Asymptotics of exponential moments of a weighted local time of a Brownian motion with small variance

Kulik, Alexei; Sobolieva, Daryna

doi:10.15559/16-VMSTA49

1 Introduction and the main result

Let

$\{W_{t},t\ge 0\}$ be a real-valued Wiener process, and μ be a σ-finite measure on

$\mathbb{R}$ such that

(1)

$\underset{x\in \mathbb{R}}{\sup }\mu \big([x-1,x+1]\big)<\infty .$

Recall that the local time

${L_{t}^{\mu }}(W)$ of the process W with the weight μ can be defined as the limit of the integral functionals

(2)

${L_{t}^{\mu _{n}}}(W):={\int _{0}^{t}}k_{n}(W_{s})\hspace{0.1667em}ds,\hspace{1em}k_{n}(x):=\frac{\mu _{n}(dx)}{dx},\hspace{2.5pt}n\ge 1,$

where

$\mu _{n}$ ,

$n\ge 1$ , is a sequence of absolutely continuous measures such that

$\int _{\mathbb{R}}f(x)\mu _{n}(dx)\to \int _{\mathbb{R}}f(x)\mu (dx)$

for all continuous f with compact support, and (1) holds for

$\mu _{n}$ ,

$n\ge 1$ , uniformly. The limit

${L_{t}^{\mu }}(W)$ exists in the mean square sense due to the general results from the theory of W-functionals; see [3], Chapter 6. This definition also applies to

$\varepsilon W$ instead of W for any positive ε. In what follows, we will treat

$\varepsilon W$ as a Markov process whose initial value may vary, and with a slight abuse of notation, we denote by

$\mathbf{P}_{x}$ the law of

$\varepsilon W$ with

$\varepsilon W_{0}=x$ and by

$\mathbf{E}_{x}$ the expectation w.r.t. this law.

In this note, we study the asymptotic behavior as

$\varepsilon \to 0$ of the exponential moments of the family of weighted local times

${L_{t}^{\mu }}(\varepsilon W)$ . Namely, we prove the following theorem.

Theorem 1.

For arbitrary finite measure μ on

$\mathbb{R}$ ,

(3)

$\underset{\varepsilon \to 0}{\lim }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}=\frac{t}{2}\underset{y\in \mathbb{R}}{\sup }\mu {\big(\{y\}\big)}^{2}.$

For arbitrary σ-finite measure μ on

$\mathbb{R}$ that satisfies (1),

(4)

$\underset{x\in \mathbb{R}}{\sup }\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}=\frac{t}{2}\underset{y\in \mathbb{R}}{\sup }\mu {\big(\{y\}\big)}^{2}.$

We note that in this statement the measure μ can be changed to a signed measure; in this case, in the right-hand side, only the atoms of the positive part of μ should appear. We also note that, in the σ-finite case, the uniform statement (3) may fail; one example of such a type is given in Section 3.

Let us briefly discuss the problem that was our initial motivation for the study of such exponential moments. Consider the one-dimensional SDE

(5)

$d{X_{t}^{\varepsilon }}=a\big({X_{t}^{\varepsilon }}\big)\hspace{0.1667em}dt+\varepsilon \sigma \big({X_{t}^{\varepsilon }}\big)\hspace{0.1667em}dW_{t}$

with discontinuous coefficients

$a,\sigma$ . In [7], a Wentzel–Freidlin-type large deviation principle (LDP) was established in the case

$a\equiv 0$ under mild assumptions on the diffusion coefficient σ. In [8], this result was extended to the particular class of SDEs such that the function

$a/{\sigma }^{2}$ has a bounded derivative. This limitation had appeared because of formula (7) in [8] for the rate transform of the family

${X}^{\varepsilon }$ . This formula contains an integral functional with kernel

${(a/{\sigma }^{2})}^{\prime }$ of a certain diffusion process obtained from

$\varepsilon W$ by the time change procedure. If

$a/{\sigma }^{2}$ is not smooth but is a function of a bounded variation, this integral function still can be interpreted as a weighted local time with weight

