The subleading order of two dimensional cover times

Abstract

The $\varepsilon$-cover time of the two dimensional torus by Brownian motion is the time it takes for the process to come within distance $\varepsilon >0$ from any point. Its leading order in the small $\varepsilon$-regime has been established by Dembo et al. (Ann Math 160:433–464, 2004). In this work, the second order correction is identified. The approach relies on a multi-scale refinement of the second moment method, and draws on ideas from the study of the extremes of branching Brownian motion.

Appendix

In the Appendix we collect some less important proofs. We first give the proof of the large deviation bound Lemma 4.6 for sums of a binomial number of geometric random variables, which was used to prove the upper bound Proposition 3.5.

Proof of Lemma 4.6

Note that

$$\begin{aligned} \mathbb {P}\left[ \sum _{i=1}^{n}J_{i}G_{i}\le \theta \right] =\mathbb {P}\left[ \sum _{i=1}^{J_{1}+\cdots +J_{n}}G_{i}\le \theta \right] . \end{aligned}$$

(9.1)

Now (since a sum of geometrics is a negative binomial distribution) we have

$$\begin{aligned} \mathbb {P}\left[ \sum _{i=1}^{m}G_{i}\le \theta \right] =\mathbb {P}\left[ I_{1}+\cdots +I_{\theta }\ge m\right] \quad \text{ for } m\ge 1, \end{aligned}$$

where $I_{1},I_{2},\ldots$ are iid Bernoulli random variables with success probability p, which can be taken to be independent of the $J_{i}-$s. Thus (by conditioning on $J_{1}+\cdots +J_{n}$ in (9.1)) we have in fact

$$\begin{aligned} \mathbb {P}\left[ \sum _{i=1}^{n}J_{i}G_{i}\le \theta \right] =\mathbb {P}\left[ I_{1}+\cdots +I_{\theta }\ge J_{1}+\cdots +J_{n}\right] . \end{aligned}$$

For any $\lambda >0$ this probability is bounded above by

$$\begin{aligned}&\mathbb {E}\left[ \exp \left( \lambda \left( I_{1}+\cdots +I_{\theta }-J_{1}-\ldots -J_{n}\right) \right) \right] \\&\quad =\left( 1+p\left( e^{\lambda }-1\right) \right) ^{\theta }\left( 1+q\left( e^{-\lambda }-1\right) \right) ^{n}\le \exp \left( \theta p\left( e^{\lambda }-1\right) +qn\left( e^{-\lambda }-1\right) \right) , \end{aligned}$$

where we have used that $1+x\le e^{x}$. Now (4.23) follows by setting $\lambda =\frac{1}{2}\log \frac{qn}{\theta p}$. $\square$

Next we derive the characterisation Lemma 7.7 of local times of continuous time random walk on $\left\{ 0,\ldots ,L\right\}$ from the generalized second Ray–Knight theorem. Recall the definition of $\tilde{\mathbb {P}}_{l}$ and $Y_{t}$ from above (7.39) and the definition of $L_{l}^{t}$ from (7.39).

Proof of Lemma 7.7

Let $\mathcal {L}=\left\{ 0,\ldots ,L\right\}$. The generalized second Ray–Knight theorem (see [16] or Theorem 8.2.2 [22]) implies that $\left( L_{l}^{\tau \left( t\right) }+\frac{1}{2}\eta _{x}^{2}\right) _{l\in \mathcal {L}}\overset{\text{ law }}{=}\left( \frac{1}{2}\left( \eta _{l}+\sqrt{2t}\right) ^{2}\right) _{l\in \mathcal {L}}$, where $\eta _{l}$ is a centered Gaussian process on $\mathcal {L}$ with covariance $\mathbb {E}\left[ \eta _{a}\eta _{b}\right] =\tilde{\mathbb {E}}_{b}\left[ L_{a}^{H_{0}}\right] =a$ for $a\le b$, independent of $L_{l}^{\tau \left( t\right) }$. Thus $\eta _{l},l\in \mathcal {L}$, is in fact Brownian motion at the integer times $l\in \mathcal {L}$. This in turn implies that $\left( \frac{1}{2}\eta _{l}^{2}\right) _{l\in \mathcal {L}}$ has the $\mathbb {Q}_{0}^{1}$-law of $\left( \frac{1}{2}X_{l}\right) _{l\in \mathcal {L}}$ and $\left( \frac{1}{2}\left( \eta _{l}+\sqrt{2t}\right) ^{2}\right) _{l\in \mathcal {L}}$ has the $\mathbb {Q}_{2t}^{1}$-law of $\left( \frac{1}{2}X_{l}\right) _{l\in \mathcal {L}}$ (recall (7.24)). By the additivity property (7.26) of Bessel processes we thus have that $\left( L_{l}^{\tau \left( t\right) }+\frac{1}{2}X_{l}^{1}\right) _{l\in \mathcal {L}}\overset{\text{ law }}{=}\left( \frac{1}{2}X_{l}^{1}+\frac{1}{2}X_{l}^{2}\right) _{l\in \mathcal {L}}$ where $\left( X_{t}^{1}\right) _{t\ge 0}$ has law $\mathbb {Q}_{0}^{1}$, $X_{t}^{2}$ haw law $\mathbb {Q}_{2t}^{0}$. Now the claim follows because we may “cancel out” $\frac{1}{2}X_{l}^{1}$ from this equality in law, since all random variables involved are non-negative (see [28, (2.56)]). $\square$

