consequence of Burkholder’s inequality and of Gronwall’s lemma. But this arguments does not guarantee that the continuity hypothesis ii) holds true. So we are not able to derive the equation veri…ed by the derivative (we may use "formal derivation" but this is not rigorous). So …nally we have to check the convergence of (Xn;DXn); n N: Another important point concerns2 the "variance of constants method" which permits to give a nice expression for DsXt: It is convenient to develop this argument for SDE0s written in Stratonovich form. So we assume that the i coe¢ cients j; j = 1; :::; d are twice di¤erentiable. Then the equation of X reads

d t t i i i j i Xt = x + j(Xs) dWs + b (Xs)ds; i = 1; :::; m j=1 0 0 X Z Z j d where dWs designees the Stratonovich integral and b = b j=1 j j: We de…ne r @Xi P Y ij = t ; i; j = 1; m: t @xj Then Y satis…es the following SDE written in matrix notation

d t t j Yt = I + @j(Xs)Ys dWs + @b(Xs)Ysds j=1 0 0 X Z Z i where @j is the m m matrix with components @kj; k; i = 1; :::; m: We also consider Z solution of the SDE

d t t j Zt = I Zs@j(Xs) dWs Zs@b(Xs)ds: j=1 0 0 X Z Z 1 Using Itô’sformula one checks that d(ZtYt) = 0 so that Zt = Yt : It follows that

DsXt = YtZs(Xs): In order to prove this one writes

d t t j Yt = Ys + @j(Xr)Yr dWr + @b(Xr)Yrdr j=1 s s X Z Z and multiplies with Zs(Xs) the above equation in order to obtain

d t j YtZs(Xs) = YsZs(Xs) + @j(Xr)YrZs(Xs) dW r j=1 Zs t X + @b(Xr)YrZs(Xs)dr: Zs Since YsZs(Xs) = (Xs) this is the equation of DsXt and so, using the uniqueness of the solutions of SED0s we get the equality.

0.2.4. The integration by parts formula An important consequence of the duality formula is the following integration by parts formula. In order to give the statement we have to de…ne the Malliavin 1;2 covariance matrix. Let F = (F1; :::; Fm) with Fi D : We de…ne 2 1 d ij k k = DFi;DFj = D Fi D Fjds: F h i s s 0 k=1 Z X This is a symmetric positive de…nite matrix. d 1 k k k 2 Notice that if Fi = (s)dW with L ([0; 1]) then the vector F k=1 0 i s i 2 is Gaussian and F coincides with the covariance matrix of F: EX. We introduce now theP followingR non-degeneracy assumption:

p (N D) E((det F ) ) < ; p N: 1 8 2 In the Gaussian case F is deterministic so this just means that det F = 0: This is 6 equivalent with the fact that F is invertible and this is a necessary and su¢ cient condition in order that the law of F is absolutely continuous with respect to the Lebesgue measure on Rm - so F is "non degenerated". 1 If (N D) holds true then F is almost surely invertible. We denote = : F F 1; 1; Theorem 14. Let F = (F1; :::; Fm) with Fi D 1 and G D 1: Suppose also i;j 1; j 2 2 that F D 1 and D Fi p N Domp; j = 1; :::; d and that (N D) holds for F: Then2 for every C1(R2m \;R2) we have 2 b E(@i(F )G) = E((F )Hi(F;G)); i = 1; :::; m with m m i;j i;j i;j Hi(F;G) = (G F DFj) = G F LFj D(G F );DFj : j=1 j=1 X X Proof. Using the chain rule

Ds(F ) = (F )DsF: r We multiply with DsF and we integrate with respect to s [0; 1] and we obtain 2 D(F );DF d = F (F ) h iH r which yields (F ) = D(F ); DF : r h F i Then we use the duality formula in order to obtain E( (F )G) = E( D(F ); G DF ) = E((F )(G DF )): r h F i F 1 d We also have the following easy generalisation. Consider i = (i ; :::; i ) (Dom ())m; i = 1; :::; m and denote 2 1 1 d ij k k = DFi; = D Fi (s)ds: F; j s j Z0 k=1 X 1; 1 d Proposition 15. Let F = (F1; :::; Fm) with Fi D 1 and let i = (i ; :::; i ) m ij 1; 2 p2 (Dom ()) ; i = 1; :::; m such that F; D 1; ij; = 1; :::; m and E(( det F; ) < 1 1; 2 1 m j j ; p N: Then for every G D 1 and C (R ;R) we have 1 8 2 2 2 p E(@i(F )G) = E((F )Hi (F;G); i = 1; :::; m with m m i;j i;j i;j Hi (F;G) = (G F;DFj) = G F;LFj D(G F;);DFj : j=1 j=1 X X Proof: Exactly the same but we multiply with j instead of DFj: 0.2.5. The Clark Ocone formula We recall the martingale representation formula: if F is a square integrable random variable which is F1 = (Ws; s 1) measurable then there exists a previsible 1 2 process = ( ; :::; ) such that E ds < and 1 d 0 j sj 1 R d 1 j F = EF + j(s)dWs : j=1 0 X Z In the case when F is di¤erentiable in Malliavin sense we may give a precise expression for j: This is done by the Clark Ocone formula:

Theorem 16. Let F D1;2: Then 2 d 1 j j F = EF + E(DsF Fs)dWs : j=1 0 j X Z Proof. We assume that EF = 0 (if not we work with F EF ): We take U = (U1; :::; Ud) a simple process which is adapted and we use the duality relation:

1 d 1 d j j E(F (U)) = E( DsF Uj(s)ds) = E(Uj(s)E(DsF Fs))ds: 0 j=1 0 j=1 j Z X Z X

Moreover, as Uj are adapted, the Skorohod integral coincides with the Itô inte- d d 1 j gral so that (U) = j=1 j(Uj) = j=1 0 Uj(s)dWs : Using the representation theorem we have F = d 1 (s)dW j so we obtain P j=1 0 j P s R

P R d 1 d 1 j j E(F (U)) = E(( j(s)dWs ) ( Uj(s)dWs )) j=1 0 j=1 0 X Z X Z 1 d = E( Uj(s)j(s)ds): 0 j=1 Z X It follows that

1 d j E( Uj(s)(E(DsF Fs) j(s))ds) = 0: 0 j=1 j Z X Since the adapted simple processes are dense (in L2 sense) in the adapted square j integrable processes we conclude that E(D F Fs) = (s) almost surely (with s j j respect to P ds): Consequences: 1. Let F D1;2: Then DF = 0 F = C (constant). 2 1;2 () 1;2 2. Let A F1: Then 1A D P (A) = 1 or P (A) = 0: In particular D is strictly included2 in L2: 2 () 2 In order to prove 2: we use the chain rule and we get D1A = D1 = 2D1A 1A: A If D1A = 0 then we may simplify and we get 1 = 2 1A which is not possible. So 6 D1A = 0 and we conclude that 1A is a constant - so zero or one.

0.2.6. Bismut-Elworthy formula and sensitivity computations Let us …rst present the problem of sensitivity computations - which is central in mathematical …nance and is known as "greeks computation". Assume that you have a …nancial asset St with the dynamics given by the Black Scholes model:

dSt = StdWt;S0 = x where W is an one dimensional Brownian motion. For simplicity we take the interest rate r = 0: The price of an option of maturity T and payo¤ on the asset S is given by = E((ST )): For some special payo¤s (as for example for call’sand put’s) one has an explicit expression for (a closed form) but generally such a formula is not available and one has to compute using a Monte Carlo method. This is a …rst stage. But in the second stage one needs to compute the sensitivity of the price with respect to certain parameters as the initial value of S; that is S0 = x (this derivative is called "delta" of the option) or the second derivative with respect to x; or the derivative with respect to ::: Each of these derivatives is denoted by a greek letter and this is why we speak about computation of Greeks. Anyway what we want to compute is @x: And there are several ways to do it - the simpler and the most common one is using …nite di¤erences. But it turns out that if the payo¤ function is singular such methods do not work very well (because they have a big variance) and then Malliavin calculus is an e¢ cient instrument in order to avoid the di¢ culty. Let us be more precise. We write

@x = @xE((ST )) = E(0(ST )@xST ): If is smooth then 0 makes sense and we may compute the above quantity using the Monte Carlo method. This is for example the case when (x) = (x K)+ which is the payo¤ of a call option. But if (x) = 1[K; )(x) (a digital option) 1 then 0 = K is a Dirac function and we are no more able to compute the above quantity. Then we have to use integration by parts in order to "regularize" 0: But if we go straight away we obtain E(0(ST )@xST ) = E((ST )(U)) where U is some stochastic process which is not adapted. So we have to deal with an anticipative integral and this is hard to simulate. We would like to get the same type of formula but with a process U which is adapted so that (U) becomes an Ito integral - which is simple to simulate. And this is what Bismut Elworthy formula produces. Now the motivation is given and we go on to the result itself. We consider an one dimensional di¤usion process

dXt = (Xt)dWt + b(Xt)dt:

1;2 We assume that and b are smooth so that Xt D : We denote Yt = @xXt;Zt = 2 1=Yt and by the variance of constants method we obtain DsXt = YtZs(Xs):

2 p Proposition 17. Suppose that ; b Cp and E( (Xt) ) < : Then for every 1; 1 2 j j 1 G D 1 and C 2 2 T T 1 Ys DsG Ys E(0(XT )YT G) = E((XT )(G dWs ds)) T (X ) (X ) Z0 s Z0 s where Yt = @xXt:

