KMT theory applied to approximations of SDE

Transcription

1 KMT theory apped to approxmatons of SD A. M. Dave September 18, 014 Abstract The dyadc method of Komós, Major and Tusnády s a powerfu way of constructng smutaneous norma approxmatons to a sequence of parta sums of..d. random varabes. We use a verson of ths KMT method to obtan order 1 approxmaton n a Vasersten metrc to soutons of vector SDs under a md non-degeneracy condton usng an easy mpemented numerca scheme. 1 Introducton The pathwse smuaton of soutons of vector stochastc dfferenta equatons s chaengng because, usng standard methods, to obtan approxmatons to order greater than 1 requres smuaton of terated ntegras of the Brownan path, whch s dffcut. One approach s to see approxmatons n a Vasersten metrc, meanng that there s a coupng between the approxmate and exact soutons wth respect to whch the error s of the desred order. [] descrbes an easy generated scheme, based on the standard order 1 Msten scheme, whch s order 1 n a Vasersten metrc, provded the SD has a nondegenerate dffuson term. Here we descrbe a modfed verson of the scheme from [] whch gves order 1 under a weaer nondegeneracy condton. The proof uses a constructon of a coupng based on the KMT method. Secton revews the bascs of SD approxmaton and states the man resut. Secton 3 brefy revews the KMT theorem and presents some requred matera from coupng and optma transport theory. The rest of the paper s devoted to the proof of the theorem and a reevant exampe. Some other wor on SD approxmaton usng coupng s descrbed n the fna chapter of voume of [6]. We aso menton [1] whch obtans an order ɛ bound n a Vasersten 3 metrc for the uer method n one dmenson. The author s gratefu to the referee for severa suggestons whch mproved the presentaton of the paper. Approxmaton of SDs Here we brefy revew the Msten scheme and formuate our new verson. Consder an Itô SD dx (t) = b (t, x(t))dw (t), x (0) = x (0), = 1,, q (1) =1 on an nterva [0, T ], for a q-dmensona vector x(t), wth a d-dmensona drvng Brownan path W (t). If the coeffcents b (t, x) satsfy a goba Lpschtz condton a (t, x) a (t, y) C x y, b (t, x) b (t, y) C x y () 1

2 for a x, y R q, t [0, T ] and a,, where C s a constant, and f a and b are contnuous n t for each x, then (1) has a unque souton x(t) whch s a process adapted to the ftraton nduced by the Brownan moton. Ths souton satsfes satsfes x(t) p < for each p [1, ) and t [0, T ]. The standard approach to the strong or pathwse approxmaton of the souton of (1), as descrbed for exampe n [4], s to dvde [0, T ] nto a fnte number N of subntervas, whch we sha usuay assume to be of equa ength h = T/N, and to approxmate the equaton on each subnterva usng a stochastc Tayor expanson. Such expansons are descrbed n deta n chapter 5 of [4]. The smpest such approxmaton, usng ony the near term n the expanson, gves the uer (aso nown as uer-maruyama) scheme x (j+1) = x (j) + =1 whe addng the quadratc terms gves the Msten scheme x (j+1) = x (j) + =1 b (t j, x (j) )V (j) + b (t j, x (j) )V (j) (3),=1 ρ (t j, x (j) )I (j) (4) where V (j) = W ((j + 1)h) W (jh), I (j) = (j+1)h {W jh (t) W (jh)}dw (t) and ρ (t, x) = q m=1 b m(t, x) b x m (t, x). Assumng () the uer scheme has order 1, n the sense that (maxn j=1 x (j) x(jh) ) = O(h) and under a stronger smoothness condton on the b the Msten scheme has order 1, ndeed ( max N j=1 x(j) x(jh) ) = O(h ) (5) (see Koeden and Paten [], Secton 10.3). These L bounds can be extended to L p for any p 1. The uer scheme s straghtforward to mpement, as the ony random varabes one has to generate are the normay-dstrbuted V (j), but for Msten one has aso to generate the area ntegras I (j) whch s non-trva f d. Order 1 s the best one can do n genera when the ony random varabes