Semiclassical analysis and a new result for Poisson - Lévy excursion measures

Transcription

1 E l e c t r o n i c J o u r n a l o f P r o b a b i l i t y Vol 3 8, Paper no 45, pages Journal URL Semiclassical analysis and a new result for Poisson - Lévy excursion measures Ian M Davies Department of Mathematics, Swansea University, Singleton Park, Swansea, SA 8PP, Wales, UK IMDavies@Swanseaacuk Abstract The Poisson-Lévy excursion measure for the diffusion process with small noise satisfying the Itô equation dx = bx tdt + dbt is studied and the asymptotic behaviour in is investigated The leading order term is obtained exactly and it is shown that at an equilibrium point there are only two possible forms for this term - Lévy or Hawkes Truman We also compute the next to leading order Key words: excursion measures; asymptotic expansions AMS Subject Classification: Primary 6H, 4A6 Submitted to EJP on March, 7, final version accepted June 5, 8 83

2 Introduction Consider a one-dimensional diffusion process defined by dxt = bxtdt + dbt, X = a, where b is a Lipschitz-continuous function and Bt is a standard Brownian motion The generator, G, of the above diffusion is G = d dx + bx d dx, and the putative invariant density is x ρ x = exp budu If ρ L R, dx the boundary {, } is inaccessible We assume this in what follows The transition density p t x, y = PXt dy X = x/dy, satisfies p t x, y t = p t x, y byp t x, y, = G yp t x, y, limp t x, y = δ x y, t G y being the L adjoint of G y, and δ being the Dirac delta function The density of the diffusion ρ t y = ρ xp t x, ydx therefore satisfies Evidently, so ρ is the invariant density ρ t t = G yρ t y G y ρ y =, ρ y t =, Crucial in what follows is the operator identity for any well-behaved f Gf = ρ / Hρ / f, or G = ρ / Hρ / and G = ρ / Hρ /, 84

3 where H is the one dimensional Schrödinger operator with potential V = b + b, This follows because Hρ / d H = dx + V x, ie ρ / is the ground state of H For convenience, we will assume V C R, V bounded below together with V, V polynomially bounded with derivatives Excursion Theory The map s Xs is continuous and so {s > : Xs a} is an open subset of R Therefore, {s > : Xs a} can be decomposed into a countable union of open intervals upward downward excursion intervals Define and the local time at a L ± t = Leb{s [, t] : Xs a}, L a t = lim h h Leb{s [, t] : Xs a h/, a + h/} L a t has inverse γ a t, the time required to wait until L a equals t It can be seen that γ a t is a stopping time with Xγ a t = a Moreover, as is intuitively obvious, Jumps in γ a t = Excursions of X from a up to L a equals t Example Lévy [954] Lévy proved that for b, for each λ >, E a exp λγ a t = exp with Poisson-Lévy excursion measure ν a [s, = { t Equating powers of λ in the above, we conclude that is Poisson with } e λs dν a s, / s /, s > π s, t= Number of excursions of duration exceeding s up to L a equals t P s, t = N = exp tν a [s, tν a [s, N /N!, for N =,,,, and so the expected number of excursions of duration exceeding s per unit local time at a is ν a [s,, the Poisson-Lévy excursion measure 85

4 Example Hawkes and Truman [99] For the Ornstein-Uhlenbeck process bx = kx, where k is a positive constant, the Hamiltonian is just H = d dx + k x k and ρ x = C exp kx This leads to { E exp λγ a t = exp t } e λs dν s, λ >, with ν [s, = k/ π / e ks / V + symm = upward downward We discuss generalisations of the above to upward and downward excursions Note that excursions can only be affected by values of bx for x a Therefore it is natural to define the symmetrised potential { V x, x > a, V a x, x < a with V symm being defined in a similar manner In an analogous manner we also define d H ± = dx + V symmx ± We now have the result due to Truman and Williams [99] Proposition Modulo the above assumptions E a exp λl ± γ a t = exp { t } e λs dν a ± s with Remarks ν a ± [t, = y a dy ρ/ y ρ / a exp th ± x, y x=a L ± γ a t are independent with L + γ a t + L γ a t = γ a t Jumps in L ± γ a t = upward downward excursions from a up to La equals t 3 ν a ± [s, is the expected number of local time at a upward downward excursions of duration exceeding s per unit 86

