Chapter 2 Application to Markov Processes

Transcription

1 Chapter 2 Application to Markov Processes 2.1 Problem Let X t be a standard Markov process with the state space. The time interval [, is denoted by T.Leta be a fixed state and σ a the hitting time for a. We impose the following four assumptions. (A-1 P b (σ a < = 1; (A-2 E b (σ a 1 asb a; (A-3 inf b U c E b (σ a 1 > for every neighborhood U of a; (A-4 a is a discontinuous exit state. We will explain the meaning of this condition. s is called an exit time from a for the path (X t (ω if, for every ε>, and if, for some ε>, {t : X t (ω = a} (s ε, s = {t : X t (ω = a} (s, s + ε =. All exit times from a for the path (X t (ω form a countable set depending on ω. An exit time s from a for the path (X t (ω is called a continuous or discontinuous exit time according as X s (ω = a or X s (ω = a. a is called a discontinuous exit state if all exit times from a for the path (X t (ω are discontinuous a.s. with respect to P a. The Author(s 215 K. Itô, Poisson Point Processes and Their Application to Markov Processes, pringerbriefs in Probability and Mathematical tatistics, DOI 1.17/ _2 19

2 2 2 Application to Markov Processes Let Xt = X t σa. ince the hitting time σa of the path (X t (ω is the same as σ a, the conditions (A-1, (A-2 and (A-3 are equivalent to (A -1 P b (σa < = 1; (A -2 E b (σa 1, as b a; (A -3 inf b U c E b (σa 1 > for every neighborhood U of a. By the strong Markov property of (X t, the probability laws of the path (X t is determined by the probability laws of the path (Xt and the probability law of the path (X t starting at a. ymbolically we have p.l. of (X t = p.l. of (X t + p.l. of (X t starting at a. (2.1 ince the path (X t starting at a behaves outside of a in the same way as the path (Xt, the union relation in (2.1 is no disjoint union. We want to extract some information I from the probability law of the path (X t starting at a to obtain a symbolic information relation p.l. of(x t = p.l. of(xt + I (disjoint union. (2.1 In the subsequent sections we will prove that I consists of two elements: jumping-in measure k(db and stagnancy rate m. 2.2 The Poisson Point Process Attached to a Markov Process at a tate a We use the same notations as in ect. 2.1 and impose the conditions (A-1, (A-2, (A-3 and (A-4. Let A(t be a local time of (X t at a. By our assumptions (A-1 and (A-2, A(t is determined up to a multiplicative constant and we have P b (A(t < for every t = 1 and P b (A(t as t = 1. We refer the reader to Blumenthal and Getoor [1] for the definition and the properties of local times. Let U be the space of all right continuous functions: T with left limits. The sample path of (X t belongs to U a.s. for every starting point. From now on we will refer to P a for the probability law of X t (ω unless the contrary is explicitly stated. Let us define a point process X : T U by D X ={A(s : s moves over all exit times from a for the path}, X ω (t = (X θ A 1 (t for t D X,

3 2.2 The Poisson Point Process Attached to a Markov Process at a tate a 21 Fig. 2.1 X t, A t and X t where θ t is a shift operator and is a stopping operator. Note that D X consists of all values of A(t corresponding to the flat t-intervals of A(t and that X ω (t is a function : T belonging to U for ω and t fixed and (Fig. 2.1 X ω (t(s = X s σa (θ τ ω(θ τ ω for s T, τ = A 1 (t. Figure 2.2 is an intuitive picture of X t. Fig. 2.2 An intuitive picture of X t

