\section{Introduction and notation}
\subsection{Introduction}
Feller (in Section XIII.7 of \cite{f1}) showed that if $p^{(k)}_n$ is the
probability that there is no run of heads of length $k$  or
more in $n$ tosses of a fair coin, then, for a suitable positive constant $c_k$
$$
p^{(k)}_n\sim c_ks_k^{n+1},
$$
where $s_k$ is the largest real root in (0,1) of the equation
\begin{equation}\label{coin}
x^k-\sum_{j=0}^{k-1}2^{-(j+1)}x^{k-1-j}=0.
\end{equation}

More generally, if the probability of a head is $p=1-q$, then the same
asympotic formula is valid, with equation (\ref{coin}) modified to
become
\begin{equation}\label{coin2}
x^k-q\sum_{j=0}^{k-1}p^{j}x^{k-1-j}=0,
\end{equation}
and $c_k=\frac{s_k-p}{q((k+1)s_k-k)}$.

The continuous-time analogue of this question is to seek the asymptotic
behaviour of the probability that $Y$, a Poisson process with rate
$r$, has no inter-jump time exceeding one unit by time $T$.
It follows, essentially  from Theorem \ref{rec1} that, denoting  by $\tau_Y$ the first time that
$Y$ waits to jump longer than one unit of time,
\begin{equation}\label{coin3}
{\mathbb P}(\tau_Y>t)\sim c_r e^{-{\phi_r} t},
\end{equation}
for a suitable constant $c_r$, where ${\phi_r}=1$ if $r=1$ and otherwise
${\phi_r}$ is the root (other than $r$ itself) of the equation
\begin{equation}\label{coin4}
xe^{-x}=r e^{-r}.
\end{equation}

A natural extension is then to seek the tail behaviour of the distribution of $\tau\equiv \tau_X$,
the first time that a Markov chain, $X$, waits longer than one unit of
time at a distinguished state,
0. In general, there has also been much interest (see \cite{JW}, \cite{JW2}, \cite{JW3}, \cite{Ber}, \cite{Ber2},
 \cite{Ber3}, \cite{Ferr}, \cite{Kes}, \cite{JR}, \cite{JR2}, \cite{JR3}, \cite{JR4}, \cite{Kyp})
in conditioning an evanescent
Markov process $X$ on
its survival time being increasingly large and in seeing whether a weak limit exists.

\goodbreak
\subsection{Notation}
We consider a continuous-time Markov chain $X$
on a countable state\-space
$S$, with a distinguished state $\partial$. We denote $S\setminus
\{\partial\}$ by $C$. For convenience, and without loss of generality, we assume henceforth that
$S={\mathbb Z}^+$ or $S=\{0,\ldots,n\}$ and
$\partial=0$ so that $C={\mathbb N}$ or $C=\{1,\ldots, n\}$.

We assume that $X$ is irreducible, and non-explosive. We denote the
transition semi-group of $X$ by $\{P(t); {t\geq 0}\}$ and its $Q$-matrix by $Q=(q_{ij})$ and set $q_i=-q_{ii}$. We define
the process $\tilde X$ as $X$ killed on first hitting 0 and we shall usually
assume that $\tilde X$ is also irreducible on $C$. We denote the substochastic
semigroup for $\tilde X$ by $(\tilde P(t))_{t\geq 0}$.
We first define the return and departure epochs as follows:
$$S_0=\inf\{t\geq 0:\; X_t=0\},\hbox{ }T_0=\inf\{t\geq 0:\; X_t\neq 0\}$$
and then for $n\geq 1$:
$$T_n=\inf\{t\geq S_n:\; X_t\neq 0\},$$
and
$$S_n=\inf\{t\geq T_{n-1}:\; X_t= 0\},$$
then we define the successive holding and return times $H^n_{n\geq 0}$
and $R^n_{n\geq 0}$ by
$$H^0=T_0\hbox{ and }R^0=S_0$$
and
$$H^n=T_n-S_n\hbox{ and }R^n=S_{n+1}-T_n\hbox{ for }n\geq 1.$$
Then we define the
current holding time as follows:
$${H}_t=t-S_n\hbox{ if }S_n\leq t<T_n\hbox{ for some }n.$$
It will be convenient in what follows to define
$${H}_t=\emptyset\hbox{ if }X_t\neq 0.$$

