Rough paths methods 3: Second order structures Samy Tindel - PowerPoint PPT Presentation

Rough paths methods 3: Second order structures Samy Tindel University of Lorraine at Nancy KU - Probability Seminar - 2013 Samy T. (Nancy) Rough Paths 3 KU 2013 1 / 46

Sketch Heuristics 1 Controlled processes 2 Differential equations 3 Aplication to fBm 4 Final remarks 5 Higher order structures Lyons theory Some projects Samy T. (Nancy) Rough Paths 3 KU 2013 2 / 46

Sketch Heuristics 1 Controlled processes 2 Differential equations 3 Aplication to fBm 4 Final remarks 5 Higher order structures Lyons theory Some projects Samy T. (Nancy) Rough Paths 3 KU 2013 3 / 46

Examples of fBm paths H = 0 . 3 H = 0 . 5 H = 0 . 7 Samy T. (Nancy) Rough Paths 3 KU 2013 4 / 46

General strategy Aim: Define and solve an equation of the type: � t y t = a + 0 σ ( y s ) dB s , where B is fBm. Properties of fBm: Generally speaking, take advantage of two aspects of fBm: Gaussianity Regularity Remark: For 1 / 3 < H < 1 / 2, Young integral isn’t suficient Levy area: We shall see that the following exists: � t � u B 2 , i , j v ∈ C 2 γ s dB i s dB j = for γ < H 2 st u Strategy: Given B and B 2 solve the equation in a pathwise manner Samy T. (Nancy) Rough Paths 3 KU 2013 5 / 46

Pathwise strategy Aim: For x ∈ C γ 1 con 1 / 3 < γ < 1 / 2, define and solve an equation of the type: � t y t = a + 0 σ ( y u ) dx u (1) Main steps: � z s dx s for z : function whose increments are Define an integral controlled by those of x Solve (1) by fixed point arguments in the class of controlled processes Remark: Like in the previous chapters, we treat a real case and b ≡ 0 for notational sake. Caution: d -dimensional case really different here, because of x 2 Samy T. (Nancy) Rough Paths 3 KU 2013 6 / 46

Heuristics (1) Hypothesis: Solution y t exists in a space C γ 1 ([ 0 , T ]) A priori decomposition for y : � t δ y st ≡ y t − y s = s σ ( y v ) dx v � t = σ ( y s ) δ x st + s [ σ ( y v ) − σ ( y s )] dx v = ζ s δ x st + r st Expected coefficients regularity: ζ = σ ( y ) : bounded, γ -Hölder, r : 2 γ -Hölder Samy T. (Nancy) Rough Paths 3 KU 2013 7 / 46

Heuristics (2) Start from controlled structure: Let z such that ζ ∈ C γ , r ∈ C 2 γ δ z st = ζ s δ x st + r st , with (2) Formally: � t � t s z v dx v = z s δ x st + s δ z sv dx v � t � t = z s δ x st + ζ s s δ x sv dx v + s r sv dx v � t z s δ x st + ζ s x 2 = st + s r sv dx v Samy T. (Nancy) Rough Paths 3 KU 2013 8 / 46

Heuristics (3) Formally, we have seen: z satisfies � t � t s z v dx v = z s δ x st + ζ s x 2 st + s r sv dx v Integral definition: z s δ x st trivially defined st well defined, if Levy area x 2 provided ζ s x 2 � t s r sv dB v defined through operator Λ if r ∈ C 2 γ 2 , x ∈ C γ 1 and 3 γ > 1 Remark: � t • We shall define s z v dx v more rigorously • Equation (1) solved within class of proc. with decomposition (2) Samy T. (Nancy) Rough Paths 3 KU 2013 9 / 46

Sketch Heuristics 1 Controlled processes 2 Differential equations 3 Aplication to fBm 4 Final remarks 5 Higher order structures Lyons theory Some projects Samy T. (Nancy) Rough Paths 3 KU 2013 10 / 46