5 Proof Outline The proof uses the result of Lévy [954] E a exp λγ a t = exp t/ p λ a, a, where p λ x, y = e λs p s x, yds and p s, is the transition density We can deduce that p λ a, a = λ ρ x ρ a Ee λτxa dx, where τ x a = inf{s > : Xs = a X = x} Here the point is that for any point a intermediate to x and y p t x, y = Pτ x a dup t u a, y Since the right hand side is a convolutional product, taking Laplace transforms and letting y a gives Ee λτxa = p λ x, a/ p λ a, a Now multiply both sides by ρ x and integrate with respect to x using the fact that ρ is the invariant density to get the desired result for p λ Some elementary computation then leads to the result in Proposition 3 The Poisson-Lévy Excursion Measure for Small Noise upward We will now consider the downward excursions from the equilibrium point for the onedimensional time-homogeneous diffusion process with small noise, X t, where dx t = bx tdt + dbt Introducing the small noise term into the Truman-Williams Law seen in the previous section, we get: Proposition The expected number of per unit local time at is given by upward downward excursions from of duration exceeding s, ν ± [s, = ± y ρ y ρ exp sh± x, ydy, x= where ρ is the invariant density and H ± is the symmetrized Hamiltonian for V = b + b One should note the form of V, in particular the presence of as a multiplier of b This rather specific dependence originates from the Shrödinger operator mentioned earlier Consequently, we are unable to resort to the usual methods for resolving such a dependence We now give a result due to Davies and Truman Davies and Truman [98] 87

6 Proposition 3 Let X min be the minimising path for the classical action Az = t ż sds + t b zsds with z = x, zt = y [ Set AX min ] = Ax, y, t Then for the self-adjoint quantum mechanical Hamiltonian H = +V, where V = b + b C R and is convex where V = V + V C 4, bounded below with V β, then for each finite time t for t π/ β exp th x, y { Ax, y, t Ax, y, t = π exp + K + O } K is a rather complicated expression with many terms involving sums and products of b and its derivatives, V and its derivatives and the Feynman-Green function Gτ, σ of the Sturm- Liouville differential operator d V X dσ min τ with zero boundary conditions ie G, τ = Gt, τ =, and discontinuity of derivative across τ = σ of For a proof of this result see Davies and Truman [98] Henceforth, for simplicity we assume that b x is an even function of x so that ν + = ν Theorem Using the notation and assumptions of Proposition 3, the leading term of the Poisson-Lévy excursion measure, for excursions away from the position of stable equilibrium, where b =, and b is given by ν + [t, π { A with the action Ax, y, t = x= dy exp { y A exp x= A,, t b t b X min s ds }, ż sds + V zs ds, where V = b } A,, t Proof As usual, the classical path X min t = Xx, y, t satisfies, correct to first order in Ẍ min V X min + V X min For V = b assumed to be convex with V =, V =, and V >, for example V x = x, bx = x, and V = b, then to leading order Ẍmin = V X min The contribution to the action Ax, y, t = ż ds+ V zs ds from V is to leading order V X min sds = = 88 b X min sds b X min s ds,

