LECTURE 1: GENERALIZED RAY KNIGHT THEOREM FOR FINITE MARKOV CHAINS

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1 LECTURE : GENERALIZED RAY KNIGHT THEOREM FOR FINITE MARKOV CHAINS We will work wih a coninuous ime reversible Markov chain X on a finie conneced sae space, wih generaor Lf(x = y q x,yf(y. (Recall ha q x,x = y x q x,y. We will be ineresed in he local ime process of X (defined below and is relaion wih cerain Gaussian fields.. Preliminaries - Io s formula We begin wih a couple of preparaory lemmas. Lemma. (Io s formula. Noaion as above. For any funcion f, f(x = f(x + where M is a maringale. Lf(X s ds + M Proof. Fix u >. Le F be he cannonical filraion. Noe ha for δ small, E[f(X u+δ F u ] = f(x u + δ y q Xu,yf(y + O(δ where he O(δ comes from he possibiliy of having wo or more jumps and is uniform. Therefore, we have E[f(X u+δ F ] = E[E[f(X u+δ F u ] F ] (The erm O(δ is uniform and deerminisic. On he oher hand, E[ u+δ Lf(X s ds F ] = E[ = E[f(X u F ] + δe[lf(x u F ] + O(δ. u Lf(X s ds F ] + δe[lf(x u F ] + O(δ. Define Z u = E[f(X u u Lf(X sds F ]. Then, from he above, Z u+δ = Z u + O(δ, i.e. dz u /du = for all u. Hence, Z u = Z, for u, which proves he claim. Lemma. (Exponenial Io s formula. Noaion as above. For any funcion f wih Lf = g and min f >, ( g(x s f(x exp f(x s ds is a maringale.

2 LECTURE : GENERALIZED RAY KNIGHT THEOREM FOR FINITE MARKOV CHAINS Proof. Wrie ( ( g(x s Z +δ = f(x +δ exp f(x s ds exp ( +δ g(x s = Z exp f(x s ds +δ g(x s f(x s ds ( g(x s +[f(x +δ f(x ] exp f(x s ds + O(δ +δ Lf(X s = Z Z f(x s ds ( Lf(X s +[f(x +δ f(x ] exp f(x s ds + O(δ. Taking condiional (on F expecaion, we ge Lf(X E[Z +δ F ] = Z δz f(x ( +δlf(x exp g(x s f(x s ds + O(δ Lf(X = Z δz f(x + δz Lf(X f(x + O(δ = Z + O(δ. From here, proceed as in he proof of Io s lemma. Exercise. Show ha he same proof works if insead of X we consider he Markov process (X, Y where Y = Y + h(x sds, replacing he generaor L by Lf(x, z = Lf(x, z + h(x f z (x, z. This remark will be imporan in wha follows.. Green funcions We now consider our Markov chains on graphs G = (V, E wih no self loops. We inerpre he q xy wih x y as conducances W x,y > on he edges, and se λ x = y x W x,y. Thus q x,x = λ x. Inroduce he Dirichle form E(f, f = W x,y (f(y f(x. x,y V Fix x V and se U = V \ x. Le τ = inf{ : X = x }. We consider he chain killed a ime τ, and inroduce he Green funcion τ g(x, y = E x X=yd =: E x l y (τ where l y ( = X s=yds is he local ime of he Markov chain a y. Lemma.. g is a symmeric, posiive definie marix. Proof. Consider he discree ime process Y n = X τn where τ n are he jump imes. This is of course a Markov chain wih jump probabiliies P x,y = W x,y /λ x. Le N = min{n : Y n = x }. Le τ y + denoe a jump ime from y (i.e., exponenial wih