4 22 2 Application to Markov Processes Theorem and Definition The point process X defined above is a Poisson point process: T U; it is called the Poisson point process attached to the Markov process (X t. Proof Let {B t } be a family of sub-σ -algebras of B in the definition of the Markov process (X t. ince A 1 (t+ and A 1 (t = sup n A 1 (t 1 + are both stopping n times, B A 1 (t and B A 1 (t+ are well-defined. Let B t (X be the σ -algebra generated by X d [, t. Then B t (X B A 1 (t B A 1 (t+. For t fixed, we have P(X (A 1 (t+ = a = 1. Thus, for B B t (X and M P, wehave P a (B ((θ t X M = P a (BP a (X M because of the additivity of A(t, the strong Markov property of (X t and the definition of X. Thus X has renewal property. This implies that X is differential and stationary. To prove the σ -discreteness of X we will introduce a map h : U T by h(u = inf{t : u(t = a}. If u is the path of (X t (ω, then h(u will be σ a (ω. Let U n be the set of all u such that h(u > 1 n. By (A-4 we have X = n X r U n a.s. ince N(X, [, t U n n A 1 (t < a.s., X ω r U n is discrete a.s. for each n. X is therefore σ -discrete.

5 2.3 The Jumping-In Measure and the tagnancy Rate The Jumping-In Measure and the tagnancy Rate Let us consider a map e : U by e(u = u(, u U. ince the path of X t has no discontinuities of the second kind and since A 1 (t < for t <, the distance ρ(x t, a between X t and a can be larger than ε(> a finite number of times during [, A 1 (t a.s. for t <. This implies that e X is a σ -discrete point process. By Theorem 1.6.1, we see that e X is a Poisson point process. Definition The characteristic measure k of e X is called the jumping-in measure of the Markov process X t from a. It is obvious that k = n X e 1. ince n X is concentrated on the paths starting at points in {a} by (A-4, k is concentrated on {a}. It is obvious that the total measure of k is the same as that of n X. ince X is σ -discrete, the total measure of k is σ -finite. ince A 1 (t is known to be an increasing homogeneous Lévy process (=a subordinator, it can be written as when J(t is a pure jump process. A 1 (t = m t + J(t, m >, Definition The coefficient m is called the stagnancy rate of the Markov process (X t. The following theorem shows that the characteristic measure n X is determined by the measure k and the probability law of the path of (X t. Theorem n X (V = k(dbp b (X V where X denotes the path (Xt (ω, t T. Proof Let i denote the set {b : ρ(a, b >1/i} for i = 1, 2,... Then i = {a}.letu i ={u U : u( i }=e 1 ( i. Then U i increases with i and the limit U is the space of all paths in U starting from points in {a}.wehave X = X r U = i X i, X i = X r U i by (A-4. The set A 1 (D Xi [, A 1 (t+] is included in the set of the time points s [, A 1 (t+] for which ρ(x s, X s >1/i. ince the sample path of (X t has no

6 24 2 Application to Markov Processes discontinuity points of the second kind, the latter set is finite and so is A 1 (D Xi [, A 1 (t+]. This implies D Xi [, t] is finite. X i is therefore a discrete Poisson point process. By Theorem 1.4.1, wehave n Xi (V i = λ i P a (X i (τ i V i, V i U i U i U, where λ i = n Xi (U i and τ i is the smallest element in D Xi. By the definition we have X i (τ i = X(τ i = (X θ σi, σ i = A 1 (τ i. ince n Xi V U, = n X U i and since σ i is a stopping time with respect to {B t },wehave,for n X (U i V = λ i P a ((X θ σi V U i = λ i i P a (X σi dbp b (X V U i et V = e 1 (B i, B i i i. Then V e 1 ( i = U i and so k(b i = λ i P a (X σi B i. Thus we have n X (U i V = k(dbp b (X V U i. i Letting i,wehave n X (V = k(dbp b (X V = k(dbp b (X V, {a} which completes the proof. The jumping-in measure k is not arbitrary. We have: Theorem k is concentrated on {a} and E b (σa 1k(db<. Proof h X is also a Poisson point process whose integrated process is the discontinuous part of the increasing homogeneous Lévy process A 1 (t. Therefore h X is summable and so (1 tn h X (dt<

