Rough paths methods 2: Young integration Samy Tindel Purdue - PowerPoint PPT Presentation

Rough paths methods 2: Young integration Samy Tindel Purdue University University of Aarhus 2016 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 1 / 75

Outline Some basic properties of fBm 1 Simple Young integration 2 Increments 3 Algebraic Young integration 4 Differential equations 5 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 2 / 75

Outline Some basic properties of fBm 1 Simple Young integration 2 Increments 3 Algebraic Young integration 4 Differential equations 5 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 3 / 75

Definition of fBm Complete probability space: (Ω , F , P ) Definition 1. A 1-d fBm is a continuous process B = { B t ; t ≥ 0 } such that: B 0 = 0 B is a centered Gaussian process 2 ( | s | 2 H + | t | 2 H − | t − s | 2 H ), for H ∈ (0 , 1) E [ B t B s ] = 1 d -dimensional fBm: B = ( B 1 , . . . , B d ), with B i independent 1-d fBm Samy T. (Purdue) Rough Paths 2 Aarhus 2016 4 / 75

fBm: variance of the increments Notation: If f : [0 , T ] → R d is a function, we shall denote: | δ f st | δ f st = f t − f s , and � f � µ = sup | t − s | µ s , t ∈ [0 , T ] Variance of the increments: for a 1-d fBm, E [ | δ B st | 2 ] ≡ E [ | B t − B s | 2 ] = | t − s | 2 H Samy T. (Purdue) Rough Paths 2 Aarhus 2016 5 / 75

FBm regularity Proposition 2. FBm B ≡ B H is γ -Hölder continuous on [0 , T ] for all γ < H , up to modification. Proof: We have δ B st ∼ N (0 , | t − s | 2 H ). Thus for n ≥ 1, � | δ B st | 2 n � � | δ B st | 2 n � = c n | t − s | 2 Hn = c n | t − s | 1+(2 Hn − 1) E i.e E Kolmogorov: B is γ -Hölder for γ < (2 Hn − 1) / 2 n = H − 1 / (2 n ). Proof finished by letting n → ∞ . Samy T. (Purdue) Rough Paths 2 Aarhus 2016 6 / 75

Some properties of fBm Proposition 3. Let B be a fBm with parameter H . Then: { a − H B at ; t ≥ 0 } is a fBm (scaling) 1 { B t + h − B h ; t ≥ 0 } is a fBm (stationarity of increments) 2 B is not a semi-martingale unless H = 1 / 2 3 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 7 / 75

Proof of claim 3 Semi-martingale and quadratic variation: If B were a semi-martingale, we would get on [0 , 1]: n ( B i / n − B ( i − 1) / n ) 2 = � B � 1 , � P − lim n →∞ i =1 were � B � is the (non trivial) quadratic variation of B . We will show that � B � is trivial (0 or ∞ ) whenever H � = 1 / 2. Samy T. (Purdue) Rough Paths 2 Aarhus 2016 8 / 75

Proof of claim 3 (2) A p -variation: Define n � | B i / n − B ( i − 1) / n | p , Y n , p = n pH − 1 V n , p . V n , p = and i =1 By scaling properties, we have: n ( d ) = ˆ ˆ Y n , p = n − 1 � | B i − B i − 1 | p . Y n , p Y n , p , with i =1 The sequence { B i − B i − 1 ; i ≥ 1 } is stationary and mixing Y n , p converges P − a . s and in L 1 towards E [ | B 1 − B 0 | p ] ⇒ ˆ ⇒ P − lim n →∞ Y n , p = E [ | B 1 | p ] ⇒ P − lim n →∞ V n , p = 0 if pH > 1, ∞ if pH < 1 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 9 / 75