Next we give the proof of Lemma 7.8, which describes the law of the local times $L_{l}^{D_{t}},l\in \left\{ 0,\ldots ,L\right\} ,$ of continuous time random walk on $\left\{ 0,1,\ldots ,L\right\}$ when conditioned on $L_{L}^{D_{t}}=0$. Recall the definition of $D_{t}$ from (7.40). For the proof let us denote by $\Gamma$ the state space of $\left( Y_{t}\right) _{t\ge 0}$, that is the space of all piecewise constant cadlag functions from $[0,\infty )$ to $\left\{ 0,\ldots ,L\right\}$.

Proof of Lemma 7.8

Define the successive returns to and departures from $\left\{ 0,\ldots ,L\right\} \backslash \left\{ 1\right\}$ of $Y_{t}$ by $\tilde{D}_{0}=H_{1}$,

$$\begin{aligned} \tilde{R}_{n}=H_{\left\{ 0,2\right\} }\circ \theta _{\tilde{D}_{n-1}}+\tilde{D}_{n-1},n\ge 1,\quad \text{ and }\quad \tilde{D}_{n}=H_{1}\circ \theta _{\tilde{R}_{n}}+\tilde{R}_{n},n\ge 1. \end{aligned}$$

Collect the excursions of $Y_{t}$ into a marked point process $\mu$ on $[0,\infty )\times \Gamma$ defined by

$$\begin{aligned} \mu =\sum _{i\ge 1}\delta _{\left( L_{1}^{\tilde{R}_{i}},Y_{\left( \tilde{R}_{i}+\cdot \right) \wedge \tilde{D}_{i}}\right) }. \end{aligned}$$

The point process $\mu$ is a Poisson point process on $\mathbb {R}_{+}\times \Gamma$ of intensity

$$\begin{aligned} \left( 2\lambda \right) \otimes \left( \frac{1}{2}\tilde{\mathbb {P}}_{0}\left[ Y_{\cdot \wedge H_{1}}\in dw\right] +\frac{1}{2}\tilde{\mathbb {P}}_{2}\left[ Y_{\cdot \wedge H_{1}}\in dw\right] \right) , \end{aligned}$$

(9.2)

where $\lambda$ is Lebesgue-measure. We can decompose this point process into

$$\begin{aligned} \mu _{1}=1_{\mathbb {R}_{+}\times \left\{ Y_{0}=0\right\} }\mu \quad \text{ and }\quad \mu _{2}=1_{\mathbb {R}_{+}\times \left\{ Y_{0}=2,H_{1}<H_{L}\right\} }\mu \text{ and } \mu _{3}=1_{\mathbb {R}_{+}\times \left\{ Y_{0}=2,H_{L}<H_{1}\right\} }\mu , \end{aligned}$$

where $\mu _{1}$ collects the excursions that start in 0, $\mu _{2}$ collects the excursions that start in 2 and avoid L, and $\mu _{3}$ has the excursions that start in 2 and hit L . Since we are restricting $\mu$ to disjoint sets, $\mu _{1}$, $\mu _{2}$ and $\mu _{3}$ are independent Poisson point processes.

