Semi-Markov approach to continuous time random walk limit processes

Transcription

1 Semi-Markov approach to continuous time random walk limit processes Joint Mathematics Meetings San Antonio, TX January 13, 2015 Mark M. Meerschaert Michigan State University East Lansing, Michigan, USA Peter Straka University of New South Wales Sydney, Australia Partially supported by NSF grants DMS , DMS , and NIH grant R01-EB

2 Abstract Continuous time random walks (CTRWs) are versatile models for anomalous diffusion processes that have found widespread application in the quantitative sciences. Their scaling limits are typically non-markovian, and the computation of their finitedimensional distributions is an important open problem. This paper develops a general semi-markov theory for CTRW limit processes in d-dimensional Euclidean space with infinitely many particle jumps (renewals) in finite time intervals. The particle jumps and waiting times can be coupled and vary with space and time. By augmenting the state space to include the scaling limits of renewal times, a CTRW limit process can be embedded in a Markov process. Explicit analytic expressions for the transition kernels of these Markov processes are then derived, which allow the computation of all finite dimensional distributions for CTRWlimits. Two examples illustrate the proposed method.

3 Continuous time random walks R d J 2 J 3 J 1 0 W 1 W 2 W 3 W 4 W 5 J 4 J 5 t The CTRW is a Markov process in space-time: (S n,t n ) = (J 1,W 1 )+ (J n,w n ).

4 CTRW limit process Assume that the CTRW at scale c > 0 converges (S c [cu],tc [cu] ) = (A 0,D 0 )+ [cu] k=1 (J c k,wc k ) (A u,d u ) in D([0, ),R d+1 ) as c. If N c t = max{k 0 : Tc k t} then X c t = Sc N c t is the location of a randomly selected particle at time t 0. Assume D u strictly increasing and unbounded. Then X c t X t := (A Et ) + in D([0, ),R d ) as c, where E t = inf{u > 0 : D u > t} is the first passage time of D u.

5 Two examples Example 1: Take P[W c n > t] = c 1 t β /Γ(1 β) iid, J c n = c 1. Then (A u,d u ) = (u,d u ), D u is the β-stable subordinator with E[e sd u] = e usβ, and X t = A Et = E t, the inverse subordinator. Example 2: Take P[W c n > t] = c 1 t β /Γ(1 β) iid, J c n = Wc n. Then (A u,d u ) = (D u,d u ), and X t = (D Et ) + is the height of D u just before it jumps over level t > 0. Note P(X t < t) = 1.

6 One sample path Sample path of D u (a pure jump process) showing E t and X t. Both E t and X t are non-markov processes. D(u) t X(t) E(t) u

7 Inverse processes Graph of E t is just the graph of D u with the axes flipped. E(t) u D(u-) t D(u)

8 Markov embedding Let V t denote the spent waiting time for E t. Extend an idea from renewal theory: Any (X t,v t ) is an F Et Markov process. V(t) E(t) t

9 Assumptions The semigroup T u f(x,t) = E x,t [f(a u,d u )] on C 0 (R d+1 ) has generator in jump-diffusion form, as in Applebaum Eq. (6.42): Af(x,t) = d i= b i (x,t) xi f(x,t)+γ(x,t) t f(x,t) 1 i,j d a ij (x,t) 2 x i x j f(x,t) + f(x+y,t+w) f(x,t) d i=1 h i (y,w) xi f(x,t) K(x,t;dy,dw) where h i (x,t) = x i 1{(x,t) [ 1,1] d+1 }, (a ij ) is non-negative definite, jump kernel K(x,t;dy,dw) on (dy,dw) R d [0, ), γ(x,t) 0, and [1 ( y 2 + w ) ] K(x,t;dy,dw) < (x,t) R d+1.

