Yaglom limits can depend on the starting state Ce travail conjoint - PowerPoint PPT Presentation

Yaglom limits can depend on the starting state Ce travail conjoint avec Bob Foley 1 est d´ edi´ e ` a Franc ¸ois Baccelli. David McDonald 2 Mathematics and Statistics University of Ottawa david.r.mcdonald@gmail.com 12 January 2015 1 Partially supported by NSF Grant CMMI-0856489 2 Partially supported by NSERC Grant 4551 / 2011

A quotation Semi-infinite random walk with absorption–Gambler’s ruin Our example Periodic Yaglom limits Applying the theory ρ -Martin entrance boundary Closing words

The long run is a misleading guide . . . The long run is a misleading guide to current affairs. In the long run we are all dead. Economists set themselves too easy, too useless a task if in tempestuous seasons they can only tell us that when the storm is past the ocean is flat again. John Maynard Keynes ◮ Keynes was a Probabilist: Keynes, John Maynard (1921), Treatise on Probability, London: Macmillan & Co. ◮ Rather than insinuating that Keynes didn’t care about the long run, probabilists might interpret Keynes as advocating the study of evanescent stochastic process: P x { X n = y | X n ∈ S } .

An evanescent process–Gambler’s ruin ◮ Suppose a gambler is pitted against an infinitely wealthy casino. ◮ The gambler enters the casino with x > 0 dollars. ◮ With each play, the gambler either wins a dollar with probability b where 0 < b < 1 / 2 . . . ◮ . . . or loses a dollar with probability a where a + b = 1 . ◮ The gambler continues to play for as long as possible. ◮ In the long run the gambler is certainly broke. ◮ What can be said about her fortune after playing many times given that she still has at least one dollar?

A quasi-stationary distribution ◮ Seneta and Vere-Jones (1966) answered this question with the following probability distribution π ∗ : � y − 1 �� π ∗ ( y ) = 1 − ρ b y for y = 1 , 2 , . . . (1) a a √ ◮ where a = 1 − b and ρ = 2 ab .

Limiting conditional distributions ◮ Let X n be her fortune after n plays. ◮ Notice that her fortune alternates between being odd and even. ◮ For n large, Seneta and Vere-Jones proved that � π ∗ ( y ) for y even, x + n even, π ∗ (2 N ) P x { X n = y | X n ≥ 1 } ≈ π ∗ ( y ) for y odd, x + n odd. π ∗ (2 N − 1) ◮ The subscript x means that X 0 = x , N := { 1 , 2 , . . . } . ◮ The probability π assigns to the even and odd natural numbers is denoted by π ∗ (2 N ) and π ∗ (2 N − 1) , respectively.

Gambler’s ruin as a Markov chain ◮ The Seneta–Vere-Jones example has a state space N 0 := { 0 } ∪ N where 0 is absorbing. ◮ The transition matrix between states in N is  0 b 0 0 0 · · ·  a 0 b 0 0 · · ·   P =  .   0 a 0 b 0 · · ·    . . . ◮ P is irreducible, strictly substochastic, and periodic with period 2.

Graphic of Gambler’s ruin b b b . . . 1 2 3 a a a a 0 Figure: P restricted to N .

Facts from Seneta and Vere-Jones ◮ The z -transform of the return time to 1 is given in Seneta and Vere-Jones: √ � � 1 − 4 abz 2 1 − F 11 ( z ) = . 2 ◮ Hence the convergence parameter of P is R = 1 /ρ where √ ρ = 2 ab . ◮ Moreover F 11 ( R ) = 1 / 2 so P is R -transient. ◮ Using Stirling’s formula as n → ∞ : for y − x even � y − 1 � x − 1 �� √ xy � n �� a b � P 2 n ( x, y ) ∼ √ πn 3 / 2 2 ab . b a ◮ Denote the time until absorption by τ so P x ( τ = n ) = f ( n ) x 0 . ◮ If n − x is even then from Feller Vol. 1 x · 2 n +1 f ( n ) 1 1 2 ( n − x ) a 2 ( n + x ) . ∼ (2 π ) 1 / 2 ( n ) 3 / 2 b x 0

Define the kernel Q ◮ It will be convenient to introduce a chain with kernel Q on N 0 with absorption at δ ◮ defined for x ≥ 0 by Q ( x, y ) = P ( x + 1 , y + 1) b b b . . . 0 1 2 a a a a δ Figure: Q is P relabelled to N 0 .