Let

$$\begin{aligned} \mu _{1}=\sum _{i}\delta _{\left( S_{i},w_{i}\right) }, \end{aligned}$$

for $S_{1}<S_{2}<\ldots$, so that $S_{t}$ is the local time at vertex 1 until the t-th jump to 0. Note that (recall (7.40))

$$\begin{aligned} L_{1}^{D_{t}}=S_{t},\quad \text{ for } t\in \left\{ 1,2,\ldots \right\} . \end{aligned}$$

We have

$$\begin{aligned} L_{l}^{D_{t}}=\sum _{\left( s,w\right) \in \mu _{2}\cup \mu _{3}:s\le S_{t}}L_{l}^{\infty }\left( w_{i}\right) \quad \text{ for } l\in \left\{ 2,3,\ldots \right\} , \end{aligned}$$

(9.3)

where $L_{l}^{\infty }\left( w\right)$ is the local time at l of the path w, i.e. $L_{l}^{\infty }\left( w\right) =d_{l}^{-1}\int _{0}^{\infty }1_{\left\{ w_{s}=l\right\} }ds$ for $d_{l}$ as in (7.38). For any $u\ge 0$ define the vector

$$\begin{aligned} V_{u}=\left( u,\sum _{\left( s,w\right) \in \mu _{2}:s\le u}L_{2}^{\infty }\left( w_{i}\right) ,\ldots ,\sum _{\left( s,w\right) \in \mu _{2}:s\le u}L_{L}^{\infty }\left( w_{i}\right) \right) \in \mathbb {R}^{L}. \end{aligned}$$

(9.4)

By (9.3) we have

$$\begin{aligned} \left( L_{l}^{D_{t}}\right) _{l\in \left\{ 1,\ldots ,L\right\} }=V_{S_{t}} \text{ on } \text{ the } \text{ event } \left\{ L_{L}^{D_{t}}=0\right\} =\left\{ \mu _{3}\left( \left[ 0,S_{t}\right] \times \Gamma \right) =0\right\} . \end{aligned}$$

Furthermore note that $L_{1}^{D_{t}}$ and $\left\{ L_{L}^{D_{t}}=0\right\}$ only depend on $\mu _{1}$ and $\mu _{3}$, while $V_{u}$ only depends on $\mu _{2}$, which is independent of $\mu _{1}$ and $\mu _{3}$. Therefore

$$\begin{aligned} \begin{array}{l} \tilde{\mathbb {P}}_{0}\left[ \left( L_{l}^{D_{t}}\right) _{l\in \left\{ 1,\ldots ,L\right\} }\in A|L_{L}^{D_{t}}=0\right] =\tilde{\mathbb {P}}_{0}\left[ f\left( L_{1}^{D_{t}}\right) |L_{L}^{D_{t}}=0\right] ,\\ \text{ where } f\left( u\right) =\tilde{\mathbb {P}}_{0}\left[ V_{u}\in A\right] . \end{array} \end{aligned}$$

(9.5)

We are thus interested in the law of $V_{u}$. Let $\tilde{Y}_{t}$ be continuous time random walk on $\left\{ 1,\ldots ,L\right\}$ with local times and inverse local time at vertex 1 given by

$$\begin{aligned} \tilde{L}_{l}^{u}=\frac{1}{1+1_{\left\{ 1<l<L\right\} }}\int _{0}^{u}1_{\left\{ \tilde{Y}_{s}=l\right\} }ds\quad \text{ and }\quad \tilde{\tau }\left( t\right) =\inf \left\{ s\ge 0:\tilde{L}_{1}^{s}>u\right\} . \end{aligned}$$

Sampling $\tilde{Y}_{t},t\ge 0,$ by “stitching together” the excursions in the point processes $\mu _{2}$ and $\mu _{3}$ we see that

$$\begin{aligned} \left( \tilde{L}_{l}^{\tilde{\tau }\left( u\right) }\right) _{l\in \left\{ 2\ldots ,L\right\} }\overset{\text{ law }}{=}\left( \sum _{\left( s,w\right) \in \mu _{2}\cup \mu _{3}:s\le u}L_{l}^{\infty }\left( w\right) \right) _{l\in \left\{ 2,\ldots ,L\right\} }. \end{aligned}$$

(9.6)