10 Main Result: Finite dimensional distributions Time-invariant case: Transition probabilities of (X t,v t ) are P t (x 0,0;dx,dv) = γ(x,t)u x 0,t 0 (x,t)dxδ 0 (dv) +K(x 0 ;R d [v, ))U x 0,t 0 (dx,t dv)1{0 v t}, P t (x 0,v 0 ;dx,dv) = δ x0 (dx)δ v0 +t(dv)k v0 (x 0 ;R d [v 0 +t, )) + y R d w [v 0,v 0 +t) P v0 +t w(x 0 +y,0;dx,dv)k v0 (x 0 ;dy,dw), where the 0-potential (mean occupation measure) [ ] f(x,v)u x 0,t 0 (dx,dv) = E x 0,t 0 f(a u,d u )du 0 has density u x 0,t 0 (x,v), and the conditional jump intensity K t (x 0 ;dx,dv) = K(x 0;dx,dv1{v > t}) K ( x 0 ;R d [t, ) ), Proof: Sample path analysis, compensation formula.

11 Example 1: FDD of E t Space drift b = 1, time drift γ = 0, diffusion a = 0. D u has PDF g(t,u), Lévy measure φ[y, ) = y β /Γ(1 β). 0-potential density u x 0,t 0 (x,t) = g(t t 0,x x 0 )1{t > t 0,x > x 0 } Jump intensity K(x,t;dy,dw) = δ 0 (dy)βw β 1 dw/γ(1 β) Conditional jump intensity K v (x,t;dy,dw) = δ 0(dy)βw β 1 dw/γ(1 β) v β /Γ(1 β) = δ 0 (dy)β(v/w) β dw/w Transition probability (for t 0 = 0 and v 0 = 0) P t (x 0,0;dx,dv) = v β Γ(1 β) g(t v,x x 0)dxdv

12 Example 1: Case v 0 = 0 (point of increase of E t ) Transition probability P t (x 0,0;dx,dv) = v β Γ(1 β) g(t v,x x 0)dxdv D(u) t v=v(t) t-v x 0 x=e(t) u

13 Inverse stable subordinator density v β Since E 0 = 0, E t has density h(x,t) = 0 Γ(1 β) g(t v,x)dv. Here β = 0.6 and t = 1. Can show h(0+,t) = t β /Γ(1 β). t h(x,t) x

14 Example 1: Case v 0 > 0 (resting point of E t ) Depends on v 0 = 0 case. We compute ( v0 +t ) β 1{v 0 > 0} P t (x 0,v 0 ;dx,dv) = δ x0 (dx)δ v0 +t(dv) v 0 + ( ) β v v0 +t v v 0 s=v 0 g((t v) (w v 0 ),x x 0 ) βw β 1 Γ(1 β) dwdxdv. The first term represents the chance that D x0 > v 0 + t given D x0 > v 0, in which case x = x 0 and v = v 0 +t. Then E t continues to rest throughout the interval of length t.

15 Example 1: Case v 0 > 0 (resting point of E t ) 2nd term ( ) v0 β v0 +t v v s=v 0 g((t v) (w v 0 ),x x 0 ) βw β 1 Γ(1 β) dwdxdv D(u) t v=v(t) (t-v)-(w-v 0 ) w v 0 x=e(t) u x 0

16 Example 1: Finite dimensional distributions Two time points: Put together cases v 0 = 0 and v 0 > 0 to get P 0,0 [E t1 dx,v t1 dv,e t2 dy,v t2 dw] = P t1 (0,0;dx,dv)P t2 t 1 (x,v;dy,dw) = v β Γ(1 β) g(t 1 v,x)dxdv1{x > 0,0 < v < t 1 } [ ( ) v +t2 t β δ x (dy)δ v+t2 t 1 (dw) 1 v + ( ) v β v+t2 t 1 w w s=v g((t 2 t 1 w) (s v),y x) βs β 1 Γ(1 β) dsdydw Integrate out v,w to get joint PDF of E t1,e t2. ]

17 Example 2: FDD of X t Recall X t = (D Et ) +, height of D u before jump over level t. Space drift b = 0, time drift γ = 0, diffusion a = 0. (D u ) has PDF g(t,u), Lévy measure φ[y, ) = y β /Γ(1 β). Jump Intensity is K(x,t;dy,dw) = δ w (dy)βw β 1 dw/γ(1 β) 0-potential density U x 0,0 (dx,dt) = δ x0 +t(dx)t β 1 dt/γ(β) Conditional jump intensity K v (x,t;dy,dw) = δ w (dy)β(v/w) β dw/w. Transition probability (for t 0 = 0 and v 0 = 0) is v β (t v) β 1 P t (x 0,0;dx,dv) = δ x0 +t v(dx) dv1{0 < v < t} Γ(1 β) Γ(β)