We denote the first time that $X$ waits in 0 for time 1 by $\tau$, i.e.
$$\tau=\inf\{t:\; H_t\geq 1\},$$
and
denote the process $X$ killed at time $\tau$ by $\hat X$. We denote the
statespace augmented by the current holding time in 0 by
$\hat S\buildrel {\mathrm{def}}\over = C\cup\{\{0\}\times[0,1)\}$. By a slight abuse of notation, we denote the (substochastic)
Markov chain $(\hat X_t, H_t)$ on the statespace
$\hat S$ by $\hat X$ also\footnote{Note that if $\hat X_t\neq
0$ then $H_t=\emptyset$ so $(\hat X_t,H_t)= \hat X_t$.}.
The associated semigroup is
denoted $(\hat P(t))_{t\geq 0}$.
Throughout the rest of the paper we denote by ${\mathbb P}_i$ the probability on
Skorokhod
pathspace $D(S,[0,\infty))$, conditional on $\hat X_0=i$, and the
corresponding filtration by $({\mathcal F}_t)_{t\geq 0}$.
Finally, we denote a typical hitting time of 0 from state $i$ by $\tau^{(i)}_0$
and its density by $\rho_i$. We denote the density of a typical return time,   ${R}^1$, by $\rho$.

\subsection{Convergence/decay parameters for evanescent chains}
\bigskip
We recall (see for example \cite{JR}) that, if $X^*$ is a
Markov chain on $C$, with substochastic transition semigroup $P^*$ and $Q$-
matrix $Q^*=(q^*_{ij})_{(i,j)\in C\times C}$, then $X^*$ is said to be evanescent if it is irreducible
and dies with probability one. In that case,
we define
$$
\alpha_{X^*}=\alpha=\inf\{\lambda\geq 0:\,\int_0^\infty P^*_{ij}(t)e^{\lambda t}dt
=\infty\}
$$
for any $i,j\in C$, and (see, for example, Seneta and Vere-Jones
\cite{Se2})  $X^*$ is classified as $\alpha$-recurrent or $\alpha$-transient
depending on whether $\int_0^\infty P^*_{ij}(t)e^{\alpha t}dt
=\infty$ or is finite. Moreover,
$X^*$ is $\alpha$-recurrent if and only if $\int_0^\infty f^*_{ii}(t)e^{\alpha t}dt
=1,$
where $f^*_{ii}$ is the defective density of the first return time to $i$
(starting in $i$).

In the $\alpha$-recurrent case, $X^*$ is
$\alpha$-positive recurrent if $$\int_0^\infty tf^*_{ii}(t)e^{\alpha
t}dt<\infty,$$
otherwise $X^*$ is
$\alpha$-null recurrent.
Defining $q^*_i=-q^*_{ii}$,
it is easy to see that $\alpha<q^*_i$ for all $i\in {\mathbb N}$ and hence
$$0\leq\alpha\leq \inf_iq^*_i.$$

Thus $\alpha$ measures the rate of
decay of transition probabilities (in $C$).
There is a second decay parameter --- $\mu^*$, which
measures the rate of dying.

We define $\tau^*$ as the death time of $X^*$ and we define
$s^*_i(t)=\sum_jP^*_{ij}(t)={\mathbb P}_i(\tau^*>t)$ and set
$$\mu^*=\inf\{\lambda:\,\int_0^\infty s^*_{i}(t)e^{\lambda t}dt
=\infty\}.$$

Notice that $\mu^*$ is independent of $i$ by the usual irreducibility
argument, moreover, since $1\geq s^*_i(t)\geq P^*_{ii}(t)$ it follows that
$$0\leq \mu^*\leq \alpha^*.$$

Note that in our current setting, we shall take $X^*=\tilde X$ and write
$\tau^*=\tau_0$, the first hitting time of 0. We shall denote the
rate of hitting 0, which is the death rate for $X^*$, by $\mu^C$
and $\alpha^*$ by $\alpha^C$ and the survival probabilities for $\tilde X$ as
$s^\cstate$, so that $s^\cstate_i(t)={\mathbb P}_i(\tau_0>t)$.

\subsection{Doob $h$-transforms}
Recall (see, for example, III.49 of Williams \cite {R+W}) that
we may form the $h$-transform of a substochastic Markovian semigroup on
$S$,
$(P(t))_{t\geq 0}$, if $h:S\rightarrow{\mathbb R}^+$ is
$P$-superharmonic (i.e. $[P(t)h](x)\leq h(x)$ for all $x\in S$ and for all $t\geq 0$).
The $h$-transform of $P$, $P^h$, is specified by its transition kernel which
is given by
$$P^h(x,dy;t)\buildrel {\mathrm{def}}\over = \frac{h(y)}{h(x)}P(x,dy;t),$$
so that if we consider the corresponding substochastic measures on path-space,
${\mathbb P}_x$ and ${\mathbb P}^h_x$ (conditional on $X_0=x$) then
$$\frac{d{\mathbb P}^h_x}{d{\mathbb P}_x}\big|_{{\mathcal F}_t}=h(X_t)
$$
and $P^h$ forms another
substochastic Markovian semigroup.
If $h$ is actually {\em space-time} $P$-superharmonic then appropriate changes
need to be made to these definitions. In particular, if $h(x,t)=e^{\phi
t}h_x$ then
$$
\frac{d{\mathbb P}^h_x}{d{\mathbb P}_x}\big|_{{\mathcal F}_t}=e^{\phi
t}h_{X_t}.
$$
As shown in \cite{JR}, in general, when a weak limit or a vague limit
exists for the problem of interest, it must be a Doob-$h$-transform
of the original process, with the state augmented
by the current waiting time in state 0 in the case we study here.