generated are the V (j). We remar here that we can wrte I (j) (j) (V V (j) hδ ) + ζ (j) V (j) ζ (j) V (j) + K (j) = 1 wth random varabes ζ (j), K(j) for 1, d a havng zero mean, varance h, satsfyng 1 K (j) = K (j), and such that the d(d + 1)/ random varabes consstng of ζ(j) : 1 d and K (j) : 1 < d are mutuay uncorreated (though not ndependent). Motvated by ths remar we consder the foowng modfcaton of Msten, whch requres the generaton of norma random varabes ony. x (j+1) where agan the z (j) = x (j) + + λ (j), where the z(j) ), and then we set λ(j) =1 b (t j, x (j) ) +,=1 are ndependent N(0, h) and J (j) for 1 d and λ (j) for < and λ (j) ρ (t j, x (j) )J (j) (6) = 1 (j) (Ṽ for 1 < d are ndependent hδ ) + z (j) N(0, h = 1 λ(j) = 0. Our man resut s that, under sutabe reguarty condtons and a fary md nondegeneracy condton, the scheme (6) has order 1 under a sutabe coupng. To formuate the nondegeneracy condton, we defne for each (t, x) [0, T ] R d a near mappng L t,x : R d S d R q by L t,x (r, s) = d =1 b (t, x)r + d,=1 ρ (t, x)s for r R d and s S d, where S d s the

3 space of sew-symmetrc d d matrces. We w requre that L x be surjectve for each (t, x) [0, T ] R d ; ths s equvaent to requrng that for each (t, x) the vectors b (x) and [b, b ](x), for 1, d, w span R q (here b s the vector whose th component s b, and [b, b ] denotes the Le bracet, regardng b and b as vector feds on R q for each t). Ths can be thought of as a strengthened Hörmander condton. In fact we need a verson wth some unformty n (t, x), whch we state precsey n the man theorem: Theorem 1. Suppose that the frst and second dervatves of b are bounded on [0, T ] R q, and that there constants δ > 0 and K > 0 such that for each (t, x) [0, T ] R q the mage under L t,x of the unt ba n R r S d contans the ba of radus δ(1 + x ) K n R q. Then there s a constant C > 0 such that f N N s gven we can fnd ndependent N(0, h ) random varabes z(j) 1 for 1 d and λ (j) for 1 < d and 0 j < N, defned on the same probabty space as the Brownan path W (t), such that, f x (j) s as gven by the scheme (6), we have x (j) x(jh) Ch for j = 1,, N. A smar resut s proved n [] for the scheme x (j+1) = x (j) + =1 b (t j, x (j) )V (j) + 1,=1 ρ (t j, x (j) )(V (j) V (j) hδ ) under a stronger nondegeneracy condton that the matrx (b ) has ran q. The proof of theorem 1 occupes much of the remander of the paper. We note here some propertes of the jont characterstc functon χ of the random varabes (z ), (λ ), regarded as a functon on R d(d+1)/. An expct expresson for χ can be found n [9]. What we requre are the foowng (tang the case h = 1, from whch the genera case can be deduced by scang): χ extends to be anaytc on a strp {x + y : x, y R d(d+1)/ n C d(d+1)/ and y < δ} for some δ > 0, and χ(x + y) < C(1 + x ) 1 on ths strp. 3 Coupng and KMT theory If we have two probabty spaces (X, F, P) and (Y, G, Q) then a coupng between P and Q s a measure on X Y whch has P and Q as ts margna dstrbutons. Theorem 1 asserts the exstence of a coupng between the probabty space of the Brownan path and that of the random varabes used n the approxmaton (6). We coect here some resuts on coupngs whch we sha need. Frst we menton the Vasersten metrcs on probabty measures on R n. If P 1 and P are such measures, we defne W p (P 1, P ) to be the nfmum of X Y p, taen over a coupngs between P 1 and P where X and Y have dstrbutons P 1 and P respectvey. For p 1 one can then show that W p s a metrc on the set of a probabty measures P on R n havng fnte pth moment (.e. satsfyng R n x p dp(x) < ). W p s nown as the p-vasersten metrc after [7]. (Note: we use the transteraton Vasersten from the Cyrc as that s the one used by Vasersten hmsef; Wassersten s aso used n the terature). We aso note the eementary resut (see e.g. proposton 7.10 n [8]) that W p (µ, ν) (p 1)/p { x p d µ ν (x)} 1/p (7) for any two probabty measures µ, ν on R n and for any p 1. Ths s qute a good bound f p = 1 but s ess good for p > 1; we sha however use t for boundng some sma remander terms. The KMT theorem [5] s a form of smutaneous Centra Lmt Theorem usng coupng. A varant of ths resut (modfed from the orgna to be coser to the type of resut we w 3