Proof. We write

Ds(XT ) = 0(XT )DsXT = 0(XT )YT Zs(Xs) which gives Ys 0(XT )YT = Ds(XT ) : (Xs) Since the equality holds true for each s [0;T ] we may integrate and obtain 2 T 1 Ys 0(X )Y = D (X ) ds: T T T s T (X ) Z0 s We use now the duality formula in order to obtain

T 1 GYs 1 GY E(0(XT )YT G) = E( Ds(XT ) ds) = E((XT )( )) T 0 (Xs) T (X ) Z T T 1 Ys DsG Ys = E((XT )(G dWs ds)): T (X ) (X ) Z0 s Z0 s Finally we give some exercises concerning sensitivity computations. EX 1. Assume that S follows the Black Scholes dynamics

dSt = StdWt;S0 = x and let = E((ST )): Compute @x and @ for = 1[K; ): EX 2. (Stochastic volatility model) Assume that S follows1 the Black Scholes dynamics

1 dSt = tStdWt ;S0 = x; d = ( )dt + dW 2 t t t where W 1 and W 2 are two correlated Brownian motions with d W 1;W 2 = h it dt; [ 1; 1]: Compute @x and @ for = 1[K; ): EX2 3. (Change option). Let S1 and S2 be two …nancial1 assets with dynamics

1 1 1 1 1 dSt = 1St dWt ;S0 = x ; 2 2 2 2 2 dSt = 2St dWt ;S0 = x :

1 2 1 2 where W and W are two correlated Brownian motions with d W ;W t = 1 h 2 i dt; [ 1; 1]: We consider an option which pays one dollar if ST ST : Compute 2 1 the sensitivity of the price with respect to 1 and to x : Solution of EX 2. The …rst step is to decorate the Brownain motions: We write W 1 = 1 2B1 + B2;W 2 = B2 t t t t t where B1;B2 are two independentp Brownian motions. Then clearly W 1 and W 2 are Brownian motions with correlation so the SDE0s become

2 1 2 dSt = St( 1 B + B );S0 = x; t t t d = ( )dt + dB2 t pt t We want to compute

@ = @E(1[K; )(ST )) = E(1[0K; )(ST ))@ST ): 1 1 This problem do not enter directly in the framework of the Bismut Elworthy formula but we will use the same strategy as there in order to solve it. First of all we solve the …rst equation:

T T T 2 1 2 1 2 ST = x exp( 1 tBt + tBt t dt): 0 0 2 0 p Z Z Z Then T T 2 1 @ST = ST ( B B ) = ST G with t t 2 t t 0 1 0 Z Z T T G = B2 p B1: t t 2 t t 0 1 0 Z Z We compute now p

1 2 D1S = S 1 2 = @ S s : s T T s T G p p It follows that (with = 1[K; )) 1 2 1 1 s 1 D (S ) = 0(S )D S = 0(S )@ S : s T T s T T T G p Then T G 1 1 1 G 0(ST )@ST = D (ST ) = D (ST ) ds: 2 s 2 s 1 T 1 0 s s Z We use now the dualityp formula with respectp to B1 and we obtain

T 1 1 G E(0(ST )@ST ) = E( D (ST ) ds) 2 s T 1 0 s Z 1 G = p E((ST ) 1( )) T 1 2 1 1 1 1 = p E((ST ) (G1( ) D G; )): T 1 2 p 1 T 1 1 Since is adapted we have 1( ) = 0 s dBs and we also have

1 R D G = s: s 1 2 We conclude that, up to an error, we havep

T 1 1 1 T E(0(ST )@ST ) = E((ST ) (G dB + )): 2 s s 2 T 1 0 1 Z p p

0.2.7. Higher order derivatives The higher order derivatives are de…ned in the same way as the …rst order derivatives: to begin one de…nes them on the simple functionals and then pass to the limit in order to obtain an extension. For F Sn we de…ne 2 @2F D(i;j) F = Di Dj F = : s1;s2 s1 s2 i;s1 j;s2 @n @n The de…nition does not depend on n: Moreover we have the following duality relation. For U1;U2 Pn we have 2 1 1 1 1 (i;j) i j E( D F U1(s1) U2(s2)ds1ds2) = E(F ( U1(s1) U2(s2)dW )dW )): s1;s2 s1 s2 Z0 Z0 Z0 Z0 In the above formula dW i and dW j designees Skorohod integrals. We dob not giveb a more explicit expression of the above double integral. But recall that U1 and U2 are simple processes; then it is clear that the above random variable is in any Lp: And using the above formula permits to check that D(i;j) is closable. Then one de…nes the domain of the second order derivative and the extension of this operator as usual. The notation is rather heavy so we prefer to give directly the from of the space of second order di¤erentiable functionals in terms of Sobolev norms. We de…ne on S the norm

1 d d p p i 2 p=2 i;j 2 p=2 F 2;p = E F +E(( DsF ds) )+E(( Ds1;s2 F ds1ds2) ): k k j j 2 Z0 i=1 Z(0;1) i;j=1 X X And we put 2;p 2;p 2; 2;p D = Sk k ;D 1 = p N D : \ 2 In order to de…ne higher order derivatives we proceed in the same way. We k consider a multi -index = ( 1; :::; k) 1; :::; d and we denote = k: Then, 2 f g j j for F Sn we de…ne 2 @kF D F = D 1 :::D k F = : s1;:::;sk s1 sk 1;s1 k;sk @n :::@n We use a duality argument in order to check that D is closable and we de…ne the extension of the operator. Finally we de…ne on S the norm

k p p 2 p=2 F k;p = E F + E(( Ds1;:::sr F ds1:::dsr) ) k k j j r r=1 Z(0;1) =r X jXj and k;p k;p k; k;p k;p D = Sk k ;D 1 = p N D ;D1 = k N p N D : \ 2 \ 2 \ 2 The space D1 is the "good" space in order to work because we will be able to iterate the integration by parts formula. It represents the analogues of C1 in the standard analysis. But notice that in Malliavin calculus (and in particular in the expression of the weight coming on in the integration by parts formula) we have two basic operators: the Malliavin derivative D and the Ornstein Uhlenbeck operator L: And for F D1 we are sure that we may use D as many times as 2 we want - but not L! It will be proved in the following section that for F D1 k 2 we may also compute L F for any k N: And moreover, D1 is an algebra. So a recurrence procedure based on the previous2 integration by parts formula gives:

Theorem 18. Let F = (F1; :::; Fm) with Fi D1;F DomL and LF D1 and 2 2 2 m let G D1: Suppose that (N D) holds true for F: Then for every Cp1(R ;R) 2 p 2 and every multi-index = ( 1; :::; p) 1; :::; m we have 2 f g E(@ (F )G) = E((F )H (F ; G)) with Hi(F;G); i = 1; :::; m given in the basic integration by parts formula and with H (F ; G) constructed by

H( 1;:::; p)(F ; G) = H p (F;H( 1;:::; p 1)(F ; G)):

Proof. We have to check that for each ; H (F ; G) D1: Once we know this we apply recursively the basic integration by parts formula.2 So we have to check that G D1 Hi(F ; G) D1: And this is true because D1 is an algebra 2 ) 2 and Fj D1 Fj DomL and LFj D1 (the last two assertions have to be 2 ) 2 2 proved - they follow from Mayer’sinequalities). Remark 4. We have the following useful estimates. For every multi-index and every p; k N there exists some constants C; q; q0; p0 (depending on and on p; k) such that 2

q q q0 q H (F ; G) C G (1 + F + LF ) (E(det F ) ) : k kk;p k kk+1;p0 k kk+2;p0 k kk;p0 0.2.8. Link between the integration by parts formula and the density of the law of a random variable We discuss this problem in an abstract framework. On a probability space ( ;F;P ) we consider some random variables F = (F1; :::; Fm) and G and a multi-index = ( 1; :::; k): We say that we have an integration by parts formula if there 1 exists some random variable H (F;G) L such that 2 m IP (F;G) E(@ (F )G) = E((F )H (F;G)) C1(R ;R): 2 b

Notice that H (F;G) is not unique. The Malliavin calculus produces such a weight, but maybe other methods produce other weights. Notice also that F and G play not a symmetric part in this formula. We will discuss the link between such integration by parts formulas and the density of the random variable F: For simplicity we consider …rst the one dimensional case m = 1: Now IPk(F;G) means

(k) IPk(F;G) E( (F )G) = E((F )Hk(F;G)) C1(R;R): 2 b

Suppose …rst that we have IP1(F; 1): Then we write down the following formal computation. If pF is the density of the law of F then

pF (x) = E(0(F x)) = E(1[00 ; )(F x)) = E(1[0; )(F x)H1(F; 1)): 1 1

So we obtain an integral representation for pF : We prove now that this computation may be done rigorously.

