Mini-course On Rough Paths (TU Wien, 2009) P.K. Friz, Last .

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Mini-course on Rough Paths (TU Wien, 2009)P.K. Friz, Last update: 20 Jan 2009

ContentsChapter 1. Rough Paths1. On control ODEs2. The algebra of iterated integrals3. Rough Path Spaces4. Rough Path Estimates for ODEs I5. Rough Paths Estimates for ODEs II6. Rough Di erential Equations11614202325Chapter 2. Applications to Stochastic Analysis1. Enhanced Brownian motion as geometric rough path2. Approximations to Enhanced Brownian Motion3. RDEs driven by EBM4. Stroock-Varadhan’s support theorem for SDEs5. Freidlin-Wentzell large deviations for SDEs6. Comments and References29293436384146iii

CHAPTER 1Rough Paths1. On control ODEs1Given 0 write b c f g with b c 2 N [ f0g and f g 2 (0; 1]. CallLip (Re ) the class of b c-times di erentiable vector elds on Re with all derivativesbounded and b cth -derivative f g-Hölder continuous.Given vector elds V1 ; :::; Vd 2 Lip1 (Re ) ( bounded & Lipschitz continuousvector elds) and x 2 C 1 [0; T ] ; Rd , at least piecewise, then basic ODE resultstell us that there exists a unique solution y (0; y0 ; x) to the (control) ODEdyt Vi (yt ) dxit(1.1)V (yt ) dxt ; t 2 [0; T ] :started at y0 :Notation 1. We write jdxr j : jx r j dr and ys;t : yt ys for path-increments.jx jAlso, jV (:)j : max fjVi (:)j : i 1; :::; dg and jxjLip;[a;b] : supa s t b jt s;tsjProposition 2. There exists a constant C2 so thatj (0; y0 ; x)jLip;[s;t]Proof. We havejys;t j ZZC2 jxjLip ;[s;t] :tV (yr ) dxrstsjV (yr )j jdxr jjV j1 jxjLip; [s;t] jtsj :The Lip-(semi-)norm here can not be replaced by sup-(semi-)norm. That is, ingeneral @C such that jys;t j C sup0 s t T jxs;t j :Exercise 3. V1 (y) 1; 0; 12 y 2 ; V2 (y) 0; 1; 21 y 1 with y y 1 ; y 2 ; y 3 2pR : Let xt n1 exp 21n2 t 2 C R2 and y0 0 2 R3 . Clearly, x ! 0uniformly on [0; 1]. Show that y13 ! area of the unit circle as n ! 1:3Lemma 4. For every Lipschitz continuous path x there exists a sequence ofpiecewise smooth xn such that xn ! x pointwise andsup jxn jLip ;[0;T ]jxjLip ;[0;T ] 1:n1Corrections and comments to

21. ROUGH PATHSProof. Fix a dissecion D (tj ) of [0; T ]. We construct an approximationxD by connecting the points xti with geodesics in Rd . In other words, xD is thepiecewise linear approximation to x. When ti s t ti 1 for some i; thenxDs;t and hence xDs;tjtt sxt ;tti 1 ti i i 1sj jxjLip ;[0;T ] . When tisti 1tjttj 1 thenDxDs;ti 1 xti 1 ;tj xtj; txDs;tjti 1jtsj jtjsj jxjLip ;[0;T ] :ti 1 j tijt jxjLip ;[0;T ]Clearly, for any sequence Dn with mesh jDn j ! 0 we have pointwise convergence.Lemma 5. Let xn be Lipschitz with xn ! x pointwise and supn jxn jLip ;[0;T ] C 1: Then x is Lipschitz, xn ! x uniformly and jxjLip ;[0;T ] lim inf n!1 jxn jLip ;[0;T ]C:Proof. Let tD denote the closest (left) element in some dissection D for agiven t 2 [0; T ]. We show that xn is Cauchy w.r.t. uniform norm (which impliesthat x is the uniform limit). Indeedjxntxmt jnxntD xmtD xt 2 2C jDjxntD xmtxmtDfor n; mN ( ) and D with mesh jDj 4C. For the second statement, notethat, along a subsequence, jxnk jLip ;[0;T ] ! lim inf n!1 jxn jLip;[0;T ] C 0 : Thenxns;tk xs;tjxs;t jfor kk ( ) and some xedjxnk jLip ;[0;T ] jt! 0 we ndjxs;t jsj 0. Sending k ! 1 we nd thatjxs;t jand sendingxns;tkC 0 jtC 0 jtsj sj ) jxjLip ;[0;T ]C 0:Lemma 6. Let x be Lipschitz and xn be (piecewise) C 1 with xn ! x pointwise and supn jxn jLip ;[0;T ] C 1: Let (f n ) be a sequence of continuous paths,uniformly convergent on [0; T ] to some (continuous) path f: ThenZ TZ Tnnf dx !f dx:00The integral on the right hand side can be viewed, equivalently, as Riemann-Stieltjesintegral or as Lebesgue integral with integrand f x,noting that Lipschitz paths areabsolutely continuous.

