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

Rough paths methods 3: Second order structures Samy Tindel Purdue University University of Aarhus 2016 Samy T. (Purdue) Rough Paths 3 Aarhus 2016 1 / 49

Outline Heuristics 1 Controlled processes 2 Differential equations 3 Additional remarks 4 Other rough paths formalisms Higher order structures Samy T. (Purdue) Rough Paths 3 Aarhus 2016 2 / 49

Outline Heuristics 1 Controlled processes 2 Differential equations 3 Additional remarks 4 Other rough paths formalisms Higher order structures Samy T. (Purdue) Rough Paths 3 Aarhus 2016 3 / 49

Examples of fBm paths H = 0 . 3 H = 0 . 5 H = 0 . 7 Samy T. (Purdue) Rough Paths 3 Aarhus 2016 4 / 49

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 , ij v ∈ C 2 γ s dB i s dB j st = for γ < H 2 u Strategy: Given B and B 2 solve the equation in a pathwise manner Samy T. (Purdue) Rough Paths 3 Aarhus 2016 5 / 49

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. (Purdue) Rough Paths 3 Aarhus 2016 6 / 49

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. (Purdue) Rough Paths 3 Aarhus 2016 7 / 49

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. (Purdue) Rough Paths 3 Aarhus 2016 8 / 49

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. (Purdue) Rough Paths 3 Aarhus 2016 9 / 49

Outline Heuristics 1 Controlled processes 2 Differential equations 3 Additional remarks 4 Other rough paths formalisms Higher order structures Samy T. (Purdue) Rough Paths 3 Aarhus 2016 10 / 49

Controlled processes Definition 1. Let 1 / 3 < κ ≤ γ z ∈ C κ 1 We say that z is a process controlled by x , if z 0 = a ∈ R , and δ z = ζδ x + r , i.e. δ z st = ζ s δ x st + r st , s , t ∈ [0 , T ] , (3) with ζ ∈ C κ 1 r is a remainder such that r ∈ C 2 κ 2 Samy T. (Purdue) Rough Paths 3 Aarhus 2016 11 / 49

Space of controlled processes Definition 2. Space of controlled processes: Denoted by Q κ, a 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 N [ ζ ; C b 1 ( V )] = sup 0 ≤ s ≤ T | ζ s | V Samy T. (Purdue) Rough Paths 3 Aarhus 2016 12 / 49

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. (Purdue) Rough Paths 3 Aarhus 2016 13 / 49

Composition of controlled processes Proposition 3. Consider z ∈ Q κ, a , ϕ ∈ C 2 b . Define ˆ z = ϕ ( z ) , ˆ a = ϕ ( a ) . Then ˆ z ∈ Q κ, ˆ a , and z = ˆ δ ˆ ζδ x + ˆ r , with ˆ ζ = ∇ ϕ ( z ) ζ r = ∇ ϕ ( z ) r + [ δ ( ϕ ( z )) − ∇ ϕ ( z ) δ z ] . and ˆ a ] ≤ c ϕ, T (1 + N 2 [ z ; Q κ, a ]). Furthermore, N [ˆ z ; Q κ, ˆ Samy T. (Purdue) Rough Paths 3 Aarhus 2016 14 / 49

Proof Algebraic part: Just write δ ˆ z st = ϕ ( z t ) − ϕ ( z s ) = ∇ ϕ ( z s ) δ z st + ϕ ( z t ) − ϕ ( z s ) − ∇ ϕ ( z s ) δ z st = ∇ ϕ ( z s ) ζ s δ x st + ∇ ϕ ( z s ) r st + ϕ ( z t ) − ϕ ( z s ) − ∇ ϕ ( z s ) δ z st = ˆ ζ s δ x st + ˆ r st Samy T. (Purdue) Rough Paths 3 Aarhus 2016 15 / 49