Proof of claim 3 (3) Recall: V n , p = � n i =1 | B i / n − B ( i − 1) / n | p Definition: P − lim n →∞ V 1 / p n , p ≡ V p ( B ) is called p -variation of B ⇒ We have seen V p ( B ) = 0 if pH > 1, ∞ if pH < 1 Property: if p 1 < p 2 , then V p 1 ( B ) ≥ V p 2 ( B ) Case H > 1 / 2: choose p < 2 such that pH > 1 ⇒ V p ( B ) = 0 ⇒ V 2 ( B ) = 0 Case H < 1 / 2: choose p > 2 such that pH < 1 ⇒ V p ( B ) = ∞ ⇒ V 2 ( B ) = ∞ Conclusion: if H � = 1 / 2, It¯ o’s type methods do not apply in order to define stochastic integrals Samy T. (Purdue) Rough Paths 2 Aarhus 2016 10 / 75

Outline Some basic properties of fBm 1 Simple Young integration 2 Increments 3 Algebraic Young integration 4 Differential equations 5 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 11 / 75

Strategy for H > 1 / 2 Generally speaking, take advantage of two aspects of fBm: ◮ Gaussianity ◮ Regularity For H > 1 / 2, regularity is almost sufficient Notation: C γ 1 = C γ 1 ( R ) ≡ γ -Hölder functions of 1 variable If H > 1 / 2, B ∈ C γ 1 for any 1 / 2 < γ < H a.s We shall try to solve our equation in a pathwise manner Samy T. (Purdue) Rough Paths 2 Aarhus 2016 12 / 75

Equation under consideration � t � t X t = a + 0 σ ( X s ) dB s + 0 b ( X s ) ds , t ∈ [0 , T ] (1) a ∈ R n initial condition b , σ coefficients in C 1 b B = ( B 1 , . . . , B d ) d -dimensional Brownian motion B i iid Brownian motions Samy T. (Purdue) Rough Paths 2 Aarhus 2016 13 / 75

Notational simplification Simplified setting: In order to ease notations, we shall consider: Real-valued solution and fBm: n = d = 1. However, we shall use d -dimensional methods b ≡ 0 Simplified equation: we end up with � t t ∈ [0 , T ] X t = a + 0 σ ( X s ) dB s , (2) a ∈ R , σ ∈ C 1 b ( R ) B is a 1-d Brownian motion Samy T. (Purdue) Rough Paths 2 Aarhus 2016 14 / 75

Pathwise strategy Aim: Let x be a function in C γ 1 with γ > 1 / 2. We wish to define and solve an equation of the form: � t y t = a + 0 σ ( y s ) dx s (3) Steps: � z s dx s for z ∈ C κ Define an integral 1 , with κ + γ > 1 Solve (3) through fixed point argument in C κ 1 with 1 / 2 < κ < γ Notation: We set � t J st ( z dx ) = ” s z w dx w ” for reasonable extensions of Riemann’s integral Samy T. (Purdue) Rough Paths 2 Aarhus 2016 15 / 75

Particular Riemann sums � 1 1 , x ∈ C γ 0 z s dx s for z ∈ C κ Aim: Define 1 , with κ + γ > 1 Dyadic partition: set t n i = i / 2 n , for n ≥ 0, 0 ≤ i ≤ 2 n Associated Riemann sum: 2 n − 1 2 n − 1 � � I n ≡ z t n i [ x t n i +1 − x t n i ] = z t n i δ x t n i +1 . i t n i =0 i =0 Question: Can we define J 01 ( z dx ) ≡ lim n →∞ I n ? Possibility: Control | I n +1 − I n | and write (if the series is convergent): ∞ � J 01 ( z dx ) = I 0 + ( I n +1 − I n ) . n =0 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 16 / 75

Control of I n +1 − I n We have: 2 n − 1 2 n − 1 � � � � I n = z t n i δ x t n i +1 = z t n +1 δ x t n +1 2 i +1 + δ x t n +1 i t n t n +1 2 i +1 t n +1 2 i 2 i 2 i +2 i =0 i =0 2 n − 1 � � � I n +1 = z t n +1 δ x t n +1 2 i +1 + z t n +1 2 i +1 δ x t n +1 t n +1 2 i +1 t n +1 2 i 2 i 2 i +2 i =0 Therefore: 2 n − 1 � � � � � | I n +1 − I n | = δ z t n +1 2 i +1 δ x t n +1 � � t n +1 2 i +1 t n +1 � � 2 i 2 i +2 � i =0 � 2 n − 1 | κ � x � γ | t n +1 � � z � κ | t n +1 2 i +1 − t n +1 2 i +2 − t n +1 2 i +1 | γ ≤ 2 i i =0 � z � κ � x � γ = 2 κ + γ 2 n ( κ + γ − 1) Samy T. (Purdue) Rough Paths 2 Aarhus 2016 17 / 75

