An Inroducion o Malliavin calculus and is applicaions Lecure 5: Smoohness of he densiy and Hörmander s heorem David Nualar Deparmen of Mahemaics Kansas Universiy Universiy of Wyoming Summer School 214 David Nualar (Kansas Universiy) May 214 1 / 31
Le F = (F 1,..., F m ) be such ha F i D 1,2 for i = 1,..., m. The Malliavin marix of F is γ F = ( DF i, DF j H ) 1 i,j m. Theorem (Crierion for absolue coninuiy) If de γ F > a.s., hen he law of F is absoluely coninuous wih respec o he Lebesgue measure on R m. Theorem (Crierion for smoohness of he densiy) If F i D and E[(de γ F ) p ] < for all p 1, hen he law of F possesses and infiniely differeniabiliy densiy. David Nualar (Kansas Universiy) May 214 2 / 31
d-dimensional Brownian moion (Ω, F, P) is he canonical probabiliy space associaed wih a d-dimensional Brownian moion {W i (), [, T ], 1 i d} : i) Ω = C ([, T ]; R d ). ii) P is he law of he d-dimensional Brownian moion. iii) F is he compleion of he Borel σ-field of Ω wih respec o P. The Hilber space here is H = L 2 ([, T ]; R d ), and for any h H, W (h) is he Wiener inegral T W (h) = hsdw i s. i i=1 The derivaive DF of a random variable F D 1,2 will be a d-dimensional process denoed by {D i F, [, T ], i = 1,..., d}. For example D i sw j = δ i,j 1 [,] (s). David Nualar (Kansas Universiy) May 214 3 / 31
Diffusion processes Consider he m-dimensional sochasic differenial equaion X = x + j=1 A j (X s )dws j + B(X s )ds, (1) where A j, B : R m R m, 1 j d are measurable funcions. We know ha under he Lipschiz condiion ( max Aj (x) A j (y), B(x) B(y) ) K x y, (2) j for all x, y R m here exiss a unique soluion X = {X, [, T ]} of Equaion (1). David Nualar (Kansas Universiy) May 214 4 / 31
Differeniabiliy of he soluion Proposiion Suppose ha he coefficiens A j, B are in C 1 b (Rm ; R m ). Then, for all [, T ] and i = 1,..., m, X i D 1, and for r m Dr j X = A j (X r ) + k A l (X s )D j r X k s dw l s + m k=1 l=1 r k=1 r k B(X s )D j r X k s ds. (3) If he coefficiens are infiniely differeniable in he space variable and heir parial derivaives of all orders are uniformly bounded, hen X i () belongs o D. David Nualar (Kansas Universiy) May 214 5 / 31
Proof : To simplify we assume B =. Consider he Picard approximaions given by X () = x and X (n+1) = x + j=1 A j (X (n) s )dw j s. if n. We will prove he following claim by inducion on n : Claim : X (n),i and for some consans c 1, c 2. D 1, for all i,, and for all p > 1 we have ( ψ n () := sup E r ) sup D r X (n) s p < (4) s [r,] ψ n+1 () c 1 + c 2 ψ n (s)ds, (5) David Nualar (Kansas Universiy) May 214 6 / 31
Clearly, he claims holds for n =. Suppose i is rue for n. Applying propery (2) of he divergence and chain rule, for any r, i = 1,..., m and l = 1,..., d, we ge Dr l X (n+1),i = Dr l [ A i (n) j (X s )dws] j = δ l,j A i l(x (n) r ) + = δ l,j A i l(x (n) r ) + r m k=1 D l r [A i j r (n) (X s )]dws j k A j (X (n) s )Dr l X (n),k s dws. j David Nualar (Kansas Universiy) May 214 7 / 31
From hese equaliies and condiion (4) we see ha X (n+1),i D 1,, and we obain using Burkholder s inequaliy ( ) [ ] ( E sup D r X (n+1) s p c p γ p + T p 1 K p E Dr j X (n) s p) ds, (6) r s r where γ p = sup E( sup A j (X (n) ) p ) <. n,j T So (4) and (5) hold for n + 1 and he claim is proved. David Nualar (Kansas Universiy) May 214 8 / 31