Definition 1. Let z ∈ C κ 1 with 1 / 3 < κ ≤ γ . We say that z is a process controlled by x , if z 0 = a ∈ R , and s , t ∈ [ 0 , T ] , (3) δ z = ζδ x + r , i.e. δ z st = ζ s δ x st + r st , with ζ ∈ C κ 1 , and r is a remainder such that r ∈ C 2 κ 2 . Controlled processes space denoted by Q κ, a , and a controlled process z ∈ Q κ, a should be considered as a couple ( z , ζ ) . Natural semi-norm on Q κ, a : N [ z ; Q κ, a ] = N [ z ; C κ 1 ] + N [ ζ ; C b 1 ] + N [ ζ ; C κ 1 ] + N [ r ; C 2 κ 2 ] with N [ g ; C κ 1 ] = � g � κ and N [ ζ ; C b 1 ( V )] = sup 0 ≤ s ≤ T | ζ s | V . Samy T. (Nancy) Rough Paths 3 KU 2013 11 / 46

Operations on controlled processes In order to solve equations, two preliminary steps: Study of transformation z �→ ϕ ( z ) for 1 ◮ Controlled process z ◮ Smooth function ϕ Integrate controlled processes with respect to x 2 Samy T. (Nancy) Rough Paths 3 KU 2013 12 / 46

Composition of controlled processes Proposition 2. Consider z ∈ Q κ, a , ϕ ∈ C 2 b . Define ˆ z = ϕ ( z ) , ˆ a = ϕ ( a ) . z ∈ Q κ, ˆ Then ˆ a , and z = ˆ δ ˆ ζδ x + ˆ r , with ˆ ζ = ∇ ϕ ( z ) ζ and ˆ r = ∇ ϕ ( z ) r + [ δ ( ϕ ( z )) − ∇ ϕ ( z ) δ z ] . a ] ≤ c ϕ, T ( 1 + N 2 [ z ; Q κ, a ]) . Furthermore, N [ˆ z ; Q κ, ˆ Samy T. (Nancy) Rough Paths 3 KU 2013 13 / 46

Composition of controlled processes (2) Remark: In previous proposition Quadratic bound instead of linear as in the Young case Due to Taylor expansions of order 2 Next step: Define J ( z dx ) for a controlled process z : Start with smooth x , z Try to recast J ( z dx ) with expressions making sense for a controlled process z ∈ C κ 1 Samy T. (Nancy) Rough Paths 3 KU 2013 14 / 46

Integration of smooth controlled processes Hypothesis: x , ζ smooth functions, r smooth increment Smooth controlled process z ∈ Q 1 , a , namely δ z st = ζ s δ x st + r st Expression of the integral: J ( zdx ) defined as Riemann integral and � t � t s z u dx u = z s [ x t − x s ] + s [ z u − z s ] dx u Otherwise stated: J ( z dx ) = z δ x + J ( δ z dx ) . Samy T. (Nancy) Rough Paths 3 KU 2013 15 / 46

Integration of smooth controlled processes (2) Levy area shows up: if δ z st = ζ s δ x st + r st , J ( z dx ) = z δ x + J ( ζδ x dx ) + J ( r dx ) . (4) Transformation of J ( ζδ x dx ) : � t s ζ s [( δ x ) su dx u ] = ζ s x 2 J st ( ζδ x dx ) = st Plugging in (4) we get J ( z dx ) = z δ x + ζ x 2 + J ( r dx ) Multidimensional case: � t � t � � s ζ ij δ x i su dx j = ζ ij s x 2 , ij s ζ s [ δ x su dx u ] ← → s u st Samy T. (Nancy) Rough Paths 3 KU 2013 16 / 46

Levy area Recall: J ( z dx ) = z δ x + ζ x 2 + J ( r dx ) → For γ < 1 / 2, x 2 enters as an additional data ֒ Hypothesis: Path x is γ -Hölder with γ > 1 / 3, and admits a Levy area, i.e x 2 ∈ C 2 γ x 2 = ” J ( dxdx )” , 2 ( R d , d ) , formally defined as and satisfying: δ x 2 = δ x ⊗ δ x , δ x 2 , ij = δ x i su δ x j i.e. ut , for any s , u , t ∈ [ 0 , T ] and i , j ∈ { 1 , . . . , d } . Remark: • If x is a regular path, Levy area in the Riemann sense. • fBm also admits a Levy area in the Stratonovich sense. Samy T. (Nancy) Rough Paths 3 KU 2013 17 / 46