4 use) states that f P s a sutaby we-behaved probabty dstrbuton on R, wth zero mean, varance 1 and zero 3rd moment, then there s a constant C > 0 such that the foowng hods: f n N and X 1,, X n are ndependent wth dstrbuton P, and f Y 1,, Y n are ndependent N(0, 1), then there s a coupng between the random vectors (X 1,, X n ) and (Y 1,, Y n ) such that { (X Y )} C =1 for = 1,, n. There are varous generasatons n the terature. nmah [3] extended the resut to vector random varabes and Zatsev [10] further extended t to non-dentca dstrbutons whch are unformy non-degenerate. What we requre s a varant of ths atter resut where the dstrbutons are themseves random. It s not cear that ths can be easy deduced from resuts n the terature so we prefer to gve a sef-contaned argument n the context we need. Ths argument w use the emma and coroary beow, on poynoma perturbatons of norma dstrbutons. We denote by φ the standard norma N(0, I) dstrbuton on R q. Lemma. Let X be an R q -vaued random varabe wth N(0, I) dstrbuton, et p : R q R q be a poynoma functon of degree 3, and defne ρ : R q R q by ρ(x) = x + p(x). Let P be the probabty dstrbuton of ρ(x) and et ν be the sgned measure on R q wth densty φ(y)(1 + y.p(y).p(y)). Then for any M 1 we have a bound R q (1 + y ) M d P ν (y) Cɛ (8) where C s a postve constant dependng ony on q and M, and ɛ s an upper bound for the absoute vaues of the coeffcents of p. Proof. We use C 1, C etc to denote postve constants whch depend ony on q and M. Frst we can fnd C 1 1 such that and max( ρ(x) x, Dρ(x) I ) C 1 ɛ(1 + x ) 3 (9) max( r(x), Dr(x) ) C 1 ɛ (1 + x ) 9 (10) for a x R q, where r(x) = p(x) p(x + p(x)). Then et R = (C 1 ɛ) 1/6 1 and et B R = {x R q : x < R} (whch w of course be empty f R 0, whch can happen f ɛ s not very sma). Now defne a measure µ as the mage under ρ of the restrcton to B R of the N(0, I) dstrbuton on R q. We aso defne ν = ν ρ(b R ). Then we have R q (1 + y ) M d P ν (y) Ω 1 + Ω + Ω 3 where Ω 1 = (1 + y ) M d µ ν (y), Ω R q = (1 + y ) M d(p µ)(y) and Ω R q 3 = (1 + R q y ) M d ν ν (y). We frst bound Ω 1. To ths end we note that, by the defnton of R, for x B R the RHS of (9) s bounded by 1 ɛ 1/. It then foows from (9) that for x B R we have Dρ(x) I 1 and so ρ s bjectve on B R. Then the densty f of ν on ρ(b R ) s gven by f(y) = det Dρ 1 (y)φ(ρ 1 (y)) and so we have Ω 1 = (1 + y ) M det Dρ 1 (y)φ(ρ 1 (y)) (1 + y.p(y).p(y))φ(y) dy ρ(b R ) 4

5 To bound the RHS, we fx x B R and set y = ρ(x), notng that x y mn(1, y 1 ) by (9). Notng that x = y p(y) + r(x) and usng the bound (10) we ready fnd that and φ(x) (1 + y.p(y).p(y))φ(y) C ɛ (1 + y C 3 )φ(y) det Dρ 1 (y) (1.p(y)) C ɛ (1 + y C 3 ). From ths we easy deduce that Ω 1 C 4 ɛ. Smar bounds for Ω and Ω 3 are aso easy proved, usng the exponenta decay of φ, and the resut foows. We requre a coroary of ths emma, for whch we frst ntroduce some notaton. Let P denote the space of a rea-vaued poynomas on R q, and P q the space of R q -vaued functons p = (p 1,, p q ) on R q such that each p s a poynoma. Let P 0 denote the subspace of S P such that R q S(y)φ(y)dy = 0. We can characterse P 0 as foows. Let L : P q P be the near mappng defned by Lp(x) =.p(x) x.p(x). Then.