7 introducing for convenience Therefore, from Proposition 3, exp th Ax, y, t Ax, y, t x, y π exp from V is and so, the contribution to term exp exp Ax,y,t b X min s ds, the Zero Point Energy term Therefore, we have using Proposition the leading order term in the Poisson-Lévy excursion measure, for upward excursions from stable equilibrium point given by ν + [t, π x= y dy exp budu [ t exp b X min s ds exp Ax, y, t A, ] Hence, for small, the leading order term is y ν + [t, π dy exp budu A, y, t [ A A t exp b X min s ] ds Comparing this to the Laplace Integral x= + O 3 I = b a e φx θx dx, where the main contribution comes from the asymptotic behaviour at points x i [a, b] with φ x i =, we can see that the main contribution to the integral in equation 3 comes from those yt, satisfying A, y, t by = If we expand φy = budu A, y, t in a Taylor series about yt, x = we get φy = φ + y yt, xφ + y yt, x φ + = φ + y yt, x φ + 89

8 Hence, ν + [s, π dy exp { A,, s + [ y ys, x budu A, y, s { t exp b X min s A ds x= ] y=ys,x= A x= } }, giving, ν + [s, π { exp and so the result follows { A,, s dy exp b X min s A ds [ A, y, s x= A ] b y x= } } y y=, 4 Poisson-Lévy Excursion Measure leading order behaviour We have seen in the previous section that in order to calculate the Poisson-Lévy excursion measure ν + [t, for a general process Xt = X[x, y, t], we require expressions for the following derivatives of the action Ax, y, t Ax, y, t where p = A x=, A Xt A where = the Van Vleck identity, p x= A where by = A x, y, t x= These expressions are evaluated in the following propositions :- Proposition 4 For p = b =, b, A, y, t b y sinh b, as y t Proof Observe that p x y, t = Ax,y,t = initial momentum at x needed to reach y in time t, satisfies du t = x p y, t + b u b x 9

9 Therefore, p y, t satisfies t = du 4 p y, t + b u Changing integration variable u = y v, t = p y,t y + b yv y since to first order V = b and V = by assumption, Therefore, Letting v = p b t = p y p y sinhw we get p, as y, as y p, t + v b, t = b sinh b p, 4 giving, for b, p, t = b sinh b t 43 Note that for b =, we get p, t = ±/t and so equation 4 has the correct limiting behaviour Proposition 5 For b, A, y, t sinh b t, as y b Proof Using the fact that p y = A/ and equation 4 we quickly get p y = p y p y + by Again, changing the variable of integration u = yv, we get du p y + bu 3 p y = p y y p y y by + y p y y 3 + byv y 9

10 Now, following the previous argument, as y, Letting v = p b p y p, t p, t + b sinhw in the above equation gives p + b v 3 p y p + b b p = sinh b t b, sinh b p cosh w dw using p = b sinh b t Proposition 6 For b, we have b A,, t = b + coth b t A is the momentum at y given that y is reached from x in time t Proof From A = p py = y, t + b y, b y A = b y p y + by + by p y p y b y b p b + p b + b p = b as y, b sinh b t + b sinh b t b + sinh b t = b b cosh b t sinh b t 9

11 Therefore { } b y A, y, t b + coth b t, y= and result follows We now come to our main result for excursions from an equilibrium point Theorem For the diffusion X with small noise satisfying dx t = b X t dt + dbt, denote the Poisson-Lévy measure for excursions from by ν ± Assuming b is continuous, b having right and left derivatives at, with b ± and b =, then if V ± = b satisfies V ± β ±, for t π, β ± ν ± [s, k± e k ±t π with k ± = b ± When b ± = the limiting behaviour is correct and yields ν ± [t, π Proof Using Theorem, and the expressions obtained in Propositions 4, 5 and 6, for the derivatives of the action as y, we get as the leading order term for excursions from the stable equilibrium position, dropping ± again for convenience, ν + [s, π s b e dy y exp t sinh b t b b sinh b t y b + coth b t = π e t b b sinh b t cosh b t + sinh b t = π b e t b et b e t b These results correspond to the Poisson-Lévy excursion measures for the examples seen earlier 93

