Yglom limits cn depend on the strting stte Ce trvil conjoint vec Bo Foley 1 est dédié à Frnçois Bccelli. Dvid McDonld 2 Mthemtics nd Sttistics University of Ottw dvid.r.mcdonld@gmil.com 12 Jnury 2015 1 Prtilly supported y NSF Grnt CMMI-0856489 2 Prtilly supported y NSERC Grnt 4551 / 2011
A quottion Semi-infinite rndom wlk with sorption Gmler s ruin Our exmple Periodic Yglom limits Applying the theory ρ-mrtin entrnce oundry Closing words
The long run is misleding guide... The long run is misleding guide to current ffirs. In the long run we re ll ded. Economists set themselves too esy, too useless tsk if in tempestuous sesons they cn only tell us tht when the storm is pst the ocen is flt gin. John Mynrd Keynes Keynes ws Proilist: Keynes, John Mynrd (1921), Tretise on Proility, London: Mcmilln & Co. Rther thn insinuting tht Keynes didn t cre out the long run, proilists might interpret Keynes s dvocting the study of evnescent stochstic process: P x {X n = y X n S}.
An evnescent process Gmler s ruin Suppose gmler is pitted ginst n infinitely welthy csino. The gmler enters the csino with x > 0 dollrs. With ech ply, the gmler either wins dollr with proility where 0 < < 1/2...... or loses dollr with proility where + = 1. The gmler continues to ply for s long s possile. In the long run the gmler is certinly roke. Wht cn e sid out her fortune fter plying mny times given tht she still hs t lest one dollr?
A qusi-sttionry distriution Senet nd Vere-Jones (1966) nswered this question with the following proility distriution π : π (y) = 1 ρ y where = 1 nd ρ = 2. ( ) y 1 for y = 1, 2,... (1)
Limiting conditionl distriutions Let X n e her fortune fter n plys. Notice tht her fortune lterntes etween eing odd nd even. For n lrge, Senet nd Vere-Jones proved tht P x {X n = y X n 1} { π (y) π (2N) π (y) π (2N 1) for y even, x + n even, for y odd, x + n odd. The suscript x mens tht X 0 = x, N := {1, 2,...}. The proility π ssigns to the even nd odd nturl numers is denoted y π (2N) nd π (2N 1), respectively.
Gmler s ruin s Mrkov chin The Senet Vere-Jones exmple hs stte spce N 0 := {0} N where 0 is soring. The trnsition mtrix etween sttes in N is 0 0 0 0 0 0 0 P = 0 0 0.. P is irreducile, strictly sustochstic, nd periodic with period 2.
Grphic of Gmler s ruin 1 0 2 3... Figure: P restricted to N.
Fcts from Senet nd Vere-Jones The z-trnsform of the return time to 1 is given in Senet nd Vere-Jones: ( 1 ) 1 4z F 11 (z) = 2. 2 Hence the convergence prmeter of P is R = 1/ρ where ρ = 2. Moreover F 11 (R) = 1/2 so P is R-trnsient. Using Stirling s formul s n : for y x even P 2n (x, y) xy ( 2 ) ( ) x 1 ( n πn 3/2 ) y 1. Denote the time until sorption y τ so P x (τ = n) = f (n) x0. If n x is even then from Feller Vol. 1 f (n) x0 x 2 n+1 (2π) 1/2 (n) 3/2 1 2 (n x) 1 2 (n+x).
Define the kernel Q It will e convenient to introduce chin with kernel Q on N 0 with sorption t δ defined for x 0 y Q(x, y) = P (x + 1, y + 1) 0 δ 1 2... Figure: Q is P relelled to N 0.
Our exmple The kernel K of our exmple hs stte spce Z. For x > 0, K(x, y) = Q(x, y), K( x, y) = Q(x, y), K(0, 1) = K(0, 1) = /2, K(0, δ) =. Folding over the two spoke chin gives the chin with kernel Q.... -2 /2 /2-1 0 1 2 δ Figure: K restricted to Z.
0 1 2 δ... -2 /2 /2-1 0 1 2 δ
Yglom limit of our exmple Define fmily σ ξ of ρ-invrint qsd s for K indexed y ξ [ 1, 1] nd given y σ ξ (0) = 1 ρ For x, y 2Z, lim n [1 + y + ξ y] σ ξ (y) = σ ξ (0) 2 K 2n (x, y) K 2n (x, 2Z) = 1 + ρ ρ σ ξ(x)(y) where where ξ(x) = x for x Z. 1 + x Notice the limit depends on x! (2) ( ) y for y Z (3) ρ 1 + ρ = σ ξ(x)(2z).