$\mu ={(a/{\sigma }^{2})}^{\prime }$ . Thus, Theorem 1 can be used in order to study the LDP for the SDE (5) with discontinuous coefficients. One of such particular results can be derived immediately. Namely, if μ is a continuous measure, then by Theorem 1 the exponential moments of

${L_{t}^{\mu }}(\varepsilon W)$ are negligible at the logarithmic scale with rate function

${\varepsilon }^{2}$ . This, after simple rearrangements, allows us to neglect the corresponding term in (7) of [8] and to obtain the statement of Theorem 2.1 of [8] under the weaker condition that

$a/{\sigma }^{2}$ is a continuous function of bounded variation. The problem how to describe in a more general situation the influence of the jumps of

$a/{\sigma }^{2}$ on the LDP for the solution to (5) still remains open and is the subject of our ongoing research. We just remark that due to Theorem 1 the respective integral term is no longer negligible, which well corresponds to the LDP results for piecewise smooth coefficients

$a,\sigma$ obtained in [1, 2, 6].

2 Proof of Theorem 1

2.1 Preliminaries

For a measure ν satisfying (1), denote by

(6)

${f_{t}^{\nu ,\varepsilon }}(x)=\mathbf{E}_{x}{L_{t}^{\nu }}(\varepsilon W)={\int _{0}^{t}}\int _{\mathbb{R}}\frac{1}{\sqrt{2\pi s{\varepsilon }^{2}}}{e}^{-\frac{{(y-x)}^{2}}{2s{\varepsilon }^{2}}}\nu (dy)\hspace{0.1667em}ds,\hspace{1em}t\ge 0,\hspace{2.5pt}x\in \mathbb{R},$

the characteristic of the local time

${L}^{\nu }(\varepsilon W)$ considered as a W-functional of

$\varepsilon W$ ; see [3], Chapter 6.

The following statement is a version of Khas’minskii’s lemma; see [9], Section 1.2.

Lemma 1.

Suppose that

(7)

$\underset{x\in \mathbb{R}}{\sup }{f_{s}^{\nu ,\varepsilon }}(x)\le \frac{1}{2}.$

Then

$\underset{x\in \mathbb{R}}{\sup }\mathbf{E}_{x}{e}^{{L_{s}^{\nu }}(\varepsilon W)}\le 2.$

Using the Markov property, as a simple corollary, we obtain, for arbitrary

$t>0$ ,

(8)

$\underset{x\in \mathbb{R}}{\sup }\mathbf{E}_{x}{e}^{{L_{t}^{\nu }}(\varepsilon W)}\le {2}^{1+t/s}=2{e}^{(\log 2)(t/s)},$

where

$s>0$ is such that (7) holds. This inequality, combined with (6), leads to the following estimate.

Lemma 2.

For a nonzero measure ν satisfying (1), denote

$N(\nu ,\gamma )=\underset{x\in \mathbb{R}}{\sup }\nu \big([x-\gamma ,x+\gamma ]\big),\hspace{1em}\gamma >0.$

For any

$\lambda \ge 1$ and

$\gamma >0$ , there exists

$\varepsilon _{\lambda ,\gamma }>0$ such that

(9)

$\underset{x\in \mathbb{R}}{\sup }\mathbf{E}_{x}{e}^{\lambda {L_{t}^{\nu }}(\varepsilon W)}\le 2{e}^{(4\log 2)c_{0}N{(\nu ,\gamma )}^{2}t{\lambda }^{2}{\varepsilon }^{-2}},\hspace{1em}\varepsilon \in (0,\varepsilon _{\lambda ,\gamma }),$

with

$c_{0}=\frac{2}{\pi }{\Bigg(1+2\sum \limits_{k=1}^{\infty }{e}^{-\frac{{(2k-1)}^{2}}{2}}\Bigg)}^{2}.$

Proof.

