Large Deviations for One Dimensional Diffusions with a Strong Drift

We derive a large deviation principle which describes the behaviour of a diffusion process with additive noise under the influence of a strong drift. Our main result is a large deviation theorem for the distribution of the end-point of a one-dimensional diffusion with drift $\theta b$ where $b$ is a drift function and $\theta$ a real number, when $\theta$ converges to $\infty$. It transpires that the problem is governed by a rate function which consists of two parts: one contribution comes from the Freidlin-Wentzell theorem whereas a second term reflects the cost for a Brownian motion to stay near a equilibrium point of the drift over long periods of time.


Introduction
The Freidlin-Wentzell theorem and its generalisations are well-known large deviation results.This theorem provides a large deviation principle (LDP) on the path space for solutions of the SDE dX = b(X)dt + √ ε dB when ε converges to 0. The related, but different, problem of the large deviation behaviour of a diffusion process under the influence of a strong drift is less studied.In this article we derive an LDP for the behaviour of the endpoint X ϑ t of solutions of the R-valued stochastic differential equation dX ϑ s = ϑb(X ϑ s )ds + dB s for all s ∈ [0, t] X ϑ 0 = z ∈ R (1.1) when the parameter ϑ converges to infinity.
For comparison with the Freidlin-Wentzell result one can convert the case of strong drift into the case of weak noise with the help of the following scaling argument: Define Xϑ s = X ϑ s/ϑ and Bs = √ ϑB s/ϑ for all s ∈ [0, ϑt].Then the process Xϑ is a solution of the SDE Xϑ 0 = z and we have P (X ϑ t ∈ A) = P Xϑ ∈ {ω | ω ϑt ∈ A} .The rescaled problem looks more similar to the situation from the Freidlin-Wentzell theory, but now the event in question depends on the parameter ϑ.Thus the Freidlin-Wentzell theorem still does not apply easily.Therefore a more sophisticated proof will be required.
The text is structured as follows: In section 2 we state our main result and two corollaries.Since the proof of the theorem is quite long we give an overview of the proof of our theorem in section 3. The proof itself is spread over sections 4, 5 and 6.
The result presented in this text was originally derived as part of my PhD-thesis [Vos04].

Results
Recall that a family (X ϑ ) ϑ>0 of random variables with values in some topological space X satisfies the LDP with rate function I : X → [0, ∞], if it satisfies the estimates for every open set O ⊆ X and lim sup for every closed set A ⊆ X .The family (X ϑ ) ϑ>0 satisfies the weak LDP if the upper bound holds for every compact (instead of closed) set A ⊆ X .For details about the theory of large deviations we refer to [DZ98].
Our main result is the following theorem together with the corollaries 2 and 4.
Note that the condition b = −Φ ′ defines Φ only up to a constant, but the rate function J t does not depend on the choice of this constant.
In the theorem V b a (Φ) can be interpreted as the "cost" for the process of going from a to b.Using b = −Φ ′ we find for any a, b ∈ R. The term (Φ ′′ (m)) − can be interpreted as the "cost" of staying near m for a unit of time.This term only occurs, if the equilibrium point m is unstable.Given the sign of b ′ (m) the rate function from the theorem can be simplified because the drift b has only one zero.The following corollary describes the case of b ′ (m) < 0, which corresponds to attracting drift.In this case the weak LDP from the theorem can be strengthend to the full LDP.
Corollary 2 Under the conditions of theorem 1 with b ′ (m) < 0 the following claims hold.a) For every t > 0 the family (X ϑ t ) ϑ>0 satisfies the weak LDP on R with rate function J t (x) = 2 Φ(x) − Φ(m) for all x ∈ R.
(2.3) b) If b is monotonically decreasing, then the family (X ϑ t ) ϑ>0 satisfies the full LDP with rate function J t .
In the situation of corollary 2 the rate function is independent of the interval length t and of the initial point z.This makes sense, because for strong drift we would expect the process to reach the equilibrium very quickly.Because we have lim inf |x|→∞ |b(x)| > 0 the potential Φ converges to +∞ for |x| → ∞ and J t is a good rate function.In fact the rate function coincides with the rate function of the LDP for the stationary distribution as given in theorem 3 (This is an easy application of the Laplace principle, see e.g.[Vos04] for details).Proof.(of corollary 2) a) Since we assume that m is the only zero of the drift b, for b ′ (m) < 0 the point m is the minimum of Φ.In this case we have V m z (Φ) = Φ(z)−Φ(m), V x m (Φ) = Φ(x)−Φ(m) and Φ ′′ (m) > 0, so the rate function simplifies to the expression given in formula (2.3).
b) To strengthen the weak LDP to the full LDP we have to check exponential tightness, i.e. we have to show that for every c > 0 there is an a > 0 with lim sup ϑ→∞ 1 ϑ log P |X ϑ t − m| > a < −c (for reference see lemma 1.2.18 from [DZ98]).We use a comparison argument to obtain this estimate.