(φp)(x) = Lp(x)φ(x) and t foows from the dvergence theorem that Lp P 0 for every p P q. In the converse drecton, we note that f u P has degree n 1 then L u = nu + r where r P has degree ess than n. If ths u s n P 0 then we have r P 0 and by nducton on n we can deduce that u s n the range of L. So P 0 s precsey the range of L. Coroary 3. Let g P have degree 4, et µ be the measure wth densty φ (.e. the standard norma probabty measure) and et λ be a probabty measure on R q such that R q (1 + y ) M d (1 + g)µ λ (y) α (11) Then W M (µ, λ) C(ɛ + (ɛ + α) 1/M ), where C s a postve constant dependng ony on q and M, and ɛ s an upper bound for the absoute vaues of the coeffcents of g. Proof. Let β = gdµ. Then (11) gves β α, and g β P 0. So by repacng g by g β we can assume g P 0. Then as descrbed above we can fnd p P q wth Lp = g, and from the constructon of p t s cear that p has degree 3 and ts coeffcents are bounded by C 1 ɛ. Let X be an N(0, I) random varabe and et Y = X + p(x). By emma we have R q (1 + y ) M d gµ ν (y) C ɛ and so R q (1 + y ) M d ν λ (y) C ɛ + α Hence by (7), W M (ν, λ) C 3 (ɛ + α) 1/M. Fnay Y X M = p(x) M C 4 ɛ M so W M (µ, ν) C 1/M 4 ɛ and the resut foows by the trange nequaty. 4 Frst reducton For the Msten approxmaton (x (j) ) gven above, we now that x (j) x(jh) Ch hods under the assumptons of the theorem. So to prove the theorem t suffces to obtan a bound x (j) x (j) Ch. We w construct a coupng between the set of random varabes V (j), I (j) used for Msten and the set of random varabes used by (6), such that ths bound hods. 5

6 We frst spt each of the random varabes V (j) where Q (j) N(0, h h ) and R (j) as the sum of two parts: V (j) = Q (j) +R(j) N(0, h ) are ndependent. (See the remars foowng the proof of Theorem 1 for dscusson of ths spttng). Now et (u (j) approxmaton defned by the recurrence reaton u (j+1) = u (j) + =1 ) be the modfed uer b (u (j) )Q (j) (1) wth u (0) = x (0). Then defne the q q matrx A (j) by A (j) = d b =1 x (u (j )Q (j), and a modfed matrx by Â(j) = A (j) f A (j) 1, and Â(j) = 0 otherwse. We aso defne a modfed verson of I (j) ζ (j) Q(j) ζ (j) Q (j) by repacng V by Q, namey I (j) + K (j). Then defne α(j) by the recurrence reaton α (j+1) = {(I + Â(j) )α (j) } + =1 b (u (j) )R (j) + = 1 (Q(j) Q(j) (h h )δ ) +,=1 ρ (u (j) )I (j) (13) wth α (0) = 0. Next defne β (j) = x (j) u (j) α (j) R q and note that then β (0) = 0. We have β (j+1) β (j) = x (j+1) x (j) (u (j+1) u (j) ) (α (j+1) α (j) ) and usng (4), (1) and (13) we fnd after some rearrangement that { } β (j+1) β (j) b q = (u (j) )β (j) Q (j) + b (x (j) ) b (u (j) b ) (u (j) )(x (j) u (j) ) x x,=1 + =1 =1 {b (x (j) ) b (u (j) )}R (j) + + {(A (j) Â(j) )α (j) } +,=1 =1 (ρ (x (j) ) ρ (u (j) ))I,=1 ρ (u (j) )(I (j) I (j) ) We now bound the RHS of (14). Frst note that, condtona on the random varabes Q (), R (), ζ (), K () for < j, each of the 6 terms on the RHS has expectaton 0. Aso the frst term has varance bounded by C 1 β (j) h. Next, we see that the scheme (1) has order 1, beng an uer scheme wth the random term scaed by 1 h = 1 + O(h), so that x (j) u (j) C h. Then we see that each of the other 3 terms on the RHS has varance bounded by C 3 h 3. Then we concude from (14) that β (j+1) (1 + C 1 h) β (j) + C 3 h 3 and hence that x (j) u (j) α (j) = β (j) C 4 h (15) for j = 1,, N. We can do a smar anayss for (x (j) z (j) R (j) ) as defned by (6) usng random varabes, and λ (j) as above. We agan wrte (j) (j) (j) = Q + R wth Q N(0, h h ) and N(0, h ). Our ntenton s to construct a coupng between the two sets of random varabes so that they a a defned on the same probabty space, on whch we can compare the two approxmatons. Our coupng w satsfy Q (j) = Q (j), so we w assume ths from now on. Then, usng the same Â(j) and u (j) as above, we defne α (j) by the recurrence reaton α (j+1) = {(I + Â(j) ) α (j) } + b (u (j) ) =1 6 R (j) +,=1 (14) ρ (u (j) )J (j) (16) Q (j)