3 LECTURE : GENERALIZED RAY KNIGHT THEOREM FOR FINITE MARKOV CHAINS 3 parameer λ y. Le U(n, x, y denoe he collecion of pahs (z = x,..., z n = y enirely inside U. Then, by he srong Markov propery, τ E x X=y = P x (Y n = y, n < N E y (τ y + = P x (Y n = y, n < N λ y n= = λ y n= z U(n,x,y n= W (x, z λ x W (z, z λ z W (z n, y λ zn Reversing he seps (using ha W x,y is symmeric, we obain ha his equals E y τ X =x, proving ha g(x, y = g(y, x. To see ha g is posiive definie, noe ha he compuaion above shows ha he Green funcion is relaed o he discree Green funcion G D (x, y = λ y n= P (Y n = y, n < N = λ y n= ˆP n (x, y, where ˆP is he ransiion marix of he discree chain resriced o U. Since G is conneced, ˆP is sub-sochasic and is op eigenvalue is sricly less han. Therefore, G D (x, y = λ y (I ˆP. I follows ha all eigenvalues are sricly posiive. We noe ha if ˆL is he generaor resriced o U hen G D = ( L. Indeed, le Λ be he diagonal marix wih Λ xx = λ x, noe ha L = Λ(I ˆP and herefore ( L = (I ˆP Λ and herefore ( L (x, y = (I + ˆP + ˆP + (x, y λ y = G D (x, y. Exercise. Show ha E(f, f = f, Lf where he inner produc is in l (V. 3. The Gaussian free field associaed on G Since g is symmeric and posiive definie, we can associae o i a cenered Gaussian field {φ x } x V by seing φ x = and Eφ x φ y = g(x, y for x, y U. Since g = ( L, we see from Exercise ha he Gaussian densiy is proporional o exp( E(φ, φ (wih φ x =. 4. The generalized second Ray-Knigh heorem Se θ u = inf{ : l x ( > u}, he inverse local ime a x. Theorem 4.. Le {X } and {φ x } be independen, as above. ( l x (θ u + ( φ d x = (φ x + u x V The heorem has a long hisory, going back o Ray and o Knigh (in he Brownian moion case, passing hrough some resuls of Dynkin ha we will describe laer, hopefully, and of Eisenbaum. This version is due o Eisenbaum, Kaspi, Marcus, Rosen and Shi. Proof. The proof we presen is due o Sabo and Tarres. Throughou, we wrie E x for expecaion wih respec o he Markov chain sared a x and E x for expecaion wih respec o boh he Markov chain and he Gaussian field. x V.

4 4 LECTURE : GENERALIZED RAY KNIGHT THEOREM FOR FINITE MARKOV CHAINS Fix h > a posiive bounded measurable funcion (on R V. noaion dϕ = δ ϕx dϕ x, C = (π U / (de g /. We wrie E x ( h((l x (θ u + φ x x V ( x U Inroduce he = CE x h(l x (θ u + R ϕ x x V e E(ϕ,ϕ dϕ V = CE x h(l x (θ u + ϕ x x V e E(σϕ,σϕ dϕ where Σ V = {σ {, } V : σ x = }. Define (wih obvious vecor noaion Φ = l(θ u + ϕ and se D u = {Φ R V + : Φ x = u, Φ x/ l x (θ u, x U.} Then Φ : R V + {ϕ x = } D u has inverse given by ϕ x = Φ x l x (θ u. This is a diagonal ransformaion wih Jacobian of he inverse given by J = x U Φ x ϕ x. Making now he change of variables, we ge ha he righ side of ( equals ( CE x h(( Φ x R U x V e E(σϕ,σϕ Φ x Φ Du dφ. ϕ + x x U (In he las display, we abused noaion by disregarding he value of Φ x, since i equals u under he inegraion. Fix now Φ R V + {Φ x = u}, and define he sopping ime T = inf{ : l x ( = Φ x, some x V }. Se, for T, Φ x ( = Φ x l x (. This is a means o inroduce dynamics (in reverse ime!. Noe ha and ha in ha case, ϕ = Φ(θ u. Now we exploi he dynamics: se M σφ Φ D u X T = x T = θ u, = e E(σΦ(,σΦ( x x σ x Φ x ( x X σ x Φ x (. Noe ha he produc in he righ side, a ime T, could be wrien as σ x Φ x / σ y ϕ y, x x y X which resembles he Jacobian of he change of variables. The hear of he proof is he following lemma, whose proof we pospone. Lemma 4.. The process (M σφ T is a uniformly inegrable maringale. We will also need he following easy lemma. Se N Φ = M σφ. Lemma 4.3. For all x x, N Φ T {X T =x} =.