So by Lemma 7.7 (with $\left\{ 1,\ldots ,L\right\}$ in place of $\left\{ 0,\ldots ,L\right\}$) we have that

$$\begin{aligned}&\tilde{\mathbb {P}}_{0}\left[ \left( {\displaystyle \sum _{\left( s,w\right) \in \mu _{2}\cup \mu _{3}:s\le u}}L_{l}^{\infty }\left( w\right) \right) _{l\in \left\{ 2,\ldots ,L\right\} }\in \cdot \right] \nonumber \\&\quad =\tilde{\mathbb {P}}_{0}\left[ \left( \tilde{L}_{l}^{\tilde{\tau }\left( u\right) }\right) _{l\in \left\{ 2\ldots ,L\right\} }\in \cdot \right] =\mathbb {Q}_{2u}^{0}\left[ \left( \frac{1}{2}X_{l}\right) _{l\in \left\{ 1,\ldots ,L-1\right\} }\in \cdot \right] . \end{aligned}$$

(9.7)

Now since

$$\begin{aligned} \left\{ \mu _{3}\left( \left[ 0,t\right] \times \Gamma \right) =0\right\} =\left\{ \sum _{\left( s,w\right) \in \mu _{2}\cup \mu _{3}:s\le t}L_{L}^{\infty }\left( w\right) =0\right\} , \end{aligned}$$

and $V_{u}$ is independent of $\mu _{3}$ we have

$$\begin{aligned}&\tilde{\mathbb {P}}_{0}\left[ V_{u}\in A\right] \\&\qquad \;\, =\tilde{\mathbb {P}}_{0}\left[ V_{u}\in A|\mu _{3}\left( \left[ 0,t\right] \times \Gamma \right) =0\right] \\&\quad \overset{(9.4),(9.6)}{=} \mathbb {P}_{0}\left[ \left( \tilde{L}_{L}^{\tau \left( u\right) }\right) _{l\in \left\{ 1,\ldots ,,L\right\} }|\tilde{L}_{L}^{\tau \left( u\right) }=0\right] \\&\qquad \overset{(9.7)}{=} \mathbb {Q}_{2u}^{0}\left[ \left( \frac{1}{2}X_{l}\right) _{l\in \left\{ 0,\ldots ,L-1\right\} }\in A|X_{L-1}=0\right] \\&\qquad \overset{(7.23)}{=} \mathbb {Q}_{2u\rightarrow 0}^{0,L-1}\left[ \left( \frac{1}{2}X_{l}\right) _{l\in \left\{ 0,\ldots ,L-1\right\} }\in A\right] . \end{aligned}$$

Plugging this into (9.5) gives the claim. $\square$

The same construction of $Y_{t}$ from the Poisson point processes $\mu _{1},\mu _{2}$ and $\mu _{3}$ can be used to prove Lemma 7.9, which gives a control on the law of $L_{1}^{D_{t}}$ conditioned on $L_{L}^{D_{t}}=0$.

Proof of Lemma 7.9

We will first show that

$$\begin{aligned}&\text{ the } \tilde{\mathbb {P}}_{0}\left[ \cdot |L_{L}^{D_{t}}=0\right] \text{-law } \text{ of } \frac{L}{L-1}L_{1}^{D_{t}}\, \text{ is } \text{ that } \text{ of } \text{ a } \text{ sum } \text{ of } t\nonumber \\&\quad \text{ independent } \text{ standard } \text{ exponential } \text{ random } \text{ variables }. \end{aligned}$$

(9.8)

In the notation of the proof of Lemma 7.8: Since $L_{1}^{D_{t}}=S_{t}$ and $\left\{ L_{L}^{D_{t}}=0\right\} =\left\{ \mu _{3}\left( \left[ 0,S_{t}\right] \times \Gamma \right) =0\right\}$ we are interested in the law of $S_{t}$ given $\left\{ \mu _{3}\left( \left[ 0,S_{t}\right] \times \Gamma \right) =0\right\}$. Since $\mu _{3}$ is independent of $S_{t}$ we have that

$$\begin{aligned} \tilde{\mathbb {P}}_{0}\left[ S_{t}=ds,\mu _{3}\left( \left[ 0,S_{t}\right] \times \Gamma \right) =0\right] =\tilde{\mathbb {P}}_{0}\left[ S_{t}=ds,\tilde{\mathbb {P}}\left[ \mu _{3}\left( \left[ 0,s\right] \times \Gamma \right) =0\right] \right] . \end{aligned}$$