$\varepsilon \sqrt{s}\le \gamma$ , then we have

$\begin{array}{r@{\hskip0pt}l}\displaystyle {f_{s}^{\nu ,\varepsilon }}(x)& \displaystyle =\sum \limits_{k\in \mathbb{Z}}{\int _{0}^{s}}\int _{|y-x-2k\gamma |\le \gamma }\frac{1}{\sqrt{2\pi v{\varepsilon }^{2}}}{e}^{-\frac{{(y-x)}^{2}}{2v{\varepsilon }^{2}}}\nu (dy)\hspace{0.1667em}dv\\{} & \displaystyle \le \sqrt{c_{0}}N(\nu ,\gamma )\sqrt{\frac{s}{{\varepsilon }^{2}}}.\end{array}$

Take

$s={\big(2N(\nu ,\gamma )\big)}^{-2}{(c_{0})}^{-1}{\lambda }^{-2}{\varepsilon }^{2}.$

Then the inequality

$\varepsilon \sqrt{s}\le \gamma$ holds, provided that

$\varepsilon \le {\big(\gamma {\big(2N(\nu ,\gamma )\big)}^{2}c_{0}{\lambda }^{2}\big)}^{1/3}=:\varepsilon _{\lambda ,\gamma }.$

Under this condition,

${f_{s}^{\lambda \nu ,\varepsilon }}(x)=\lambda {f_{s}^{\nu ,\varepsilon }}(x)\le \frac{1}{2}.$

Now the required inequality follows immediately from (8). □

In what follows, we will repeatedly decompose μ into sums of two components and analyze separately the exponential moments of the local times that correspond to these components. We will combine these estimates and obtain an estimate for

${L_{t}^{\mu }}(\varepsilon W)$ itself using the following simple inequality. Let

$\mu =\nu +\kappa$ and

$p,q>1$ be such that

$1/p+1/q=1$ . Then

${L_{t}^{\mu }}(\varepsilon W)={L_{t}^{\nu }}(\varepsilon W)+{L_{t}^{\kappa }}(\varepsilon W)=(1/p){L_{t}^{p\nu }}(\varepsilon W)+(1/q){L_{t}^{q\kappa }}(\varepsilon W),$

and therefore by the Hölder inequality we get

(10)

$\mathbf{E}{e}^{{L_{t}^{\mu }}(\varepsilon W)}\le {\big(\mathbf{E}{e}^{{L_{t}^{p\nu }}(\varepsilon W)}\big)}^{1/p}{\big(\mathbf{E}{e}^{{L_{t}^{q\kappa }}(\varepsilon W)}\big)}^{1/q}.$

We will also use another version of this upper bound, which has the form

(11)

$\mathbf{E}{e}^{{L_{t}^{\mu }}(\varepsilon W)}1_{A}\le {\big(\mathbf{E}{e}^{{L_{t}^{p\mu }}(\varepsilon W)}\big)}^{1/p}{\big(\mathbf{P}(A)\big)}^{1/q},\hspace{1em}A\in \mathcal{F}.$

We denote

$\varDelta =\underset{x\in \mathbb{R}}{\sup }\mu \big(\{x\}\big).$

We will prove Theorem 1 in several steps, in each of them extending the class of measures μ for which the required statement holds.

2.2 Step I: μ is a finite mixture of δ-measures

$\mu =a\delta _{z}$ is a weighted δ-measure at the point z, then we have

${L_{t}^{\mu }}(\varepsilon W)=a{\varepsilon }^{-1}{L_{t}^{(z)}}(W),$

where

${L_{t}^{(z)}}(W)=\underset{\eta \to 0}{\lim }\frac{1}{2\eta }{\int _{0}^{t}}1_{|W_{s}-z|\le \eta }\hspace{0.1667em}ds$

is the local time of a Wiener process at the point z. The distribution of

${L_{t}^{(z)}}(W)$ is well known; see, e.g., [5], Chapter 2.2 and expression (6) in Chapter 2.1. Hence, the required statement in the particular case

$\mu =a\delta _{z}$ is straightforward, and we have the following:

(12)

$\underset{\varepsilon \to 0}{\lim }{\varepsilon }^{2}\underset{x}{\sup }\log \mathbf{E}_{x}{e}^{a{\varepsilon }^{-1}{L_{t}^{(z)}}(W)}=\frac{t{a}^{2}}{2}.$

Note that in this formula the supremum is attained at the point

$x=z$ .