7 2.3 The Jumping-In Measure and the tagnancy Rate 25 by virtue of Theorem ince n h X = n X h 1, this can be written (1 t k(dbp b (σa dt< by the previous theorem, namely k(dbe b (σa 1 <. Remark By this theorem and the condition (A-3, we have for every neighborhood U of a. k(u c < If k( <, then X is discrete. Then the set {t : X t (ω = a} is a sequence of disjoint intervals ordered linearly and A(t,ωis the sojourn time at a singleton {a} up to a multiplicative constant. Thus we have m > in the decomposition: A 1 (t = mt + J(t, J(t being the discontinuous part of A 1 (t. Therefore we obtain: Theorem m in general, and m > in case k( <. A(t is determined up to a multiplicative constant and m and k depend on which version of A(t we take. Let A i (t, i = 1, 2 be two versions of A(t and write the corresponding m and k as m i and k i, i = 1, 2. Then we have a constant c > such that A 2 (t = ca 1 (t. Consider the decompositions Then A 1 i (s = m i s + J i (s, i = 1, 2. A 1 2 (cs = A 1 1 (s, m 2 cs + J 2 (cs = m 1 s + J 1 (s and so m 2 = 1 c m 1.

8 26 2 Application to Markov Processes Writing # A for the number of points in A, wehave ε k 2 (B = E a [#{s : s ε, e(x s B}] = E a [#{s : s ε, X (A 1 2 (s B}] = E a [#{t : A 2 (t ε, X (t B}] = E a [#{t : ca 1 (t ε, X (t B}] = E a [# {t : A 1 (t 1 }] c ε, X (t B = 1 c εk 1(B and so k 2 = 1 c k 1. Thus we have: Theorem If A 2 (t = ca 1 (t, then m 2 = 1 c m 1 and k 2 = 1 c k 1. Therefore m and k are determined up to a common multiplicative constant. To have m and k determined uniquely, we have to take a standard version of the local time A(t. Definition A(t is called standard if ( E a e t da(t = 1, in which case E b ( e t da(t = E b (e σ a for every b. The m and k that correspond to the standard A(t are called the standard stagnancy rate and the standard jumping-in measure. Theorem The standard stagnancy rate m and the stagnancy jumping-in measure k satisfy the following conditions. (a m in general and m > in case k( <. (b k is concentrated on {a} and (i k(dbe b(σa 1 < ; (ii m + k(dbe b(1 e σ a = 1. Proof By Theorems and it is enough to prove (b-(ii. ince m and k are standard, the corresponding A(t satisfies

9 2.3 The Jumping-In Measure and the tagnancy Rate 27 ( E a e t da(t = 1. But the left side is ( E a e A 1 (t dt = E a (e A 1 (t dt = e mt t (1 e s k(dbp b(σa ds dt (see the proof of Theorem and use the Lévy Khinchin formula ( = m + (1 e s 1 k(dbp b (σa ds ( 1. = m + k(dbe b (1 e σ a This proves (ii. 2.4 The Existence and Uniqueness Theorem uppose that X t is a standard Markov process with the state space and that a is a fixed state. We assume (A-1, (A-2, (A-3 and (A-4 in ect Let Xt = X t σa, m the standard stagnancy rate and k the jumping-in measure for X t. Then we have proved: (i Xt is a standard Markov process which satisfies (A -1, (A -2, (A -3. (ii m and k satisfy (a and (b in Theorem Now we want to construct X t for Xt, m and k given. Theorem uppose that Xt, m and k satisfy (i and (ii. Then there exists a standard Markov process X t satisfying (A-1, (A-2 and (A-3 such that X t σa is equivalent to Xt and that the standard stagnancy rate and the standard jumping-in measure are respectively equal to m and k. uch X t is unique up to equivalence. Proof of existence First we will construct the Poisson point process X attached to the Markov process X t that is to be constructed. Let U be the space of all right continuous functions: T with left limits. Define a σ -finite measure n on U by n(v = k(dbp b (X V and construct a Poisson point process X : T U with n X = n by Theorem 1.3.5, or by Theorem