\subsection{Main results}
Denoting by $\hat P(t)$ the substochastic transition semigroup for $\hat X$, we define
$$s_i(t)\buildrel {\mathrm{def}}\over = {\mathbb P}_i(\tau>t)=\hat P(i,\hat S;t)\hbox{ for }i\in\hat S.$$
Our first result is
\begin{theorem}\label{trans}
Suppose that $X$ is transient. Denote ${\mathbb P}_i(X\hbox{ never hits 0})$ by
$\beta_i$ and
define $\Delta=\sum_{j\in C}\frac{q_{0,j}\beta_j}{{q_0}}.$
Set
\begin{equation}\label{trans1}
p_{(0,0)}\buildrel {\mathrm{def}}\over = p_0=\frac{(1-e^{-{q_0}})\Delta}{e^{-{q_0}}+(1-e^{-
{q_0}})\Delta},
\end{equation}

\begin{equation}\label{trans2}
p_{(0,u)}=\frac{1-e^{-{q_0} (1-u)}}{1-e^{-{q_0}}}p_0,
\end{equation}
and
\begin{equation}\label{trans3}
p_i=\beta_i+(1-\beta_i)p_0\hbox{ for }i\inC.
\end{equation}
Then
\begin{equation}\label{translimit}
s_i(t)\buildrel {t\rightarrow \infty}\over\longrightarrow p_i\hbox{ for all }i\in\hat S.
\end{equation}
Hence, if we condition $X$ on $\tau=\infty$ we obtain a new Markov
process, $X^\infty$, on $\hat S$  with honest semigroup $P^\infty$ given
by
\begin{equation}\label{cond1}
P^\infty_{i,j}(t)=\frac{p_j}{p_i}\hat P_{i,j}(t)\hbox{ for }j\in C
\end{equation}
and
\begin{equation}\label{cond2}
P^\infty_i((0,du);t)=\frac{p_{(0,u)}}{p_i}\hat P_i((0,du);t),
\end{equation}
so that
$X^\infty$ looks like a Markov chain with $Q$-matrix given by
$q^\infty_{i,j}=\frac{p_j}{p_i}q_{i,j}$ on $C$, whilst $X^\infty$ has a
holding time in 0 with density $d$ given by
$$d(t)=\frac{e^{-{q_0} t}}{\int_0^1 e^{-{q_0} s}ds}1_{(t<1)}$$
and a tine-homogeneous jump probability out of state 0 to state $j$ of
$\frac{q_{0,j}p_j}{{q_0} p_0}$.
\end{theorem}

In the case where $X$ is recurrent, it is clear that
$s_i(t)\buildrel {t\rightarrow \infty}\over\longrightarrow 0$ for each $i\in\hat S$.

Now let $W\buildrel {\mathrm{def}}\over = {H}^1+{R}^1$ (so
that $W$ is the first return time of $X$ to 0 from 0) and let $g$ be the
(defective)
density of $W 1_{({H}^1<1)}$ on $(0,\infty)$.
Our first result under these conditions is as follows. It is a
generalisation to our more complex setting of Seneta and Vere-Jones'
result in the $\alpha$-positive case.

\begin{theorem}\label{rec1}
Let
$$I(\lambda)\buildrel {\mathrm{def}}\over = \int_0^\infty e^{\lambda t}g(t)dt={\mathbb E} e^{\lambda W}1_{({H}^1<1)},$$
then if
\begin{equation}\label{condn2}
\hbox{there exists a $\phi$ such that }I(\phi)=1,\hbox{ and }I'(\phi-
)<\infty,
\end{equation}
then for each $i\in\hat S$,
$$e^{\phi t}s_i(t)\buildrel {t\rightarrow \infty}\over\longrightarrow p_i>0,$$
where the function $p$ is given by
\begin{equation}\label{p1}
p_{(0,0)}=\frac{e^{\phi-{q_0}}}{\phi I'(\phi-)}\buildrel {\mathrm{def}}\over = \kappa,
\end{equation}
\begin{equation}\label{p2}
p_{(0,u)}=\frac{\int_0^{1-u}e^{(\phi-{q_0})s}ds}{\int_0^{1}e^{(\phi-
{q_0})s}ds}\kappa,
\end{equation}
and, for $i\neq 0$,
\begin{equation}\label{p3}
p_i=F_{i,0}(\phi)\kappa,
\end{equation}
where
\begin{equation}\label{F}
F_{i,0}(\lambda)\buildrel {\mathrm{def}}\over = {\mathbb E} e^{\lambda \tau^{(i)}_0}=\int_0^\infty e^{\lambda t}\rho_i(t)dt.
\end{equation}
\end{theorem}