7 wth α (0) = 0, where J (j) before we obtan a bound = 1 (Q(j) Q(j) (h h )δ ) + z (j) Q(j) z (j) Q (j) + λ (j) and just as x (j) u (j) α (j) = β (j) C 5 h (17) From the bounds (15) and (17) we see that to prove the theorem t s enough to obtan a bound α (j) α (j) Ch (18) We prove ths n the next secton. As preparaton we note some propertes of the process (u (j) ). We et G denote the σ- agebra generated by Q (0),, Q (N 1), so that the u (j) and Â(j) are G-measurabe. As u(j) s an uer approxmaton to (1), wth the random term scaed by 1 h, standard bounds appy and we have u (j) p C(p) for any p 1. We aso defne B (r) = (I + Â(1) ) 1 (I + Â (r) ) 1 and we ready obtan B (r) p C(p) and (B (r) ) 1 p C(p). 5 Proof of theorem Throughout we use C to denote a constant whch may depend on the SD but s ndependent of N; each occurrence may be dfferent. Wth B (r) as defned above we set γ (r) = B (r) { =1 γ (r) = B (r) { =1 R (r) b (u (r) ) + R (r) b (u (r) ) +,=1,=1 σ (u (r) )Q (r) ζ (r) + } σ (u (r) )K (r), 1 < d σ (u (r) )Q (r) z (r) + } σ (u (r) )λ (r), 1 < d where σ (x) s the vector n R q whose th component s ρ (x) ρ (x), and we see that j 1 α (j) α (j) = (B (j) ) 1 (γ (r) γ (r) ) (19) It s convenent to reformuate the above expressons for γ (r) and γ (r) usng random varabes scaed to have varance 1. We et m = d(d + 3)/ and defne random vectors X (r) = (X (r) 1,, X m (r) ) by X (r) = h 1 R (r) for = 1,, d; X (r) = (1/h) 1/ ζ (r) d for = d + 1, d; X (+1)(d /)+ = 1 1/ h 1 K (r). Then (condtona on G), X(r) has mean 0 and covarance matrx I. We can then wrte h 1 γ (r) = G r X (r) where G r s a q m matrx defned n terms of B (r), b (u (r) ), σ (u (r) ), Q (r). In the same way we have h 1 γ (r) = G X(r) r where X (r) s N(0, I). We have nequates ( ) G r B (r) b (r) (u(r) + σ (u (r) (h 1/ Q (r) + 1) =1,=1 and G r G t r B (r) F (u (r) )B (r)t where F (x) = d =1 b (x)b (x) t + 1 r=0 1 < σ (x)σ (x) t. We note that the nondegeneracy hypothess n the theorem mpes that (F (x)f (x) t ) 1 C(1 + x ) K. From these bounds and those at the end of the ast secton we deduce that G r p C(p) and (G r G t r) 1 p C(p) for a p 1. We remar that, condtona on G, γ (r) and γ (r) have the same covarance matrx h G r G t r. 7

8 From now on we assume for convenence that N s a power of, N = κ (ths can aways be arranged by extendng the SD to the nterva [0, κ h] where κ s the smaest nteger such that κ N). Let 0 = {0, 1,, κ 1}. We ca a subset of 0 dyadc f t s of the form = {m n, m n + 1,, (m + 1) n 1} for some n {0, 1,, κ} and m {0, 1,, κ n 1}. We see then that, for each n, the dyadc sets of sze n form a partton of 0, and each dyadc set of sze n+1 s the unon of two dyadc sets of sze n. For each dyadc set of sze n we then defne γ = r γ(r), γ = r γ(r) and H = n r G rg t r. Note that snce, condtona on G, the random varabes γ (0),, γ (N 1) are ndependent, H s the (condtona) covarance matrx of Y := n/ h 1 γ. The same appes to γ. The dea s to construct coupngs between γ and γ recursvey, startng wth 0 and proceedng by successve bsecton. For ths purpose we use the foowng emma, whch s a verson of the Centra Lmt Theorem sayng that the densty of γ s cose to the (Gaussan) densty of γ. Lemma 4. Let be a dyadc set of sze n, and et f be the densty functon of Y, condtona on G. Fx η wth 0 < η < 1. Then, provded G 1 r < ηn and (G r G t r) 1 < ηn for each r, we have for v < ηn that f (v) 1 p (v) f < C(16η )n where f (v) = (π) q/ (det H ) 1/ exp( 1 vt H 1 v) s the densty of Ỹ and p (v) s a 4th degree poynoma whose coeffcents are bounded by C (4η 1)n. Proof. Note frst that the bounds on G r mpy H nη and H 1 nη. Let ψ be the characterstc functon of the random varabe X (r) (whch s ndependent of r). ψ s rea-vaued and even on R m, and extends to a compex-anaytc functon on a strp {x + y : x, y R