10 28 2 Application to Markov Processes et Ã(s = ms + α s α D X h(x(α where h(u = inf{α T : u(α = a}. Define Y (t as follows. Y (t = { X(s(t Ã(s a if Ã(s t < Ã(s if Ã(s = t = Ã(s. Now define the probability law P a of the path of X t starting at a by P a (X V = P(Y ( V and the probability law P b of the path of X t starting at a general state b by P b (X σa V 1, X θ σa V 2 = P b (X V 1P a (X V 2. It is needless to say that the definition of P a is suggested by Fig. 2.1 and that the definition of P b is suggested by the strong Markov property. First we will prove that P(Ã(s < for every s and Ã( = = 1, (2.2 so that Y (t is well-defined for every t.ifk( =, then m > and Ã(s = ms < and A( =. If k( >, then h X is a Poisson point process with n h X = nh 1 = k(dbp b (σa. ince (t 1n h X (dt = k(dbe b (σa 1 <, h X is summable and so J(s = α s α D X h(x(α < for every s < and J(s is a homogeneous Lévy process with increasing paths. ince n h X ([, = k( >, we have

11 2.4 The Existence and Uniqueness Theorem 29 P(J( = = 1. This proves (2.2. Now we will prove that the process X t defined above is a standard Markov process with (A-1, (A-2, (A-3 and (A-4. Case 1. k( <. In this case we have m >. ince n X (U = k(dbp b (X U = k(, X is discrete. et and set D X ={τ 1 <τ 1 + τ 2 <τ 1 + τ 2 + τ 3 < } ξ i = X(τ 1 + τ 2 + +τ i, i = 1, 2,... Then τ 1,τ 2,..., ξ 1,ξ 2,...are independent and P(τ i > t = e tk(, P(ξ i V = 1 k(dbp b (X ( V, V U. k( In other words the probability law of ξ i is the probability law of the path of X with the initial distribution k(db/k(. By the definition of Y (t we have Y (t = a for mτ 1 + h(ξ 1 + +mτ i 1 + h(ξ j 1 t < mτ 1 + h(ξ 1 + +mτ i 1 + h(ξ i 1 + mτ i and Y (t = ξ i (t mτ 1 h(ξ 1 mτ i 1 h(ξ i 1 mτ i for mτ 1 + h(ξ 1 + +mτ i 1 + h(ξ i 1 + mτ i t < mτ 1 + h(ξ 1 + +mτ i + h(ξ i. ince P(mτ i > t = P(τ i > t/m = e tk(/m, X (t can be described as follows. If it starts at a, it stays at a for an exponential holding time with the parameter = k(/m, then jumps into db with probability

12 3 2 Application to Markov Processes Fig. 2.3 Markov process from a point process k(db/k( and moves in the same way as Xt does until it hits a, it will repeat the same motion afterwards independently of its past history. If it starts at b = a, it performs the same motion as Xt until it hits a and then it will act as above. We can verify the strong Markov property of this motion by routine. It is easy to check the other properties of X t stated above (see Fig Case 2. k( =. Everything can be verified by routine except the fact that the sample path of Y (t belongs to U a.s. ince it is obvious that Y (t is right continuous and has left limits as far as it is in {a}, the only fact that needs proof is that the set of s such that σ ε (X(s <, σ ε (u = inf{t : ρ(a, u(t ε} forms a discrete set a.s. for every ε>. ince X(s(t = a for t h(x(s a.s., σ ε (X(s < is equivalent to a.s. It is therefore enough to prove that σ ε (X(s < h(x(s X r V ε, V ε ={u : σ ε (u <h(u} is discrete a.s., namely that n X (V ε <. et Then δ>by(a -3. δ = inf{e b (σ a 1 : ρ(b, a ε}.