We know ha as n ends o infiniy. E ( ) sup X (n) s X s p s T By Gronwall s lemma applied o (5) we deduce ha he derivaives of he sequence X (n),i are bounded in L p (Ω; H) uniformly in n for all p 2. This implies ha he random variables X i belong o D 1,. Finally, applying he operaor D o Equaion (1) we deduce he linear sochasic differenial equaion (3) for he derivaive of X i. This complees he proof of he proposiion. David Nualar (Kansas Universiy) May 214 9 / 31
Consider he m m marix-valued process defined by Y = I + l=1 A l (X s )Y s dws l + B(X s )Y s ds, where A l denoes he m m Jacobian marix of he funcion A l, ha is, ( A l ) i j = j A i l. In he same way, B denoes he m m Jacobian marix of B. If he coefficiens of Equaion (1) are of class C 1+α, α >, hen here is a version of he soluion X (x ) o his equaion ha is coninuously differeniable in x, and Y is he Jacobian marix X x : Y = X x. David Nualar (Kansas Universiy) May 214 1 / 31
Recall ha Y = I + l=1 A l (X s )Y s dws l + B(X s )Y s ds. Proposiion For any [, T ] he marix Y is inverible. Is inverse Z saisfies Z = I l=1 Z s A l (X s )dw l s Z s [ B(X s ) ] A l (X s ) A l (X s ) ds. l=1 David Nualar (Kansas Universiy) May 214 11 / 31
Proof : By means of Iô s formula, one can check ha Z Y = Y Z = I, which implies ha Z = Y 1. In fac, Z Y = I + l=1 l=1 Z s A l (X s )Y s dws l + Z s B(X s )Y s ds Z s A l (X s )Y s dw l s Z s [ B(X s ) and similarly we show ha Y Z = I. ] A l (X s ) A l (X s ) Y s ds l=1 ( ) Z s A l (X s ) A l (X s ) Y s ds = I, l=1 David Nualar (Kansas Universiy) May 214 12 / 31
Lemma The m d marix (D r X ) i j = D j r X i can be expressed as D r X = Y Y 1 r A(X r ). (7) Proof : I suffices o check ha he process Φ,r := Y Yr 1 A(X r ), saisfies In fac, Φ,r = A(X r ) + l=1 A(X r ) + + r l=1 r A l (X s )Φ s,r dws l + B(X s )Φ s,r ds. r = A(X r ) + [Y Y r ] Y 1 r r A l (X s ) { Y s Yr 1 A(X r ) } dws l B(X s ) { Y s Yr 1 A(X r ) } ds A(X r ) = Y Yr 1 A(X r ). David Nualar (Kansas Universiy) May 214 13 / 31
Consider he Malliavin marix of X, denoed by γ X := Q, given by ha is Equaion (7) leads o where C = Y 1 s Q i,j = Q = l=1 D l sx i D l sx j ds, (D s X )(D s X ) T ds. Q = Y C Y T, (8) A(X s )A T (X s ) ( Y 1 ) T ds = and σ is he m m diffusion marix σ = AA T. s Y 1 s σ(x s ) ( Y 1 ) T ds, Taking ino accoun ha Y is inverible, he nondegeneracy of he marix Q will depend only on he nondegeneracy of he marix C, which is called he reduced Malliavin marix. s David Nualar (Kansas Universiy) May 214 14 / 31
Absolue coninuiy under ellipiciy condiions Consider he sopping ime defined by S = inf{ > : de σ(x ) } T. Theorem (Bouleau and Hirsch 85) Le {X(), [, T ]} be a diffusion process wih C 1 and Lipschiz coefficiens. Then for any < T he law of X() condiioned by { > S} is absoluely coninuous wih respec o he Lebesgue measure on R m. David Nualar (Kansas Universiy) May 214 15 / 31