The following simple condition ensures that condition
(\ref{condn2}) holds.
\begin{lemma}\label{rec2}
Suppose that $\tilde X$ is $\alpha$-recurrent, and both $N_0\buildrel {\mathrm{def}}\over = \{i:\,
q_{i,0}>0\}$ and $N^*_0\buildrel {\mathrm{def}}\over = \{i:\, q_{0,i}>0\}$ are finite, then (\ref{condn2}) holds.
\end{lemma}

\begin{corollary}\label{cor1}
Let $X^T$ denote the chain on $\hat S$ obtained by conditioning $\hat X$
on the event $(\tau >T)$, then, if condition (\ref{condn2}) holds,
for each $s>0$, the restriction of the law of $X^T$ to
${\mathcal F}_s$ converges weakly as $T\rightarrow \infty$ to that of $X^\infty$ restricted to ${\mathcal F}_s$, where
the transition semigroup of $X^\infty$ is given by equations (\ref{cond1})
and  (\ref{cond2}).
\end{corollary}

In the case where $I(\phi)<1$ or $I'(\phi-)=\infty$, Theorems \ref{subexp1},
\ref{subexp2} and \ref{subexp3} (may)  apply,
giving some sufficient conditions for weak or vague convergence to take
place.
In Theorem \ref{subexp4} and Corollary \ref{bd2}, we give an application to
the case of a recurrent birth and death process conditioned not to wait
too long in state 0.
\goodbreak

\section{Proof of the transient and $\alpha$-positive cases}
To prove Theorem \ref{trans} is straightforward.

{\em Proof of Theorem \ref{trans}}
It is trivial to establish the equations
\begin{equation}
p_i=\beta_i+(1-\beta_i)p_0
\end{equation}
and
\begin{equation}
p_0=(1-e^{-{q_0}})\sum_{j\in C}\frac{q_{0,j}}{{q_0}}p_i.
\end{equation}
Equations (\ref{trans1})-(\ref{trans3}) follow immediately. Then the
conditioning result follows straightforwardly
\hfill$\square$

\begin{example}
\label{eg1}
We take a transient nearest-neighbour random walk with
reflection at 0 and with up-jump rate of $b$ and down-jump rate of $d$. Note that $1-\beta$
is the minimal positive solution to
$P(t)h=h$ with $h(0)=1$, and that $1-\beta_i=(\frac{d}{b})^i$.
\end{example}

The main tool in the proof of Theorem \ref{rec1} is the Renewal Theorem.

{\em Proof of Theorem \ref{rec1}}
Recall that state $(0,u)$ denotes that the killed chain is at 0 and
its current holding time is $u$.
First note that $s_{(0,0)}$ satisfies the renewal equation
\begin{equation}\label{renew1}
s_{(0,0)}(t)=\bigl(1-\int_0^\infty g(u)du\bigr)1_{(t<1)}+\int_t^\infty g(u)du+\int_0^t g(u)s_{(0,0)}(t-u)du.
\end{equation}
If we define
$$f(t)=e^{\phi t}s_{(0,0)}(t),$$
it follows immediately from (\ref{renew1}) that
\begin{equation}\label{renew2}
f(t)=e^{\phi t}\bigl((1-\int_0^\infty g(u)du)1_{(t<1)}+\int_t^\infty g(u)du\bigr)+\int_0^t \tilde g(u)f(t-u)du,
\end{equation}
where
$\tilde g(t)\buildrel {\mathrm{def}}\over = e^{\phi t}g(t).$ Now, it is easy to check that the conditions of Feller's alternative
formulation of the Renewal Theorem (see XI.1 of \cite{Feller}, p.363) are satisfied, so we conclude that
\begin{equation}\label{lim1}
f(t)\buildrel {t\rightarrow \infty}\over\longrightarrow \mu^{-1}\int_0^\infty e^{\phi t}((1-\int_0^\infty g(u)du)1_{(t<1)})+\int_t^\infty g(u)du)dt,
\end{equation}
where
$$
\mu=\int_0^\infty t\tilde {g}(t)dt=I'(\phi-).
$$
It is trivial to establish, by changing the order of integration, that
$$
\int_0^\infty e^{\phi t}\int_t^\infty g(u)dudt=\int_0^\infty g(u)\int_0^u
e^{\phi t}dt du=\frac{I(\phi)-\int_0^\infty g(u)du}{\phi}=\frac{1-
\int_0^1 {q_0} e^{-{q_0} u}du}{\phi}=\frac{e^{-{q_0}}}{\phi},
$$
and hence (\ref{p1}) follows.