Using the assumption lim inf |x|→∞ |b(x)| > 0 we find that exp(−2ϑΦ) is integrable and SDE (2.1) has a stationary distribution with density proportional to exp(−2ϑΦ).Let X ϑ be a solution of (2.1) with start in z and Y ϑ be a stationary solution, both with respect to the same Brownian motion.Then we get the deterministic differential equation for the difference between the processes.First assume X ϑ 0 − Y ϑ 0 ≥ 0. Because for X ϑ t − Y ϑ t = 0 the right hand side vanishes, the process X ϑ t − Y ϑ t can never change its sign and stays positive.Since b is decreasing we have b(X ϑ t ) − b(Y ϑ t ) ≤ 0 and we can conclude For the case X ϑ 0 − Y ϑ 0 ≤ 0 we can interchange the roles of X and Y to obtain the estimate Combining these two cases gives Using Now let c > 0. Then using theorem 3 we can find an a > 0 with and using the above estimate we get Since this is the required exponential tightness condition, the proof is complete.
The case of repelling drift, i.e. of b ′ (m) > 0, is described in the following corollary.
Corollary 4 Under the conditions of theorem 1 with b ′ (m) > 0, for every t > 0 the family (X ϑ t ) ϑ>0 satisfies the weak LDP on R with constant rate function (2.4) Proof.(of corollary 4) In the case b ′ (m) > 0 the point m is the maximum of Φ and because of for all x ∈ R.
The corollary shows that in the case of repelling drift the rate function does not depend on x.In particular it is not a good rate function.Here it is impossible to strengthen the weak LDP to the full LDP because we have

Overall Structure of the Proof
The remaining part of this text contains the proof of theorem 1.Since the proof is quite long, we use this section to give an overview of the proof.All the technical details are contained in sections 4, 5, and 6.Let X ϑ be a solution of the SDE (1.1).From the Girsanov formula we know the density of the distribution of X ϑ t w.r.t. the Wiener measure W: assuming X ϑ 0 = 0 and b = −∇Φ we get where For large values of ϑ the ϑ 2 G term dominates over the ϑF term and we show that the only paths which contribute for the large deviations behaviour of X ϑ t are those, which correspond to very small values of G.These paths run quickly to the equilibrium point m of the drift b, stay close to this point for most of the time, and shortly before time t move quickly into the set A. Assuming for the moment A = B(a, δ) with a small δ > 0, we get and thus (3.2) Lemma 26 in section 6 resolves the technical details which are hidden in the ≈-signs here and also gives the required upper and lower limits for (3.2).
To evaluate the integral on the right hand side of (3.2) we use the following result about upper and lower limits in Tauberian theorems of exponential type.The theorem is proved in [Vos04].It is a generalisation of de Bruijn's theorem (see theorem 4.12.9 in [BGT87]).
Theorem 5 Let X ≥ 0 be a random variable and A an event with P (A) > 0. Define the upper and lower limits Then −r 2 /4 = s and for the lower limits we have the sharp estimates −r ¯2 ≤ s ¯≤ −r ¯2/4.Using theorem 5 we can reduce the original problem to the calculation of exponential rates like In section 4 we examine the situation that during a short time interval the process runs from 0 to m or from m to a respectively while still keeping b 2 (ω s ) ds small.This will be used for the initial and the final section of the path.As indicated in section 1 we can rescale the problem in these domains and apply the known results for weak noise.The problem here is to identify the infimum of the rate function.
In section 5 we examine the situation that b 2 (ω s ) ds is small over a long interval of time.This will be used to study the middle section of the path.We will use theorem 5 again to deduce the probability for this case from the known Laplace transform of t 0 B 2 s ds.Finally, in section 6, we fit these two results together to complete the proof of theorem 1.This part of the proof is modelled after the proof of proposition 6 which we give below.We want to use X 1 , X 2 , X 3 = b 2 (B s ) ds where the integral is taken over the initial, middle, and final section of the path respectively.Since these random variables are not independent, we cannot directly apply proposition 6 but have to use an enhanced version of the proof.This is provided in lemma 27.
We give the full prove of proposition 6 here, because we will need the proposition itself in the proof of lemma 23, and also because we hope that reading the proof of proposition 6 might make it easier to follow the proof of lemma 27 below.
Proposition 6 Let X 1 , . . ., X n be independent, positive random variables with Proof.Let δ > 0. Since the simplex for all ε > 0. This gives and for the individual terms in the sum we can use the relation where equality holds if and only if there is a λ ∈ R with λα k = c k for k = 1, . . ., n.Thus we get for every α ∈ D δ n .Using lemma 1.2.15 of [DZ98] we can conclude lim sup ε↓0 From (3.4) we know that we should choose α k proportional to b k in order to get the optimal lower bound.This leads to the estimate which completes the proof.