m, y < a} for some a > 0. In a neghbourhood of 0 n C, og ψ has a convergent expanson og ψ(z) = 1 z + c 4 (z) + c 6 (z) + where c (z) s a homogeneous poynoma of degree, and c (z) (C z ) for even 4. Then ψ(z) = exp zt z + χ(z) where χ(z) = c 4 (z) + c 6 (z) +. From ths t foows that there exsts δ > 0 such that f x, y R wth y x < δ then ψ(x + y) e x /6 (0) Then usng the decay of ψ as x and the fact that ψ(x) < 1 for x R wth x 0, we can fnd γ wth 0 < γ < 1 and δ > 0 so that f x, y R wth x δ and y δ then ψ(x + y) mn(γ, C x 1 ) (1) Now et Ψ be the characterstc functon of Y ; then Ψ(u) = r ψ( n/ G t ru) and f (v) = (π) q/ R q e utv Ψ(u)du, whch by transatng the subspace of ntegraton n C q by H 1 v we can wrte as f (v) = (π) q e vt H 1 R v e utv Ψ(u H 1 v)du () q If u 4ηn+1 we can wrte Ψ(u H 1 v) = r ψ( n/ G t ru n/ G t rh 1 v). Now usng (0) and (1) we see that each term n the product s bounded by ether mn(γ, (C n(η+1/) u 1 ) or exp( (1+η)n u /6), and we deduce that Ψ(u H 1 v) mn(γ, (Cn(η+1/) u 1 ) n + exp( ηn u /6) for u 4ηn+1. It then foows easy that Ψ(u H 1 v) du C { nm γ n + exp( 6ηn 1 ) } (3) u 4ηn+1 8

9 To get a bound for u 4ηn+1 we wrte w = u H 1 v and note that e utv Ψ(w) = exp( 1 vt H 1 v 1 ut H u + Λ(w) where Λ(w) = r χ( n/ G t rw) = = S (w) where S (w) = n r c (G t rw). We see that S s a homogeneous poynoma of degree and satsfes S (w) C (1 +η)n w. We fnd that e Λ(w) 1 S 4 (w) C (8η )n w 6 and hence that e 1 vt H 1 u v e utv Ψ(u H 1 v) (1+S 4(u H 1 H u v))e u du C (16η )n (4) 4ηn+1 We aso have 1+S u 4ηn+1 4 (u H 1 v) e 1 ut H u du Ce ηn and combnng these bounds the emma foows, wth p (v) = S R q 4 (u H 1 v)e 1 u H u du whch s a poynoma of degree 4 whose coeffcents are bounded by C (4η 1)n. Inta step. We start the constructon by fndng a coupng between Ỹ 0 and Y 0. Let 0 be the event that condton (7) beow hods wth = 0. Then provded 0 hods, emma 4 gves f 0 (y)/ f 0 (y) 1 p 0 (y) < C (16 η)n for y nη/3. To appy Coroary 3 we wrte y = H 1/ 0 u and g(u) = (det H 0 ) 1/ f 0 (H 1/ 0 U), and deduce that { ( )} (1 + u ) p 0 H 1/ φ(u) g(u) du < C (16η )n (5) A 0 u where A = {u R q : H 1/ 0 u < nη/3 }. One can easy see that the ntegra over A c s bounded by C n so that (5) hods wth the ntegra over R q. And the poynoma p 0 (H 1/ u) has coeffcents bounded by C (4η 1)n so from Coroary 3 we have W 3 (g, φ) C (16η )n/3. Then W 3 (f 0, f 0 ) H 1/ 0 W 3 (g, φ) C (17η )n/3. So can can fnd a coupng between Y 0 and Y so that Y 0 Ỹ 0 3 C (17η )n (6) Recursve step. Let be a dyadc set of sze n where n 1. We can wrte n a unque way as the unon of two dsjont dyadc sets F and G of sze n 1 and note that Y F + Y G = 1/ Y and ỸF + ỸG = 1/ Ỹ. We suppose a coupng between Ỹ and Y has been defned, condtona on G. In other words, for each choce of Q (0),, Q (N 1), we have a jont dstrbuton of Y and Ỹ wth the correct condtona margna dstrbutons. We wsh to extend ths coupng to a coupng between (Y F, Y G ) and (ỸF, ỸG). For each x R q, et f x be the densty of Y F condtona on Y = x and on G, and et f x be the densty of ỸF condtona on Ỹ = x and on G. We note that the condtona dstrbuton H G. So f x s the densty functon of N(Jx, H). We need to fnd a coupng between Y F and ỸF, condtona on Y = x and Ỹ = x. To do ths we need a coupng between the dstrbutons wth denstes f x and f x. We sha n fact construct a coupng between f x and f x, then use the fact that f x s just f x transated by J( x x). of ỸF, gven Ỹ = x and G, s N(Jx, H) where J = H F H 1 and H = 1 H F H 1 Frst we note that f x (y) = 1/ f F (y)f G ( 1/ x y) f. Then the provded the condton (x) G r < ηn/6 and (G r G t r) 1 < ηn/3 for each r (7) hods, appyng emma 4 to each of, F, G gves f x (y) f x (y) 1 p x(y) < C(16η )n (8) for x, y nη/3, where p x (y) = p F (y) + p G ( 1/ x y) p (x). 9