5 LECTURE : GENERALIZED RAY KNIGHT THEOREM FOR FINITE MARKOV CHAINS 5 Proof. Le σ x equal σ excep ha he spin in he x locaion is flipped. Noe ha Φ x (T = if X T = x, by definiion. On he laer even we herefore have ha σ x Φ(T = σφ(t. On he oher hand, (σ x ϕ x = σ x Φ x ( = σφ x (. Hence, M σx Φ T = MT σφ on he even X T = x. Since = = x Φ N Φ T MT σφ we obain ha NT Φ {XT =x} = (MT σφ MT σ, + M σx Φ T {XT =x} =. We can now complee he proof. Noe ha he expression in ( equals C h(( Φ x x V E x [NT Φ XT =x ]dφ = C h(( Φ x x V E x [NT Φ ]dφ = C h(( Φ x x V N Φ dφ, where he firs equaliy is due o Lemma 4.3 and he second o Lemma 4.. The las expression hen equals C h(( Φ x R U x V e E(σΦ,σΦ dφ = C h(( Φ x + σ Σ R x V e E(Φ,Φ dφ. U V Since Φ x = u, we see ha Φ under he las measure is disribued like φ + u. This complees he proof. I remains o prove Lemma 4.. Proof of Lemma 4.. We inroduce he process X = (X, l wih generaor L = (L + l x. Se f(x, l =. y x σ y Φ y l y Recall ha (3 M σφ = e E(σΦ(,σΦ( x x σ x Φ x ( x X σ x Φ x (. Noe ha, since Φ x ( = Φ x l x (, we have dφ x (/d = {X=x}/Φ x ( and herefore d d E(σΦ(, σφ( = d W x,y (σ x Φ x ( σ y Φ y ( d = x,y (σφ( X L(σΦ((X.

6 6 LECTURE : GENERALIZED RAY KNIGHT THEOREM FOR FINITE MARKOV CHAINS On he oher hand, Lf(X, l f(x, l = = y X W X,y y X W X,y(f(y, l f(x, l f(x, l ( f(y, l f(x, l = y X W X,y Combining he las wo displays we ge ha d d E(σΦ(, σφ( = Lf(X, l f(x, l = Lf(X, l f(x, l, ( (σφ(y = L(σΦ((X (σφ( X (σφ((x. where he second equaliy is due o he fac ha f(x, l does no depend on l x. Subsiuing in (3, we conclude ha M σφ M σφ = f(x, l f(x, e Lf f (Xs,lsds. Applying Lemma., we conclude ha M T σφ n is a maringale, where T n = inf{ : l x ( Φ x n for some x V }. Clearly, T n T. We will prove ha M Tn is uniformly (in n and inegrable, which hen implies ha M T is a uniformly inegrable maringale. Toward he laer, a direc compuaion similar o he above gives ha Lf (x, l f = σ y Φ y l y σ x Φ x l x W x,y σ x Φ x l x. Therefore, y x Lf f (X, l C(W, Φ Φ X (. Thus, Lf f (X s, l s ds C(W, Φ = x V C(W, Φ Φ Xs (s ds lx( Φ x l dl = Φ x ( C(W, Φ dl x V Φ x l = x V C(W, Φ( Φ x Φ x( C(W, Φ, V. Anoher observaion is ha since V < and x V l x( =, we have ha T C(Φ, V. In paricular, combined wih he las display we conclude ha, for T, (4 M σφ C(Φ, V, W f(x, l. So i only remains o show ha f(x T, l T is uniformly inegrable. Toward his end, run he chain X unil ime S a which l x (S Φ x/ for all x. Clearly, S < since V is conneced. Le T x = inf{ : l x ( = Φ x/} and le s x denoe he ime accumulaed in he las visi o x before. Noe ha if x X hen

7 LECTURE : GENERALIZED RAY KNIGHT THEOREM FOR FINITE MARKOV CHAINS 7 Φ x l x ( T s x. On he oher hand, he variables s x are independen, exponenially disribued wih parameer λ x. We hen ge ha f(x, l =: Q. sx sx x x X Combined wih (4 we conclude ha M σφ C(Φ, V, W Q. Noe ha Q does no depend on or T. On he oher hand, by he independence of he s x s, and he square-roo ogeher wih he posiive densiy of s x a, we obain ha compleing he proof. EQ ( C(W C(W r V e <, r

Math 334 Fall 2011 Homework 11 Solutions

Math 334 Fall 2011 Homework 11 Solutions Dec. 2, 2 Mah 334 Fall 2 Homework Soluions Basic Problem. Transform he following iniial value problem ino an iniial value problem for a sysem: u + p()u + q() u g(), u() u, u () v. () Soluion. Le v u. Then