Proposition 19. Suppose that we have IP1(F; 1) Then the law of F is absolutely continuous with respect to the Lebesgue measure and the density veri…es

pF (x) = E(1[0; )(F x)H1(F; 1)): 1

In particular x pF (x) is contiguous. ! Moreover suppose that IP1(F;G) with G = H1(F; 1) holds true also and 1+ H1(F;H1(F; 1)) L for some > 0: Then x pF (x) is continuously dif- ferentiable and we2 have !

pF0 (x) = E(1[0; )(F x)H1(F;H1(F; 1))): 1

Proof. We take C1(R;R+) which is symmetric, non negative and null 2 1 1 outside ( 1; 1): We also assume that = 1: Then we de…ne (x) = " (x" ) " which converge weakly to the Dirac function. Moreover we denote "(x) = x R 1 "(y)dy which converge to 1(0; )(x) = 1(0; )(x) + 2 0(x): Take now f Cb: 1 1 1 2 Since f" = f n f we have E(f(F )) = lim" E(f"(F )): We use the integration byR parts formula in! order to obtain

E(f"(F )) = E( f(y) (F y)dy) = f(y)E(0 (F y))dy " " ZR ZR = f(y)E("(F y)H1(F; 1))dy f(y)E(1(0; )(F y)H1(F; 1))dy ! 1 ZR ZR This proves that the law of F is absolutely continuos. But then we may change 1(0; ) with 1(0; ) because the Lebesgue measure does not charge points. So we have1 the integral1 representation. And the continuity follows from the Lebesgue convergence theorem. Now we use one more integration by parts and we obtain

pF (x) = E(1[0; )(F x)H1(F; 1)) = E((F x)+H1(F;H1(F; 1))): 1 1+ And if we now that H1(F;H1(F; 1)) L then we may di¤erentiate under the 2 integral. So pF is di¤erentiable and pF (x) = E(1[0; )(F x)H1(F;H1(F; 1))): Let us now give a proposition which is somehow1 the converse of the above one.

Proposition 20. Suppose that the law of F is absolutely continuous with respect tot the Lebegue measure and that x pF (x) is continuous and E pF0 (F ) < :. 1 ! j j 1 Then for every g Cb and G = g(F ) we have the following integration by parts formula 2 E(0(F )G) = E((F )H1(F;G)) with H1(F;G) = (g0 + g@ ln pF ): Proof. We use standard integration by parts on R in order to get

E(0(F )G) = 0(x)g(x)pF (x)dx = (x)(g0(x)pF (x) + g(x)p0 (x))dx F ZR ZR = (x)(g0(x)pF (x) + g(x)@(ln pF (x))pF (x)dx ZR = E((F )(g0 + g@ ln pF )(F )): The multi-dimensional statement is the following.

Proposition 21. Let F = (F1; :::; Fm) be a random variable such that for each k multi-index = ( 1; :::; k) 1; :::; m the integration by parts formula I (F; 1) 2 f g 1 given in the beginning of this section, holds true. Then P F (x) = pF (x)dx and one has m

pF (x) = E( 1[0; )(Fi xi)H(1;:::;m)(F; 1)): 1 i=1 Y Moreover x pF (x) is of class C1 and ! m

@ pF (x) = E( 1[0; )(Fi xi)H(1;:::;m; 1;:::; k)(F; 1)): 1 i=1 Y The proof follows the same idea as above so we skip it.

0.3. Wiener chaos decomposition

For notational convenience we will work with an one dimensional Brownian motion 2 k in this section. We denote by L the space of the functions fk : [0; 1] R which s;k ! are symmetric ( we mean that f(t1; :::; tk) = f(t1 ; :::; tk ) for any permutation ) 2 2 and such that [0;1]k fk ds1:::dsk =: fk k < : For such a function we de…ne the iterated integral j j k k 1 R 1 t1 tk 1 ::: fk(t1; :::; tk)dWtk :::dWt1 Z0 Z0 Z0 and the "multiple stochastic integral"

1 t1 tk 1 Ik(fk) := k! ::: fk(t1; :::; tk)dWtk :::dWt1 : Z0 Z0 Z0 If we take fk to be constant on rectangles we may see that the iterated integral is the integral on a simplex - and since the function is symmetric, we have to multiply with k! (the number of simplexes) in order to obtain the integral on the whole cube [0; 1]k: EX: k = 2:::: We have the following basic relations:

i) E(Ik(fk) Ip(fp)) = 0 if k = p; 6 2 2 ii) E(Ik(fk)Ik(gk)) = k! fk; gk so E(I (fk)) = k! fk : h ik k k kk The second relation is obtained using the isometry property k times (after replac- ing the multiple integral by iterated integrals). In order to obtain the …rst relation suppose that k < p: We use the isometry property k times and the stochastic integrals in Ik(fk) disappear - we get a deterministic quantity. But we have still p k 1 stochastic integrals from Ip(fp): And they are of expectation zero. We de…ne the "chaos of order k" as

2 2 Hk = Ik(fk): fk L L ( ;F1;P ): f 2 k;sg 2 This is a linear closed subspace of L ( ;F1;P ): In order to see that Hk is closed n one employes the isometry property ii) : so if Fn = Ik(fk ); n N is a Cauchy 2 n 2 2 sequence in L ( ;F1;P ) then f ; n N is a Cauchy sequence in L and we get k 2 k;s a limit fk: Moreover, using i) we see that Hk and Hp are orthogonal for k = p: And we 6 have (with the convention H0 = R = constants)

2 L ( ;F1;P ) = 1 Hk: k=0 This is the chaos decomposition theorem (which I do not prove). So for every F 2 2 2 2 L ( ;F1;P ) there exist some kernels fk Lk;s; k N such that k1=0 k! fk k < and 2 2 k k 1 P 1 2 1 2 F = Ik(fk); F = k! fk : k k2 k kk k=0 k=0 X X We denote Prk(F ) = Ik(fk).

1;2 Proposition 22. A. For every k N one has Hk D and 2

DsIk(fk) = k Ik 1(fk(s; )): p;2 Moreover Hk p N D and for p k \ 2 p Ds ;:::;s Ik(fk) = k(k 1):::(k p + 1) Ik p(fk(s1; :::; sp; )): 1 p 2 B. Let fk Ls;k: Then (Ik 1(fk(s; )))s [0;1] Dom2 and 2 2 2 1 Ik 1(fk(s; ))dWs = (Ik 1(fk(s; )) = Ik(fk): Z0 C. Hk DomL and LIk(fk) = kIk(fk):

Proof. A. We denote

T t1 tk 1 Jk(fk)(T ) = ::: fk(t1; :::; tk)dWtk :::dWt1 Z0 Z0 Z0 1;2 and we prove that Jk(fk)(T ) D and that DsJk(fk)(T ) = Jk 1(fk(s; ))(T ): 2 Since Ik(fk) = Jk(fk)(1) A follows. We proceed by recurrence on k: Suppose that the assertion is true for k 1 and let us prove it for k: We will use the rule of derivation of stochastic integrals given in Ex 4. In order to do it we need to check some continuity properties but we postpone this for the end of the proof. For the moment we assume that we are allowed to use that rule. We have

Jk(fk)(T ) = Jk 1(fk(t1; ))(t1)dWt1 Z0 so we take derivatives and we obtain

DsJk(fk)(T ) = Jk 1(fk(s; ))(s) + DsJk 1(fk(t1; ))(t1)dWt1 : Zs

We use the recurrence hypothesis and we develop Jk 1(fk(s; ))(s) in order to get s T

DsJk(fk)(T ) = Jk 2(fk(s; t1; ))(t1)dWt1 + Jk 2(fk(t1; s; ))(t1)dWt1 : Z0 Zs

Now we use symmetry and get fk(t1; s; ) = fk(s; t1; ) so that T

DsJk(fk)(T ) = Jk 2(fk(s; t1; ))(t1)dWt1 = Jk 1(fk(s; )): Z0 So the formula is proved. We will now check that we have the needed continuity properties for Ut = Jk 1(fk(t; ))(t): T T T 2 2 E Ut U (t) dt + E DsUt DsU (t) dtds 0: n n ! Z0 Z0 Z0

It is easy to check that this is true if fk is continuos. So we have the result for continuous kernels. 2 Take now some general f Ls;k and approximate it by a sequence of continuos n 2 2 k kernels fk (because the continuous functions are dense in L [0; 1] : But these functions may be not symmetric. Then we take the symmetrization:

n 1 f (t ; :::; t ) = f n(t ; :::; t ) k 1 k k! k 1 k X n n where the sum is over all the permutation : Denote f (t1; :::; tk) =: f (t ; :::; t ): k k 1 k Since fk is symmetric we have fk = fk so that

n 1 n n f k fk fk fk k = fk fk k 0: k k! k k k k ! X n n n Now we have Ik(f k ) Ik(fk) and DsIk(f k ) = kIk(f k (s; )) Ik(fk(s; )) so ! 1;2 ! we conclude that Ik(fk) D and DsIk(fk) = Ik(fk(s; )): In order to obtain the2 formula for higher order derivatives we iterate the procedure. B. Suppose for a moment that we know that Ik 1(fk(s; )) Dom2 and let us prove the formula . We will taste against multiple integrals. 2 Using the duality relation we obtain

1 E(Ip(gp)(Ik 1(fk(s; ))) = E( DsIp(fp) Ik 1(fk(s; ))ds) 0 Z1 = p E(Ip 1(fp(s; )) Ik 1(fk(s; ))ds Z0 and this is zero if k = p and is k gk; fk if k = p: Moreover we have E(Ip(gp)(Ik(fk)) = 6 h i 0 if k = p and E(Ip(gp)(Ik(fk)) = k gk; fk if k = p: So the two coincides and 6 h i this proves that (Ik 1(fk(s; )) = Ik(fk): Let us now prove that Ik 1(fk(s; )) Dom2: Suppose …rst that fk is a step 2 functuion. Then Ik 1(fk(s; )) is a simple process so it belongs to the domain of : Then by approximation we get the result for fk continuos and then for a general fk (the fact that we know that (Ik 1(fk(s; )) = Ik(fk) is useful in order to prove convergence). C. Just composition L = D: We are now able to characterize the domains of the operators.