In this section, we will extend this result to the case where μ is a finite mixture of δ-measures, that is,

$\mu =\sum \limits_{j=1}^{k}a_{j}\delta _{z_{j}}.$

Let

$j_{\ast }$ be the number of the maximal value in

$\{a_{j}\}$ , that is,

$\varDelta =a_{j_{\ast }}$ . Then

${L_{t}^{\mu }}(\varepsilon W)\ge \varDelta {\varepsilon }^{-1}{L_{t}^{(z_{j_{\ast }})}}(W)$ , and it follows directly from (12) that

(13)

$\underset{\varepsilon \to 0}{\liminf }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}\ge \frac{t{\varDelta }^{2}}{2}.$

In what follows, we prove the corresponding upper bound

(14)

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}\le \frac{t{\varDelta }^{2}}{2},$

which, combined with this lower bound, proves (3).

Observe that, for

$\gamma >0$ small enough,

$N(\mu ,\gamma )=\varDelta .$

Then by Lemma 2, for any

$\lambda \ge 1$ ,

(15)

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{\lambda {L_{t}^{\mu }}(\varepsilon W)}\le c_{1}{\lambda }^{2}t{\varDelta }^{2}$

with

$c_{1}=(4\log 2)c_{0}=\frac{8\log 2}{\pi }{\Bigg(1+2\sum \limits_{k=1}^{\infty }{e}^{-\frac{{(2k-1)}^{2}}{2}}\Bigg)}^{2}.$

In particular, taking

$\lambda =1$ , we obtain an upper bound of the form (14), but with a worse constant

$c_{1}$ instead of required

$1/2$ . We will improve this bound by using the large deviations estimates for

$\varepsilon W$ , the Markov property, and the “individual” identities (12).

Denote

$\mu _{j}=a_{j}\delta _{z_{j}},j=1,\dots ,k$ . Then

${L_{t}^{\mu }}(\varepsilon W)=\sum \limits_{j=1}^{k}{L_{t}^{\mu _{j}}}(\varepsilon W).$

Fix some family of neighborhoods

$O_{j}$ of

$z_{j},j=1,\dots ,k$ , such that the minimal distance between them equals

$\rho >0$ , and denote

${O}^{j}=\mathbb{R}\setminus \bigcup \limits_{i\ne j}O_{i}.$

For some

$N\ge 1$ whose particular value will be specified later, consider the partition

$t_{n}=t(n/N)$ ,

$n=0,\dots ,N$ , of the segment

$[0,t]$ and denote

$B_{n,j}=\big\{f\in C(0,t):f_{s}\in {O}^{j},s\in [t_{n-1},t_{n}]\big\},\hspace{1em}j\in \{1,\dots ,k\},\hspace{2.5pt}n\in \{1,\dots ,N\},$

$C_{j_{1},\dots ,j_{N}}=\bigcap \limits_{n=1}^{N}B_{n,j_{n}},\hspace{1em}j_{1},\dots ,j_{N}\in \{1,\dots ,k\}.$

Observe that if the process

$\varepsilon W$ does not visit

$O_{j}$ on the time segment

$[u,v]$ , then

${L}^{\mu _{j}}(\varepsilon W)$ on this segment stays constant. This means that, on the set

$\{\varepsilon W\in C_{j_{1},\dots ,j_{N}}\}$ , we have

${L_{t}^{\mu }}(\varepsilon W)=\sum \limits_{n=1}^{N}\big({L_{t_{n}}^{\mu _{j_{n}}}}(\varepsilon W)-{L_{t_{n-1}}^{\mu _{j_{n}}}}(\varepsilon W)\big).$