To establish (\ref{p3}), notice that (by conditioning on the time of the first hit
of 0),
$$
s_i(t)=\int_t^\infty \rho_i(u)du+\int_0^t \rho_i(u)s_{(0,0)}(t-u)du,$$
and so, denoting $e^{\phi t}s_i(t)$ by $f_i(t)$, we obtain
$$
f_i(t)=e^{\phi t}\int_t^\infty \rho_i(u)du +\int_0^t \tilde {\rho}_i(u)f(t-u)du,$$
where $\tilde {\rho}_i(t)\buildrel {\mathrm{def}}\over = e^{\phi t}\rho_i(t)$. Now $f$ is continuous and
converges to $\kappa$ so, by the Dominated Convergence Theorem,
$$
\int_0^t \tilde {\rho}_i(u)f(t-u)du\buildrel {t\rightarrow \infty}\over\longrightarrow \int_0^\infty \kappa \tilde {\rho}_i(u)du=\kappa
F_{i,0}(\phi).
$$
Moreover, since $F_{i,0}(\phi)=\int_0^\infty \tilde {\rho}_i(u)du<\infty$, it follows that
$e^{\phi t}\int_t^\infty \rho_i(u)du \leq \int_t^\infty \tilde {\rho}_i(u)du\buildrel {t\rightarrow \infty}\over\longrightarrow
0$, and hence
$$f_i(t)\buildrel {t\rightarrow \infty}\over\longrightarrow \kappa F_{i,0}(\phi)$$
as required.

To establish (\ref{p2}), observe that
$$s_{(0,u)}(t)=e^{-{q_0} t}1_{(t<1-u)}+\int_0^t\int_0^{(1-u)\wedge v}{q_0} e^{-{q_0} v}\rho(w-
v)s_{(0,0)}(t-w)dvdw$$
and hence
$$f_{(0,u)}(t)\buildrel {\mathrm{def}}\over = e^{\phi t}s_{(0,u)}(t)=e^{(\phi-{q_0}) t}1_{(t<1-u)}+\int_{w=0}^t\int_{v=0}^{(1-u)\wedge w}{q_0} e^{(\phi-{q_0}) v}\tilde {\rho}(w-
v)f(t-w)dvdw,$$
and hence, by the Dominated Convergence Theorem,
\begin{displaymath}
\begin{array}{rcl}
f_{(0,u)}(t)&\buildrel {t\rightarrow \infty}\over\longrightarrow &\kappa \int_0^\infty \int_0^{w\wedge (1-u)}{q_0} e^{(\phi-
{q_0}) v}\tilde {\rho}(w-v)dv\\
&=&\kappa\int_0^\infty\tilde {\rho}(t)dt\int_0^{1-u}{q_0}
e^{(\phi-{q_0}) v}dv=\kappa
\frac{\int_0^{1-u}e^{(\phi-{q_0})s}ds}{\int_0^{1}e^{(\phi-
{q_0})s}ds},\\
\end{array}
\end{displaymath}
as required \hfill$\square$
\begin{remark}
Note that the case mentioned in the introduction, where $Y$ is a
Poisson($r$) process and we let $\tau_Y$ be the first time that an
interjump time is one or larger, can be addressed using the proof of Theorem
\ref{rec1}. In this case, if we consider that the chain \lq\lq returns
directly to 0" at each jump time $Y$ then
$$
I(\lambda)=\int_0^1re^{(\lambda-r)t}dt,
$$
and so $\phi$ satisfies $r\frac{e^{\phi-r}-1}{\phi-r}=1$ which
establishes (\ref{coin4}), and
$$
e^{\phi t}{\mathbb P}(\tau>t)\buildrel {t\rightarrow \infty}\over\longrightarrow \frac{e^{\phi-r}}{\phi I'(\phi-)}=\frac{\phi -
r}{r(\phi -1)},
$$
for $r\neq 1$.
The case $r=1$ gives $\phi=1$ and $c_1=2$.
\end{remark}
Now we give the