Proof : I suffices o show ha de C > a.s. on he se {S < }. Suppose > S. For any u R m wih u = 1 we can wrie u T C u = u T Y 1 s σ(x s )(Y 1 ) T uds ( ) inf v T σ(x s )v v =1 s (Ys 1 ) T u 2 ds. ( Noice ha inf v =1 v T σ(x s )v ) is he smalles eigenvalue of σ(x s ) which is sricly posiive in an open inerval conained in [, ] by he definiion of he sopping ime S and because > S. On he oher hand, (Ys 1 ) T u u Y s 1. Therefore we obain u T C u k u 2, for some posiive random variable k >, which implies ha he marix C is inverible. This complees he proof. David Nualar (Kansas Universiy) May 214 16 / 31
Regulariy of he densiy under Hörmander s condiions Assume ha he coefficiens are infiniely differeniable wih bounded derivaives of all orders. Then, X i belong o D. Consider he following vecor fields on R m : A j = B = m i=1 m i=1 A i j (x) x i, j = 1,..., d, B i (x) x i. The Lie bracke beween he vecor fields A j and A k is defined by where [A j, A k ] = A j A k A k A j = A j A k A k A j, A j A k = m A l j la i k. x i i,l=1 David Nualar (Kansas Universiy) May 214 17 / 31
Se A = B 1 2 A l A l. The vecor field A appears when we wrie he sochasic differenial equaion (1) in erms of he Sraonovich inegral insead of he Iô inegral : X = x + A j (X s ) dws j + A (X s )ds. j=1 Hörmander s condiion : The vecor space spanned by he vecor fields A 1,..., A d, [A i, A j ], i d, 1 j d, [A i, [A j, A k ]], i, j, k d,... a poin x is R m. For insance, if m = d = 1, A 1 1 (x) = a(x), and A1 (x) = b(x), hen Hörmander s condiion means ha a(x ) or a n (x )b(x ) for some n 1. l=1 David Nualar (Kansas Universiy) May 214 18 / 31
Theorem Assume ha Hörmander s condiion (H) holds. Then for any > he random vecor X has an infiniely differeniable densiy. This resul can be considered as a probabilisic version of Hörmander s heorem on he hypoellipiciy of second-order differenial operaors. David Nualar (Kansas Universiy) May 214 19 / 31
Lemma Le {Z, } be a real-valued, adaped coninuous process such ha Z = z. Suppose ha here exiss α > such ha for all p 1 and [, T ], ( E ) sup Z s z p C p,t pα. s Then, for all p 1 and (, T ], [ ( ) p ] E Z s ds <. David Nualar (Kansas Universiy) May 214 2 / 31
Proof : For any < ɛ < z 2 we have ( P ) Z s ds < ɛ P which implies he desired resul. ( ) 2ɛ/ z Z s ds < ɛ ( ) P sup Z s z > z s 2ɛ/ z 2 ( ) 2p C p,t 2ɛ pα z p, z David Nualar (Kansas Universiy) May 214 21 / 31
Lemma (Norris) Consider a coninuous semimaringale of he form where Y = y + a(s)ds + a() = α + β(s)ds + i=1 i=1 u i (s)dw i s, γ i (s)dw i s, and c = E ( sup T ( β() + γ() + a() + u() ) p) < for some p 2. Fix q > 8. Then, for all r < q 8 27 here exiss ɛ such ha for all ɛ ɛ we have ( T ) T P Y 2 d < ɛ q, ( a() 2 + u() 2 )d ɛ c 1 ɛ rp. David Nualar (Kansas Universiy) May 214 22 / 31
Skech of he proof of Hörmander s heorem : Sep 1 We need o show ha for all > and all p 2, E[(de Q ) p ] <, where Q is he Malliavin marix of X. Taking ino accoun ha ( de ) E Y 1 p + de Y p <, i suffices o show ha E[(de C ) p ] < for all p 2. Sep 2 have Fix >. Then he problem is reduced o show ha for all p 2 we sup P{v T C v ɛ} ɛ p v =1 for any ɛ ɛ (p), where he quadraic form associaed o he marix C is given by v T C v = v, Ys 1 A j (X s ) 2 ds. (9) j=1 David Nualar (Kansas Universiy) May 214 23 / 31