Because

${L}^{\mu _{j}}(\varepsilon W)$ is a time-homogeneous additive functional of the Markov process

$\varepsilon W$ , we have

$E_{x}\big[{e}^{{L_{t_{n}}^{\mu _{j_{n}}}}(\varepsilon W)-{L_{t_{n-1}}^{\mu _{j_{n}}}}(\varepsilon W)}\big|\mathcal{F}_{t_{n-1}}\big]=E_{y}{e}^{{L_{t/N}^{\mu _{j_{n}}}}(\varepsilon W)}\Big|_{y=\varepsilon W_{t_{n-1}}}.$

Then by (12), for any

$j_{1},\dots ,j_{N}\in \{1,\dots ,k\}$ ,

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}1_{\varepsilon W\in C_{j_{1},\dots ,j_{N}}}\le \frac{t}{2N}\sum \limits_{n=1}^{N}{(a_{j_{n}})}^{2}\le \frac{t{\varDelta }^{2}}{2}.$

Because we have a fixed number of sets

$C_{j_{1},\dots ,j_{N}}$ , this immediately yields

(16)

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}1_{\varepsilon W\in C}\le \frac{t{\varDelta }^{2}}{2}$

with

$C=\bigcup \limits_{j_{1},\dots ,j_{N}\in \{1,\dots ,k\}}C_{j_{1},\dots ,j_{N}}.$

Hence, to get the required upper bound (14), it suffices to prove an analogue of (16) with the set C replaced by its complement

$D=C(0,t)\setminus C$ . Using (11) with

$p=2$ ,

$A=\{\varepsilon W\in D\}$ , and (15) with

$\lambda =2$ , we get

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}1_{\varepsilon W\in D}\le 2c_{1}t{\varDelta }^{2}+\frac{1}{2}\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{P}_{x}(\varepsilon W\in D).$

By the LDP for the Wiener process ([4], Chapter 3, §2),

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{P}_{x}(\varepsilon W\in D)=-\underset{f\in \mathrm{closure}(D)}{\inf }I(f),$

where

$I(f)=\left\{\begin{array}{l@{\hskip10.0pt}l}(1/2){\textstyle\int _{0}^{t}}{({f^{\prime }_{s}})}^{2}\hspace{0.1667em}ds,& f\hspace{2.5pt}\text{is absolutely continuous on}\hspace{2.5pt}[0,t];\\{} +\infty & \text{otherwise.}\end{array}\right.$

For any trajectory

$f\in D$ , there exists n such that f visits at least two sets

$O_{j}$ on the time segment

$[t_{n-1},t_{n}]$ . Therefore, any trajectory

$f\in \mathrm{closure}(D)$ exhibits an oscillation

$\ge \rho$ on this time segment. On the other hand, for an absolutely continuous f,

$|f_{u}-f_{v}|=\Bigg|{\int _{u}^{v}}{f^{\prime }_{s}}\hspace{0.1667em}ds\Bigg|\le |u-v{|}^{1/2}{\Bigg({\int _{0}^{t}}{\big({f^{\prime }_{s}}\big)}^{2}\hspace{0.1667em}ds\Bigg)}^{1/2}.$

This means that, for any

$f\in \mathrm{closure}(D)$ ,

$I(f)\ge \frac{{\rho }^{2}N}{2t},$

which yields

If in this construction, N was chosen such that

$N\ge (4c_{1}-1){\rho }^{-2}{t}^{2}{\varDelta }^{2},$

then the latter inequality guarantees the analogue of (16) with D instead of C. This completes the proof of (14).

