A limit theorem for the moments in space of Browian local time increments

Transcription

1 A limit theorem for the moments in space of Browian local time increments Simon Campese LMS EPSRC Durham Symposium July 18, 2017

2 Motivation Theorem (Marcus, Rosen, 2006) As h 0, L x+h t L x t h q dx a.s. C q (L x t) q/2 dx 2 / 21

3 CLT for second moment in space Theorem (Chen, Li, Marcus, Rosen, 2010) For h 0, ( 1 ( ) ) 2 h 3/2 L x+h t L x t dx 4ht where Z N (0, 1), independent of (L x t) x R. d c 2 (L x t) 2 dx Z, 3 / 21

4 CLT for second moment in space Theorem (Chen, Li, Marcus, Rosen, 2010) For h 0, ( 1 ( ) ) 2 h 3/2 L x+h t L x t dx 4ht where Z N (0, 1), independent of (L x t) x R. Proof by Method of Moments d c 2 (L x t) 2 dx Z, 3 / 21

5 CLT for second moment in space Theorem (Chen, Li, Marcus, Rosen, 2010) For h 0, ( 1 ( ) ) 2 h 3/2 L x+h t L x t dx 4ht where Z N (0, 1), independent of (L x t) x R. Proof by Method of Moments d c 2 (L x t) 2 dx Z, Alternative proof by Hu/Nualart (2009) using Malliavin calculus 3 / 21

6 CLT for second moment in space Theorem (Chen, Li, Marcus, Rosen, 2010) For h 0, ( 1 ( ) ) 2 h 3/2 L x+h t L x t dx 4ht where Z N (0, 1), independent of (L x t) x R. Proof by Method of Moments d c 2 (L x t) 2 dx Z, Alternative proof by Hu/Nualart (2009) using Malliavin calculus Another proof by Rosen (2011) using self-intersection local times 3 / 21

7 CLT for third moment in space Theorem (Rosen, 2011) For h 0, 1 ( ) 3 h 2 L x+h t Lt x d dx c3 where Z N (0, 1), independent of (L x t) x R. (L x t) 3 dx Z, 4 / 21

8 CLT for third moment in space Theorem (Rosen, 2011) For h 0, 1 ( ) 3 h 2 L x+h t Lt x d dx c3 where Z N (0, 1), independent of (L x t) x R. Again, proof by Method of Moments (L x t) 3 dx Z, 4 / 21

9 CLT for third moment in space Theorem (Rosen, 2011) For h 0, 1 ( ) 3 h 2 L x+h t Lt x d dx c3 where Z N (0, 1), independent of (L x t) x R. Again, proof by Method of Moments Malliavin calculus proof by Hu/Nualart (2010) (L x t) 3 dx Z, 4 / 21

10 Conjecture for fourth moment in space Conjecture (Rosen, 2011) For h 0, ( 1 h 5/2 ( h L x t) 4 dx 24h ( h L x t) 2 L x t dx + 48h 2 (L x t) 2 dx where Z N (0, 1), independent of (L x t) x R. ) d c 4 (L x t) 4 dx Z, 5 / 21

11 Main result: Limit theorem for any moment Theorem (C., 2016) For h 0, ( 1 h (q+1)/2 ( h L x t) q dx q 2 ) + h k a q,k ( h L x t) q 2k (4L x t) k dx k=1 where Z N (0, 1), independent of (L x t) x R. d c q (L x t) q dx Z, 5 / 21

12 Key ingredient: Kailath-Segall identity For continuous L 2 -martingale (M x ) x R, define iterated integrals I 0 (x) = 1, I 1 (x) = x dm u and I q (x) = x I q 1 (u) dm u. 6 / 21

13 Key ingredient: Kailath-Segall identity For continuous L 2 -martingale (M x ) x R, define iterated integrals I 0 (x) = 1, I 1 (x) = x dm u and I q (x) = x I q 1 (u) dm u. Theorem (Kailath, Segall, 1976) q I q (x) = I q 1 (x) M x I q 2 (x) M, M x 6 / 21