12 5 Poisson-Lévy Excursion Measure higher order behaviour In calculating higher order terms in the Poisson-Lévy excursion measure ν + [s,, we obtain the surprising result that the next order term is identically zero We now write the leading term as ν + Once again we must emphasise the particular dependence of V on and how this requires us to follow a rather complicated route in determining the higher order dependencies on This arises due to our study originating from stochastic mechanics where the Schrödinger equation and operator hold sway Theorem 3 For the diffusion process with small noise, assuming bx for all x, the Poisson-Lévy excursion measure is given by ν + = ν + i e the second order term is identically zero + O 3 Proof - First part For the derivation of the next order term of the Poisson- Levy excursion measure ν + [s, about x =, we must include the second order term in the expression for kernel exp x, y given in Proposition 3 Hence, from Propositions and 3 th ν + [s, π exp ρ y Ax, y, t ρ [exp x= b ] A X min s ds [ + K] dy, y< where K, recall, is a very complicated expression involving the Feynman-Green function 94

13 Therefore, up to order, we have assuming A ν + [s, = π dy exp budu { [ A A, y, t exp x= [ + A A x= x= b X min sds A x= + A A, y, t exp exp + O x= We now use the result Olver [974] Proposition 7 exp = fy exp fy Γ g f gy dy exp exp x= [g y + g y + A, y, t exp b A X min s ds t exp b X min s A, y, t exp b X min s ds A x= ] ds b X min s ds x= K ]} 5! g y + ]dy 5 + Γ 3 g f g + Γ g f 3 f f Proof For a proof of this standard result on asymptotic approximations see Olver [974] Comparing equation 5 with our expression for ν + [s, to second order in equation 5, we get and g y = π { Ax, y, t t exp b Xs ds g y = π A, y, t t exp b Xs ds { t b Xs ds + Ax, y, t Ax, y, t } Ax, y, t, x= K } Ax, y, t x= 95

14 Since we know for x = A = p y, so A = p y >, and A = p y x= x= x= giving A = p y >, we can write the expressions for g y and g y as { p y g y = π t exp and g y = π {, A = p y, } b Xs ds p y x= p y t exp b Xs ds b Xs ds + p y } p y Kp y If we now expand each term in the expression for g y in a Taylor series we get { p g y = π + yp + t exp b Xs ds + y } t b Xs ds + p + yp + y= Therefore, we can now see that in order to obtain the first order expressions for g y and g y, the following terms need to be evaluated :, y= x= p y p y b Xs ds b Xs ds p y b Xs ds Let us be very thankful that an evaluation of K is not needed in this rather complicated computation Each of these terms is evaluated in the following Propositions Recall that we have already seen in equation 43 p = p y = b sinh b t 96

15 Proposition 8 For p = b =, as y, p y = b cosh b t sinh 3 54 b t Proof In order to calculate p y we return to the identity, t = du p y + b u Differentiating the equation above wrt y, and then using the change of variable u = yv, gives dropping some inessential modulus signs for ease of presentation p y = p y p y + b y = p y y p y by + y y p p + b du p y + b u 3 p y y p + b v 3 + byv yv v 3 as y, 55 using the same argument as seen in Proposition 4 Expanding the rhs of equation 55 in a Taylor Series, using for simplicity the notation p = p and b = b, gives p + y p + p + y p + + b + y b + = p + y p + + b + yv 3 b + v + y p p + b b p + y p + p + b p + b v 3 p + y p + p + b + y p + b v 3 3 y + p + b + y p p + b b v 3 p + b v + p p + b b p + b p p + b b v 3 p + b v 5 = zero order term + fy + higher order termsy

16 The zero order term is p p + b p + b v 3 Now the integral term in the equation above can be written as b 3 p 3 + v b Using the change of variable v = p b sinhw in the equation above gives b 3 sinh b p = b p dw sinh b p p p b cosh w b + p b sinh w Therefore, the zero order term is because of equation 43 p p + b dw 3 cosh w = p 3 cosh b t p 3 cosh b t = p The coefficient of y is given by f = p p + b 3 p p + b p + b v 3 + p p p + b b p + b p p + b b v 3 p + b v 5 p + b v 3 Therefore, we have that p = p + yf + = p + yp f + = p yp f 56 Now, by letting p = b a giving a = sinh b t, we can write p f = p + p 3 p p + b b p 3 + a p + a { 3 p 3 p + a p p p a a 3 } + b b p 5 3a + 3a, 3a 4 + a 3 98

