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

The Poisson-Lévy excursion measure for the diffusion process with small noise satisfying the Itô equation dXε = b(Xε(t))dt + √ ε dB(t) is studied and the asymptotic behaviour in ε 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.


Introduction
Consider a one-dimensional diffusion process defined by where b is a Lipschitz-continuous function and B(t) is a standard Brownian motion.The generator, G, of the above diffusion is and the putative invariant density is , where H is the one dimensional Schrödinger operator with potential This follows because Hρ 1/2 0 ≡ 0, i.e. ρ 1/2 0 is the ground state of H.For convenience, we will assume V ∈ C 2 (R), V bounded below together with V ′′ , V polynomially bounded with derivatives.
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 X(γ 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 1 Lévy [1954] Lévy proved that for b ≡ 0, for each λ > 0, (1 − e −λs ) dν a (s) , with Poisson-Lévy excursion measure s −1/2 , s > 0. (2.1) Equating powers of λ in the above, we conclude that ♯(s, t)= Number of excursions of duration exceeding s up to L a equals t is Poisson with for N = 0, 1, 2, . . ., 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.
Example 2 Hawkes and Truman [1991] For the Ornstein-Uhlenbeck process b(x) = −kx, where k is a positive constant, the Hamiltonian is just We discuss generalisations of the above to upward and downward excursions.Note that upward downward excursions can only be affected by values of b(x) for x ≷ a. Therefore it is natural to define the symmetrised potential symm being defined in a similar manner.In an analogous manner we also define We now have the result due to Truman and Williams [1991] Proposition 1. Modulo the above assumptions 2. Jumps in L ± (γ a (t)) = upward downward excursions from a up to L a equals t.
3. ν ± a [s, ∞) is the expected number of upward downward excursions of duration exceeding s per unit local time at a.

Proof. (Outline)
The proof uses the result of Lévy [1954] E a exp(−λγ a (t)) = exp (−t/p λ (a, a)) , where pλ (x, y) = ∞ 0 e −λs p s (x, y) ds and p s (•, •) is the transition density.We can deduce that Ee −λτx(a) dx, where τ x (a) = inf{s > 0 : Here the point is that for any point a intermediate to x and y Since the right hand side is a convolutional product, taking Laplace transforms and letting y → a gives Ee −λτx(a) = pλ (x, a)/p λ (a, a).Now multiply both sides by ρ 0 (x) and integrate with respect to x (using the fact that ρ 0 is the invariant density) to get the desired result for p−1 λ .Some elementary computation then leads to the result in Proposition 1.

The Poisson-Lévy Excursion Measure for Small Noise
We will now consider the upward downward excursions from the equilibrium point 0 for the onedimensional time-homogeneous diffusion process with small noise, X ε (t), where Introducing the small noise term into the Truman-Williams Law seen in the previous section, we get: Proposition 2. The expected number of upward downward excursions from 0 of duration exceeding s, per unit local time at 0 is given by where ρ 0 is the invariant density and H ± is the symmetrized Hamiltonian for One should note the form of V , in particular the presence of ε 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 [1982].
Proposition 3. Let X min (•) be the minimising path for the classical action and is convex (where 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 G(τ, σ) of the Sturm-Liouville differential operator d 2 dσ 2 − V ′′ (X min (τ )) with zero boundary conditions i.e.G(0, τ ) = G(t, τ ) = 0, and discontinuity of derivative across τ = σ of 1.
For a proof of this result see Davies and Truman [1982].
Henceforth, for simplicity we assume that b 2 (x) is an even function of x so that ν + 0 = ν − 0 .
Theorem 1.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 0, where b(0) = 0, and b ′ (0) ≤ 0 is given by Proof.As usual, the classical path X min (t) = X(x, y, t) satisfies, correct to first order in ε , and so, the contribution to term exp Therefore, we have using Proposition 2 the leading order term in the Poisson-Lévy excursion measure, for upward excursions from stable equilibrium point 0 given by .
Hence, for small ε, the leading order term is Comparing this to the Laplace Integral where the main contribution comes from the asymptotic behaviour at points x i ∈ [a, b] with φ ′ (x i ) = 0, we can see that the main contribution to the integral in equation 3.1 comes from those y(t, 0) satisfying b(y) = ∂A(0, y, t) ∂y .
If we expand φ(y) = y 0 b(u)du − A(0, y, t) in a Taylor series about y(t, x) = 0 we get and so the result follows.