1;2 2 Proposition 23. We have D = F : 1 k k! fk < and f k=1 k kk 1g

1 P 1 2 2 2 DsF = k Ik 1(fk(s; )); F 1;2 = E(F ) + k k! fk k : k k k k k=1 k=1 X X p;2 k! 2 Moreover D = F : k1=p (k p)! k! fk k < and f k k 1g P 1 p k! Ds ;:::;s F = Ik p(fk(s1; :::; sp; )); 1 p (k p)! k=p X p 1 2 2 k! 2 F = E(F ) + k! fk : k kp;2 (k p )! k kk p =1 k=p 0 X0 X0 2 Proof. Let A := F : k1=1 k k! fk k < : If F A then we approxi- n f k k 1g 1;2 2 1;2 mate it by Fn = k=0 Ik(fk) and we obtain that F D : So A D and it is easy to check that it is a closedP linear subspace. In order2 to see that A = D1;2 we P 1;2 will check that every F D which is orthogonal (with respect to ; 1;2) on all the multiple integrals,2 is null. Indeed h i

0 = Ik(fk);F = E(Ik(fk)F ) + E( DIk(fk);DF ) = E((Ik(fk) + (DIk(fk))F ) h i1;2 h i = (1 + k)E(Ik(fk)F ):

And this implies F = 0:

Corollary 24. (Strook’sformula) If F Dp;2 then 2 1 f (s ; :::; s ) = E(Dp F ): p 1 p p! s1;:::;sp Proof. We take expectation in the formula of the derivative operator and we get 1 p k! E(Ds ;:::;s F ) = E(Ik p(fk(s1; :::; sp; ))): 1 p (k p)! k=p X For k > p we the expectation is null. And for k = p we get k!fk(s1; :::; sk): 3 EX. Write down the chaos decomposition of F = W1 using two methods: 1. Strook formula. 2. Clark Ocone formula.

2 2 Proposition 25. Dom2(L) = F : k1=1 k k! fk k < : And we have f 2 2 k 2k 1g LF = 1 k Ik(fk):and E LF = 1 k k! fk : k=1 j j P k=0 k kk TheP proof is the same as for theP derivatives, but we work with the norm 2 2 2 F L = F 2 + LF 2 : k k k k k k 2;2 An easy computation shows that F L = F 2;2 so Dom2(L) = D : One may de…ne other norms in thek k followingk k way. We have (I L)(F ) = s s k1=1(1 k) Ik(fk) so we may de…ne (I L) (F ) = k1=1(1 k) Ik(fk); s > 0: Then one de…nes P P p F = (I L)q=2(F ) = (E( (I L)q=2(F ) ))1=p: kj jkq;p Lp

Notice that for q = 2 we have F 2 = E((I L)(F ))2 = E(F 2) 2E(F LF ) + E(LF )2 kj jk2;2 = F 2 + 2E DF;DF = F 2 + 2E DF;DF k kL h i k k2;2 h i so we obtain F 2 F 2 3 F 2 : k k2;2 kj jk2;2 k k2;2 This inequalities may be generalized (completely non trivial) for every p 1 and q N: 2 Theorem 26. (Meyer’sinequalities [7]) For every p 1 and q N there exists 2 some cq;p;Cq;p such that

cq;p F F Cq;p F : k kq;p kj jkq;p k kq;p We will now discuss the domain of : Let u L2( ;L2([0; 1]): Then for almost every t [0; 1] we have u(t; ) L2( ) 2and consequently u(t; ) = 2 2 2 2 1 Ik(fk(t; )) and the sum converges in L ( ;L ([0; 1]): Having in mind a re- k=0 sult given previously we believe that Ik(fk(t; )) Dom2 and that (Ik(fk(t; ))) = P 2 Ik+1(fk): But the trouble is that now fk is a function of k + 1 variables which is not symmetric with respect to the …rst variable. So we have to come back and extend the previous result in this case. We de…ne the symmetrization of fk: 1 f (t ; ::::; t ) := f (t ; ::::; t ): k 1 k+1 (k + 1)! k 1 k+1 X Notice that if gk+1 is any symmetric function then we have

gk+1fk = gk+1f k: k+1 k+1 Z[0;1] Z[0;1]

Then we have the following result: the process (Ik(fk(t; ))t [0;1] is in the domain of (this is obtained by approximation and we do not insist 2 on it) and we have 1 (Ik(fk(t; )) = ( 0 Ik(fk(t; )dWt = Ik+1(f k): In order to prove the last equality we test against I (g ): pR p b 1 E((Ik(fk(t; ))Ip(gp)) = pE Ik(fk(t; )Ip 1(gp(t; )))dt: Z0 This quantity is zero if p = k + 1 and if p = k + 1 we get 6 1 (k + 1) E(Ik(fk(t; )Ik(gk+1(t; ))dt 0 Z 1 = (k + 1)k! fk(t; )gk+1(t; ))dt = (k + 1)! f kgk+1 k k+1 Z0 Z[0;1] Z[0;1] = E(Ik(f k)Ik+1(gk+1)) and we are done. The general result is now the following.

2 2 Proposition 27. Let u L ( ;L ([0; 1]) such that u(t; ) = k1=0 Ik(fk(t; ): 2 2 Then u Dom2() if and only if (k + 1)! f < (which is equivalent 2 k k k+1 1 P with the fact that I (f ) converges in L2): In this case (u) = I (f ): k k+1 k P k k+1 k

P P n Proof. Suppose that Ik+1(f ) converges. Then un(t; ) = Ik(fk(t; ) k k k=0 ! u(t; ) and (un) = k Ik+1(f k) k Ik+1(f k) so we have the result. Assume P ! P now that u Dom2() and let us prove that the above sum is convergent. Since 2 2 P P (u) L we have (u) = k Ik(fk) for some kernels fk and the above sum is convergent2 in L2: We write now P b b 1 E(Ip(fp)Ip(gp)) = E((u)Ip(gp)) = pE u(t)Ip 1(gp(t; ))dt 0 1 Z b = pE Ip 1(fp(t; )Ip 1(gp(t; ))dt Z0 and the same calculus as above gives E(Ip(f p)Ip(gp)): So we conclude that f p = fp and we are done. b 0.4. Di¤usion processes

We consider the di¤usion process

d t t i i i j i Xt = x + j(Xs)dWs + b (Xs)ds; i = 1; :::; m: j=1 0 0 X Z Z i i m We assume that j; b Cb1(R : R): A …rst result concerns the regularity of Xt in Malliavin sense. 2

i i m i Proposition 28. Suppose that j; b Cb1(R : R): Then Xt DompL; p N i i 2 2 8 2 and X ; LX D1 and one has t t 2 i i X + LX Ck;p(1 + x ) k; p N: t k;p t k;p j j 8 2 where Ck;p depends on the bounds of ; b and of their derivatives up to order k+2:

The proof is strightforward but rather long so we skip it.

Consider now the Malliavin covariance matrix Xt : We have seen in a previous section that we have DsXt = YtZs(Xs) where

d t t j Yt = I + @j(Xs)Ys dWs + @b(Xs)Ysds and j=1 0 0 X Z Z d t t j Zt = I Zs@j(Xs) dWs Zs@b(Xs)ds: j=1 0 0 X Z Z d j Here b = b j=1 j j and dWs designs the Stratonovich integral. Now the integration rby parts formula from Malliavin calculus reads: P i i m Theorem 29. Suppose that ; b C1(R : R) and moreover j 2 b 1 E( p ) < ; p N: (det Xt ) 1 2

m Then for every G D1 and every C1(R ;R) we have 2 2 p

E(@ (Xt)G) = E((Xt)H (Xt;G)) (1)

k where = ( 1; :::; k) 1; :::; m is a multi-index, @ is the derivative corre- 2 f g sponding to and the weights H (Xt;G) are de…ned by recurrecne:

m ij j ij j Hi(Xt;G) = ( G Xt LXt D(G Xt ;DXt ); j=1 X H (Xt;G) = H k (Xt;H( 1;:::; k 1 (Xt;G)) 1 with Xt = Xt :

We also have the following useful estimation:

Lemma 30. Let k; p N: There exists some constants C = Ck;p; q = qk;p; q0 = 2 qk;p0 (depending on the bounds of ; b and their derivatives up to order k + 2) such that for every multi-index with k one has j j

p q0 q q0 E H (Xt;G) C(1 + x ) G (E((det X ) )) : (2) j j j j k kk;q t This is another strightforward but long computation which we will skip. In fact, in a …rst step one proves the general inequality

p q0 q q0 E H (Xt;G) C(1 + F + LF ) G E((det X ) )) j j k kk;q k kk;q k kk;q t and then employs the estimation of F k;q ; LF k;q given above. k k k k q So the problem we have now is to estimate E((det Xt ) ): A …rst step is to use the variance of constants method in order to obtain

q 2q E((det X ) ) CqE((det Ut) ) (3) t where the matrix Ut is given by t i;j i;j Ut = (Zs(Xs)Zs) ds: Z0