Reaching the Final Point
The results of this section help to estimate the probability that the path travels quickly between the equilibrium point of the drift and the final resp.initial point.Here Schilder's theorem (see theorem 5.2.1 in [DZ98]) can be applied and we will reduce the evaluation of the rate function to a variational problem.
The main result of this section is the following proposition which describes the large deviation behaviour of the event 1 2 when ε ↓ 0, where the final point B tε stays in a fixed, compact set.Evaluating the rate for fixed t > 0 is difficult, but it transpires that there is an explicit representation for the limit of the rate as t tends to infinity.The modulus of the integrals is taken to properly handle the cases m < z and a < m.The proof of proposition 7 is based on the following two lemmas.Lemma 8 evaluates the infimum of the rate function from Schilder's theorem.Since the proof of lemma 8 is quite long, we defer the proof until the end of the section.We will write C 0 ([0, t], R) = ω ∈ C([0, t], R) ω 0 = 0} as an abbreviation.
Lemma 8 Let v : R → [0, ∞) be a positive C 2 -function with lim inf |x|→∞ v(x) > 0 and m ∈ R with v(x) = 0 if and only if x = m and v ′′ (m) > 0. For a, z ∈ R and β ≥ 0 define Consider the rate function uniformly over all a ∈ K 2 and z ∈ K 1 .
Lemma 9 Let M a,z,β t be as in lemma 8. Then for every pair Proof.By definition of the sets M a,z,β t we have because K 2 is compact and v and the integral are continuous.In both cases we can find an ε > 0, such that the ball B(ω, ε) also lies in With these preparations in place we can now give the proof for proposition 7.
Proof.(of proposition 7) We want to apply Schilder's theorem [DZ98, theorem 5.2.1] and to evaluate the rate function using lemma 8. Let K 1 , K 2 ⊆ R be compact.Define the process B by setting Br = (B rε − z)/ √ ε for every r > 0. Then B is a Brownian motion with start in 0 and we get M a,z,β t and thus Since from lemma 9 we know that the set z∈K1 a∈K2 β≤1 M a,z,β t is closed in the path space C 0 [0, t], • ∞ , we can apply Schilder's theorem to get First assume m ∈ K 1 ∩ K 2 .Define the path ω by ω s = 0 for all s ∈ [0, t].Then clearly we have ω ∈ M m,m,0 t for every t and since we find I t (ω) = 0 we have for all t ≥ 0. On the other hand we have J(m, m) = 0.
Otherwise the evaluation of the infimum is done in lemma 8. Using v(x) = b 2 (x) we get v ′′ (m) = 2(b ′ (m)) 2 > 0 and for every η > 0 we can find a t 0 > 0, such that for every η > 0. Together with the relation (4.2) this proves the upper bound.
For the lower bound we follow the same procedure.Without loss of generality we can assume that O is bounded.Here we get M a,z,β t where the set • ∞ .So we can use the lower bound from Schilder's theorem and lemma 8 to complete the proof.
Corollary 10 Under the assumptions of proposition 7 we have Because O is open and b and the integral are continuous we can find an E > 0, such that for every η < E the ball and using Schilder's theorem and the relation Now we can evaluate the infimum on the right hand side as we did in proposition 7. We get for every η < E. Taking the limit δ ↓ 0 completes the proof.
The only thing which remains to be done in this section is to give a proof for lemma 8. Before we do so we need some preparations.For the remaining part of this section we assume throughout that v is non-negative and two times continuously differentiable and that a, z ∈ R are fixed.
Notation: For x, y ∈ R we will write [x, y] for the closed interval between x and y; in the case x < y this is to be read as [y, x] instead.
As a first step towards the proof of lemma 8 we get rid of the parameter β.
Then we have ω0 = 0, ωt/β = ω t , and Thus ω → ω is a one-to-one mapping from M a,z,β t onto M a,z,1 t/β .Because of the set M a,z,0 t is empty and we find for all t > t 0 , z ∈ K 1 , and a ∈ K 2 .Then for every t > t 0 sup B and every β > 0 we have for all t > t 0 sup B, z ∈ K 1 , and a ∈ K 2 .Because η was arbitrary, this completes the proof.
Because of I t (ω + z) = I t (ω) we can shift every path from M a,z,1 t by z and get For the moment assume that there is a path ω with I t (ω) = inf I t (ω) ω ∈ M a,z,1 t .Later we will show that such an ω in fact does exist.In order to evaluate the rate function I t for this path ω, we solve the Euler-Lagrange equations (see section 12 of [GF63]) for extremal values of I t under the constraint and with the boundary conditions Because of v ∈ C 2 (R) we can use theorem 1 from section 12.1 of [GF63] to find that for every extremal point ω of I, under the given constraints, there is a constant λ, such that ω solves the equations ωs = λv ′ (ω s ) for all s ∈ (0, t], and ω 0 = z (4.3a) Existence of solutions: the autonomous second order equation (4.3a) describes the motion of a classical particle on the real line in the potential −λv.The differential equation can be reduced to an autonomous first order equation in the plane with the usual trick: defining x(s) = (ω s , ωs ) and F (x 1 , x 2 ) = x 2 , λv ′ (x 1 ) the equation becomes See e.g.section 5.3 of [BR89] for details.Because v ′ and thus F is locally Lipschitz continuous, for every pair ω 0 = z, ω0 = v 0 of initial conditions and every bounded region we find a unique solution of the ODE at least up to the boundary of that region (see theorem 8 in section 6.9 of [BR89]).