Our example ◮ The kernel K of our example has state space Z . ◮ For x > 0 , K ( x, y ) = Q ( x, y ) , K ( − x, − y ) = Q ( x, y ) , ◮ K (0 , 1) = K (0 , − 1) = b/ 2 , K (0 , δ ) = a. ◮ Folding over the two spoke chain gives the chain with kernel Q . b/ 2 b/ 2 b b b b . . . -2 -1 0 1 2 a a a a a a a δ Figure: K restricted to Z .

b b b 0 1 2 a a a a δ b/ 2 b/ 2 b b b b . . . -2 -1 0 1 2 a a a a a a a δ

Yaglom limit of our example ◮ Define a family σ ξ of ρ -invariant qsd’s for K ◮ indexed by ξ ∈ [ − 1 , 1] and given by σ ξ (0) = 1 − ρ (2) a � | y | �� σ ξ ( y ) = σ ξ (0)[1 + | y | + ξ y ] b for y ∈ Z (3) 2 a ◮ For x, y ∈ 2 Z , K 2 n ( x, y ) K 2 n ( x, 2 Z ) = 1 + ρ ρ lim σ ξ ( x ) ( y ) where 1 + ρ = σ ξ ( x ) (2 Z ) . ρ n →∞ x ◮ where ξ ( x ) = 1 + | x | for x ∈ Z . ◮ Notice the limit depends on x !

Definition of Periodic Yaglom limits ◮ For periodic chains, define k = k ( x, y ) ∈ { 0 , 1 , 2 , . . . d − 1 } so that K nd + k ( x, y ) > 0 for n sufficiently large. ◮ We can partition S into d sets labeled S 0 , . . . , S d − 1 so that the starting state x ∈ S 0 and that K nd + k ( x, y ) > 0 for n sufficiently large if y ∈ S k . ◮ Theorem A of Vere-Jones implies that for any y ∈ S k , [ K nd + k ( x, y )] 1 / ( nd + k ) → ρ . ◮ We say that we have a periodic Yaglom limit if for some k ∈ { 0 , . . . , d − 1 } P x { X nd + k = y | X nd + k ∈ S } = K nd + k ( x, y ) K nd + k ( x, S ) → π k x ( y ) (4) where π k x is a probability measure on S with π k x ( S k ) = 1 .

Asymptotics of Periodic Yaglom limits Proposition ◮ If π k x is the periodic Yaglom limit for some k ∈ { 0 , 1 , . . . , d − 1 } , then there are periodic Yaglom limits for all k ∈ { 0 , 1 , . . . , d − 1 } . ◮ Moreover, there is a ρ invariant qsd π x such that π k x ( y ) = π x ( y ) /π x ( S k ) for y ∈ S k for each k ∈ { 0 , 1 , . . . , d − 1 } . ◮ We conclude K nd + k ( x, y ) K nd + k ( x, S ) → π x ( y ) π x ( S k ) for all k ∈ { 0 , 1 , . . . , d − 1 } where x ∈ S 0 by definition and y ∈ S k .