The intensity of $\mu _{3}$ is $\left( 2\lambda \right) \otimes \frac{1}{2}\tilde{\mathbb {P}}_{2}\left[ \cdot ,H_{L}<H_{1}\right]$ (recall (9.2)), so that

$$\begin{aligned}&\tilde{\mathbb {P}}_{0}\left[ \mu _{3}\left( \left[ 0,s\right] \times \Gamma \right) =0\right] =e^{-s\tilde{\mathbb {P}}_{2}\left[ H_{L}<H_{1}\right] }=e^{-\frac{s}{L-1}}, \text{ and }\\&\tilde{\mathbb {P}}_{0}\left[ S_{t}=ds,\mu _{3}\left( \left[ 0,S_{t}\right] \times \Gamma \right) =0\right] =e^{-\frac{s}{L-1}}\tilde{\mathbb {P}}_{0}\left[ S_{t}=ds\right] . \end{aligned}$$

The $\tilde{\mathbb {P}}_{0}$-law of $S_{t}$ is the gamma distribution with shape t and scale 1. Thus

$$\begin{aligned} \tilde{\mathbb {P}}_{0}\left[ S_{t}=ds,\mu _{3}\left( \left[ 0,S_{t}\right] \times \Gamma \right) =0\right] =e^{-\frac{s}{L-1}}\frac{s^{t-1}e^{-s}}{\left( t-1\right) !}=e^{-\left( \frac{1}{L-1}+1\right) s}\frac{s^{t-1}}{\left( t-1\right) !}, \end{aligned}$$

so that the $\tilde{\mathbb {P}}_{0}\left[ \cdot |\mu _{3}\left( \left[ 0,S_{t}\right] \times \Gamma \right) =0\right]$-law of $S_{t}$ is the gamma distribution with shape t and scale $\left( 1+1/\left( L-1\right) \right) ^{-1}=\left( L-1\right) /L$. This proves (9.8).

Now the central limit theorem shows that

$$\begin{aligned} 0<c\le \tilde{\mathbb {P}}_{0}\left[ \left| \frac{L}{L-1}L_{1}^{D_{t}}-t\right| \le \sqrt{t}|L_{L}^{D_{t}}=0\right] , \end{aligned}$$

(9.9)

and a standard large deviation bound shows that for $x\ge 0$,

$$\begin{aligned} \tilde{\mathbb {P}}_{0}\left[ \left| \frac{L}{L-1}L_{1}^{D_{t}}-t\right| \ge x\sqrt{t}|L_{L}^{D_{t}}=0\right] \le e^{-cx^{2}}. \end{aligned}$$

(9.10)

From the assumption $t\le 10L^{2}$ we see that $\frac{L}{L-1}$ can be replaced by $\left( \frac{L}{L-1}\right) ^{2}$, and then (7.41) follows from (9.9) and (7.42) follows from (9.10). $\square$

It remains to prove Lemma 1, giving a large deviation bound for the number of traversals $\tilde{T}_{l}^{t}$ (recall (7.43)) given the continuous local times $L_{l}^{D_{t}}$. For this we will need the following computation of the conditional distribution of $\tilde{T}_{l}^{t}$ (which can be seen as a special case of the results of [14, Section 4]). To prove it we use the following fact about the modified Bessel function of the first kind $I_{1}\left( \cdot \right)$:

$$\begin{aligned} \sum _{m\ge 1}\frac{z^{m}}{m!\left( m-1\right) !}=\sqrt{z}I_{1}\left( 2\sqrt{z}\right) \quad \text{ for } \text{ all } z\in \mathbb {R}. \end{aligned}$$

(9.11)

Lemma 9.1

For all $u_{0},u_{1},u_{2},\ldots ,u_{L}\in [0,\infty )$ such that $u_{i}=0\implies u_{i+1}=0$, and any $l\in \left\{ 1,\ldots ,L-1\right\}$ such that $u_{l+1}>0$ we have for $m\in \left\{ 1,2,\ldots \right\}$

$$\begin{aligned} \mathbb {\tilde{P}}_{0}\left[ \tilde{T}_{l}^{t}=m|L_{l}^{D_{t}}=u_{l},l=0,\ldots ,L\right] =\frac{\left( u_{l}u_{l+1}\right) ^{m}/\left( m!\cdot \left( m-1\right) !\right) }{\sqrt{u_{l}u_{l+1}}I_{1}\left( 2\sqrt{u_{l}u_{l+1}}\right) }. \end{aligned}$$