There are two degrees of freedom in (4.3a) because we can choose ω0 and λ.In the following we will show, that the two additional conditions (4.3b) and (4.3c) guarantee the existence of a unique solution to the system (4.3).
For λ = 0 the only solution of (4.3a) and (4.3c) is given by ω s = z + (a − z)s/t for 0 ≤ s ≤ t and consequently in this case we have In the following assume t > 1/c.Then we know from (4.3b) that every solution of (4.3) has λ = 0.
The interpretation as the motion of a classical particle helps us to determine the behaviour of the solutions.We can use conservation of energy: Because of This conservation law describes the speed for any point of the path: the speed of the path at point ω s is Thus the rate function I t can be expressed as a function of E and λ as follows.
where λ and E are determined by equations (4.3b) and (4.3c).Because of relation (4.4) we find that whenever ω is a solution of (4.3a) we have E ≥ −λv(ω s ) for all s ∈ [0, t] and the path can only stop and turn at points x with −λv(x) = E. Let x ∈ R be such a point and assume v ′ (x) = 0. Then η with η s = x for all s ≥ 0 is the unique solution of (4.3a) with η 0 = x and η0 = 0. Now assume that ω s = x for some s > 0. Then (ω s−r ) r∈[0,s] is also a solution of (4.3a) with start in x and initial speed 0, so we have ω s−r = η r = x for all r ∈ [0, s].This shows that a point x = z with E = −λv(x) and v ′ (x) = 0 cannot be reached by a solution ω of (4.3a).Thus whenever a non-constant path reaches an x ∈ R with E = −λv(x) then we have ωs = λv ′ (ω s ) = 0 and the path always changes direction there.Figure 1 illustrates two different kinds of solution, one where ω s moves monotonically and one where the path reaches a point b with −λv(b) = E and turns there.
Since the differential equation (4.3a) is autonomous and since a solution ω changes direction every time is reaches a point x with −λv(x) = E, the path can reach at most two distinct points of these nature.In this case the solution oscillates between these points periodically.Thus every solution of (4.3a) changes direction only a finite number of times before time t.
In order to find the path which minimises the rate function I t we need to keep track of the different possible traces of the path.For the remaining part of this section we use the following notation.The path (ω s ) 0≤s≤t is said to have trace T = (x 0 , x 1 , . . ., x n ) when ω 0 = x 0 , ω t = x n , and the path ω moves monotonically in either direction from x i−1 to x i for i = 1, . . ., n in order and changes direction only at the points x 1 , . . ., x n−1 .We use the abbreviation for the length of the trace and sometimes identify T with the set n i=1 [x i−1 , x i ] of covered points to write min T , max T , v| T , or inf x∈T v(x).For positive functions f : R → R we use the notation The absolute values are taken to make the integral positive even when x i < x i−1 .If a solution ω of (4.3a) has trace T = (x 0 , x 1 , . . ., x n ), this then implies that v(x 1 ) = • • • = v(x n−1 ) = −E/λ and each of the x 1 , . . ., x n−1 is either min T or max T .Between the points x i the path is strictly monotonic, i.e. after the start in z it oscillates zero or more times between min T and max T before it reaches a at time t.Using this notation we can formulate the following Lemma.
Lemma 12 Let λ, E ∈ R and a trace T = (x 0 , . . ., x n ) be given.Then the following two conditions are equivalent.
for all min T < x < max T , and the pair (λ, E) solves and Proof.Assume the conditions from (j).Then ω is a solution of (4.3a), there are times t 0 , t 1 , . . ., t n with ω ti = x i for i = 0, . . ., n, and between the times t i the process moves monotonically.For any integrable, positive function g : R → R substitution using (4.5) yields 2(E + λv(x)) dx. (4.8) Applying (4.8) to the function g = v gives This is equation (4.7a).Applying (4.8) to the constant function g = 1 gives which is equation (4.7b).Now assume condition (ij).For i = 1, . . ., n define the function F i by for all x between x i−1 and x i .Then F i is finite because of (4.7b), strictly monotonic (increasing if x i > x i−1 and decreasing else), and has F i (x i−1 ) = 0. Further define Equation (4.7b) gives t n = t.Because the functions F i are monotonic they have inverse functions We will prove that ω satisfies all the conditions from (j).