This follows in the following way: we have Xt = Yt Ut Yt and then det Xt = 2 1 (det Yt) det Ut: One has YtZt = I so that (det Yt) = det Zt: And since Zt has 4q 4q …nite moments of any order we obtain E(det Yt) = E(det Zt) C Then we q 2q use Schwartz inequality and we obtain E((det Xt ) ) CqE((det Ut) ): So now on we will focus on U: We will …rst discuss the elliptic case - which is much simpler then the general case given by Hörmander’s theorem. We …x the starting point x Rm and we assume that for some c > 0 we have 2 (x) c I: (4) Then we have the following estimate:

Proposition 31. Suppose that (x) c I: Then q 4 mq+1 Cq E((det Ut) ) 1 + ( + 1) + (5) tc c2(mq+2) where Cq is a constant which depends on q; m and on the bounds of and b and of their derivatives of order one and two. Proof. Step 1. We compute …rst

t t Ut; = Zs(Xs)Z; ds = (Xs)Z; (Xs)Z ds h i h s i h s s i Z0 Z0 d t 2 = j(Xs)Zs; ds: j=1 Z0 X Then the smallest eigenvalue t of Ut is

d t 2 t := inf Ut; = inf j(Xs)Zs; ds: =1 h i =1 j j j j j=1 Z0 X m q mq Since det Ut t we have E((det U) ) E(t ) so we estimate this last quantity. Step 2. Using the elementary inequality (a + b)2 1 a2 b2 we obtain 2 d d d 2 1 2 2 (X )Z; (x); (X )Z (x); j s s 2 j j s s j j=1 j=1 j=1 X X X d 1 2 c j(Xs)Zs j(x) : 2 j=1 X

The second inequality hods true if = 1: In order to dominate the …rst term we j j d 2 2 have used the ellipticity assumption: j=1 j(x); = (x); c : Step 3 We take now " > 0 and a > 0 and we write h i j j P d a" 2 P (t ") P ( inf j(Xs)Zs; ds ") =1 j j j=1 Z0 X a" d 1 2 P ( ( c j(Xs)Zs j(x) )ds ") 2 Z0 j=1 X d ac 2 P ( " sup j(Xs)Zs j(x) a" "): 2 s a" j=1 X ac We take a = 4=c such that = 2 and we conclude that 2 d 2 P (t ") P (sup j(Xs)Zs j(x) a" ") (6) s a" j=1 X d p d 2 c 4 2p = P (sup j(Xs)Zs j(x) ) p E(sup j(Xs)Zs j(x) ) s a" 4 c s a" j=1 j=1 X X Cp Cp (a")p = "p cp c2p where Cp is a constant which depends on p and on the bounds of and b and of their derivatives of order one and two. mq Step 4. We estimate now E(t ): We write

mq mq 1 mq E(t ) = E(t 1 t>1 ) + E(t 1 1=k t>1=(k+1) ) f g f g k=1 X 1 mq 1 1 + (k + 1) P (t ): k k=1 X We will now use (6). Notice that this inequality holds true only if a" < t (because t a" 1 4 we have used the inequality 0 0 ): So, if " = k ; we need c k < t which gives 4 k > tc : So we come back and write R R

1 mq mq mq 1 E(t ) 1 + (k + 1) + (k + 1) P (t ) 4 4 k k tc k> tc X X 1 4 mq+1 Cp mq 1 1 + ( + 1) + 2p (k + 1) p : tc c 4 k k> tc X In order to get a convergent sum we take p = mq + 2 and the above inequality gives mq 4 mq+1 Cmq+2 E(t ) 1 + ( + 1) + : tc c2(mq+2) We turn no to the application of the above result to the study of the density of the law of Xt:

i;j i m Theorem 32. Suppose that ; b Cb1(R ;R) and (x) c I: Then the 2 law of Xt(x) is absolutely continuous with respect to the Lebesgue measure and m the density pt(x; y) is of class C1 with respect to y R : Moreover 2 2 1 C0 x y pt(x; y) C( + 1) exp( j j ) (c t)p t where C;C0; p are constants which depend on m and on the bounds of ; b and of their derivatives . The same inequality holds true for the derivatives of the density with respect to y: Proof. We have the integration by parts formula

E(@ (Xt)) = E((Xt)H (Xt; 1)) with

1 qp H (Xt; 1) Cp(1 + x )( + 1) ; p N: k kp j j c t 8 2

Let us …rst study the di¤erentiability of pt(x; y): We will use a formal calculus but it is clear that one may get a rigorous presentation using a regularization procedure. We take m 1 i m (y) = 1[0; )(y ) (y )+ 1 i=1 Y and = (1; 2; :::; m 1; m; m) so that @ = 0: Then using integration by parts one obtains

pt(x; y) = E(0(Xt y)) = E(@ (Xt y)) = E((Xt y)H (Xt; 1)): Then it is easy to check that we may di¤erentiate the term in the right hand side and we obtain

m i i @ym pt(x; y) = @ym E((Xt y)H (Xt; 1)) = E( 1[0; )(Xt y )H (Xt; 1)): 1 i=1 Y For higher order derivatives one proceeds in a similar way. Let us now prove the estimates for the tails of the density. If we work with Xt directly there will be some (1 + x ) coming on in our estimate (because this j j quantity appears in the estimate of H (Xt; 1) p): This is why it is more clever k k qp to look to F := Xt x: Notice that we have H (Xt x; 1) p Cp(1 + c t) k k without 1 + x : This is because Xt x Cp: Let us denote qt(x; y) the j j k kp density of F = Xt x: We have pt(x; y) = qt(x; y x): So if we prove that 2 2 C0y C0 y x qt(x; y) C exp( jt j ) then we get pt(x; y) C exp( j t j ): We take y such that yi > xi; i = 1; :::; m and we write

m i i m i i 1=2 qt(x; y) = E( 1[0; )(F y )H (F; 1)) (P ( i=1 F > y )) H (Xt x; 1) 2 1 i=1 \ f g k k Y i i 1=2 q2 ( min P (F > y )) C2(1 + (c t) ): i=1;m We use now Bernstein inequality in order to prove that

d t t i 2 i i i;j j i i C0(y ) =t P (F > y ) = P ( (Xs)dWs + b (Xs)ds > y ) Ce : j=1 0 0 X Z Z Then

i i i 2 2 min P (F > y ) C exp( C0 max(y ) =t) C exp( C0 y =t): i=1;m j j We present now the Hörmander condition. For two di¤erentiable functions f; g : Rm Rm we de…ne the Lie bracket [f; g] := f@g g@f which reads on ! i m j i j i 1 d components [f; g] := j=1 f @jg g @jf : Then we denote j = (j ; :::; j ); j = 1 m 1; :::; d and 0 = (b ;P :::; b ) so that the equation of our di¤usion process reads 0 (with Stratonovich integrals and with the convention Wt = t):

d j dXt = j(Xt) dWt : j=0 X We de…ne now recursively

S0 = 1; :::; d ;Sk+1 = [; j]; j = 0; :::; d ;S = k N Sk: f g f g 1 [ 2 Then, given x Rm; the Hörmander condition in x is 2 m (Hx) Span (x): S = R : f 2 1g

Notice that the ellipticity condition (x) c I is equivalent with SpanS0(x) = Rm: We explain now how this condition appears. Recall that we deal with the t matrix 0 Zs(Xs)Zsds with Z solution of the equation R d j dZt = Zt@j(Xt) dWt : j=0 X Using Itô’sformula we obtain for every f C2(Rm;Rm): 2 d j d(Ztf(Xt)) = Zt[j; f](Xt) dWt : j=0 X Then we use this formula in order to develop Ztj(Xt) in "stochastic series". We take k = 1; :::; d and we write

d t j Ztk(Xt) = k(x) + Zs[j; k](Xs) dWs j=0 0 X Z d d t j j = k(x) + [j; k](x)Wt + (Zs[j; k](Xs) [j; k](x)) dWs : j=0 j=0 0 X X Z If we continue to develop Zs[j; k](Xs) [j; k](x) then we obtain, for every N N : 2 ; Ztk(Xt) = k(x) + T (k)(x)Wt 1 N jXj d t t1 tN 1 N N + ::: (T (k)(Xt ) T (k)(x)) dW ::: dW N s =N j=0 0 0 0 j Xj X Z Z Z ; where T (k) and Wt are de…ned recursively in the following way. If = p ( 1; :::; p) 0; :::; d is a multi-index, we denote = ( 1; :::; p 1) and ! = 2 f g p: Then we de…ne

j T (f) = [f; j]; j = 0; :::; d; T (f) = [T (f); p ] = [[:::[f; 1 ]::::]; p ] and t ; ; p W = W dW : t s s Z0

Notice that SN = T (1); :::; T (d); = N : So the vectors which appear in f j j g Hörmander’sconditions are exactly the coe¢ cients of the development of Ztk(Xt); k = 1; :::; d in stochastic series. Notice that S0 contains only 1; :::; d and not 0 because in the covariance matrix 0 does not appear. In the following section we t will discuss the non degeneracy of a matrix of the form 0 Vsds where V is a matrix valued process which may be developed in Taylor sums. As an immediate consequence of Theorem 32 given there we obtain the following.R

Theorem 33. Suppose that ; b Cb1 and that (Hx) holds true. Then the law of x 2 X is absolutely continuous with respect to the Lebesgue measure and y pt(x; y) t ! is of class C1: Remark 5. One may also prove that pt(x; y) is of class C1 with respect to x m 2 R and t > 0: And one may obtain very precise exponential bounds for pt(x; y) and its derivatives with respect to t; x; y: This is a rather heavy computation - see Kusuoka-Strook.