Sep 3 Fix a smooh funcion V and use Iô s formula o compue he differenial of Y 1 V (X ) d ( ) Y 1 V (X ) = Y 1 +Y 1 [A k, V ](X )dw k k=1 { [A, V ] + 1 2 } [A k, [A k, V ]] (X )d. (1) k=1 David Nualar (Kansas Universiy) May 214 24 / 31
Sep 4 We inroduce he following ses of vecor fields : Σ = {A 1,..., A d }, Σ n = {[A k, V ], k =,..., d, V Σ n 1 } if n 1, Σ = n=σ n, and Σ = Σ, Σ n = {[A k, V ], k = 1,..., d, V Σ n 1; [A, V ] + 1 [A j, [A j, V ]], V Σ 2 n 1} if n 1, Σ = n=σ n. j=1 David Nualar (Kansas Universiy) May 214 25 / 31
i) We denoe by Σ n (x) (resp. Σ n(x)) he subse of R m obained by freezing he variable x in he vecor fields of Σ n (resp. Σ n). ii) Clearly, he vecor spaces spanned by Σ(x ) or by Σ (x ) coincide, and under Hörmander s condiion his vecor space is R m. iii) Therefore, here exiss an ineger j such ha he linear span of he se of vecor fields j j= Σ j (x) a poin x has dimension m. iv) As a consequence here exis consans R > and c > such ha j v, V (y) 2 c, (11) j= V Σ j for all v and y wih v = 1 and y x < R. David Nualar (Kansas Universiy) May 214 26 / 31
Sep 5 For any j =, 1,..., j we pu m(j) = 2 4j and we define he se E j = v, Ys 1 V (X s ) 2 ds ɛ m(j). V Σ j Noice ha {v T C v ɛ} = E because m() = 1. Consider he decomposiion E (E E c 1 ) (E 1 E c 2 ) (E j 1 E c j ) F, where F = E E 1 E j. Then for any uni vecor v we have P{v T C v ɛ} = P(E ) P(F) + We are going o esimae each erm of his sum. j j= P(E j E c j+1 ). David Nualar (Kansas Universiy) May 214 27 / 31
Sep 6 Le us firs esimae P(F ). By he definiion of F we obain j P(F ) P v, Ys 1 V (X s ) 2 ds (j + 1)ɛ m(j ). j= V Σ j Then, aking ino accoun (11) we can apply he previous lemma o he process j Z s = v, Ys 1 V (X s ) 2, j= V Σ j we obain E j j= V Σ j v, Y 1 s V (X s ) 2 ds p <. Therefore, for any p 1 he exiss ɛ such ha for any ɛ < ɛ. P(F ) ɛ p David Nualar (Kansas Universiy) May 214 28 / 31
Sep 7 For any j =,..., j he probabiliy of he even E j Ej+1 c is bounded by he sum wih respec o V Σ j of he probabiliy ha he wo following even happens and k=1 + v, Ys 1 [A k, V ](X s ) 2 ds v, Y 1 s v, Ys 1 V (X s ) 2 ds ɛ m(j) [A, V ] + 1 2 [A j, [A j, V ]] (X s ) j=1 where n(j) denoes he cardinaliy of he se Σ j. 2 ds > ɛm(j+1) n(j), David Nualar (Kansas Universiy) May 214 29 / 31
Consider he coninuous semimaringale { v, Y 1 s V (X s ), s }. From (1) we see ha he quadraic variaion of his semimaringale is equal o k=1 s v, Yσ 1 [A k, V ](X σ ) 2 dσ, and he bounded variaion componen is s v, Yσ 1 [A, V ] + 1 [A j, [A j, V ]] 2 (X σ) dσ. Taking ino accoun ha 8m(j + 1) < m(j), from Norris Lemma applied o he semimaringale Y s = v T Ys 1 V (X s ) we ge ha for any p 1 here exiss ɛ > such ha P(E j Ej+1 c ) ɛp, for all ɛ ɛ. The proof of he heorem is now complee. j=1 David Nualar (Kansas Universiy) May 214 3 / 31
Example Consider he following example. dx 1 = dw 1 + sin X 2 dw 2, dx 2 = 2X 1 dw 1 + X 1 dw 2 wih iniial condiion x =. In his case he diffusion marix [ 1 + sin 2 ] x σ(x) = 2 x 1 (2 + sin x 2 ) x 1 (2 + sin x 2 ) 5x1 2 [ degeneraes along ] he line x 1 =. The Lie bracke [A 1, A 2 ] is equal o 2x1 cos x 2. Therefore, he vecor fields A 1 2 sin x 1 and [A 1, A 2 ] a x = span R 2 2 and Hörmander s condiion holds. So X has a C densiy for any >. David Nualar (Kansas Universiy) May 214 31 / 31