13 2.4 The Existence and Uniqueness Theorem 31 Observe that h(u 1 n X (du h(u 1 n X (du U V ε (h(u σ ε (u 1 n X (du V ε = (h(u σ ε (u 1 k(dbp b (X du V = k(dbe b [(σa σ ε(x 1, σa >σ ε(x ] = k(dbe b [E X (σε (X (σa 1, σ a >σ ε(x ] δ k(dbp b (σa >σ ε(x = δ k(dbp b (X V ε = δn X (V ε and that U h(u 1 n X (du = h(u 1 k(dbp b (X du U = k(dbe b (h(x 1 = k(dbe b (σa 1. Thus we have n X (V ε <. The proof of uniqueness is easy, because the probability law of the path of (Xt and k determine n X and so the probability law of X, which, combined with m determines the probability law of the path of X t. 2.5 The Resolvent Operator and the Generator of the Markov Process Constructed in ect. 2.4 The generator of a Markov process is defined in many ways which are not always equivalent to each other. We will adopt the following definition due to E.B. Dynkin. Let X t be a Markov process with right continuous paths. The transition probability p(t, b, E is defined by p(t, b, E = P b (X t E,

14 32 2 Application to Markov Processes and the transition operator p t is defined by p t f (b = p(t, b, dc f (c = E b ( f (X t. p t carries the space B( of all bounded real Borel measurable functions into itself. It has the semi-group property: p t+s = p t p s, p = I (=identity operator. The resolvent operator (potential operator of order α R α (α > is defined by R α = e αt p t dt i.e., R α f (b = e αt p t f (b dt = E b ( e αt f (X t dt. It satisfies the resolvent equations: R α R β + (α βr α R β =. The Dynkin subspace L of B( is defined by L ={f B( : lim t p t f (b = f (b for every b}. L is a linear subspace of B(. Because of the right continuity of the path of X t we have C( L B(, C( being the space of all bounded continuous real functions on. It is easy to see that p t L L, R α L L. In view of this fact we will regard p t and R α as operators : L L, unless the contrary is stated explicitly. By virtue of the resolvent equation R = R α L is independent of α. R α : L R is 1 1 and so Rα 1 is well-defined.

15 2.5 The Resolvent Operator and the Generator 33 Definition The generator G of (X t is defined by { D(G = f L : 1 t (p t f f converges boundedly as t } to a function L and 1 G f (b = lim t t (p t f (b f (b, f D(G. Theorem D(G = R = R α L, G f = α f Rα 1 f, f D(G. Let X t be a standard Markov process and a be a fixed state. We assume that (A-1, (A-2 and (A-3 are satisfied. Let Xt = X t σa. Then Xt is also a standard Markov process with (A -1, (A -2 and (A -3. We will denote the transition operator, the resolvent operator and the generator of X t respectively by p t, R α and G and the corresponding operators for Xt are denoted by pt, R α and G. Theorem D(G D(G and G f (b = G f (b, b = a, G f (a =. Proof If f D(G, then f = R α g, g L. By Dynkin s formula we have f (b = E b ( = E b ( σ a = E b ( σ a e αt g(x t dt e αt g(x t dt + E b (e ασ a f (X σa e αt g(x t dt + E b (e ασ a f (a. et Then g (b = { g(b b = a, αr α g(a = α f (a b = a. ( Rα g (b = E b e αt g (Xt dt = E b ( σ a e αt g(x t dt + E b (e ασ a Rα g (a.

16 34 2 Application to Markov Processes ince we have R α g (a = e αt α f (a dt = f (a, f (b = R α g (b. To complete the proof of D(G D(G, we need only prove that g belongs to the Dynkin space L of X t. ince X t = a for t σ a,wehave p t g (a = g (a g (a as t. uppose b = a. Then P b (σ a > = 1 and so Therefore lim P b (σ a t =. t pt g (b g (b = E b (g (Xt g (b = E b (g(x t, t <σ a + E b (g (a, t σ a g(b = E b (g(x t E b (g(x t, t σ a + E b (g (a, t σ a g(b = E b (g(x t g(b +( g + g (a P b (t σ a, where g =sup c g(c. ince f = R α g = R α g, we have and so G f = α f g, G f = α f g G f (b = G f (b for b = a and G f (a = α f (a g (a =. Let X t be the Markov process constructed from X t, m and k in ect The resolvent and the generator for X t are denoted respectively by R α and G and the corresponding operators for X t are denoted respectively by R α and G.