2.3 Step II: μ is finite

Exactly the same argument as that used in Section 2.2 provides the lower bound (13). In this section, we prove the upper bound (14) for a finite measure μ and thus complete the proof of the first assertion of the theorem. For finite μ and any

$\chi >0$ , we can find

$\gamma >0$ and decompose

$\mu =\mu _{0}+\nu$ in such a way that

$\mu _{0}$ is a finite mixture of δ-measures and

$N(\nu ,\gamma )<\chi$ . Let

$p,q>1$ be such that

$1/p+1/q=1$ . The measure

$p\mu _{0}$ has the maximal weight of an atom equal to

$p\varDelta$ . Since we have already proved the required statement for finite mixtures of δ-measures, we have

(17)

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log {\big(\mathbf{E}_{x}{e}^{{L_{t}^{p\mu _{0}}}(\varepsilon W)}\big)}^{1/p}\le \frac{t}{2}p{\varDelta }^{2}.$

On the other hand, we have

$N(\nu ,\gamma )<\chi$ and then by Lemma 2

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log {\big(\mathbf{E}_{x}{e}^{{L_{t}^{q\nu }}(\varepsilon W)}\big)}^{1/q}\le c_{1}qt{\chi }^{2}.$

Hence, by (10),

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}\le \frac{t}{2}p{\varDelta }^{2}+c_{1}qt{\chi }^{2}.$

Now we can finalize the argument. Fix

$\varDelta _{1}>\varDelta :=\max _{x\in \mathbb{R}}\mu {(\{x\})}^{2}$ and choose

$p,q>1$ such that

$1/p+1/q=1$ and

$p{\varDelta }^{2}<{\varDelta _{1}^{2}}$ . Then there exists

$\chi >0$ small enough such that

$p{\varDelta }^{2}+2c_{1}qt{\chi }^{2}<{\varDelta _{1}^{2}}.$

Taking the decomposition

$\mu =\mu _{0}+\nu$ that corresponds to this value of χ and applying the previous calculations, we obtain an analogue of the upper bound (14) with Δ replaced by

$\varDelta _{1}$ . Since

$\varDelta _{1}>\varDelta$ is arbitrary, the same inequality holds for Δ.

2.4 Step III: μ is σ-finite

In this section, we prove the second assertion of the theorem. As before, the lower bound can be obtained directly from the case

$\mu =a\delta _{z}$ , and hence we concentrate ourselves on the proof of the upper bound

(18)

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}\le \frac{t{\varDelta }^{2}}{2},\hspace{1em}x\in \mathbb{R}.$

We will use an argument similar to that from the previous section and decompose μ into a sum

$\mu =\mu _{0}+\nu$ with finite

$\mu _{0}$ and ν, which is negligible in a sense. However, such a decomposition relies on the initial value x, and this is the reason why we obtain an individual upper bound (18) instead of the uniform one (14).

Namely, for a given x, we define

$\mu _{0},\nu$ by restricting μ to

$[x-R,x+R]$ and its complement, respectively. Without loss of generality, we assume that for each R, the corresponding ν is nonzero. Since we have already proved the required statement for finite measures, we get (17).

Next, denote

$M=\sup _{x\in \mathbb{R}}\mu ([x-1,x+1])$ and observe that

$N(\nu ,1)\le M$ . Then by Lemma 2 with

$\gamma =1$ and the strong Markov property, for any stopping time τ, the exponential moment of

${L_{t}^{q\nu }}(\varepsilon W)$ conditioned by

$\mathcal{F}_{\tau }$ is dominated by

$2{e}^{c_{1}{M}^{2}t{q}^{2}{\varepsilon }^{-2}}$ . This holds for

$\varepsilon \le {\varepsilon _{q,1}^{x,R}}$ , where we put the indices

$x,R$ in order to emphasize that this constant depends on ν, which, in turn, depends on

$x,R$ . Since we have assumed that, for any

$x,R$ , the respective ν is nonzero, the constants

${\varepsilon _{q,1}^{x,R}}$ are strictly positive.