Integration of smooth controlled processes (3) Analysis of J ( r dx ) : we have seen J ( r dx ) = J ( z dx ) − z δ x − ζ x 2 Apply δ on each side of the identity: [ δ ( J ( r dx ))] sut = δ z su δ x ut + δζ su x 2 ut − ζ s δ x 2 sut = ζ s δ x su δ x ut + r su δ x ut + δζ su x 2 ut − ζ s δ x su δ x ut = r su δ x ut + δζ su x 2 ut . 2 , x 2 ∈ C 2 γ 2 , δ x ∈ C γ Expression with Λ : If r ∈ C 2 κ 2 , δζ ∈ C κ 2 , with κ + 2 γ > 2 κ + γ > 1, then: δ ( J ( r dx )) = r δ x + δζ x 2 J ( r dx ) = Λ( r δ x + δζ x 2 ) ⇒ Samy T. (Nancy) Rough Paths 3 KU 2013 18 / 46

Integration of smooth controlled processes (4) Conclusion: We have seen: z δ x + ζ x 2 + J ( r dx ) J ( z dx ) = Λ( r δ x + δζ x 2 ) J ( r dx ) = Thus, if m , x are smooth paths: J ( z dx ) = z δ x + ζ x 2 + Λ( r δ x + δζ x 2 ) Substantial gain: This expression can be extended to irregular paths! Samy T. (Nancy) Rough Paths 3 KU 2013 19 / 46

Proposition 3. Let x ∈ C γ 1 , with 1 / 3 < κ < γ , and Levy area x 2 . Let z ∈ Q κ, b , such that δ z st = ζ s ( δ x ) st + r st , with ζ ∈ C κ 1 , r ∈ C 2 κ 2 Define ℓ by z 0 = a ∈ R , and δℓ ≡ J ( z dx ) = z δ x + ζ · x 2 − Λ( r δ x + δζ · x 2 ) . Then ℓ is an element of Q κ, a and: The semi-norm of ℓ in Q κ, a satisfies: 1 � � 1 + T γ − κ N [ z ; Q κ, b ] N [ ℓ ; Q κ, a ] ≤ c x We have 2 n � � z t i ( δ x ) t i , t i + 1 + ζ t i · x 2 � J st ( z dx ) = lim t i , t i + 1 | π st |→ 0 i = 0 Samy T. (Nancy) Rough Paths 3 KU 2013 20 / 46

Sketch Heuristics 1 Controlled processes 2 Differential equations 3 Aplication to fBm 4 Final remarks 5 Higher order structures Lyons theory Some projects Samy T. (Nancy) Rough Paths 3 KU 2013 21 / 46

Pathwise strategy Hypothesis: x is a function of C γ 1 with 1 / 3 < γ ≤ 1 / 2. It x admits a Levy area x 2 Aim: We wish to define and solve an equation of the form: � t y t = a + 0 σ ( y s ) dx s (5) Meaning of the equation: y ∈ Q a ,κ , and δ y = J ( σ ( y ) dx ) Samy T. (Nancy) Rough Paths 3 KU 2013 22 / 46

Fixed point: strategy A map on a small interval: Consider an interval [ 0 , τ ] , with τ to be determined later Consider κ such that 1 / 2 < κ < γ < 1 In this interval, consider Γ : Q a ,κ ([ 0 , τ ]) → Q a ,κ ([ 0 , τ ]) defined by: z 0 = a , and for s , t ∈ [ 0 , τ ] : Γ( z ) = ˆ z , with ˆ � t δ ˆ z st = s σ ( z r ) dx r = J st ( σ ( z ) dx ) Aim: See that for a small enough τ , the map Γ is a contraction → our equation admits a unique solution in C κ ֒ 1 ([ 0 , τ ]) Remark: Same kind of computations as in the Young case ֒ → but requires more work (quadratic estimates, patching)! Samy T. (Nancy) Rough Paths 3 KU 2013 23 / 46