Definition of Periodic Yglom limits For periodic chins, define k = k(x, y) {0, 1, 2,... d 1} so tht K nd+k (x, y) > 0 for n sufficiently lrge. We cn prtition S into d sets leled S 0,..., S d 1 so tht the strting stte x S 0 nd tht K nd+k (x, y) > 0 for n sufficiently lrge if y S k. Theorem A of Vere-Jones implies tht for ny y S k, [K nd+k (x, y)] 1/(nd+k) ρ. We sy tht we hve periodic Yglom limit if for some k {0,..., d 1} P x {X nd+k = y X nd+k S} = Knd+k (x, y) K nd+k (x, S) πk x(y) (4) where π k x is proility mesure on S with π k x(s k ) = 1.
Asymptotics of Periodic Yglom limits Proposition If π k x is the periodic Yglom limit for some k {0, 1,..., d 1}, then there re periodic Yglom limits for ll k {0, 1,..., d 1}. Moreover, there is ρ invrint qsd π x such tht π k x(y) = π x (y)/π x (S k ) for y S k for ech k {0, 1,..., d 1}. We conclude Knd+k (x, y) K nd+k (x, S) π x(y) for ll π x (S k ) k {0, 1,..., d 1} where x S 0 y definition nd y S k.
Periodic rtio limits We sy tht we hve periodic rtio limit if for x, y S 0 K nd (y, S 0 ) h(y) lim n K nd = λ(x, y) = (x, S 0 ) h(x). Proposition If we hve oth periodic Yglom nd rtio limits on S 0 then for ny k, m {0, 1,..., d 1}, u S k nd y S m, K nd+d m+k (u, y)/k nd+d m+k (u, S k ) π u (y)/π u (S m ).
Theory pplied to our exmple Let S 0 = 2Z nd let x S 0. We check tht for y 2Z, lim n K 2n (x, y) K 2n (x, 2Z) = 1 + ρ 1 σ ξ(x)(y) where σ ξ(x) (2Z) = 1 1 + ρ. From Proposition 1 we then get for y 2Z 1, lim n K 2n+1 (x, y) K 2n (x, 2Z 1) = 1 + ρ ρ σ ξ(x)(y) where σ ξ(x) (2Z 1) = ρ 1 + ρ.
Checking the periodic Yglom limit I Assume x, y 1. Similr to the clssicl llot prolem, there re two types of pths from x to y: those tht visit 0 nd those tht do not. From the reflection principle, ny pth from x to y tht visits 0 hs corresponding pth from x to y with the sme proility of occurring. Thus, if {0} K n (x, y) denotes the proility of going from x to y without visiting zero, we hve K n (x, y) = {0} K n (x, y) + K n ( x, y) = {0} K n (x, y) + K n (x, y). From the coupling rgument, {0} K n (x, y) = P n (x, y).
Checking the periodic Yglom limit II For x, y 0, Q n (x, y) =K n (x, y ) := K n (x, y) + K n (x, y). Hence, K n (x, y) = K n (x, y ) K n (x, y) = K n (x, y ) (K n (x, y) {0} K n (x, y)) = 1 2 ( {0} Kn (x, y) + K n (x, y )). Similrly, K n (x, y) = 1 2 (Kn (x, y ) {0} K n (x, y)).
Checking the periodic Yglom limit III For x, y > 0 nd oth even, from (35) in Vere-Jones nd Senet {0} K2n (x, y) = P 2n (x, y) xy ( 2 ) ( ) x 1 ( ) y 1 2n. πn 3/2 Moreover, K 2n (x, y )) = Q 2n (x, y) + Q 2n (x, y) = P 2n (x + 1, y + 1) + P 2n (x + 1, (y + 1)) ( ) x ( ) y 1 (4) n (x + 1) (y + 1) π n 3/2.
Checking the periodic Yglom limit IV Let τ δ e the time to sorption for the chin X. so P x (τ δ = n) = P x+1 (τ = n) nd P x (τ δ > 2n) = v=n+1 f 2v 1 x+1,0. (5) P x (τ > 2n) (x + 1) 2 2v (2π) 1/2 (2v 1) 3/2 1 2 (2v 1 (x+1)) 1 2 (2v 1+(x+1)) v=n+1 ( ) x (x + 1) (4) n 4 (2π) 1/2 (2n) 3/2 1 4.
Checking the periodic Yglom limit V Hence, for x, y > 0, K 2n (x, y) P x (τ > 2n) = 1 K 2n (x, y )) + {0} K 2n (x, y) 2 P x (τ > 2n) ) ( x (y + 1) + 1 2 (x + 1) ( 1 xy 2 πn 3/2 (x+1) ) x (4) n ( (2π) 1/2 (2n) ( 3/2 ) 2n ( ) x/2 ( y/2 ) (x+1) (2π) 1/2 ( 1 4 ( ) x (4) n 4 (2n) 3/2 1 4 1 + y + ξy ) 2 ( ) y 1 (4) n π n 3/2 4 1 4 ) y = (1 + ρ)σ ξ(x)(y).