Now we take by τ the first time moment when

$|\varepsilon W_{\tau }-x|=R$ . Observe that

${L_{t}^{\nu }}(\varepsilon W)$ equals 0 on the set

$\{\tau >t\}$ and it is well known that

$\mathbf{P}_{x}(\tau <t)\le 4\mathbf{P}_{x}(\varepsilon W_{t}>R)\le C{e}^{-t{R}^{2}{\varepsilon }^{-2}/2}.$

Summarizing the previous statements, we get

$\mathbf{E}_{x}{e}^{{L_{t}^{q\nu }}(\varepsilon W)}\le 1+2C{e}^{t{\varepsilon }^{-2}(c_{1}{M}^{2}{q}^{2}-{R}^{2}/2)},\hspace{1em}\varepsilon \le {\varepsilon _{\lambda ,1}^{x,R}},$

which implies

(19)

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\log {\big(\mathbf{E}_{x}{e}^{{L_{t}^{q\nu }}(\varepsilon W)}\big)}^{1/q}\le t\big(c_{1}{M}^{2}q-{R}^{2}/(2q)\big)_{+},$

where we denote

$a_{+}=\max (a,0)$ . By (10) inequalities (17) and (19) yield

$\underset{\varepsilon \to 0}{\limsup }{\varepsilon }^{2}\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu _{0}}}(\varepsilon W)}\le \frac{t}{2}p{\varDelta }^{2}+t\big(c_{1}{M}^{2}q-{R}^{2}/(2q)\big)_{+}.$

Now we finalize the argument in the same way as we did in the previous section. Fix

$\varDelta _{1}>\varDelta$ and take

$p>1$ such that

$p{\varDelta }^{2}\le {\varDelta _{1}^{2}}$ . Then take R large enough so that, for the corresponding q,

$c_{1}{M}^{2}q-{R}^{2}/(2q)\le 0.$

Under such a choice, the calculations made before yield (18) with Δ replaced by

$\varDelta _{1}$ . Since

$\varDelta _{1}>\varDelta$ is arbitrary, the same inequality holds for Δ.

3 Example

Let

$\mu =\sum \limits_{k=1}^{\infty }(\delta _{{k}^{2}}+\delta _{{k}^{2}+{2}^{-k}}).$

Then μ satisfies (1) and

$\varDelta =1$ . However, it is an easy observation that when the initial value x is taken in the form

$x_{k}={k}^{2}$ , the respective exponential moments satisfy

$\mathbf{E}_{x_{k}}{e}^{{L_{t}^{\mu }}(\varepsilon W)}\to \mathbf{E}_{0}{e}^{{L_{t}^{\nu }}(\varepsilon W)},\hspace{1em}k\to \infty ,$

with

$\nu =2\delta _{0}$ . Then

$\underset{\varepsilon \to 0}{\liminf }{\varepsilon }^{2}\underset{x\in \mathbb{R}}{\sup }\log \mathbf{E}_{x}{e}^{{L_{t}^{\mu }}(\varepsilon W)}\ge \underset{\varepsilon \to 0}{\liminf }{\varepsilon }^{2}\log \mathbf{E}_{0}{e}^{{L_{t}^{\nu }}(\varepsilon W)}=2t>\frac{t}{2},$

and therefore (3) fails.

Abstract

1 Introduction and the main result

(1)

(2)

Theorem 1.

(3)

(4)

(5)

2 Proof of Theorem 1

2.1 Preliminaries

(6)

Lemma 1.

(7)

(8)

Lemma 2.

(9)

Proof.

(10)

(11)

2.2 Step I: μ is a finite mixture of δ-measures

(12)

(13)

(14)

(15)

(16)

2.3 Step II: μ is finite

(17)

2.4 Step III: μ is σ-finite

(18)

(19)

3 Example

Acknowledgments

References

Authors

Abstract

1 Introduction and the main result

(1)

(2)

Theorem 1.

(3)

(4)

(5)

2 Proof of Theorem 1

2.1 Preliminaries

(6)

Lemma 1.

(7)

(8)

Lemma 2.

(9)

Proof.

(10)

(11)

2.2 Step I: μ is a finite mixture of δ-measures

(12)

(13)

(14)

(15)

(16)

2.3 Step II: μ is finite

(17)

2.4 Step III: μ is σ-finite

(18)

(19)

3 Example

Acknowledgments

References

Export citation

Copy and paste formatted citation

Download citation in file