{\em Proof of Lemma \ref{rec2}}
It follows from Theorem 3.3.2
of \cite{JR3} that if $N_0$ is finite then $\alpha^C=\mu^C$.
Now since $\tilde X$ is $\alpha$-recurrent it follows that
$$
\int_0^\infty e^{\lambda t}\tilde P_{ii}(t)dt<\infty\hbox{ iff
}\lambda<\alpha^C.
$$
Since $s^\cstate_i(t)\geq \tilde P_{ii}(t)$, it follows that
$$
\int_0^\infty e^{\lambda t}s^\cstate_{i}(t)dt=\infty\hbox{ if
}\lambda\geq\alpha^C.
$$
Conversely,
since $\alpha^C=\mu^C$
we see that
$$
\int_0^\infty e^{\lambda t}s^\cstate_{i}(t)dt<\infty\hbox{ if
}\lambda<\alpha^C,
$$
and so we conclude that
$$
\int_0^\infty e^{\lambda t}s^\cstate_{i}(t)dt<\infty\hbox{ iff
}\lambda<\alpha^C.
$$
Now
$$I(\lambda)=\int_0^\infty e^{\lambda t}g(t)dt=\int_0^1e^{(\lambda-
{q_0})t}dt\bigl(\sum_{i\in N^*_0}q_{0,i}F_{i,0}(\lambda)\bigr)=\int_0^1e^{(\lambda-
{q_0})t}dt\bigl(\sum_{i\in N^*_0}q_{0,i}(\frac{F_{i,0}(\lambda)-
1}{\lambda})\bigr).
$$
Now $F_{i,0}(\lambda)<\infty$ if and only if $\lambda<\alpha^C$
and so, since $N^*_0$ is finite by assumption,  $I(\lambda)<\infty$ iff $\lambda<\alpha^C$. It now follows
trivially that $\phi<\alpha^C$ and that (\ref{condn2}) is satisfied
\phantom{p}\hfill$\square$

Now we give the

{\em Proof of Corollary \ref{cor1}} This follows immediately from
Theorem \ref{rec1} and Theorem 4.1.1 of
\cite{JR} provided that we can show that $h$, given by $h:(i,t)\mapsto
e^{\phi t}p_i$, is $\hat P$-harmonic.
This is easy to check by considering the chain at the epochs when it
leaves and returns to 0, i.e. we show that, defining $\sigma$ as the
first exit time from 0,
${\mathbb E}_{(0,u)}h(\hat X_{t\wedge\sigma},t\wedge\sigma)=h((0,u),0)$ and
${\mathbb E}_ih(\hat X_{t\wedge\tau_0},t\wedge\tau_0)=h(i,0)$ for $i\in C$.
This is sufficient since $\hat X$ is non-explosive
\hfill$\square$

\section{The $\alpha$-transient case}
We seek now to consider the $\alpha$-transient case. In particular, we
shall focus on the case where $\phi=0$. This is not so specific as one
might think since one can (at the cost of a slight extra difficulty)
reduce the general case to that where $\phi=0$.

\subsection{Reducing to the case where $\phi=0$}
We discuss briefly how to transform the problem to this case.

The essential technique is to note that if, for any $\lambda\leq \phi$, we $h$-transform $\hat P$ using
the space-time $\hat P$-superharmonic function $h^\lambda$ given by
$$
h^\lambda(i,t)=F_{i,0}(\lambda)e^{\lambda t}\hbox{ for }i\in C
$$
and
$$h^\lambda((0,u),t)=(1-I(\lambda)\frac{J^\lambda(u)}{J^\lambda(1)})e^{-(\lambda-{q_0})u}e^{\lambda t}\hbox{ for }u\in [0,1),
$$
where
$$J^\lambda(x)\buildrel {\mathrm{def}}\over = \int_0^xe^{(\lambda-{q_0} )v}dv,$$
then we obtain a new chain $\overline X$ on $\hat S$, with $\phi_{\overline
X}=\phi-\lambda$ and satisfying
$g_{\overline X}(t)=e^{\lambda t}g(t)$, which dies only from state $(0,1-)$.

\begin{proof}
It is a  standard result that $h^\lambda$ is space-time harmonic for $\hat P$
off $\{0\}\times [0,1)$, while, since $I(\lambda)<1$, it is easy to see that $h^\lambda$ is
superharmonic on
$\{0\}\times [0,1)$, by conditioning on the time of first exit from 0.
Now it is easy to check that $\overline X$ dies only from state $(0,1-)$ and dies
on a visit to 0 with
probability $1-I(\lambda)$ so  the result follows
immediately.
\end{proof}
\begin{remark}
Note that, in the $\alpha$-null-recurrent case, where $I(\phi)=1$ but
$I'(\phi -)=\infty$, the transform above produces a null-recurrent $h$-transform
when $\lambda=\phi$, whereas the transform is still evanescent
in the $\alpha$-transient case.

It will follow from L'H\^opital's Theorem in the
$\alpha$-transient cases that if $\psi_i$ denotes the density (on (1,$\infty$)) of $\tau$
when starting from state $i$, then, if
$\frac{\psi_i(t-v)}{\psi_j(t)}$ has a limit as $t\rightarrow \infty$ then
it is the common limit of $\frac{s_i(t-v)}{s_j(t)}=\frac{\int_{t-v}^\infty \psi_i(u)du}
{\int_t^\infty \psi_j(u)du}$ and $\frac{h^\phi_i}{h^\phi_j}\frac{s^{h^\phi}_i(t-v)}{s^{h^\phi}_j(t)}=
\frac{\int_{t-v}^\infty e^{\phi u}\psi_i(u)du}
{\int_t^\infty e^{\phi u}\psi_j(u)du}$.