1. ON CONTROL ODES3Proof. Consider f n f as a rst case. Fix 0. Since f is continuous onthe compact [0; T ] it is uniformly continuous. Therefore, 9 0 and nitely manyintervals Ij (tj ; tj 1 ] of length or less so that [Ij (0; T ] and so that theoscillation of the continuous function f in each Ij is less than : De ne the stepfunctionXh f (tj ) 1(tj ;tj 1 ]and note jf hj1: From Lemma 5; x is Lipschitz and jxjLip ;[0;T ]implies jdxt j Cdt andZ TZ TZ Tjft ht j jdxt j CT :f dxhdx000Similarly, for any n;ZTf dxn0ZThdxnCT :0From Lemma 5, xn ! x uniformly and we haveZ TZ TXftj xntj; tj 1hdxnhdx0C. This0jxtj; tj 1 ! 0as n ! 1. By the triangle inequality and the two preceding estimates,Z TZ Tnlim supf dxf dx2CTn!100and this hold for any 0. Sending ! 0 proves the rst case.The case of asequence f n is another (simple) application of the triangle inequality, simply insert& subtract the integral of f dxn .Proposition 7. Let x be a Lipschitz path and xn be (piecewise) C 1 on [0; T ].Assume xn ! x pointwise and supn jxn jLip ;[0;T ] 1. Set y n : (0; y0 ; xn ) : Thensupn jy n jLip;[0;T ] 1 and y n converges pointwise to a unique limit, say y yt :Moreover yt satis esZ tZ tiy t y0 Vi (ys ) dxs Vi (ys ) x is ds:0We also write0(0; y0 ; x) for this unique limit y.Proof. From Proposition 2, supn j (0; y0 ; xn )jLip C2 supn jxn jLip ;[0;T ] 1and hence (y n ) is equicontinuous. It is also bounded in uniform normsup jytn jt2[0;T ]jy0 j T sup j (0; y0 ; xn )jLip :nArzela-Ascoli’s theorem tells us that, possibly after passing to a sub-sequence, yn (0; y0 ; xn ) converges uniformly on [0; T ] to some continuous path y. By de nitionof y n ;Z tny t y0 V (ysn ) dxns :0

41. ROUGH PATHSUsing Lipschitz regularity of V we see that V (y n ) ! V (y) uniformly. Since byhypothesis xn ! x pointwise with jxn jLip ;[0;T ]supn jxn jLip ;[0;T ] 1 lemma 6shows that as n ! 1Z tyt y 0 V (ys ) dxs0Z tV (ys ) x s ds: y0 0Now assume y is another limit point of the sequence f (0; y0 ; xn )gand set y y . ThenZ tj t j kV kLip1 (Re )j s j jdxs j0and from Gronwall’s lemma, y y : The result follows.Exercise 8. Let x; xn be a Lipschitz paths on [0; T ]. Assume xn ! x pointwise and supn jxn jLip ;[0;T ] 1. Set y n (0; y0 ; xn ) and y (0; y0 ; x) ; as constructed in the previous proposition. Show that supn jy n jLip;[0;T ] 1 and y n ! yuniformly on [0; T ]. Hint: use Lemma 4.It is convenient (and standard in di erential geometry) to identify a vector eldV (y) with the rst order di erential operator V j (y) @y@ jV j (y) @j :Lemma 9. Let x be a Lipschitz path on [0; T ] and y C 1 -function on Re . Thend Z tXf (yt ) f (ys ) Vi f (yr ) dxir :i 1(0; y0 ; x). Let f be asProof. When x is piecewise smooth then this is just the fundamental theoremof calculus. When x is only Lipschitz, approximate using the results above. Thedetails are left as exercise.Remark 10. Vi1ViN makes sense as N th order di erential operator providedthe vector elds are (N 1) times di erentiable.Exercise 11. (i) Check that [V1 ; V2 ] : V1 V2 V2 V1 is in fact a 1st orderdi erential operator (and hence a vector eld). It is called the Lie bracket of thevector elds V1 and V2 :(ii) Compute [V1 ; V2 ] for the vector elds given in exercise 3.(iii) Give a geometric interpretation of the Lie bracket.1.1. Euler scheme of order N . De ne T N Rd d2dNk 00Rdk. For in-stance, Ris just a (d d)-matrix. By convention RR. Let x be anRd -valued Lipschitz path and de ne k th iterated integrals of the path segment xj[s;t]asZ ZZtgk;i1 ;;ik:::sand so that gk gk;i1 ;0set g 1 2 Rd0u2uk: ss;iki1 ;;ik 2f1;:::;dgdxiu11 :::dxikuk :2 Rdk. For later convenienceR. We then de ne the (step-N ) signature of the path