Because we have t the function ω is well-defined on the connection points at times t i and is continuous.This also shows ω ti = x i for i = 0, 1, . . ., n and especially ω 0 = x 0 = z and ω t = x n = a.
Because the F i are differentiable at all points x strictly between x i−1 and x i , the function ω is differentiable on the intervals (t i−1 , t i ) with derivative Because ω is continuous and the limits lim s→ti ωs exist, we see that ω is even differentiable on [0, t] with ω0 = sgn(x 1 − x 0 ) 2(E + λv(0)) and ωti = 0 for i = 1, . . ., n − 1.Using the same kind of argument again, we find first between the t i and then on the whole interval [0, t].Thus ω really solves the differential equation from (j).
Using the substitution as in the first part, we also get back (4.3b) from (4.7a).
Now we have reduced the problem of minimising I t (ω) over the solutions ω of the system (4.3) to the problem of minimising over the solutions (E, λ) of the system (4.7).
For a trace T define and furthermore define the functions f, g : Figure 2 illustrates the domain H T .Both functions are finite in the interior of the domain, but can be infinite at the boundary.The equations (4.7) are equivalent to f (E λ , λ) = √ 2t and g(E, λ) = √ 8.For paths which change direction at some point we will find solutions (E, λ) of (4.7), which lay on the boundary of H T .For paths which go straight from z to a we will find solutions (E, λ) in the interior of H T .
Lemma 13 Let t > 0 and T be a trace from z ∈ R to a ∈ R such that v| T is not constant.Then there is at most one solution (E, λ) of (4.7).
Proof.For E > − inf x∈T λv(x) we can choose an E * between − inf x∈T λv(x) and E. Then v(x)/(E * + λv(x)) 3/2 is an integrable upper bound of v(x)/(e + λv(x)) 3/2 for all e in a (E − E * )-Neighbourhood of E. So we can use the theorem about interchanging the Lebesgue-integral with derivatives to get So for every λ the map E → g(E, λ) is strictly decreasing and there can be at most one E λ with g E λ , λ = √ 8.With the help of the implicit function theorem we can calculate the derivative of E λ .Interchanging the integral with the derivative as above we get where µ is the probability measure, with density and the normalisation constant is Furthermore for (E, λ) ∈ (H T ) • we have and thus Equality would only hold for the case of constant v| T .So the map λ → f (E λ , λ) is strictly increasing and there can be at most one λ with f E λ , λ = √ 2t.This completes the proof.
Lemma 14 Let T a trace with m ∈ T and t ≥ 2|T |/ T v(x) dx.Then equation (4.7) has a solution (E, λ) with with E, λ > 0.
Proof.Define λ * = ( T v(x) dx) 2 /8 and assume 0 < λ ≤ λ * .Then we have and the dominated convergence theorem gives Thus for all 0 < λ ≤ λ * there exists an E λ ≥ 0 with g(E λ , λ) = √ 8.Because of g(0, λ * ) = √ 8 we have E λ * = 0. Fatou's lemma then gives Because v is positive and v(m) = 0, we have v ′ (m) = 0 and v ′′ (m) ≥ 0. Then by Taylor's theorem there exists a c > 0 and a closed interval for all x ∈ I. Therefore we find and thus λ → f (E λ , λ) is a continuous function with On the other hand because of g(E 0 , 0) = √ 8 we have E 0 = ( T v(x) dx) 2 /8.So for λ = 0 we get Together this shows that for all Lemma 15 There are numbers ε, c 1 , c 2 > 0 such that the following holds: For every trace T starting in K 1 , ending in K 2 , and visiting the ball B ε (m) there is a non-empty, closed interval A ⊆ R, such that A ⊆ T , |A| = ε and we have c 1 ≤ v(x) ≤ c 2 for every x ∈ A.
Proof.Because m / ∈ K 1 ∩ K 2 either K 1 or K 2 has a positive distance from m.Let ε be one third of this distance.Define Each trace starting in K 1 , ending in K 2 , and visiting the ball Let A be the crossed interval.Then clearly |A| = ε and and because of A ⊆ A ′ the estimates for v hold on A.