17 which simplifies to p f = p + p ab + a 3 p a 3 + a ab 3a 4 + 3a 3a 4 + a Hence, from equation 56 p = p + p ab + a p p + a + 3ab + a 3a 4 + 3a 3a 4 + a, giving = p ab + a + b + a a a a 3 + a b Now since a = b p and p = b sinh b t, and again using the obvious notation s = sinh b t and c = cosh b t, we can write the equation above as = p + b s c + c s 3 + 3s s 3 c Hence, we get the result = p + b s4 s 3 = p + b s 3 = p + b s 3 p + 3s + c c c c4 c + c c 4 + c 3 c c + c3 c + 3c c = b cosh b t sinh 3 b t Proposition 9 As y, p b cosh b t sinh 3 b t = p Proof We begin with t = Differentiating both sides wrt x gives, = px, y du x x du p x, y + b u b x p p bx bx p x, y + b u b x 3 99

18 Therefore, as x = p y + p y p x= du p y + b u 3 Hence, p = x= p y p y + b u 57 3/ du If we consider the quotient term on the rhs of equation 57, with a change of variable u = yv and again letting y, we get rhs = p p + b v 3/ Expanding the denominator of equation 57 in a Taylor series [p =, p ], using the same notation as in the previous Proposition p + y p + p + y p + + b + yv b + v 3 = p + y p p + p + b v + yp p + v 3 b b + 3 = p + y p p + { 3 y p p + v 3 b b } p + b v 3 p + b v Therefore, p = zero order term + fy +, x= where f is the coefficient of the y term as shown below p x= = p + y p p p + b v 3 p + b v 3 3 p p + b b v 3 p + p + b v 5 Inverting this equation gives { p = p y x= p + b v 3 p p p +b v 3 p 3 p p +b v 3 p p +b b v 3 p +b v 5 } 3

19 Now since px, y Comparing y-terms yields = px, x= + y px, x= + x= Letting a = b p, px, = x= p p p +b v 3 p 3 p p +b v 3 p p +b b v 3 p +b v 5 p = x= p p = 3 p p +b b v 3 +a v 3 p 3 +a v 5 p +a v 3 p p + a b b + a p a + Now, substituting for p and p gives the result b b p a + a Remark A by product of the above is Proposition As y, with X min satisfying p y p cosh b t b X min s ds t b Ẍ min s = bx min sb X min s and X min = x, X min t = y Proof For u = X min s, du = Ẋminsds, and considering since, b X min s ds = s b X min s t t + sb X min sẋminsds = t b y + = t b y + tu = x x x sub udu 58 du u du p x, y + b u b x x b u p x, y + b v b x Therefore, the integral term on the rhs of equation 58 as y, and so the result follows recall y > x > 3

20 Proposition b X min s ds b p p x= = b b x= du u p + b v 3 cosh b t sinh b, as y 59 t Proof Using equations 57 and 58, a calculation along the lines of the proof of Theorem 4, using integration by parts, yields y b X min s ds = du p y p u x= x= p y + b v 3 b u b p p du u y=x= p + b v 3 as y and result follows Proposition b Xs ds b p du = b b Proof The proof of this is similar to that of Proposition u p + b v 3 cosh b t sinh b, as y 5 t Proof - Second part Therefore, by substituting equations 43, 54, 57, 59, 5 and Proposition 9 into the expressions obtained for g y and g y, and using Proposition 7, we can complete the proof of Theorem 3 g y π p + p y + exp b X t + y b cosh b t b sinh b + t = π p 3 t b exp y + p y p b b = π p 3 exp t b p + yp + y p + + p p y + Similarly, g y π p t b exp b cosh b t b sinh b + b t b cosh b t sinh b y b p cosh b t t 4 b p sinh b t [y + y p p b cosh b t b sinh b t 3, as y y cosh b t sinh b t + y +