1. ON CONTROL ODES5segment xj[s;t] asSN (x)s;txs;t1 NXk 1g k 2 T N Rd :Let I be the identity function on Re :Definition 12. Let N 2 N. Given LipN -vector elds V (V1 ; :::; Vd ) on Re ,g 2 T N Rd and y 2 Re we callE(V ) (y; g) : NXXVik I (y) gk;i1 ;V i1;ik;k 1 i1 ;:::;ik2f1;:::;dgthe (increment of ) the step-N Euler scheme. In a typical application, g is the(step-N ) signature of a path segment xj[s;t] and we callE(V ) ys ; SN (x)s;tthe (inrement of ) the step-N Euler scheme for dy V (y) dx over the time-interval[s; t].Lemma 13 (Euler ODE estimate). Let V (Vi )1 i d be a collection of vector elds in LipN (Re ) and x : [s; t] ! Rd be a Lipschitz path. Then there exists aconstant C13 C13 (N ) such that for all s t,Z tN 1:Cjdxjy;S(x)jVj(1.2)(s;y;x)E113rsNs(V )(V )s;tLips;tsProof. We rst show thatys;t E(V ) ys ; SN (x)s;tX Z [Vi1i1 ;:::;iN2f1;:::;dgViN I (yr1 )V i1s r1 ::: rN tViN I (ys )] dxir11dxirNN :To this end, consider a smooth function f and note that for any kNV i1Vik f 2 C 1 . By iterated use of the fundamental theorem of calculus,NX1 X ZV i1Vik f (ys ) dxir11dxirkkf (yt ) f (ys ) Xk 1 i1 ;:::;ik2f1;:::;dgi1 ;:::;ik2f1;:::;dgZ1;s r1 ::: rk tV i1s r1 ::: rN tViN f (yr1 ) dxir11dxirNN :and the claim follows from specializing to f I, the identity function. For theproof of (1.2) we momentarily allow all constants to depend on V . Clearly,Z tZ tjys;t j V (y) dxc1jdxr j :ss

61. ROUGH PATHSFrom LipN -regularity of the vector elds, Vi1 ::ViN I ( ) is Lipschitz continuous.Hence, for all r 2 [s; t],Z tjdxr j :jVi1 ::ViN I (yr ) Vi1 ::ViN I (ys )j c2sand after integration,Z[Vi1 ::ViN I (yr1 )Vi1 ::ViN Is r1 ::: rN t(ys )] dxir11dxirNNc3ZN 1tsjdxr j:Summation over the indices nishes the estimate. At last, a scaling argument showsthatN 1c3 c4 jV jLipNfor some constant c4 independent of V . Indeed it su ces to rewritedy V (y) dx V (y) d x x jV jLipN . In particular, V with V V jV jLipN and xLipN1 which is enoughto rerun the above estimates without picking up any dependence on V .2. The algebra of iterated integralsThere are algebraic relations between higher iterated integrals.dExample 14. Consider a Lipschitz pathR t xi startedR tatj x0 i 0 i 2j R . Sincejjii jjx dxr x dxr d x x integration yields 0 xr dxr 0 xr dxr xt xt : Similarly,for arbitrary x0 ;Z tZ t2;i;j2;j;iijxs;t xs;t xs;r dxr xjs;r dxir xis;t xjs;t :issLife is simpler without indices! Recall that the symmetric part of a matrix A isgiven by Sym (A) 21 A AT . ThenSym x2s;t 1 1x2 s;tx1s;t :Note that turns two vectors (1-tensors) v; w into a matrix v w (2-tensor).Similarly, an i-tensor x and a (k i)-tensor y give us a k-tensor x y. Let j jk. Namely, for z z i1 ; ;ik i1 ; ;i 2f1;:::;dgj j(Rd ) k denote Euclidean norm on Rdkwe haveX22jzj(Rd ) k z i1 ; ;ik :;ik 2f1;:::;dgi1 ;The resulting family of norms j j(Rd )jxyj( Rd )kk; k 1; 2; 3; ::: is compatible in the sense thatjxj( Rd )ijyj( Rd )(ki):Usually, there is no confusion in which tensor space computations take placeand we write j j instead of j j(Rd ) k : These norms easily lead to a norm on T N Rd .We setNXi22gi where gi 2 Rd:jgj : i 0