Proof

The law of $T_{1}$ under $\mathbb {G}_{a}$ can be written down explicitly as

$$\begin{aligned} \mathbb {G}_{a}\left[ T_{1}=b\right] ={a+b-1 \atopwithdelims ()a-1}\left( \frac{1}{2}\right) ^{a+b}\quad \text{ for } a\in \left\{ 1,2,\ldots \right\} ,b\in \left\{ 0,1,2,\ldots \right\} , \end{aligned}$$

since there are ${a+b-1 \atopwithdelims ()a-1}$ ways to write b as a sum of a non-negative integers, and since the probability that a geometric random variable with support $\left\{ 0,1,\ldots \right\}$ and mean 1 takes on the value k is $\left( \frac{1}{2}\right) ^{k+1}$. By Lemma 7.10 we therefore have for all $t=t_{0},t_{1},t_{2},\ldots ,t_{L-1}\in \left\{ 0,1,2,\ldots \right\}$ such that $t_{i}=0\implies t_{i+1}=0$

$$\begin{aligned} \mathbb {\tilde{P}}_{0}\left[ \tilde{T}_{i}^{t}=t_{i},i=0,\ldots ,L-1\right] =\prod _{i\in \left\{ 1,\ldots ,L-1\right\} :t_{i-1}>0}{t_{i-1}+t_{i}-1 \atopwithdelims ()t_{i-1}-1}\left( \frac{1}{2}\right) ^{t_{i-1}+t_{i}}. \end{aligned}$$

Conditioned on the number of visits to each vertex the total holding times at the vertices are independent and gamma distributed, so we have for such $t_{i}$ and any $u_{0},u_{1},\ldots ,u_{L}\in [0,\infty )$ such that $t_{l-1}=0\iff u_{l}=0$ that

$$\begin{aligned}&\tilde{\mathbb {P}}_{0}\left[ \tilde{T}_{l}^{t}=t_{l},l=1,\ldots ,L-1,L_{l}^{D_{t}}=u_{l},l=0,\ldots ,L\right] \\&\quad =\left\{ {\displaystyle \prod _{i\in \left\{ 1,\ldots ,L-1\right\} :t_{i-1}>0}}{t_{i-1}+t_{i}-1 \atopwithdelims ()t_{i-1}-1}\left( \frac{1}{2}\right) ^{t_{i-1}+t_{i}}\right\} \\&\qquad \times \left\{ \left( \frac{e^{-u_{1}}u_{1}^{t_{0}-1}}{\left( t_{0}-1\right) !}\right) \left( {\displaystyle \prod _{i\in \left\{ 1,\ldots ,L-1\right\} :t_{i-1}>0}}\frac{e^{-2u_{i}}\left( 2u_{i}\right) ^{t_{i-1}+t_{i}}}{u_{i}\left( t_{i-1}+t_{i}-1\right) !}\right) \left( \frac{e^{-u_{L}}u_{L}^{t_{L-1}-1}}{\left( t_{L-1}-1\right) !}\right) \right\} , \end{aligned}$$

where the quantity in the last parenthesis is interpreted as 1 if $t_{l-1}=0$ or $u_{L}=0$. Exploiting two cancellations the right-hand side equals

$$\begin{aligned}&\left\{ {\displaystyle \prod _{i\in \left\{ 1,\ldots ,L-1\right\} :t_{i-1}>0}}\frac{1}{\left( t_{i-1}-1\right) !t_{i}!}\right\} \\&\quad \times \left\{ \left( \frac{e^{-u_{1}}u_{1}^{t_{0}-1}}{\left( t_{0}-1\right) !}\right) \left( {\displaystyle \prod _{i\in \left\{ 1,\ldots ,L-1\right\} :t_{i-1}>0}}\frac{e^{-2u_{i}}u_{i}^{t_{i-1}+t_{i}}}{u_{i}}\right) \left( \frac{e^{-u_{L}}u_{L}^{t_{L-1}-1}}{\left( t_{L-1}-1\right) !}\right) \right\} . \end{aligned}$$