0.5. Stochastic series

We are on a probability space and W = (W 1; :::; W d) is a standard Brownian motion. We introduce some notation. For notational convenience we denote 0 k Wt = t: We work with multi-indexes = ( 1; :::; k) 0; 1; :::; d and we denote 2 f g

= k; p( ) = card i : i 1; :::; d + 2card i : i = 0 : j j f 2 f gg f g Given an adapted and square integrable process X we de…ne the iterated stochastic integral

t tk 1 k 1 It (X) = ::: Xtk dWtk :::dWt1 ; and Z0 Z0 t tk 1 k 1 Wt = It (1) = ::: 1dWtk :::dWt1 : Z0 Z0 An important point concerns the scaling property (Scaling k) W "k=2W with k = p( ): t t" p( )=2 The speci…c property of Wt is that it behaves as t as t 0: In order to give a precise de…nition of this fact we introduce some notation. ! p A function f : R R will be called ‡at if lim" 0 " f(") = 0; p N: The nice thing about this! de…nition is the following simple! remark.j j Let8F 2be a non p negative random variable. Then EF < ; p N if and only if the function " P (F < ") is ‡at. 1 8 2 !Moreover we say that a process X has order larger then N and we write N=2 O(X) N if " P (X" > " ) is ‡at: Here we use the standard notation ! N=2+ X" := supt " Xt : And we say that O(X) N if " P (X" < " ); > 0 is ‡at. We say thatj j O(X) = N if O(X) N and O(X)! N: Notice that we8 do not assume that we may de…ne O(X) for any process X. We need just inequalities. But when we say that "O(X) = k" this means that we are able to prove both inequalityes. Let us list some basic properties. Lemma 34. i) If a process X has the Scailing k property then O(X) N (respectively O(X) N) is equivalent to the fact that " P (supt 1 Xt < ") is 1 ! j j ‡at (respectively " P (supt 1 Wt > " ) is ‡at). ii) Let b be a standard! one dimensionalj j Brownian motion. Then O(b) = 1 iii) For each X;Y we have

O(X) + O(Y ) O(X Y ) in the sense that O(X) k; O(Y ) p O(X Y ) k + p: k ) iv) If X" p Cp" for every p N and every " > 0 then O(X) k: v) If Ok(X)k< O(Y ) then O(X+Y2) = O(X) (in the sense: if O(X) k < O(Y ) then O(X + Y ) k): Proof. The only non -trivial property is ii): And this property is crucial because it permits to handle all the other processes coming on in the calculus. The fact that O(b) 1 is trivial - one employs the scaling property and iv): Let us prove that O(b) 1: So we have to check that " P (supt 1 bt < ") is ‡at. We write ! j j

P (b < ") P (b < "; b1 b1=2 < 2") = P (b < ")P ( b1 b1=2 < 2") 1 1=2 1=2 C"P (b1=2 < "): p And if we continue like this we obtain P (b1 < ") Cp" : We introduce now a class of stochastic processes which may be developed in stochastic series; We denote

p C0 = X : continuous and adapted process such that E(X ) < ; p N; T > 0 : f T 1 8 2 g

Then we de…ne by recurrence Ck to be the class of the stochastic processes of the form d t i i ( ) Xt = X0 + XsdWs i=0 0 X Z i with X Ck 1: It is easy to check by recurrence that Ck Ck 1: We denote 2 C = kCk: 1 \ i Moreover if X has the representation ( ) we de…ne Pri(X) = X and we iterate this de…nition: for a multi-index = ( 1; :::; k) and X C we de…ne 2 1 Pr (X) = Pr k (Pr( 1;:::; k 1)(X)): Finally we put pr (X) = Pr (X)(0): In order to get unitary notation we put pr (X) = X0: ; Let us now take X C and develop it in stochastic series. We write 2 1 d t d d t t i i i i i 0 0 Xt = X0 + XsdWs = X0 + pri(X)Wt + (Xs pri(X))dWs + Xs dWs i=0 0 i=1 i=1 0 0 X Z X X Z Z

= pr (X)Wt + I (Pr(X) pr (X)) + I( ;0)(Pr(X)): p( ) 1 p( )=1 p( )=1 1=0 X X X We wrote the last expression in order to understand what we will obtain in the general case. And we obtain

Xt = pr (X)Wt + RN (X)(t) with RN (X)(t) = RN0 (X)(t) + RN00 (X)(t); p( ) N X RN0 (X)(t) : = I (Pr(X) pr (X));RN00 (X)(t) := I( ;0)(Pr(X)): p( )=N p( )=N 1 X X The interest of this decomposition is that it permits us to compute O(X): We have:

Proposition 35. O(X) = min p( ): pr (X) = 0 f 6 g

Proof. Suppose that pr (X) = 0 if p( ) < N and there exists with pr (X) = 0 and p( ) = N: Then we have X = p( )=N pr (X)W + RN (X): 6 (N+1)=2 It is easy to check that RN (X)(") p Cp" for every p N so that k k P 2 O(RN (X)) N + 1: So it su¢ ce to prove that O( p( )=N pr (X)W ) = N: And by the scaling property this amounts to check that " P (( pr (X)W )(1) P! p( )=N ") is ‡at. And this is a consequence of (7) in the appendix. The followning lemma which we prove in the appendix,P is crucial.

Lemma 36. Let

Dq(c)(s; t) = c ; Ws Wt and p( )+Xp( )=q t 2 Yq(t) = Dq(c)(s; t) ds: j j Z0 with c ; R: If there exists at least one c ; = 0 then O(Y ) = 2q + 2: 2 6 As a consequence we obtain

n t 2 Proposition 37. Let U; V C and Yt := 0 Us;Vt ds: Then O(Y ) = 2q + 2 2 1 h i where q = min p( ) + p( ): pr (U); pr (V ) = 0 : f h Ri 6 g Proof. Take N = q + 1 and write

U = SN (U) + RN (U);SN (U) = pr (U)W p( ) N X

V = SN (V ) + RN (V );SN (V ) = pr (V )W : p( ) N X Then

Us;Vt = SN (U)(s);SN (V )(t) + RN (s; t) with h i h i RN (s; t) = SN (U)(s);RN (V )(t) + RN (U)(s);SN (V )(t) + RN (U)(s);RN (V )(t) : h i h i h i t 2 It is easy to see that O( 0 RN (s; t)ds) 2(N + 1) + 2 > 2q + 2 so we may ignore this term. Moreover we write R

SN (U)(s);SN (V )(t) h i

= pr (U); pr (V ) W W h i s t p( ) N p( ) N X X

= pr (U); pr (V ) W W + pr (U); pr (V ) W W h i s t h i s t p( )+Xp( )=q p( )+Xp( )>q = : Dq(s; t) + Dq0 (s; t):

t 2 We have O( 0 Dq0 (s; t) ds) 2(q + 1) + 2 > 2q + 2 so we ignore this term also. t 2 And by the previous lemma O( Dq(s; t) ds) = 2q + 2 so we are done. R 0 j j We consider now a matrix valued process U Mn m(C ) and we denote by R 21 n 1 n Uj; j = 1; :::; m the columns of U; that is Uj = (Uj ; :::; Uj ) C : We de…ne 2 1 n Ek = Span pr (Uj): j = 1; :::; m; p( ) k R : f g And we consider the matrix

AU (t) = UsUsds: Z0 Lemma 38. The following are equivalent:

n i) Eq = R n ii) O( AU X;X ) 2q + 2 + 2 min O(Xi); X = (X1; :::; Xn) C : h i i=1;:::;n 8 2 1

n n Proof. Suppose that Eq = R : Take X C and let p = mini=1;:::;n O(Xi): 2 1 Then there exists some with p( ) = p such that pr (X) = 0: And then there exist 6 some j 1; :::; m and some with p( ) q such that pr (Uj); pr (X) = 0: We have2 f g h i 6 m t 2 AU (t) Xt;Xt = Uk(s);Xt ds h i h i k=1 0 X Z so that, by the previous lemma

t 2 O( AU (t) Xt;Xt ) O( Uj(s);Xt ds) 2(p( ) + p( )) + 2 2(q + p) + 2 h i h i Z0 so ii) is proved. n n Suppose now that Eq R : Then we take x R with x = 0: We have 2 6 mini=1;:::;n O(xi) = 0 so that, if ii) is true, then O( AU x; x ) 2q + 2: But if we look to the coe¢ cients of the development in stochastich seriesi we see that all the coe¢ cients with p( ) q are null. Then O( AU x; x ) 2(q +1)+2 > 2q +2: We need also the following algebra lemma.h i

Lemma 39. Let I be a comutative ring and let O : I R+ be a function such that O(0) = ;O(x) = O( x) and ! [ f1g 1 (a) O(x) + O(y) O(xy); x; y I: 8 2

Consider a matrix a Mn n(I) which veri…es 2 n (b) ax; x k + 2 min O(xi); x I : h i i=1;:::;n 8 2 Then O(det a) nk: Proof. We procced by recurrence on n: Suppose …rst that n = 1: Then a I and by (a) …rst and then by (b) 2