2. THE ALGEBRA OF ITERATED INTEGRALS7g 2 T N Rd : g0 1 is a group under truncated tensorThe set T1N Rdmultiplication which we now introduce. If g 1 g1 ::: gN 1 g and similarfor h then for k 0; :::; N(gkh) kXgihk i :i 0N1(Note that, a priori, gh is a (N 1)-tensor whereas our de nition sets (N 1)and all higher tensor to zero. Hence why we call it truncated.)The neutral element is e 1 1 0 ::: 0 and the inverse is given by theusual power series calculus(1 g )1 1g g 2:::(with truncation beyond level N ). All group operations are smooth (after all, theyare given by polynomials in the coordinates) and we see that T1N Rd is actuallya Lie group2 .If g happens to be the signature of a Lipschitz path, say g SN (x)0;1 we mayask for the signature of the scaled path, say SN ( x)0;1 with 2 R. This inducesthe dilatation map: gk 7!k kg;2 R:3The Lie algebra of T1N Rd can be identi ed withhi gT0N Rdg 2 T N Rd : g 0 0 , g ; h h hg and the exponential map with exp : T0N Rd ! T1N Rd ,1 2g :::2!(again, with truncation beyond level N ). We de ne log : T1N Rd ! T0N Rd byexp ( g) 1 g 1 2g :::2Due to truncation, log is globally de ned and it is easily checked that log expThusexpLie algebra T0N RdLie group T1N Rd .log (g) log (1 g ) g 1.log1Exercise 15. (i) Show that exp ( g) exp ( g ) : (ii) Given a straight pathsegment: x : [a; b] 7! x0 bt aa v with x0 ; v 2 Rd show thatSN (x)a;b exp (v) :1(iii) Conclude that SN ( x )a;b SN (x)a;b where x is x run backwards in time.2 A Lie group is a set which is both a group and a smooth manifold such that the groupoperations are smooth.3 A Lie algebra is a vector space L for which is given a bilinear function from L L to L, calledbracket, and denoted by [ ; ] which satis es (1) [a; a] 0 and (2) [a; [b; c]] [c; [a; b]] [b; [c; a]] 0for all a; b; c 2 L.To every Lie group G there is an associated Lie algebra. As vector space, it is usuallyidenti ed with Te G, the tangent space of the Lie group at the unit element.

81. ROUGH PATHSProposition 16. Let x : [0; T ] ! Rd be Lipschitz continuous with (step-N )signatures xs;t SN (x)s;t . Then(2.1)SN (x)s;tSN (x)t;u SN (x)s;u :kProof. SN (x)s;u is de ned as integral over(k)s;ufs r1 ::: rk ug.For xed t 2 (s; u), ignoring sets of zero k-dimensional Lebesgues measure,equals the disjoint union A0 [ ::: [ Ak withAj(k)s;u fs r1 ::: rj t rj 1 :::: rk ug (j)s;t(k j)t;uThenk;i ;:::;ikSN (x)s;u1 k ZXj 0 k ZXj 0 kXdxi1 :::dxikAji1(j)s;tijdx :::dxjSN (x)s;tZ(k j)t;udxij 1 :::dxikk jSN (x)t;u :j 0Remark 17. Di erently put, the signature of the path segment xj[s;t] tensormultiplied by the signature of the path segment xj[t;u] equals the signature of theirconcatenation ( the path segment xj[s;u] ).It is a general fact from abstract Lie group theory that the Lie algebra describes the Lie-group in a neighbourhood of the unit element. More precisely, theexponential map is a di eomorphism of a neighbourhood of 0 in the Lie algebraonto a neighbourhood of e in the Lie group. Even better, the group multiplicationis (locally) expressed in terms of the brackets. This is the famous Campbell-BakerHausdor formula. In our setting the CBH-formula holds globally. 2 T0N Rd . ThenProposition 18. Let g ; h(2.2)exp g hi ::: exp g 1 g ; hexp

Mini-course on Rough Paths (TU Wien, 2009) P.K. Friz, Last update: 20 Jan 2009. Contents Chapter 1. Rough Paths 1 1. On control ODEs 1 2. The algebra of iterated integrals 6 3. Rough Path Spaces 14 4. Rough Path Estimates for ODEs I 20 5. Rough Paths Estimates for ODEs II 23 6. Rough Di erential Equations 25 Chapter 2. Applications to Stochastic Analysis 29 1. Enhanced Brownian motion as .

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