In the $\alpha$-null recurrent case, we see that this is not of much
help. It is not hard to generalise Lemma 3.3.3 of \cite{JR2} to prove
that in this case $(i,t)\mapsto e^{\phi t}h^\phi_i$ is the unique $\hat P$-superharmonic
function of the form $e^{\lambda t}k_i$ and so gives the only possible
weak or vague limit.
\end{remark}

\subsection{Heavy and subexponential tails}
All the results quoted in this subsection, apart from the last, are taken from Sigman \cite{Sigman}.

Recall first that a random variable (normally taking values in ${\mathbb R}^+$)
$Z$, with distribution function $F_Z$,
is said to be {\em heavy-tailed}, or to have a heavy tail, if
$$
\frac{\overline{F_Z}(t+s)}{\overline{F_Z}(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow 1\hbox{ for all }s\geq 0,
$$
where $\overline{F_Z}\buildrel {\mathrm{def}}\over = 1-F_Z$, is the complementary distribution function.

Denoting the $n$-fold convolution of $F_Z$ by $F^{n}_Z$, $Z$ is said to
have a {\em subexponential tail}, or just to be subexponential, if
\begin{equation}\label{subdef}
\frac{\overline{F^{n}_Z}(t)}{\overline{F_Z}(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow n \hbox{ for all }n,
\end{equation}
and (\ref{subdef}) holds iff
\begin{equation}\label{subdef2}
\limsup_{t\rightarrow \infty}\frac{\overline{F^{n}_Z}(t)}{\overline{F_Z}(t)}\leq n \hbox{ for
some }n\geq 2.
\end{equation}
A subexponential random variable always has a heavy tail.

Two random variables, $X$ and $Y$, are said to have comparable tails, or to be tail equivalent, if
$$
\overline{F_Y}(t)\sim c\overline{F_X}(t)$$
for some $c>0$.

$Y$ is said to have a lighter tail than $X$ if
$$
\frac{\overline{F_Y}(t)}{\overline{F_X}(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow 0.
$$
\begin{lemma}\label{subexp14}
If $X$ and $Y$ are independent, $Y$ is lighter tailed than $X$ and $X$ has a subexponential tail then
$X+Y$ has a subexponential tail and
$$
\overline{F_{X+Y}}(t)\sim \overline{F_X}(t).
$$
\end{lemma}

\begin{lemma}\label{subexp15}
If $X$ and $Y$ are independent and subexponential and tail-equivalent
with
$$
\overline{F_Y}(t)\sim c\overline{F_X}(t),$$
then $X+Y$ is subexponential and
$$
\overline{F_{X+Y}}(t)\sim (1+c)\overline{F_X}(t).
$$
\end{lemma}

This generalises to the following random case:
\begin{lemma}\label{subexp16}
Suppose that $X_1,\ldots$ are iid with
common d.f $F$ which is subexponential and $N$ is an independent geometric random variable, then if
$$
S\buildrel {\mathrm{def}}\over = \sum_1^N X_i,
$$
then $S$ is subexponential and
$$
\overline{F_S}(t)\sim ({\mathbb E} N)\overline{F_X}(t).
$$
\end{lemma}

Finally, we have the following
\begin{lemma}\label{mix}
Suppose that $X_1,\ldots$ are independent and tail-equivalent with
$$F_{X_i}\buildrel {\mathrm{def}}\over = F_i,
$$
and $J$ is an independent random variable taking values in ${\mathbb N}$.
Let
$$Y=X_J,$$
(so that $Y$ is a mixture of the $X_i$s)
and denote its distribution function by $F$ (so $F(t)=\sum_{i\in
{\mathbb N}}{\mathbb P}(J=i)F_i(t)$).

Now suppose that
$$\overline{F_{i}}(t)\sim a_i\overline{F_{1}}(t):
$$
if the collection
$\{\frac{\overline{F_J}(t)}{\overline{F_{1}}(t)};\; t\geq 0\}$ is uniformly
integrable then,
\begin{equation}\label{subexp9}
\overline F(t)\sim ({\mathbb E} a_J)\overline{F_1}(t).
\end{equation}
In particular, if $J$ is a bounded r.v. then (\ref{subexp9}) holds.
\end{lemma}
\begin{proof}
It follows from the assumptions that
$$
\frac{\overline{F_J}(t)}{\overline{F_{1}}(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow a_J\hbox{ a.s.}
$$
Thus if the collection is u.i. then convergence is also in $L^1$ and so,
since ${\mathbb E} \overline{F_J}(t)=\overline{F}(t)$, we see that
$$
\frac{\overline{F}(t)}{\overline{F_{1}}(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow {\mathbb E} a_J.
$$
In particular, if $J\leq n$ a.s. then
$$
\limsup_{t\rightarrow \infty}\frac{\overline{F_J}(t)}{\overline{F_{1}}(t)}\leq
\max_{1\leq i\leq n} a_i \hbox{ a.s.},
$$
and so the collection is indeed u.i.
\end{proof}