Lemma 16 For every η > 0 there is a t 1 > 0, such that whenever t ≥ t 1 , T is a trace from z ∈ K 1 to a ∈ K 2 with m ∈ [z, a] and (E, λ) solves (4.7), then we have Proof.This case is illustrated in the left hand image of figure 1.Because of m ∈ [z, a], any path from z to a must visit m and thus we find E > −λv(m) = 0. Thus the only possible trace in this case is T = (z, a), because the process could only turn at points x where −λv(x) = E.
and thus So we can find a t 1 > 0 with Et < η (4.9) whenever t ≥ t 1 .Choosing A, c 1 , and c 2 as in lemma 15 we get So we can choose a small c 3 > 0 and increase t 1 to achieve λ > c 3 whenever t ≥ t 1 .Because of we can find a c 4 > 0 with and thus for all t ≥ t 1 .Solving this for λ we get 2λ ≥ (1 − η/J(z, a))J(z, a) = J(z, a) − η. (4.10) Because E is positive we also find and thus 2λ ≤ J(z, a).(4.11) For the rate function I t equation (4.10) gives and equations (4.9) and (4.11) give Lemma 17 For every η > 0 there is a t 2 > 0, such that whenever t ≥ t 2 , T is a trace from z ∈ K 1 to a ∈ K 2 with m / ∈ [z, a], and (E, λ) solves (4.7), then we have Proof.This case is illustrated in the right hand image of figure 1.Because the path has to change direction we will have E < 0 in this case.Without loss of generality we can assume that m < a, z.We call a value b ∈ R admissible if it lies in the interval (m, min(a, z)) and if additionally v(x) > v(b) for all x > b holds.For admissible values b consider the trace T = (z, b, a) and define dx .
Using Taylor approximation as in lemma 14, one sees that for b → m the numerator converges to +∞ and by dominated convergence the denominator converges to (0,m,a) v(x) dx.So h is a continuous function with h z,a (b) → ∞ for b → m.Let ε, c 1 , and c 2 and A be as in lemma 15.We would like to find a b ∈ B ε (m) with h z,a (b) = t, so we need an upper bound on inf b∈(m,m+ε) which is uniform in a and z.We find Using Taylor's theorem again we get (4.14) The right hand side of (4.14) is independent of a and z.So we can take the infimum over all b ∈ (m, m + ε) and use (4.13) to get the uniform upper bound on (4.12).Call this bound t 2 .Now let t > t 2 .Then for every z ∈ K 1 and a ∈ K 2 we can find a b ∈ (m, m + ε) with h z,a (b) = t.Further define λ > 0 by Then for the trace T = (z, b, a) these values E and λ solve For t → ∞ we have b → m uniformly in a and z, and again E → 0 (this time from below).This gives With all these preparations in place we are now ready to calculate the asymptotic lower bound from lemma 8.
Proof.(of lemma 8) Because of lemma 11 we can restrict ourselves to the case β = 1, i.e. we have to prove lim Assume first the case m ∈ [z, a].From lemma 16 we get a t 0 > 0, such that for every t > t 0 there exists a solution (E, λ) of (4.7) for the trace T = (z, a) with I t (E, λ) − J(a, z) ≤ η.This t 0 only depends on K 1 and K 2 , but not on z and a.Now assume the case m / ∈ [z, a].From lemma 17 we again get a t 0 > 0, such that for every t > t 0 there exists a solution (E, λ) of (4.7) for a trace T = (z, x 1 , a) with I t (E, λ) − J(a, z) ≤ η and t 0 only depends on K 1 and K 2 , but not on z and a.
In either case we can use lemma 12 to conclude, that there exists an ω, which solves (4.3a), (4.3b), and (4.3c).Because of (4.6) this path has t is closed and I t is a good rate function, the sets M n are compact, non-empty, and satisfy M n ⊇ M n+1 for every n ∈ N.So the intersection M = n∈N M n is again non-empty.Because every ω ∈ M has I t (ω) = c, we see that there in fact exists a path ω for which the infimum is attained.From the Euler-Lagrange method we know that ω also solves equations (4.3a), (4.3b), and (4.3c).From lemmas 12 and 13 we know that the solution is unique, so ω must coincide with our path ω constructed above and we get inf I t (ω) ω ∈ M a,z,1 t − J(a, z) ≤ η for all z ∈ K 1 , a ∈ K 2 and t ≥ t 0 .Since η > 0 was arbitrary this completes the proof of lemma 8.

Staying Near the Equilibrium
In this section we study the event that for some drift function b the integral 1 2 t 0 b 2 (B s ) ds is small.In contrast to the previous section, here we are considering long time intervals but have no conditions on the final point.The main result of this section are the following two propositions.
Proposition 18 Let b : R → R be a differentiable function with b(0) = 0, b ′ (0) = 0 and lim inf |x|→∞ |b(x)| > 0. Then for every η > 0 we have Proposition The rest of this section is devoted to the proof of these two propositions.The main idea of the proof is to use Taylor approximation around the zero of b to reduce the problem to the case of linear b.We start by proving a result for the case b(x) = x.
Lemma 20 Let B be a one-dimensional Brownian Motion.Then for every x ∈ R and every set A with P (B t ∈ A) > 0 and in particular Proof.Formula (1-1.9.7) from [BS96] gives By definition of cosh and sinh there are constants 0 < c 1 < c 2 with c 1 e −tϑ/2 ≤ 1 2π sinh(tϑ) ≤ c 2 e −tϑ/2 for all ϑ > 1.