14 Key ingredient: Kailath-Segall identity For continuous L 2 -martingale (M x ) x R, define iterated integrals I 0 (x) = 1, I 1 (x) = x dm u and I q (x) = x I q 1 (u) dm u. Theorem (Kailath, Segall, 1976) q I q (x) = I q 1 (x) M x I q 2 (x) M, M x Recursively, q 2 q! I q (x) = a q,k M q 2k x M, M k x k=0 6 / 21

15 Isolation of dominating term Perkins, 1981: (L x t) x R is continuous semimartingale with quadratic variation 4 x Lu t du 7 / 21

16 Isolation of dominating term Perkins, 1981: (L x t) x R is continuous semimartingale with quadratic variation 4 x Lu t du Write h L x t = h M x + h V x. 7 / 21

17 Isolation of dominating term Perkins, 1981: (L x t) x R is continuous semimartingale with quadratic variation 4 x Lu t du Write h L x t = h M x + h V x. Binomial theorem and stochastic analysis: ( h L x t) q dx = ( h M x ) q dx + o (h (q+1)/2) 7 / 21

18 Isolation of dominating term Perkins, 1981: (L x t) x R is continuous semimartingale with quadratic variation 4 x Lu t du Write h L x t = h M x + h V x. Binomial theorem and stochastic analysis: Kailath-Segall: ( h L x t) q dx = ( h M x ) q dx + o (h (q+1)/2) q 2 ( M x ) q = q! I q (x) a q,k ( M x ) q 2k ( M, M x ) k k=1 7 / 21

19 Isolation of dominating term Perkins, 1981: (L x t) x R is continuous semimartingale with quadratic variation 4 x Lu t du Write h L x t = h M x + h V x. Binomial theorem and stochastic analysis: ( h L x t) q dx = ( h M x ) q dx + o (h (q+1)/2) Kailath-Segall: ( consider martingale ( y x dm u at y = x + h) )y x q 2 ( h M x ) q = q! h I q (x) a q,k ( h M x ) q 2k ( h M, M x ) k k=1 7 / 21

20 CLT for dominating term Theorem (C., 2016) For h 0, 1 h (q+1)/2 q! h I q (x) dx d c q (L x t) q dx Z, where Z N (0, 1), independent of (L x t) x R. 8 / 21

21 Key ingredient: asymptotic Ray-Knight theorem (M h x), (N h x) sequences of continuous L 2 -martingales (β h x ), (γ h x ) Dambis-Dubins-Schwarz Brownian motions Theorem (Revuz, Yor, 1999) If sup x [,a] M h, N h prob. 0 x as h 0, then (β h x, γ h x ) converges in distribution to a two-dimensional standard Brownian motion. 9 / 21

22 Proof of CLT for dominating term Iterated stochastic Fubini: h I q (x) dx = = = x+h u1 x u1 u 1 h u2 u 1 h uq 1 K q,h (u 1 ) dm u1 uq u 1 h dm uq... dm u2 dm u1 dx dx dm uq... dm u2 dm u1 10 / 21

23 Proof of CLT for dominating term Iterated stochastic Fubini: h I q (x) dx = = = x+h u1 x u1 u 1 h u2 u 1 h uq 1 K q,h (u 1 ) dm u1 Define martingale ( M h x) x by M h x = uq u 1 h q! x h (q+1)/2 K q,h (u 1 ) dm u1 dm uq... dm u2 dm u1 dx dx dm uq... dm u2 dm u1 10 / 21

24 Proof of CLT for dominating term Use Burkholder-Davis-Gundy-type arguments to show that sup x (,x 0 ] M h, M 0 L1 x for x 0 and M h, M h x x L 1 c 2 q (L u t ) q du for x R {, }. 11 / 21

25 Proof of CLT for dominating term Asymptotic Ray-Knight: ( β, β h, M h, M h ) ( ) d β, β, c 2 q (L u t ) q du, where β, β h DDS-Brownian motions of M and M h, respectively 12 / 21

26 Proof of CLT for dominating term Asymptotic Ray-Knight: ( β, β h, M h, M h ) ( ) d β, β, c 2 q (L u t ) q du, where β, β h DDS-Brownian motions of M and M h, respectively Consequently: M h x = β h M h, M h x d βc 2 q x (Lu t )q du. 12 / 21