Checking the periodic Yglom limit VI K 2n (x, y) P x (τ > 2n) = 1 (K 2n (x, y ) {0} K 2n (x, y) 2 P x (τ > 2n) ( ) y ( ) y 1 4 (y + 1) xy 1 4 2 x + 1 2 ( ) y = 1 4 1 + y ξy ( ) = (1 + ρ)σ 2 ξ(x)( y). Finlly, for y = 0, K 2n (x, 0) = P 2n x+1,1 so K 2n (x, 0) P x (τ > 2n) = P 2n x+1,1 P x (τ > 2n) = 1 4 = (x + 1) (x+1) (2π) 1/2 ( = (1 + ρ)σ ξ(x) (0). ( ) x 1 (4) n π n ) 3/2 x (4) n 4 1 4 (2n) 3/2
Checking the periodic Yglom limit VII Therefore strting from x even we hve periodic Yglom limit with density (1 + 2 )σ ξ ( ) on S 0 = 2Z with ξ = x/( x + 1) [0, 1]. Similrly, for x, y > 0 even, K 2n ( x, y) = K 2n (x, y) nd K 2n ( x, y) = K 2n (x, y); hence, strting from x even we get Yglom limit (1 + 2 )σ ξ ( ) on 2Z with ξ = x/( x + 1) so ξ [ 1, 0].
Checking the periodic rtio limit Agin tking S 0 = 2Z, K 2n (y, 2Z) K 2n (x, 2Z) = P y(τ > 2n) P x (τ > 2n) ( / ) y ( y + 1) ( / ) x = h 0(y) h ( x + 1) 0 (x) In fct h 0 is the unique ρ-hrmonic function for Q in the fmily of ρ-hrmonic functions for K h ξ (y) := [1 + y + ξy] ( ) y for y Z. (6)
Checking the periodic Yglom limit VIII Applying Proposition 2, strting from u odd we hve periodic Yglom limit on the evens with density (1 + 2 )σ ξ(u) ( ) on S 0 = 2Z with ξ = u/( u + 1) [0, 1]. Similrly, strting from u odd we hve periodic Yglom limit on the odds: 1 + 2 2 σ ξ(u) ( )
Cone of ρ-invrint proilities The proilities σ ξ with ξ [ 1, 1] form cone. The extreml elements re ξ = 1 nd ξ = 1 since σ ξ (y) = 1 + ξ 2 σ 1(y) + 1 ξ 2 σ 1(y). Define the potentil G(x, y) = R n K n (x, y) nd n=0 the ρ-mrtin kernel M(y, x) = G(y, x)/g(y, 0). As mesure in x, M(y, x) B re the positive excessive mesures of R K normlized to e 1 t x = 0; i.e. µ RµK if µ B. Ech point y Z is identified with the mesure M(y, ) B, which y the Riesz decomposition theorem is extreml in B.
The ρ-mrtin entrnce oundry As y +, M(y, ) M(+, ) = σ 1 ( )/σ 1 (0). We conclude + is point in the Mrtin oundry of Z. We hve therefore identified + in the Mrtin oundry with the ρ-invrint mesure σ 1 ( )/σ 1 (0), which is identified with the point +1 in the topologicl oundry of { ξ = x 1 + x : x Z }. By similr rgument we see is lso in the Mrtin oundry of Z. As y +, M(y, ) M(, ) = σ 1 ( )/σ 1 (0). Agin we hve identified in the Mrtin oundry with the ρ-invrint mesure σ 1 ( )/σ 1 (0) which is identified with the point 1 in the topologicl oundry of { ξ = x 1 + x : x Z }.
Hrry Kesten s exmple Kesten (1995) constructed n mzing exmple of su-mrkov chin possessing most every nice property including hving ρ-invrint qsd tht fils to hve Yglom limit. Kesten s exmple hs the sme stte spce nd the sme structure s ours. The only difference is tht t ny stte x there is proility r x of holding in stte x nd proilities (1 r x ) nd (1 r x ) of moving one step closer or further from zero. If α = (1 r 0 ), then our chin is exctly Kesten s chin wtched t the times his chin chnges stte. It is pretty cler Hrry could hve derived our exmple with moment s thought, ut he focused on the non-existence of Yglom limits. His exmple is orders of mgnitude more sophisticted nd complicted thn ours.