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 + 0 [t, ∞) for a general process X(t) = X[x, y, t], we require expressions for the following derivatives of the action A(x, y, t).
Proof.Using the fact that p 0 (y) = −∂A/∂x and equation 4.1 we quickly get .
Again, changing the variable of integration u = yv, we get . Now, following the previous argument, as y → 0, b ′ (0) sinh w in the above equation gives ( ∂A ∂y is the momentum at y given that y is reached from x in time t.) Proof.From and result follows.
We now come to our main result for excursions from an equilibrium point 0.
Theorem 2. For the diffusion X ε with small noise satisfying denote the Poisson-Lévy measure for excursions from 0 by ν ± 0 .Assuming b is continuous, b having right and left derivatives at 0, with b(0 ± ) ≶ 0 and b(0 Proof.Using Theorem 1, and the expressions obtained in Propositions 4, 5 and 6, for the derivatives of the action as y → 0, we get as the leading order term for excursions from the stable equilibrium position 0, (dropping ± again for convenience), These results correspond to the Poisson-Lévy excursion measures for the examples seen earlier.
5 Poisson-Lévy Excursion Measure -higher order behaviour In calculating higher order terms in the Poisson-Lévy excursion measure ν + 0 [s, ∞), we obtain the surprising result that the next order term is identically zero.We now write the leading term as ε .
Once again we must emphasise the particular dependence of V ε on ε and how this requires us to follow a rather complicated route in determining the higher order dependencies on ε.This arises due to our study originating from stochastic mechanics where the Schrödinger equation operator hold sway.
Theorem 3.For the diffusion process with small noise, assuming b(x) ≤ 0 for all x, the Poisson-Lévy excursion measure is given by i. e. the second order term is identically zero.
Proof -First part.For the derivation of the next order term of the Poisson-Levy excursion measure ν + 0 [s, ∞) about x = 0, we must include the second order term in the expression for kernel exp − tH(ε) ε (x, y) given in Proposition 3. Hence, from Propositions 2 and 3 where K, recall, is a very complicated expression involving the Feynman-Green function.
Therefore, up to order ε, we have assuming (5.1) We now use the result Olver [1974].
Proof.For a proof of this standard result on asymptotic approximations see Olver [1974] .
Comparing equation 5.2 with our expression for ν + 0 [s, ∞) to second order in equation 5.1, we get , y, t) ∂x∂y Since we know for x = 0 = − ∂p 0 (y) ∂y , giving we can write the expressions for g 0 (y) and g 1 (y) as If we now expand each term in the expression for g 0 (y) in a Taylor series we get Therefore, we can now see that in order to obtain the first order expressions for g 0 (y) and g 1 (y), the following terms need to be evaluated : 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 4.3 Proposition 8.For p 0 (0) = b(0) = 0 , as y → 0, (5.4) Proof.In order to calculate ∂ 2 p 0 (y) we return to the identity, .
Differentiating the equation above w.r.t.y, and then using the change of variable u = yv, gives dropping some inessential modulus signs for ease of presentation as y → 0, using the same argument as seen in Proposition 4 Expanding the r.h.s. of equation 5.5 in a Taylor Series, using for simplicity the notation p = p 0 (0) and b = b(0), gives = zero order term + f.y + higher order terms(y 2 • • • ).
The zero order term is Now the integral term in the equation above can be written as .
Using the change of variable v = p ′ b ′ sinh w in the equation above gives Therefore, the zero order term is (because of equation 4.3) The coefficient of y is given by .
Therefore, we have that , which simplifies to .
Hence, from equation 5.6 |t , and again using the obvious notation we can write the equation above as we get the result Proposition 9.As y → 0, Proof.We begin with (5.7) If we consider the quotient term on the r.h.s. of equation 5.7, with a change of variable u = yv and again letting y → 0, we get Expanding the denominator of equation 5.7 in a Taylor series [p(0) = 0, p ′ (0) = 0], using the same notation as in the previous Proposition Therefore, ∂p ∂x where f is the coefficient of the y term as shown below Comparing y-terms yields . Now, substituting for p ′ and p ′′ gives the result.
Remark.A by product of the above is Proof.For u = X min (s), du = Ẋmin (s) ds, and considering .
Therefore, the integral term on the r.h.s. of equation 5.8 → 0 as y → 0, and so the result follows (recall y > x > 0).Therefore, and Now, in our case, the expression for f (y) in Proposition 7 is If ρ 0 ∈ L 1 (R, dx) the boundary {−∞, ∞} is inaccessible.We assume this in what follows.The transition densityp t (x, y) = P(X(t) ∈ dy|X(0) = x)/dy, y)p t (x, y) , = G * y p t (x, y), lim t↓0 p t (x, y) = δ x (y), G *y being the L 2 adjoint of G y , and δ being the Dirac delta function.The density of the diffusion ρ t (y) = ρ 0 (x)p t (x,