27 Proof of CLT for dominating term Asymptotic Ray-Knight: ( β, β h, M h, M h ) ( ) d β, β, c 2 q (L u t ) q du, where β, β h DDS-Brownian motions of M and M h, respectively Consequently: M h x = β h M h, M h x d βc 2 q x (Lu t )q du. Letting x finishes proof 12 / 21

28 Comparison with literature For q = 2 and q = 3: new proofs of CLT by Rosen et al. 13 / 21

29 Comparison with literature For q = 2 and q = 3: new proofs of CLT by Rosen et al. For q = 4, conjecture by Rosen confirmed 13 / 21

30 Generalizations Time variable t can be replaced by suitable stopping time 14 / 21

31 Generalizations Time variable t can be replaced by suitable stopping time In principle, Brownian motion should be replaceable with more general square-integrable martingale 14 / 21

32 Generalizations Time variable t can be replaced by suitable stopping time In principle, Brownian motion should be replaceable with more general square-integrable martingale analogously, one expects to be able to prove analogous results for more general functions than powers 14 / 21

33 Generalizations (cont.) Theorem (C., 2016) ( 1 h ( h L x t h ) q Y t,x dx q 2 + a q,k k=1 (Y t,x ) x R non-negative and nice. ( h L x t h ) q 2k (4 L x t) k Y t,x dx d c q ) (L x t) q Y t,x dx Z, 15 / 21

34 Generalizations (cont.) In particular, for n N 0 : ( 1 h ( h L x t h ) q (L x t) n dx q 2 + a q,k k=1 ( h L x t h ) q 2k (4L x t) k (L x t) n dx d c q ) (L x t) q+n dx Z, 16 / 21

35 Question Recursive argument should yield: ( 1 h ( h L x t h ) q dx C q with C 2p given by Marcus/Rosen (2006). ) (L x t) q/2 dx d C q (L x t) q dx Z, 17 / 21

36 Question Recursive argument should yield: ( 1 h ( h L x t h ) q dx C q with C 2p given by Marcus/Rosen (2006). ) (L x t) q/2 dx d C q Unfortunately, didn t succeed to make this rigorous using this approach. (L x t) q dx Z, 17 / 21

37 small/big blocks approach (w/ M. Podolskij) Theorem (C., Podolskij, 2017+) For f with polynomial growth, ( h L x ) [ ( t f dx P E f 2 )] L x t N dx, h where N N (0, 1), independent of (L x t) x R. 18 / 21

38 small/big blocks approach (w/ M. Podolskij) Theorem (C., Podolskij, 2017+) For f with polynomial growth, ( h L x ) [ ( t f dx P E f 2 )] L x t N dx, h where N N (0, 1), independent of (L x t) x R. Heuristic for proof as before: h L x t h 2L x t h B x t h 18 / 21

39 small/big blocks approach (w/ M. Podolskij) Theorem (C., Podolskij, 2017+) For f with polynomial growth, ( h L x ) [ ( t f dx P E f 2 )] L x t N dx, h where N N (0, 1), independent of (L x t) x R. Heuristic for proof as before: h L x t h 2L x t h B x t h small/big blocks technique to break asymptotic dependence of increments 18 / 21

40 small/big blocks approach (w/ M. Podolskij) In particular: with C 2p+1 = 0. ( h L x t h ) q dx P C q (L x t) q/2 dx, 19 / 21

41 small/big blocks approach (w/ M. Podolskij) Theorem (C., Podolskij, 2017+) For f C 1 (R) such that f, f have polynomial growth, ( 1 h ( h L x ) [ ( t f dx E f 2 ) ] ) L x t N L x t dx h where Z, N N (0, 1), independent of (L x t) x R. d U f (2 L x t ) dx Z 20 / 21

42 small/big blocks approach (w/ M. Podolskij) In particular: ( 1 h ( h L x t h ) q dx C q ) (L x t) q/2 dx d C q (L x t) q dx Z. 21 / 21