18 Example 2: Case v 0 = 0 (point of increase of E t ) v β (t v) β 1 P t (x 0,0;dx,dv) = δ x0 +t v(dx) dv1{0 < v < t} Γ(1 β) Γ(β) D(u) t v=v(t) x=x(t) t-v x 0 E(0) E(t) u

19 Example 2: Probability density of X t (given X 0 = 0) Probability density f(x,t) = (t x) β Γ(1 β) x β 1 Γ(β) 1{0 < x < t} D(u) t x=x(t) t-x x E(0) E(t) u

20 Example 2: Case v 0 > 0 (resting point of E t ) Depends on v 0 = 0 case. We compute P t (x 0,v 0 ;dx,dv) = δ x0 (dx)δ v0 +t(dv) + ( v0 +t v w=v 0 v 0 ( ) β v0 +t v 0 ) β δ x0+v0+t v(dx) (v 0 +t w v) β 1 1{0 < v < v 0 +t w} βw β 1 Γ(1 β) dwdv Γ(β) The first term represents the chance that D x0 > v 0 + t given D x0 > v 0, in which case x = x 0 and v = v 0 +t (as in Ex. 1). Then X t remains constant throughout the interval of length t.

21 δ x0 +v 0 +t v(dx) Example 2: Case v 0 > 0 (2nd term) ( ) v0 β v0 +t v w=v 0 ((t v) (w v 0 )) β 1 Γ(β) βw β 1 Γ(1 β) dwdv D(u) t x=x(t) v=v(t) (t-v)-(w-v 0 ) w v 0 E(t) u x 0

22 Summary CTRW limit process is not Markov Embed in a Markov process Can include coupled space-time jumps Can be space-time inhomogeneous All FDD can be computed Solves an important problem in physics

23 References 1. D.D. Applebaum, Lévy Processes and Stochastic Calculus. 2nd Ed. Cambridge Studies in Advanced Mathematics 116, Cambridge University Press, M.M. Meerschaert and P. Straka, Semi-Markov approach to continuous time random walk limit processes. The Annals of Probability 42 (2014), No. 4, pp M.M. Meerschaert and P. Straka, Inverse stable subordinators. Mathematical Modeling of Natural Phenomena 8 (2013), No. 2, pp

24 Example 1: Computing the 0-potential Recall that A u = u and D u has PDF g(t,u) given D 0 = 0. Make a change of variables y = u+x 0, w = t+t 0 to see that [ ] [ ] E x 0,t 0 f(a u,d u )du = E 0,0 f(a u +x 0,D u +t 0 )du 0 0 = = = 0 0 f(u+x 0,t+t 0 )g(t,u)dtdu y=x 0 y=x 0 w=t 0 f(y,w)g(w t 0,y x 0 )dwdy w=t 0 f(y,w)u x 0,t 0 (y,w)dwdy so that u x 0,t 0 (y,w) = g(w t 0,y x 0 )1{w > t 0,y > x 0 }.

25 Example 2: Computing the 0-potential Recall that A u = D u has PDF g(t,u) given D 0 = 0. WLOG t 0 = 0. Then [ ] [ ] E x 0,0 f(a u,d u )du = E 0,0 f(d u +x 0,D u )du 0 0 so that = = = 0 x R 0 x R 0 f(t+x 0,t)g(t,u)dtdu 0 f(x,t)g(t,u)duδ x 0 +t(dx)dt t=0 f(x,t)ux 0,0 (dx,dt) U x0,0 (dx,dt) = δ x0 +t(dx) g(t,u)dudt = δ x u=0 0 +t(dx) tβ 1 dt Γ(β) since 0 e st g(t,u)du = 0 0 e usβ du = s β = L [ t β 1 /Γ(β) ]