Definition of the integral We have seen: for α ≡ κ + γ − 1 > 0 and n ≥ 0: | I n +1 − I n | ≤ c x , z 2 α n Series convergence: Obviously, � ∞ n =0 ( I n +1 − I n ) is a convergent series → yields definition of J 01 ( z dx ), and more generally: J st ( z dx ) ֒ Remark: One should consider more general partitions π , with | π | → 0 ֒ → C.f Lejay (Séminaire 37) Samy T. (Purdue) Rough Paths 2 Aarhus 2016 18 / 75

Young integral, version 1 Proposition 4. 1 ([0 , T ]) , x ∈ C γ Let z ∈ C κ 1 ([0 , T ]), with κ + γ > 1, and 0 ≤ s < t ≤ T . Let ( π n ) n ≥ 0 a sequence of partitions of [ s , t ] such that lim n →∞ | π n | = 0 I n corresponding Riemann sums Then: I n converges to an element J st ( z dx ) 1 The limit does not depend on the sequence ( π n ) n ≥ 0 2 Integral linear in z , and coincides with Riemann’s integral 3 for smooth z , x If 0 ≤ s < u < t ≤ T , we have 4 J st ( z dx ) = J su ( z dx ) + J ut ( z dx ) Samy T. (Purdue) Rough Paths 2 Aarhus 2016 19 / 75

Outline Some basic properties of fBm 1 Simple Young integration 2 Increments 3 Algebraic Young integration 4 Differential equations 5 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 20 / 75

Notations: increments Simplex: For k ≥ 2 and T > 0 we set S k , T = { ( s 1 , . . . , s k ); 0 ≤ s 1 < · · · < s k ≤ T } ( k − 1)-increment: Let T > 0, a vector space V and k ≥ 1: � � C k ( V ) ≡ g ∈ C ( S k , T ; V ); t i → t i +1 g t 1 ··· t k = 0 , i ≤ k − 1 lim Remark: We mostly consider V = R for notational sake ֒ → We write C k = C k ([0 , T ]; R ) Samy T. (Purdue) Rough Paths 2 Aarhus 2016 21 / 75

Notations: operator δ Operator δ : k +1 � ( − 1) k − i g ˆ δ : C k → C k +1 , δ g t 1 ··· t k +1 = , t k +1 i i =1 where t k +1 = ( t 1 , . . . , t k +1 ) ˆ t k +1 = ( t 1 , . . . , t i − 1 , t i +1 , . . . , t k +1 ) i Examples: if g ∈ C 1 and h ∈ C 2 we have, for s , u , t ∈ S 3 , T , δ g st = g t − g s , and δ h sut = h st − h su − h ut . Samy T. (Purdue) Rough Paths 2 Aarhus 2016 22 / 75

First properties of δ Proposition 5. δδ : C k → C k +2 satisfies δδ = 0 Notation: ZC k = [ C k ∩ Ker δ ] Proposition 6. Let k ≥ 1 h ∈ ZC k +1 There exists a (non unique) f ∈ C k such that h = δ f . Samy T. (Purdue) Rough Paths 2 Aarhus 2016 23 / 75

Proofs Proposition 5, easy case: If k = 1, g ∈ C 1 and h ≡ δ g , then: δ h sut = h st − h su − h ut ( δδ g ) sut = = [ g t − g s ] − [ g u − g s ] − [ g t − g u ] = 0 Samy T. (Purdue) Rough Paths 2 Aarhus 2016 24 / 75