17 2.5 The Resolvent Operator and the Generator 35 We will discuss the relation between (R α, G and (R α, G. Let us make three cases. Case 1. k( =. In this trivial case a is a trap for X t and (X t is equivalent to (X t, so that R α = R α and G = G. Case 2. < k( <.(m > in this case. a is an exponential holding state with the rate k(/m. Theorem If < k( <, then R α g(b = Rα g(b + E b(e ασ a Rα g(a for b = a; (2.3 mg(a + k(dbrα g(b R α g(a = ; αm + k(dbe b (1 e ασ a (2.4 G f (b = G f (b for b = a; (2.5 mg f (a = k(db( f (b f (a. (2.6 Proof Equation (2.3 is obvious by Dynkin s formula. To prove (2.4, set f (a = R α g(a and f (a = R α g(a. The Poisson point process X attached to (X t is discrete. Let σ be the first point in D X and τ be the first exit time from a for (X t.lety t be the process derived from X in ect By Dynkin s formula, we have ( f (a = E a e αt g(x t dt ( τ = E a e αt g(x t dt + E a (e ατ f (X τ ( mσ = E e αt g(a dt + E[e αmσ f (X σ (] [ 1 e αmσ ] = g(ae + E[e αmσ ]E[ f (X σ (]. α

18 36 2 Application to Markov Processes Observe E(e αmσ = = e αmt e tk( k( dt k( αm + k( and Therefore we have E[ f (X σ (] = 1 k(db f (b. k( mg(a + k(db f (b f (a =, (2.7 αm + k( which, combined with (2.3, implies mg(a + k(db f (b + k(dbe b (e ασ a f (a f (a =. αm + k( olving this for f (a, wehave mg(a + k(db f (b f (a =, αm + k(dbe b (1 e ασ a which proves (2.4. Equation (2.5 is obvious by Theorem It follows from (2.7 that m(α f (a g(a = k(db( f (b f (a, which proves (2.6. Case 3. k( =. a is an instantaneous state for (X t. Theorem Theorem holds also in case k( = : ( k(db( f (b f (a = lim k(db( f (b f (a ε ρ(b,a>ε with the following proviso. If m >,(2.4 holds for g with lim g(b = g(a (2.8 b a

19 2.5 The Resolvent Operator and the Generator 37 and (2.6 holds for f = R α g with g satisfying the same condition. Proof (2.3 and (2.5 are obvious. Let ε> and set 1,ε ={b : ρ(b, a ε}, 2,ε = 1,ε, U i,ε ={u U : u( i,ε }, i = 1, 2, X i,ε = X r U i,ε, i = 1, 2. Let Y 2,ε (t be the process derived from X 2,ε in the same way as Y t was derived from X in ect ince we fix ε for the moment, we omit ε in i,ε, U i,ε etc. Let J(t, X = s t s D X h(x s. imilarly for J(t, X i. X 1 is discrete. Let σ be the first element in D X 1. Noticing that s D X, s <σ = s D X 2, we have J(σ, X = J(σ, X 2, X σ = X 1 σ and Y t = Y 2 t for t < mσ + J(σ, X = mσ + J(σ, X 2. f (a R α g(a ( = E e αt g(y t dt ( mσ +J(σ,X = E e αt g(y t dt h(xσ + E [e αmσ α J(σ,X [ + E e αmσ α J(σ,X αh(x σ ] e αt g(x σ (t dt ] e αt g(y (t,θ σ X d (, dt ( mσ +J(σ,X 2 = E e αt g(yt 2 dt [ h(x 1 + E e αmσ α J(σ σ ],X2 e αt g(xσ 1 (t dt ] + E [e αmσ α J(σ,X2 αh(xσ 1 e αt g(y (t,θ σ X d (, dt = I 1 + I 2 + I 3.