(5.1) (The value 1 is arbitrary, any positive number would do.)Also we can use the relation |2xy| ≤ x 2 + y 2 to get we can then find a ϑ 0 > 0, such that whenever ϑ > ϑ 0 the estimate holds for all x, z ∈ R. Thus we can conclude The exponential Tauber theorem [BGT87, theorem 4.12.9]now gives the first equality of the claim.The second claim follows by taking x = 0 and A = R.
We will also need a version of lemma 20 which holds uniformly in the initial condition x.This is given in the following lemma.The following lemma gives a set of conditions under which dominated terms can be neglected when calculating large deviation rate functions.The proof is elementary and we omit it here.
Lemma 22 Let f, g : R + → R + be two functions and assume that either one of the two conditions lim sup ε↓0 ε log g(ε) ≤ lim inf ε↓0 ε log f (ε) or lim sup ε↓0 ε log g(ε) < lim inf ε↓0 ε log f (ε) + g(ε) holds.Then we have for all x ∈ R. The basic large deviation result for the standard normal distribution on R now gives In this situation we can apply proposition 6 to get (5.3) Now consider the case T J > t.Choose n ∈ N with n > 2J and ε > 0 with 4ε/a 2 < t/n.Define ∆t = t/n, the intervals I 1 = [0, ∆t] and and the event A ε (k1,...,k ℓ ) = A ε ∩ S j ∈ I kj for j = 1, . . ., ℓ and S ℓ+1 > t .Then we have Choose (k 1 , . . ., k ℓ ) ∈ Q.As we have seen above the condition (5.4) for j = 1, . . ., ℓ − 1, where we use the convention k 0 = 0.If k ℓ < n then we use 5.4 also for j = ℓ and we have For k ℓ = n it will turn out that we need to treat the right endpoint of the interval specially, here we define else.Then we have Assume first the case k ℓ < n.Then we get Now we use the strong Markov property of Brownian motion for the stopping times S j and T j .
Because |B Tj | = a/2 and |B Sj | = a are deterministic and the Brownian motion is symmetric we get Repeating these two steps for j = ℓ − 1, . . ., 0 finally gives In order to use inequality (3.4) we have to calculate the individual rates for the factors on the right-hand side.Using lemma 21 we get (5.5) Using the reflection principle and the basic scaling property of Brownian motion we find The large deviation principle for the standard normal distribution on R now gives (5.6) Now we can apply inequality (3.4) to get the combined rate.The result is where n 1 = j = 1, . . ., ℓ d j > 0 .Because each of the intervals [S j , T j ] can have a non-empty intersection with at most two of the n intervals I k we have ℓ j=0 d j ≥ n − 2J and thus n 1 ≥ 1.So we find for all α ∈ D δ 2ℓ+1 and all δ > 0. Now assume k ℓ = n.This case is similar, but needs an additional argument to take care of the case t ∈ [S ℓ , T ℓ ).Here we can no longer use (5.6) for the interval [S ℓ , T ℓ ).To work around this we define a stopping time R by Given the event A αε (k1,...,k ℓ ) the process cannot have |B s | > a/2 for a period of time of length ∆t and using the special definition of d ℓ−1 for this case we get Similar to the other case we get then Using the strong Markov property for the stopping time R first gives Now we can continue splitting of terms as in the first case to get Using equations (5.5), (5.6) and inequality (3.4) as in the first case we get for all α ∈ D δ 2ℓ+1 and all δ > 0. Note that in this case n 1 = 0 is possible, this occurs in the case ℓ = 1 and S 1 ∈ I n , because I n was the interval we treated specially.
To estimate the upper exponential rate of A ε ∩ {T J > t} we need to compare all the rates from (5.7) and (5.8).We get for all δ > 0 and large enough n, where the largest bound came from the case ℓ = 1, k 1 = n.
Letting first δ ↓ 0 and then n → ∞ shows lim sup ε↓0 (5.9) This gives the upper bound for P (A ε ).Using the estimates (5.3) and (5.9) we find This completes the proof of the lemma 23.
Lemma 24 For every a > 0 and every x ∈ (−a/ √ 2, +a/ √ 2) we have Proof.The second equality is proved in lemma 20.Applying lemma 23 to the function v(x) = x 2 we see that Thus we can use lemma 22 to prove the first equality.Now we can combine the results of the previous lemmas to give the proofs of proposition 18.
Proof.(of proposition 18) Choose some 0 < δ < |b ′ (0)|.Using the Taylor formula b(x) = b ′ (0)x + o(x) we find an a > 0 with (5.10) Without loss of generality we may assume that a is smaller than η and also small enough to permit |b(x)| ≥ a |b ′ (0)| − δ for all x ∈ R with |x| > a.
We have to calculate the exponential rates of (5.11) Whenever sup 0≤s≤t |B s | ≤ a we can approximate b(x) by b ′ (0)x as in (5.10).This gives Both bounds of this estimate can be handled using which is a consequence of lemma 24.