Periodic ratio limits ◮ We say that we have a periodic ratio limit if for x, y ∈ S 0 K nd ( y, S 0 ) K nd ( x, S 0 ) = λ ( x, y ) = h ( y ) lim h ( x ) . n →∞ ◮ Proposition If we have both periodic Yaglom and ratio limits on S 0 then for any k, m ∈ { 0 , 1 , . . . , d − 1 } , u ∈ S k and 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 applied to our example ◮ Let S 0 = 2 Z and let x ∈ S 0 . ◮ We check that for y ∈ 2 Z , K 2 n ( x, y ) K 2 n ( x, 2 Z ) = 1 + ρ 1 lim σ ξ ( x ) ( y ) where σ ξ ( x ) (2 Z ) = 1 + ρ. 1 n →∞ ◮ From Proposition 1 we then get for y ∈ 2 Z − 1 , K 2 n +1 ( x, y ) K 2 n ( x, 2 Z − 1) = 1 + ρ ρ lim σ ξ ( x ) ( y ) where σ ξ ( x ) (2 Z − 1) = 1 + ρ. ρ n →∞

Checking the periodic Yaglom limit I ◮ Assume x, y ≥ 1 . Similar to the classical ballot problem, there are two types of paths from x to y : those that visit 0 and those that do not. From the reflection principle, any path from x to y that visits 0 has a corresponding path from − x to y with the same probability of occurring. ◮ Thus, if { 0 } K n ( x, y ) denotes the probability of going from x to y without visiting zero, we have 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 argument, { 0 } K n ( x, y ) = P n ( x, y ) .

Checking the periodic Yaglom 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 } K n ( x, y ) + K n ( x, | y | )) . ◮ Similarly, K n ( x, − y ) = 1 2( K n ( x, | y | ) − { 0 } K n ( x, y )) .

Checking the periodic Yaglom limit III ◮ For x, y > 0 and both even, from (35) in Vere-Jones and Seneta { 0 } K 2 n ( x, y ) P 2 n ( x, y ) = � y − 1 � x − 1 �� √ � 2 n �� a xy b � ∼ √ πn 3 / 2 2 ab . b a ◮ Moreover, K 2 n ( x, | y | )) Q 2 n ( x, y ) + Q 2 n ( x, − y ) = P 2 n ( x + 1 , y + 1) + P 2 n ( x + 1 , − ( y + 1)) = � y � � x �� (4 ab ) n �� a b 1 ∼ ( x + 1) ( y + 1) n 3 / 2 . b a π

Checking the periodic Yaglom limit IV ◮ Let τ δ be the time to absorption for the chain X . so P x ( τ δ = n ) = P x +1 ( τ = n ) and ∞ � f 2 v − 1 P x ( τ δ > 2 n ) = x +1 , 0 . (5) v = n +1 P x ( τ > 2 n ) ∞ ( x + 1) · 2 2 v 1 1 � 2 (2 v − 1 − ( x +1)) a 2 (2 v − 1+( x +1)) ∼ (2 π ) 1 / 2 (2 v − 1) 3 / 2 b v = n +1 � x (4 ab ) n �� a ( x + 1) 4 a ∼ 1 − 4 ab. (2 π ) 1 / 2 b (2 n ) 3 / 2

Checking the periodic Yaglom limit V ◮ Hence, for x, y > 0 , K 2 n ( x, | y | )) + { 0 } K 2 n ( x, y ) K 2 n ( x, y ) P x ( τ > 2 n ) = 1 2 P x ( τ > 2 n ) � y � �� �� a � x ( y + 1) (4 ab ) n 1 b 1 2 ( x + 1) b a π n 3 / 2 ∼ �� a � x (4 ab ) n ( x +1) 4 a (2 π ) 1 / 2 b (2 n ) 3 / 2 1 − 4 ab � √ � 2 n � a � x/ 2 � b � y/ 2 xy 1 ab √ πn 3 / 2 2 b a + ( x +1) � x (4 ab ) n �� a 4 a (2 π ) 1 / 2 b (2 n ) 3 / 2 1 − 4 ab � y �� 1 − 4 ab (1 + | y | + ξy b ∼ ) = (1 + ρ ) σ ξ ( x ) ( y ) . a 2 a