20 38 2 Application to Markov Processes X 1 and X 2 are independent. σ and Xσ 1 are B(X1 measurable and independent of each other. X 2, σ and Xσ 1 are therefore independent of each other. Thus we have [ I 2 = E[e αmσ α J(σ,X2 h(xσ 1 ] ]E e αt g(xσ 1 (t dt = P(σ dte αmt E[ α J(t,X2 ] 1 k(db ( σ k( 1 E a b e αt g(xt. dt ince J(t, X 2 is a Lévy process increasing with jumps whose Lévy measure is equal to n h X 2(dt = k(dbp b (σa dt, 2 we have E[e α J(t,X2 ]=e t (1 e αs n h X 2 (ds = e t 2 k(dbe b(1 e ασ a. It is obvious that Therefore I 2 = P(σ dt = e k(1 t k( 1 dt. 1 k(dbe b ( σ a e αt g(x t dt. αm + k( 1 + k(dbe b (1 e ασ a 2 By the strong renewal property of X we have [ I 3 = E[e αmσ α J(σ,X2 αh(xσ 1 ]E Thus we have = E[e αmσ α J(σ,X2 ]E[e αh(x1 σ ] f (a f (a k(dbe b (e ασ a = 1. αm + k( 1 + k(dbe b (1 e ασ a 2 ] e αt g(y t dt ( k(db [E σa ] b e αt g(xt dt + E b (e ασ a f (a f (a = I (2.9 αm + k( 1 + k(dbe b (1 e ασ a 2

21 2.5 The Resolvent Operator and the Generator 39 To evaluate I 1, consider ( f 2 (a E e αt g(yt 2 dt This implies = I 1 + E = I 1 + ( E ( e αmσ α J(σ,X2 P(σ ds e αms αj(s,x2 = I 1 + P(σ ds ( E(e αms αj(s,x2 E (by the renewal property ofx 2 = I 1 + From (2.9 wehave e αt g(y (t,θ σ X 2 d (, dt e αt g(y (t,θ s X 2 d (, dt e αt g(y (t, X 2 dt P(σ dse(e αms α J(s,X2 f 2 (a = I 1 + E(e αmσ α J(σ,X2 f 2 (a. I 1 = f 2 (a[1 E(e αmσ α J(σ,X2 ] [ = f 2 k( 1 ] (a 1 αm + k( 1 + k(dbe b (1 e ασ a 2 αm + k(dbe b (1 e ασ a = f 2 (a 2. αm + k( 1 + k(dbe b (1 e ασ a 2 αm + f (a = f 2 (a αm + k( k(dbe b (1 e ασ a 2 k(dbe b (1 e ασ a (2.1 ( k(db [E σa ] b e αt g(xt dt + E b (e ασ a f (a + 1. αm + k( 1 + k(dbe b (1 e ασ a 2

22 4 2 Application to Markov Processes olving this for f we have f (a = ( f 2 (a αm + k(dbe b (1 e ασ a 2 αm + ( σa + k(dbe b 1 k(dbe b (1 e ασ a e αt g(x t dt (2.11 Let ε, then ( σa ( k(dbe b e αt g(xt σ dt a k(dbe b e αt g(xt ; dt 1. notice that ( σa k(db E b e αt g(x t dt g k(dbe b (1 e ασ a <, = sup. norm by virtue of k(dbe b(σa 1 <. It is obvious that k(dbe b (1 e ασ a as ε. 2 It follows from (2.1 and (2.3 that αm + f (a = f 2 (a αm + k( k(dbe b (1 e ασ a 2 k(dbe b (1 e ασ a k(db f (b + 1 αm + k( 1 + k(dbe b (1 e ασ a 2 so that m(α f (a α f 2 (a + f (a k(dbe b (1 e ασ a 2 = f 2 (a k(dbe b (1 e ασ a + k(db( f (b f (a. (