Proof (2) a ( R n )], strategy: get bound on Bound for N [ˆ z ; Q κ, ˆ N [ˆ z ; C κ 1 ( R n )] N [ˆ ζ ; C κ 1 L d , n ] N [ˆ ζ ; C b 1 L d , n ] r ; C 2 κ 2 ( R n )] N [ˆ Decomposition for ˆ r : We have r 1 + ˆ r 2 ˆ r = ˆ with r 1 r 2 st = ∇ ϕ ( z s ) r st st = ϕ ( z t ) − ϕ ( z s ) − ∇ ϕ ( z s )( δ z ) st . ˆ and ˆ (4) Samy T. (Purdue) Rough Paths 3 Aarhus 2016 16 / 49

Proof (3) r 1 : ∇ ϕ is a bounded L k , n -valued function. Therefore Bound for ˆ r 1 ; C 2 κ 2 ( R n )] ≤ �∇ ϕ � ∞ N [ r ; C 2 κ 2 ( R k )] . N [ˆ (5) r 2 : Bound for ˆ st | ≤ 1 2 �∇ 2 ϕ � ∞ | ( δ z ) st | 2 ≤ c ϕ N 2 [ z ; C κ r 2 1 ( R k )] | t − s | 2 κ , | ˆ which yields r 2 ; C 2 κ 2 ( R n )] ≤ c ϕ N 2 [ r ; C 2 κ 2 ( R k )] , N [ˆ (6) r 1 + ˆ r 2 , we get from (5) and (6) Bound for ˆ r : Since ˆ r = ˆ � � r ; C 2 κ 1 + N 2 [ r ; C 2 κ 2 ( R n )] ≤ c ϕ 2 ( R k )] N [ˆ Samy T. (Purdue) Rough Paths 3 Aarhus 2016 17 / 49

Proof (4) Other estimates: We still have to bound z ; C κ 1 ( R n )] N [ˆ N [ˆ ζ ; C κ 1 L d , n ] N [ˆ ζ ; C b 1 L d , n ] Done in the same way as for ˆ r Conclusion for the analytic part: We obtain � � 1 + N 2 [ z ; Q κ, a ] N [ˆ z ; Q κ, ˆ a ] ≤ c ϕ, T Samy T. (Purdue) Rough Paths 3 Aarhus 2016 18 / 49

Composition of controlled processes (ctd) 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. (Purdue) Rough Paths 3 Aarhus 2016 19 / 49

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 ( z dx ) 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. (Purdue) Rough Paths 3 Aarhus 2016 20 / 49

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 ) . (7) 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 (7) we get J ( z dx ) = z δ x + ζ x 2 + J ( r dx ) Multidimensional case: � t � t � � s ζ ij δ x j su dx i = ζ ij s x 2 , ji s ζ s [ δ x su dx u ] ← → s u st Samy T. (Purdue) Rough Paths 3 Aarhus 2016 21 / 49

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 4. 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 sut = δ x i su δ x j i.e. ut , for any s , u , t ∈ S 3 , T and i , j ∈ { 1 , . . . , d } . Samy T. (Purdue) Rough Paths 3 Aarhus 2016 22 / 49

Levy area: particular cases Levy area defined in following cases: x is a regular path 1 ֒ → Levy area defined in the Riemann sense x is a fBm with H > 1 2 4 → Levy area defined in the Stratonovich sense ֒ Samy T. (Purdue) Rough Paths 3 Aarhus 2016 23 / 49

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 . Samy T. (Purdue) Rough Paths 3 Aarhus 2016 24 / 49

Integration of smooth controlled processes (4) Recall: We have found δ ( J ( r dx )) = r δ x + δζ x 2 Regularities: We have r ∈ C 2 κ 2 δ x ∈ C γ 2 δζ ∈ C κ 2 x 2 ∈ C 2 γ 2 Since κ + 2 γ > 2 κ + γ > 1, Λ can be applied Expression with Λ: We obtain δ ( J ( r dx )) = r δ x + δζ x 2 J ( r dx ) = Λ( r δ x + δζ x 2 ) = ⇒ Samy T. (Purdue) Rough Paths 3 Aarhus 2016 25 / 49