O(a) + 2O(x) O(axx) = O( ax; x ) k + 2O(x): h i So O(a) k: We suppose now that the assertion is true for n 1 and we prove it for n: We i p i i+j i i denote Aj = (ak)k=j;p=i and j = ( 1) det Aj: We also denote a = (j)i;j=1;n: And we have a a6 =6 det a Id: n We prove now that An Mn 1;n 1 satis…es (b): We take x = (bx1; :::; xn 1) n 1 2 n n 2 I and we denote x = (x1; :::; xn 1; 0) I : Then Anx; x = ax; x so that b 2 h i h i n O( Anx; x ) = O( ax; x ) k + min O(xi) j=1;:::;n 1 hb i h i b b the last inequality being a consequenceb ofb (b) for a and of the fact that O(0) = : n 1 So be the recurrence hypothesis we get O(det An) k(n 1): We take now x = (0; :::; 0; 1) In and we denote y = ax = (1 ; :::; n): Then 2 n n ay; y = a(ax); ax = det(a)x; ax = det(a)n: h i h i h i b n Using (a) b b b n n O(det(a)) + O(n) O(det(a)n) = O( ay; y ) k + 2 min O(yi) h i i=1;:::;n n k + 2O(yn) = k + O( ): n And then O(det(a)) k + O(n) = k + O(An) k + k(n 1) = kn: n n n Proposition 40. Let h(U) = min k : Ek = R and h(U) = is the set is void. Then f g 1 O(det AU ) (2h(U) + 1)n: n Proof. Suppose that h(U) = q < : Then Eq = R and then by the previous 1 lemma O( AU X;X ) 2q +2+2 mini=1;:::;n O(Xi): Now, using the algebra lemma h i we get O(det AU ) (2q + 1)n: Corollary 41. If h(U) < then 1 p E( det AU (T ) ) < ; p N: j j 1 8 2

Proof. Let U (t) be the smallest eigenvalue of AU (t): Since det(AU (t)) = U (t) Vt for some process V C0 we have O(U ) O(det(AU )) < : It 2 q 1 follows that there exists q such that " P (U (") " ) is ‡at. It follows that q ! " P (U (T ) " ) is also ‡at and since t U (t) is non decreasing we get that ! q ! p " P (U (T ) " ) is ‡at. And this implies E( U (T ) ) < ; p N: Since ! n j j 1 8 2 det AU (T ) (T ) the proof is compleated. U 0.5.1. Appendix. We give now a several of lemmas which precise the asymptotic behavior of linear combinations of iterated stochastic integrals. In order to do it we need some more k notation concerning multi-indexes. Let = ( 1; ::::; k) 0; ::::; d : We denote 2 f g

= ( 1; ::::; k 1) and = k: ! Moreover we denote = ( k; k 1; ::::; 1; 0; 0; ::::): On the set of the multi-indexes we consider the following order relation. Let b = ( 1; ::::; k) and = ( 1; ::::; p): We say that if the following is true. Case 1. We have p( ) < p( ): Case 2. We have p( ) = p( ) but is larger than in the lexicographic order. This means that there exists some i N such that 2 j = j for 1 j i 1 and i > i: It is easy to see thatb this relation is complete,b that is: for each and we have or : And if the two inequalitiesb b hold true then = :bThe ideab of this de…nition is the following: we prefer multi-indexes with small order. And if we have the same order we prefer the one which has larger components. But we begin to compare them starting from the last component. Moreover we denote by the set of all the multi-indexes and we consider a 2 1=2 set of numbers c = (c ) : We denote c = ( c ) : Clearly c < if and only if c has a …nite number2 of non nullj j components.2 j j For c with cj j< 1we P j j 1 de…ne cmax in the following way. We de…ne max(c) = max : ; c = 0 : The maximum is taken with respect to the order de…ned abovef and since2 this6 orderg relation is complete and the set is …nite, there exists a unique element which is strictly larger then all the other ones. So max(c) is well de…ned. Finally we put

cmax = c max(c) :

Lemma 42. A. Let SN (c; t) = p( ) N c Wt : For every p N; there exists some 2 universal constants C = Cp;N ; q = qp;N such that P p 1=p i)(E( SN (c; 1) )) C c (7) j jp 1=p j j 1=2 ii)(E( sup SN (c; t) SN (c; ) )) C c " stopping time. t +" j j j j 8

B. Let QN (c; t) = c W : For every p N; there exist some universal p( )=N t 2 P constants C = Cp;N ; q = qp;N such that

N 1 q 8 p p iii) P (QN (c; 1) < ") C( c + cmax ) " and (8) 1 j j 2 q 8N p p iv) P ( QN (c; s) ds < ") C( c + c ) " : j j j j max Z0 Proof. The properties i); ii) are standard. Let us prove iii) and iv) by recur- d i rence. If N = 1 then Q1(c; t) = i=1 ciWt has the same law as c bt where b is a j j p standard Brownian motion. So P (Q1(c; 1) < ") = P (b1 "= c ) Cp("= c ) p P 1 j j j j Cp("=cmax) : We treat now the second term. We take u = 8 and we de…ne

u u T" = inf t : Q1(c; s) 2" and T 0 = inf t T" : Q1(c; s) < " : f j j g " f j j g 1 Then we take v = 2 and we write

1 2 1 1 v P ( Q1(c; s) ds < ") P (T" > ) + P (T" ;T 0 T" " ) j j 2 2 " Z0 1 3 1 v 2 i +P (T ;T 0 T > " ; Q (c; s) ds < ") =: I : " 2 " " N " 0 j j i=1 Z X

We take p0 = 8p and we write 1 1 u u p0 P (T" > ) = P (Q(c; ) < " ) = P (Q(c; 1) < "p2) Cp (" =cmax) 2 1 2 1 0

Cp0 p = 8p " : cmax

u u u Since Q1(c; T") Q1(c; T 0) = 2" " = " we have j j j " j

1 v u P (T" ;T"0 T" " ) P ( sup Q1(c; t) Q1(c; T") " ) v 2 T" t T"+" j j 1 1 p0 p0 p0v=2 8p p E( sup Q1(c; t) Q1(c; T") ) c " = c " : p0u v p0u " T" t T"+" j j " j j j j v u For T" t T" + " T 0; we have QN (c; s) " so that " j j v 1 T"+" 2 2 v+2u QN (c; s) ds QN (c; s) ds " > " j j j j Z0 ZT" 3 so that I" = 0: Suppose now that the property holds true for N 1: Let ci; i = 0; :::; d be the vector of the coe¢ cients corresponding to multi-indexes = ( 1; :::; k) with i i k = i; that is: c = c is = i and c = 0 is = i: Then we have ! ! 6 i QN 1(c ; s) = c Ws ; for i = 1; :::; d; p( )=XN;! =i 0 QN 1(c ; s) = c Ws : p( )=XN;! =0 Then

d t t i i 0 QN (c; t) = QN 1(c ; s)dWs + QN 2(c ; s)ds =: mt + dt: i=1 0 0 X Z Z Case 1. Suppose that there exists with = i 1; :::; d such that c = 0: ! 2 f g 6 We take i = max ! : c = 0; ! 1; :::; d : Notice that we have max(c) = i f 6 2i f gg 1 max(c ) and consequently cmax = cmax: We take u = 8 and we de…ne

i u i u T" = inf t : QN 1(c ; s) 2" and T"0 = inf t T" : QN 1(c ; s) < " : f g f g

We take p0 = 8p and we use the recurrence hypothesis in order to obtain

1 i 1 u i q 1 up P (T > ) = P (Q (c ; ) < " ) C( c + )" 0 " N 1 i 8N 2p 2 2 (cmax ) 0 1 C( c q + )"p: 8N 1p j j (cmax)

3 8p Moreover, we take v = 2 and p00 = 5 and we use the same reasoning as above and obtain 1 v i i u P (T" ;T"0 T" " ) P ( sup QN 1(c ; t) QN 1(c ;T") " ) v 2 T" t T"+" 1 p C p i i 00 i 00 p00v=2 p00 p E( sup QN 1(c ; t) QN 1(c ;T") ) c " = C c " : p00u v p00u " T" t T"+" " j j 1 Finally we take r = 2 and p000 = 2p we write

p 0 r 0 000 p000r p000 p P (QN 2(c ; 1) " ) C c " C c " : j j

We conclude that

q 1 p P (Q (c; 1) < ") C( c + )" + I with N 8N 1p " j j (cmax)

1 v 0 r I" = P (T" ;T"0 T" > " ;QN 2(c ; 1) < " ;QN (c; 1) < "): 2 v We take T" t T" + " and we write

QN (c; t) QN (c; t) QN (c; T") QN (c; T") j j j j j j mt mT dt dT QN (c; T") j " j j " j j j which gives

sup mt mT" 2QN (c; 1) + sup dt dT" : v v T" t T"+" j j T" t T"+" j j 0 r v r v If we are on the set QN 2(c ; 1) < " then supT" t T"+" dt dT" " = ": j j v And if we also have QN (c; 1) < " the above inequality becomes supT" t T"+" mt mT" 3": So j j 1 v I" P (T" ;T"0 T" > " ; sup mt mT" 3"): v 2 T" t T"+" j j

We write now m(t) = b( m t) where b is a standard Brownian motion. Notice that h i

v v T"+" d T"+" 2 i 2 i v+2u m T +"v m T = QN 1(c ; s)ds QN 1(c ; s)ds " " " h i h i T" i=1 T" Z X Z