10 Let Ω = {x R q : ( Y F 3 χ YF ηn/3 Y = x & G) > n }, and et p = 60/η. Then we see that, provded (7) hods, P(Y Ω G) n ( Y F 3 χ YF ηn/3 G) 18n ( Y F p+3 G) C 18n H (p+3)/ C n (9) Let denote the event that (7) hods, Y nη/6 1 and Y / Ω. Wrte x = Y. In order to appy Coroary 3 to the condtona dstrbuton of Y F, we mae the change of varabe y = Jx + H 1/ u, notng that then f x (y) = (det H) 1/ φ(u). We defne g x (u) = (det H) 1/ f x (Jx + H 1/ u). Then, provded hods, (8) gves (1 + u ) 3 { 1 + p x (Jx + H 1/ u) } φ(u) g x (u) du < C (16η )n (30) A where A = {u R q : Jx + H 1/ u < nη/3 }. Aso, wrtng y = Jx + H 1/ u, f y nη/3 we have H 1/ u y so u 1+nη/6 y and then, usng x / Ω, we fnd (1 + u ) 3 g A c x (u) (η )n. We aso easy get (1 + A u )3 1 + p x (Jx + H 1/ u)φ(u) du < C n. Puttng these bounds together, we obtan (1 + u ) 3 { 1 + p x (Jx + H 1/ u) } φ(u) g x (u) du < C (16η )n (31) R q The poynoma p x (Jx + H 1/ u) has coeffcents bounded by C (5η 1)n and then appyng Coroary 3 we deduce that W 3 (g x, φ) C (16η )n/3. Then W 3 (f x, f x ) H 1/ W 3 (g x, φ) C (17η )n/3. In other words, condtona on Y = x and assumng, we can fnd a random varabe YF wth densty f x such that YF Y F 3 C (17η )n. If fas then we fnd an arbtrary varabe Y wth densty f x. One easy fnds that P() C 6n and then tang expectaton over G and Y we fnd that uncondtonay We can now compete the recursve step by defnng Y F Y F 3 C (17η )n (3) Ỹ F = Y F + H F H 1 (Ỹ Y ) (33) whch has the correct condtona densty f x wth x = Ỹ. Then we must have ỸG = 1/ Ỹ ỸF. It s natura to defne YG = 1/ Y YF ; then one sees that (3) and (33) both hod wth F repaced by G. Concuson of proof. We appy the recursve procedure as descrbed above, startng wth 0 (nta step), then usng the recursve step to proceed from dyadc sets of sze n to dyadc sets of sze n 1, to generate a coupng for every dyadc set. The resut can be summarsed as foows: condtona on G we have constructed a coupng between the sets of random varabes (Y ) and (Ỹ), each rangng over the dyadc sets, such that (3) and (33) hod whenever F s a dyadc set of sze n 1 contaned n a dyadc set of sze n. Now consder a gven dyadc set of sze n. We can wrte n a unque way = 1 0 where = κ n and, for each j, j s a dyadc set of sze κ j. Then from 0 (Ỹ 0 ) Y 0 ). From ths, usng (33) we obtan Ỹ Y = j=1 H H 1 j (Y j Y j ) + H H 1 (6) and (3) aong wth the L p bounds for H 1 j obtan Ỹ Y 5/ C 3(17η )n/10. Thus and H, and usng Höder s nequaty, we γ γ 5/ C (51η 1)n/10 h (34) hods whenever s a dyadc set of sze n. We now appy ths to (19). If 1 j N we can wrte {0, 1,, j 1} as a unon of dyadc sets 1 where 1,, have 10

11 dfferent szes. Then (19) gves α (j) α (j) = (B (j) ) 1 =1 ( γ γ ). Provded η < 1 51, (34) then gves (18) usng Höder s nequaty and an L 10 bound for (B (j) ) 1. Ths competes the proof of the theorem. Remars. A natura queston s whether the theorem s true wthout the nondegeneracy condton. Wthout ths condton the KMT-type argument faces consderabe technca dffcutes, but I woud conjecture that the theorem s st true. The spttng V (j) = Q (j) + R (j) s ntroduced n order to aow the vectors b as we as the Le bracets be ncuded n the nondegeneracy condton. If we have the nondegeneracy condton wth the bracets ony (.e. the term n r s omtted from the defnton of L t,x ) then we can prove the theorem wthout ths spttng - but ths condton s consderabe stronger (e.g. t cannot hod f q = d = ). Note that our resut s sghty weaer than than the bound (5) for Msten whch has a max over j. (5) s deduced from the bound for ndvdua j usng Doob s martngae nequaty; however we cannot appy ths to our scheme because our coupng does not preserve the ftratons, so the error x (j) x(jh) s not a martngae. The foowng exampe shows that the anaogue of (5) fas for scheme (6), whatever coupng s used. xampe. We consder the SD wth q = 3 and d = gven by dx 1 = dw 1, dx = dw, dx 3 = x 1 dw x dw 1 on the tme nterva [0, 1], wth nta condton x (0) = 0. Ths SD has souton x 1 = W 1, x = W, x 3 (t) = t (W 0 1(s)dW (s) W (s)dw 1 (s)). We fnd that ρ 31 (x) = 1, ρ 31 (x) = 1 and a other ρ are zero. It s then easy to chec that the hypotheses of Theorem 1 are satsfed. We aso note that the Msten approxmaton s exact, n that x (j) = x(jh) for each j. We cam that there s a constant c > 0 such that, for any N N the approxmaton usng scheme (6) wth h = 1 (j) and any coupng between the random varabes Ṽ N, z (j), λ(j) 1 used by (6) and the Brownan path W, we have P( max 0 j<n x(j) x(jh) cn 1 og N) > 1 (35) and To prove ths cam we frst note that x (j+1) 3 x (j) 3 = x (j) 1 V (j) x (j) V (j) 1 + I (j) 1 I (j) 1 (36) x (j+1) 3 x (j) 3 = x (j) 1 x (j) 1 + (z (j) 1 z (j) 1 + λ (j) 1 ) (37) We aso defne random varabes M = max 0 j<n x (j) x (j), K = max 1 j N W (jh) and K = max 1 j N ( x (j) 1, x (j) ). And we set X j = h 1 (I (j) 1 I (j) 1 ), Y (j) = h (z(j) 1 z (j) 1 ) and Z (j) = h λ(j) 1. Then subtractng (36) from (37) and usng the above defntons we fnd that h X (j) Y (j) Z (j) M(1 + K + K) (38) The dea s to use (38) to get a ower bound for M. For ths we need the dstrbutons of the random varabes on the LHS of (38). Frst note that, from the nown dstrbuton of the Lévy area, X (j) has densty 1sech(πx/) so P( X(j) λ) C 1 e πλ/ for λ > 0. And Y (j) 1 can be expressed as (P Q +R S ) where P, Q, R, S are ndependent N(0, 1), so that 3 P +R and Q +S have exponenta dstrbutons, and then a smpe cacuaton shows that Y (j) has a symmetrc exponenta dstrbuton wth P( Y (j) > λ) = e 3λ. Moreover Z (j) has N(0, 1) dstrbuton, from whch one fnds easy that P( Y (j) + Z (j) > λ) C 3 e 5λ/3 (usng 5 < 3). 3 11

12 and Now fx α and β wth 3 < β < α <. Then we have 5 π P( max 0 j<n X(j) α og N) (1 C 1 e πα og N/ ) N πα 1 exp(c 1 N ) P( max Y (j) + Z (j) β og N) C N 1 3β 5 0 j<n So f N s arge enough we have P(max 0 j<n X (j) α og N) 1 and P(max 8 0 j<n Y (j) + Z (j) β og N) 1. Moreover we can fnd a constant C 8 3 so that P(K C 3 ) 1 and 8 P( K C 3 ) 1. Then, wth probabty at east 1, we have 8 max 0 j<n X(j) α og N, max Y (j) + Z (j) β og N, K C 3, and K C 3 (39) 0 j<n Fnay, usng (38), (39) mpes M(1 + 4C 3 ) (α β)h og N, gvng the requred resut. References [1] A. Afons, B. Jourdan and A. Kohatsu-Hga, Pathwse optma transport bounds between a one-dmensona dffuson and ts uer scheme, Ann. App. Probab. 4 (014), [] A. M. Dave, Pathwse approxmaton of stochastc dfferenta equatons usng coupng, preprnt. [3] U. nmah, xtensons of resuts of Komós, Major and Tusnády to the mutvarate case, J. Mutvarate Ana. 8 (1989), [4] P.. Koeden and. Paten, Numerca Souton of Stochastc Dfferenta quatons, Sprnger-Verag [5] J. Komós, P. Major and G. Tusnády, An approxmaton of parta sums of ndependent RV s and the sampe DF. I, Z. Wahr. und Wer. Gebete 3 (1975), [6] S. T. Rachev and L. Ruschendorff, Mass Transportaton Probems, Voume 1, Theory; Voume, Appcatons. Sprnger-Verag [7] L. N. Vasersten, Marov processes over denumerabe products of spaces descrbng arge system of automata, Probemy Peredac Informac 5 (1969), 64-7 (Russan). Transaton n Probems of Informaton Transmsson 5 (1969), [8] C. Van, Topcs n Optma Transportaton, AMS 003. [9] M. Wtorsson, Jont characterstc functon and smutaneous smuaton of terated Itô ntegras for mutpe ndependent Brownan motons, Ann. App. Probab. 11 (001), [10] A. Zatsev, Mutdmensona verson of a resut of Sahaneno on the nvarance prncpe for vectors wth fnte exponenta moments. I,II,III Teor. Veroyatnost. Prmenen, 45 (000), ; 46 (001), and Transatons n Theory Probab. App. 45 (00), ; 46 (003), and Schoo of Mathematcs Unversty of dnburgh 1