For the lower bound this gives whenever δ > 0. For the upper bound we find Then by our choice of a we have v(x) ≥ x 2 ∧ a 2 and lemma 23 gives ( and thus Because ζ < a/2 < δ we can use lemma 24 to get for all sufficiently small κ > 0. Letting ζ ↓ 0 completes the proof of the first claim.
For the second claim first note that again by proposition 18.Let κ > 0 and choose δ > 0 with Using Taylor approximation we can find an a > 0 with Using the coupling argument and proposition 18 again, we get for all κ > 0. Taking the limit κ ↓ 0 completes the proof of proposition 19.
6 The LDP for the Endpoint In this section we use the results of the previous section to complete the proof of theorem 1.
Notation.To avoid complicated and hard to read expressions in small print we sometimes write (A) for the indicator function of the event A during this section.
for every compact set K ⊆ R. For ϑ > 0 let X ϑ be a solution of the SDE (1.1) with start in X ϑ 0 = 0. Then for ϑ → ∞ the family (X ϑ t ) ϑ satisfies the weak LDP with rate function J, where J is defined by Let F and G be as in (3.1).Then we find By definition of F * (x) we have Thus whenever ω t ∈ B η (x) and F (ω) − F * (x) ≥ 2δ we find Because Φ ′′ is bounded the above estimate implies that we can find an ε > 0 with i.e. we can find a g > 0 with G(ω) > g for all paths ω with ω t ∈ B η (x) and F (ω) − F * (x) ≥ 2δ.Together this gives So we can use lemma 22 to conclude and taking the supremum over all x ∈ O on the right hand side proves the lower bound.Now let K ⊆ R be compact and δ > 0. For each x ∈ K we can find an η > 0 with |Φ(y) − Φ(x)| ≤ δ whenever y ∈ B η (x).Because I is lower semi-continuous we can assume I(y) ≥ I(x) − δ for every y ∈ B η (x) by choosing η small enough.Using the compactness of K we can cover K with a finite number of such balls: there are x 1 , . . ., x n ∈ K and 0 < η 1 , . . ., η n < δ with and the above assumption on Φ and I hold for each k.For k = 1, . . ., n consider F * (x k ) as defined above.This time we find Since for fixed η this rate become arbitrarily negative when a becomes large, we can choose a large enough that the rate of p 1 (α, ε) + p 2 (α, ε) is dominated by p 2 .To treat the p 2 -term we apply inequality (3.4) as we did in the proof of proposition 6.From proposition 7 we know the individual rates lim sup This completes the proof of theorem 1.

2
and for every z ∈ R and every open set O ⊆ R we have B s ) ds ≤ ε, B tε ∈ O

Figure 1 :
Figure 1: This figure illustrates two types of solution for equation (4.3a).Here we only consider the case λ > 0. The curved line is the graph of the function x → −λv(x).The bold part of the lines corresponds to the points visited by the path ω.The thick dots are ω 0 , −λv(ω 0 ) and ω t , −λv(ω t ) .Both solutions start at z ∈ K 1 , head towards a neighbourhood of the zero m, and finally reach a point a ∈ K 2 .The left hand image shows a free solution, i.e. one with E > 0, the right hand image shows a bound solution, i.e. one with E ≤ 0 where the path ω turns at the point b with −λv(b) = E.

Figure 2 :
Figure 2: This figure illustrates the domain H T of the functions f and g.The domain is unbounded in directions λ → ∞ and E → ∞.It is bounded from below by λ → − inf x∈T λv(x), which is equal to −λ sup x∈T v(x) for λ ≤ 0 and to −λ inf x∈T v(x) for λ ≥ 0.

Lemma 26
Let Φ : R → R be a C 2 -function with bounded Φ ′′ and let b = −Φ ′ .Assume that there is an m ∈ R with b(x) = 0 if and only if x = m and lim inf |x|→∞ |b(x)| > 0. Further assume that there is a rate function I : R → [0, ∞] ω s ) ds)1 O (B t ) ≥ − inf x∈O I(x) for every open set O ⊆ R and lim sup ϑ→∞
, and d k = c k /p k for k = 1, . . ., n. Applying Jensen's inequality to the random variable which takes value d k with probability p k gives Proposition 7 Let P z be the distribution of a Brownian motion with start in z and B be the canonical process.Let b : R → R be a C 2 -function with lim inf |x|→∞ |b(x)| > 0. Assume that there is an m ∈ R with b(x) = 0 if and only if x = m and with b ′ (m) = 0. Then for every pair of compact sets K 1 , K 2 ⊆ R we have .13) Let ζ < a/2 and z ∈ [−ζ, +ζ].Then we can use lemma 25 to choose two Brownian motions B ζ and B z with B ζ 0 = ζ, B z 0 = z, and |B ζ t | ≥ |B z t | for all t ≥ 0. We find