23 2.5 The Resolvent Operator and the Generator 41 If m =, we can derive (2.4 and (2.6 from (2.11 and (2.12, letting ε and noticing that f 2 (a f 2,ε (a g /α. If m >, we need only prove that in order to derive (2.4 and (2.6 from (2.11 and (2.12. Let η> and set lim f 2,ε (a = g(a ε α, (2.13 V 1 = V 1,η ={u U : sup ρ(u(t, a η}, t V 2 = V 2,η = U V 1, Y i = Y i,ε,η = X 2,ε r V i,η, i = 1, 2. By the argument in the last step of the existence proof of Theorem 2.4.1, wehave λ λ ε,η n Y 1(V 1 = n X 2,ε(V 1,η ( 1 inf E b(σa 1 k(dbe b (σa 1 ρ(b,a>η 2,ε, ε forη fixed. Y 1 is a discrete Poisson point process. Let τ = τ ε,η be the first element in D Y 1. Then τ is exponentially distributed with rate = λ ε,η. Using the same argument as in deriving (2.9, we obtain f 2,ε (a g(a α ( E ( mτ+j(τ,y 2 = E e αt g (Yt 2,ε dt, g (b = g(b g(a e αt g (Y (t, Y 2 dt ( + E(e αmτ α J(τ,Y 2 h(yτ 1 E e αt g (Yτ 1 (t dt ( + E(e αmτ α J(τ,Y 2 αh(yτ 1 E e αt g (Yt 2,ε dt. ince ρ(y (t, Y 2, a <ηfor < t < mτ + J(τ, Y 2,wehave f 2,ε (a g(a δ(η 1 α α + E(e αmτ g α + E(e αmτ g α

24 42 2 Application to Markov Processes where δ(η = sup{g (b, ρ(b, a <η} (η by (2.8. ince τ is exponentially distributed with rate λ ε,η,wehave E(e αmτ = λ ε,η e αmt e λε,ηt λ ε,η dt = αm + λ ε,η, ε by m >. Thus we have lim sup ε f 2,ε (a g(a δ(η 1 α α, η. This completes the proof. 2.6 Examples Example 1 Let =[, and X be a diffusion in stopped at such that the generator of X is G = d d dm dx. Let be an exit or regular (i.e., exit in Feller s new terminology boundary, i.e., 1 m(ξ,1 dξ <. Then X satisfies (A -1, (A -2 and (A -3 in ect. 2.1; notice that inf E b(σ 1 = E ε(σ 1 >. ρ(b,>ε We will investigate the condition (i in Theorem 2.3.8: k(dbe b (σ 1 <. (2.14 This is equivalent to k(dbe b (1 e σ <.

25 2.6 Examples 43 ince u(b = E b (e σ is a decreasing positive solution of d d u = u, u( = 1, dm dx 1 u (1 u (ξ = u(ξm(dξ m(ξ, 1 (ξ, u( u(b ξ b (m(ξ, 1 u (1 dξ (α(ξ β(ξ (ξ means that we have c 1, c 2 > independent of ξ such that c 1 β(ξ < α(ξ < c 2 β(ξ near ξ =. Case 1. (regular case If is a regular (i.e., exit and entrance in Feller s new terminology boundary, i.e., m(, 1 <, then E b (1 e σ = u( u(b b (b. ince E b (1 e σ 1asb, Eb (1 e σ b 1in< b <. Therefore our condition (2.14 turns out to be k(db(b 1 <. Case 2. (exit case If is an exit (i.e., exit and non-entrance in Feller s new terminology boundary, i.e., m(, 1 =, then (2.14 turns out to be [ b ] k(db m(ξ, 1 dξ 1 <. Example 2 Let =[, and X be a deterministic motion with constant speed 1. Then P b (σ = b = 1 and so (2.14 is written as k(db(b 1 <. Reference 1. Blumenthal, R., Getoor, R.: Markov Processes and Potential Theory. Academic Press, New York (1968

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