Considering only the terms that depend on $t_{l}$ we have that if $u_{0},u_{1},\ldots ,u_{l+1}>0$

$$\begin{aligned} \tilde{\mathbb {P}}_{0}\left[ \tilde{T}_{l}^{t}=m|L_{l}^{D_{t}}=u_{l},l=0,\ldots ,L\right] =\frac{1}{\tilde{Z}}\frac{\left( u_{l}u_{l+1}\right) ^{m}}{\left( m-1\right) !m!},m\ge 1,l\in \left\{ 1,\ldots ,L-1\right\} pg{\!}, \end{aligned}$$

for a normalizing constant $\tilde{Z}$ depending only on $t,u_{0},\ldots ,u_{L}$. Using (9.11) we can identify the constant as

$$\begin{aligned} \tilde{Z}=\sum _{m\ge 1}\frac{\left( u_{l}u_{l+1}\right) ^{m}}{\left( m-1\right) !m!}=\sqrt{u_{l}u_{l+1}}I_{1}\left( 2\sqrt{u_{l}u_{l+1}}\right) . \end{aligned}$$

$\square$

We now prove the large deviation result Lemma 7.12 for the traversal process $\tilde{T}_{l}^{t}$ conditioned on $L_{l}^{D_{t}},l=0,\ldots ,L$.

Proof of Lemma 7.12

Denote $\tilde{\mathbb {P}}_{0}\left[ \cdot |\sigma \left( L_{l}^{D_{t}}:l=0,\ldots ,L\right) \right]$ by $\tilde{\mathbb {Q}}$. By Lemma 1,

$$\begin{aligned} \tilde{\mathbb {Q}}\left[ \exp \left( \lambda \tilde{T}_{l}^{t}\right) \right] =\sum _{m\ge 1}\frac{\left( e^{\lambda }\mu ^{2}\right) ^{m}/\left( m!\cdot \left( m-1\right) !\right) }{\mu I_{1}\left( 2\mu \right) }\overset{(9.11)}{=}\frac{e^{\lambda /2}\mu I_{1}\left( 2e^{\lambda /2}\mu \right) }{\mu I_{1}\left( 2\mu \right) } \text{ for } \lambda \in \mathbb {R}. \end{aligned}$$

Thus for all $\lambda >0$

$$\begin{aligned} \tilde{\mathbb {Q}}\left[ \tilde{T}_{l}^{t}\ge \mu +\theta \right] \le e^{\lambda /2}\frac{I_{1}\left( 2e^{\lambda /2}\mu \right) }{I_{1}\left( 2\mu \right) }\exp \left( -\lambda \left( \mu +\theta \right) \right) . \end{aligned}$$

Using the standard estimate $I_{1}\left( z\right) =\frac{e^{z}}{\sqrt{2\pi z}}\left( 1+O\left( z^{-1}\right) \right)$ we have that

$$\begin{aligned} I_{1}\left( 2e^{\lambda /2}\mu \right) /I_{1}\left( 2\mu \right) \le ce^{\lambda /4}e^{2\left( e^{\lambda /2}-1\right) \mu }, \end{aligned}$$

so that for all $\lambda >0$

$$\begin{aligned} \tilde{\mathbb {Q}}\left[ \tilde{T}_{l}^{t}\ge \mu +\theta \right] \le ce^{c\lambda }\exp \left( 2\left\{ e^{\lambda /2}-1\right\} \mu -\lambda \left\{ \mu +\theta \right\} \right) \le ce^{c\lambda }\exp \left( c\lambda ^{2}\mu -\lambda \theta \right) . \end{aligned}$$

Setting $\lambda =c\theta /\mu$ for a small enough c the right-hand side is bounded above by $ce^{c\theta /\mu -c\theta ^{2}/\mu }$, giving one half of (7.45). By estimating $\tilde{\mathbb {Q}}\left[ \exp \left( -\lambda \tilde{T}_{l}^{t}\right) \right]$ one can similarly show that $\tilde{\mathbb {Q}}\left[ \tilde{T}_{l}^{t}\le \mu -\theta \right] \le ce^{c\theta /\mu -c\theta ^{2}/\mu }$, giving the other half. $\square$