21 Therefore, and g = π p 3 t b exp, g = π p 3 t b exp p p b b cosh b t sinh b t Now, in our case, the expression for fy in Proposition 7 is so Also, since pt = p y + b y, so f y = A, y, t fy = A, y, t f = A,, t f = bu du, budu = budu = p t by = p + b b =, since p =, the momentum required to go from x = to y = p y + b y by, Now, f y = p, y, t b y, so differentiating pt = p y + b y twice with respect to y, and using the fact that b = and pt as y, we get { pt As y, + pt pt } x= = p y + p yp y + b y + byb y 5 p t since p = b =, and p t p t p + b,, as y Hence, p t = p + b x= = b sinh b t + b = b + sinh b t sinh b t = b coth b t Therefore, f y b coth b t b = b coth b t +, 33

22 since we are dealing with bx < Similarly, f y = p, y, t b y Differentiating equation 5 again wrt y gives, { 3 pt 3 pt + 3 pt } pt = 3p yp y + p yp y x= + 3b yb y + byb y And again, since p = b =, and p t as y, we get { p t } p t = p p + b b, y= giving, after a little calculation Therefore, as y, f y = b { } p t = p p + b b b coth b t y= p p + b b b coth b t b cosh b t cosh b tsinh b t 3 sinh b t cosh b t Finally, we conclude the proof of Theorem 3 by substituting into equation 53 to get y ν + [t, = dy exp budu A, y, t g y + g y + dy π p 3 t b exp f { π π + f 3 p 3 t b 4 exp 3 f 6 π 4 f p 3 t b } exp f π + f 3 π p exp t b b cosh b t b sinh b + b t b p p b b cosh b t sinh b t cosh b t sinh b t 34

23 Substituting for p and p eventually gives, using an obvious shorthand notation ν + [t, π p 3 t b exp f + 3 π p 3 t b { [b π c exp f p s 3 b c 4 b s b c s 4 cs 3 ] } s 4 b cs c + s b c + s b c p b + b c s p b s = π p 3 t b t b exp [ π f b c c b s s + + c s c + s + c cs 3 s ] c s = π p 3 t b exp f + p t b exp b b f A tedious calculation shows that the terms within the f + 3 π p exp c s [ ] c s 3 [ ] cancel leaving the result, ν + [t, = π p 3 t b exp f + O 3, b = e b t + O 3 π Acknowledgements The author wishes to thank Aubrey Truman for his crucial input into an earlier joint collaboration concentrating on the leading order behaviour of the Poisson - Lévy excursion measure Davies and Truman [998] I am, as always, indebted to David Williams for his help and never-ending enthusiasm for probability References P Lévy, Processus Stochastiques et Mouvement Brownien, Gauthier Villars, Paris, 954 MR

24 J Hawkes and A Truman, Statistics of Local Time and Excursions for the Ornstein-Uhlenbeck Process, Lecture Notes of the London Mathematical Society 67, 9 99 MR6648 A Truman and D Williams, Excursions and Itô Calculus in Nelson s Stochastic Mechanics, Recent Developments in Quantum Mechanics, Kluwer Academic Publishers, Netherlands, 99 I M Davies and A Truman, Laplace asymptotic expansions of conditional Wiener Integrals and generalized Mehler kernel formulas, J Math Phys 3, MR68 I M Davies and A Truman, Semiclassical analysis and a new result in in excursion theory, in Probability Theory and Mathematical Statistics, Proceedings of the Seventh Vilnius Conference 998, edited by B Grigelionis et al, Utrecht: VSP, Vilnius: TEV, 999, 7 76 F W J Olver, Asymptotics and Special Functions, Academic Press, New York and London, 974 MR