2 i 2u v the last inequality being true because QN 1(c ; s) > " for T" < s < T" +" T"0: This gives

1 (u+v=2) 1=8 p I" P ( sup b(t) < ") = P (b(1) < " ) = P (b(1) < " ) Cp" 0

1 q 1 q 1 0 up0 0 p P (T" > ) C( c + N 3 )" C( c + N 2 )" : p08 p8 2 cmax cmax 1 We take v = 2 and p00 = 8p: Then, the same argument as above gives

1 v 0 v q p00( u) q p P (T" ;T T" " ) C c " 2 = C c " : 2 " j j j j It follows that

0 q 1 p P (QN (c; 1) < ") C( c + N 2 )" p8 cmax 1 0 v +P (T" ;T T" > " ;Q (c; 1) < "): 2 " N v 0 Notice that for T" t T" + " T"0; the function t QN 2(c ; t) keeps the 0 u ! same sign and QN 2(c ; s) > " : It follows that j j v v T"+" T"+" 0 0 v+u 3=4 QN 2(c ; t)dt = QN 2(c ; t) dt " = " : ZT" ZT"

But we also have v v T"+" T"+" T" 0 0 0 QN 2(c ; t)dt QN 2(c ; t)dt + QN 2(c ; t)dt 2": ZT" Z0 Z0

Since 2" < "3=4 this is impossible so the proof …nishes.

Let us now prove iv): The proof is identical with the one in the case N = 1 so we skip it. We consider now the double sum

Dk(c; s; t) = c ; W W ; 0 s t 1: s t p( )+Xp( )=k We de…ne

max(c) = max : such that c ; = 0 ; (c) = max : c (c); = 0 ; f 9 6 g max f max 6 g cmax = c : max(c); max(c) Lemma 43. For every p; k N; there exists some universal constants C = Cp;k 2 and q = qp;k such that

t 2 q p8k p P (sup Dk(c; s; t)ds ") C( c + cmax ) " : (9) t 1 0 j j Z Proof. For every l; r 1; :::; k we de…ne 2 f g

Ql;r(c; s; t) = c ; Ws Wt pX( )=l p(X )=r with the convention c ; = 0 if p( ) + p( ) = k: We take N = p( max(c)) and 6 we notice that for p( ) < N we have c ; = 0; : So Ql;k l = 0 for l < N: So k 8 Dk = l=N Ql;k l: We also denote M = k N: v We take = " with v > 0 (to be chosen latter on) and for = ( 1; :::; k) we de…neP

t tk 1 t tk 1 ; k 1 ; k 1 Wt = :::: 1 tk> dWt :::dWt ; W t = :::: 1 tk dWt :::dWt f g k 1 f g k 1 Z0 Z0 Z0 Z0 ; ; so that Wt = Wt + W t : Moreover, for with p( ) = N we denote (recall that M = k N) ; ; d (t) = c ; Wt ; d (t) = c ; W t p(X )=M p(X )=M so that QN;M (c; s; t) = QN;M (c; s; t) + QN;M (c; s; t) with

QN;M (c; s; t) = d (t)Ws and QN;M (c; s; t) = d (t)Ws : p(X )=N p(X )=N 2 1 2 Then, for t > we write (we use the elementary inequalities (a + b) 2 a and (a + b)2 2a2 + 2b2) t 2 2 Dk(c; s; t)ds Dk(c; s; t)ds 0 0 Z Z 1 2 2 Q (c; s; t) ds C Q (c; s; t) ds 2 N;M N;M Z0 Z0 k 2 C Ql;k l(c; s; t) ds j j l=N+1 0 X Z where C is an universal constant. We obtain

t k 2 l P (sup Dk(c; s; t)ds ") I"; + I"; + J" t 1 0 l=N+1 Z X with

2 2 I"; = P ( sup QN;M (c; s; t) ds C"); I"; = P ( sup QN;M (c; s; t) ds "); 0 and we de…ne T = inf t > : d (t) "u : " f g This random time is F;1 measurable so it is independent of Ws; s : We write 1 1 2 Id;" P (T > ) + P (T ; QN (d (T ); s) ds C") =: I0 + I00 : " 2 " 2 " d;" d;" Z0

We treat …rst

u u I0 = P ( sup d (t) < " ) = P ( sup Q (c; t) < " ) d;" M

Wt : By de…nition we have max(c) = max(c) so that cmax = cmax: We take now p0 = p=u and we use the point iii) of the previous lemma and we obtain

q 8M 1p up q 8M 1p=u p I0 C( c + (c ) 0 ) " 0 C( c + c ) " : d;" j j max j j max We estimate now Id;"00 : We write

2 Id;"00 = E(1 T 1 P ( QN (d; s) ds C" d (T" ) = d)): f " 2 g j j j Z0

Notice that is the largest multi-index for which there exists with c ; = 0: 6 So for we have c ; = 0; and consequently d (T" ) = 0: So (d(T )): By the de…nition of T 8we have d (t) = "u > 0 so we conclude that max " " u = max(d (T )): And consequently if d = d (T ) then dmax = d (T ) = " : " " " We take some p00 N and we use the scaling property and the point iv) from the previous lemma and2 we obtain

1 2 N+1 2 P ( QN (d; s) ds C" d (T ) = d) = P ( QN (d; s) ds C" d (T ) = d) j j j " j j j " Z0 Z0 q 8N u (N+1) p C( d + " )(" ) 00 : j j q q0 Since E( d (T ) ) C0 c we obtain " j j N q 8 u (N+1) p00 I00 C( c + 1) " (" ) : d;" j j

We estimate now I";: We write

2 2 2 2 ; QN;M (c; s; t) ds C c sup Ws sup W t 0 j j s< j j t 1 Z p( )=N p( )=M X X so that 2 I"; = P ( sup QN;M (c; s; t) ds ")

We have

4p ; 4p000 000 1=2 Np000 1=2 p000 (E(sup W t )) C ; (E(sup W t )) C s< t 1 so, using Chebyshev inequality …rst and Schwarz inequality then we obtain C c 2 N+1 p000 I"; ( j j ) : " k Let us now estimate J" : We have

2 2 2 2 Ql;k l(c; s; t) ds C c sup Ws sup Wt : 0 j j j j s< j j t 1 Z p( )=l p( )=k l X X

Notice that for p( ) = l N + 1 we have

4p000 4p000 1=2 (N+1)p000 1=2 (E(sup Wt )) C ; (E(sup Wt )) C s< j j t 1 so the same estimation as above gives

C c 2 J k ( j j N+1)p000 : " " We choose now our parameters. We take now = "v with 1 1 1 2N + 3 v = ( + ) = : 2 N + 1 N + 2 2(N + 1)(N + 2)

Then v(N + 2) 1 = 1=2(N + 1) so we take p000 = 2(N + 1)p and we obtain k k 2p000 p000=2(N+1) 4(N+1)p p I"; + J C c " = C c " : " j j j j l=N+1 X N We take now u = 1 and p00 = 2(p + 8 )(N + 2): Since 1 v(N + 1) = 1=2(N + 2) we obtain N 8 u + p00(1 v(N + 1)) = p: This gives

q 8N u (N+1) p q p I00 C( c + 1) " (" ) 00 = C( c + 1) " d;" j j j j and …nally

q 8M p=u p q 8M p p I0 C( c + c ) " = C( c + c ) " : d;" j j max j j max We put all this together and we obtain (9). 0.6. References

[1] Bally V. On the connection between the Malliavin covarience matrix and Hör- mander condition. J. Func. Ann. 96, pp 219-255, (1991). [2] Bell D. R. The Malliavin calculus. Pitman Monographs and Surveys in Pure and Applied Math. 34, Longman and Wiley (1987) [3] Hörmander L : Hypoeliptic second order di¤erential equations. Acta math. 119, pp 147-171 (1967) [4] Ikeda N. and Watanabe S. Stochastic di¤erential equations and di¤usion processes. North Holand, second ed. (1989). [5] Kusuoka R. and Stroock D.W.: Application of the Malliavin calculus II: J. Fac. Sci. Univ. Tokyo Sect IA Math. 32pp. 391-442 (1987). [6] Malliavin P. Stochastic calculus of variations and hypoelliptic operators. In: Proc. Inter. Symp. on Stoch. Di¤e. Equations, Kyoto (1976), Wiley, 313, pp. 195-263. [7] Meyer P-A. Transformation de Riez pour les lois gaussiennes. In Séminaire de Probabilités XVIII. Lecture Notes in Math. 1059, pp 179-193. Springer Verlag (1984). [8] Nualart D. Malliavin Calculus and related topics. Springer Verlag (1995). [9] Ocone D. Malliaivin Calculus and Stochastic Integral Representation of Di¤usion Processes. Stochastics and Stochastic Repports 12, pp 161-185 (1984). [10] Sanz-Solé M. Malliavin cCalculus with applications to Stochastic Di¤erential Equations EPFL Press, Fundamental Sciences, Mathematics, (2006). [11] Shigekawa I. Stochastic Analysis. Translations of Mathematical Mono- graphs, Volume 224, American Math? Soc. (2004). [12] Watanabe S. Lectures on stochastic di¤erential equations and Malliavin calculus. Tata institute of fundamental research, Springer Verlag, Berlin (1984).