\subsection{Results for heavy tails}
Suppose first that $0=\phi<\mu^{C}$.
\begin{theorem}\label{subexp1}
If $0=\phi<\mu^{C}$ and $\tau$ is subexponential, then $\frac{s_i(t-
v)}{s_j(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow 1$ for all $v\geq 0$ and $s_{(0,u)}(t-v)/s_{(0,0)}(t)\buildrel {t\rightarrow \infty}\over\longrightarrow \frac{1-e^{-
{q_0}(1-u)}}{1-e^{-{q_0}}}.$
\end{theorem}
\begin{proof}
Notice first that, since $\mu^{C}>0$, ${\mathbb P}_i(\tau_0>t)\leq k_i e^{-\mu^{C}
t/2}$, so that $\tau^{(i)}_0$ has a lighter tail than $\tau$ so, by Lemma \ref{subexp14},
$$s_i(t-v)={\mathbb P}_{(0,0)}(\tau^{(i)}_0+\tau>t-v)\sim
{\mathbb P}_{(0,0)}(\tau>t-v)\sim {\mathbb P}_{(0,0)}(\tau>t)=s_{(0,0)}(t).$$
Similarly,
$$s_{(0,u)}(t-v)=\int_0^{1-u}{q_0} e^{-{q_0} w}{\mathbb P}({R}^1+\tau>t-v-w)dw\sim  (1-e^{-
{q_0} (1-u)}){\mathbb P}({R}^1+\tau>t)$$
and so $s_{(0,u)}(t-v)/s_{(0,0)}(t)$ converges
to the desired limit.
\end{proof}
It is easy to see that $h$, defined by $h_i=1$, for $i\in C$ and
$h_{(0,u)}=\frac{1-e^{-{q_0}(1-u)}}{1-e^{-{q_0}}}$ is strictly $\hat P$-
superharmonic and is harmonic on $C$:
the following theorem then follows easily from a mild
adaptation of Theorem 4.1.1 of \cite{JR}.
\begin{theorem}\label{subexp2}
Under the conditions of Theorem \ref{subexp1}, the restriction of the law of $\tilde X^T$ to
${\mathcal F}_s$ converges {\em vaguely} to that of $X^\infty$ restricted to ${\mathcal F}_s$, where
$P^\infty$ is the (substochastic) $h$-transform of $\tilde P$ (which dies from state $(0,u)$ with hazard rate
$\lambda(u)=\frac{{q_0} e^{-{q_0}}}{1-e^{-{q_0}(1-u)}}$).
\end{theorem}

\begin{example}
Consider the case where $\sum_{j\in C}q_{0,j}F_{j,0}(\lambda)=\infty$ for all
$\lambda>0$ but $\mu^{C}>0$. For example, we may take the nearest-neighbour random
walk on ${\mathbb N}$ with up-jump rate $b$ and down-jump rate $d$ (with $b<d$) and then set
$$q_0=1;\; q_{0,i}=\frac{6}{\pi^2i^2}\hbox{ for }i\in {\mathbb N}.
$$
It is well-known that
$$\mu^C=b+d-2\sqrt{bd},$$
and
$$
F_{i,0}(\lambda)=\gamma_\lambda^i,
$$
where
$$\gamma_\lambda=\frac{b+d-\lambda-\sqrt{(b+d-\lambda)^2-4bd}}{2b}>1
$$
for $0<\lambda\leq\mu^C$.
So, for any $\lambda>0$, $\sum_{i\in {\mathbb N}}q_{0,i}F_{i,0}(\lambda)={\mathbb E} e^{\lambda {R}^1}=\infty$
and hence $\phi=0$.
\end{example}

Now we consider the case where $\mu^{C}=0$ (and hence $\phi=0$ also).

\begin{theorem}\label{subexp3}
Denote by $\tau^{(i)}$ a generic random variable having the distribution
of the $\tau$ conditional on $X_0=i$.
Suppose that $\tau^{(i)}$ have comparable heavy tails, so that
${\mathbb P}(\tau^{(i)}>t)={\mathbb P}_i(\tau>t)\sim c_i{\mathbb P}(\tau^{(0)}>t)=c_i{\mathbb P}_{(0,0)}(\tau>t)$ and $\frac{{\mathbb P}_i(\tau>t+s)}{{\mathbb P}_i(\tau>t)}\buildrel {t\rightarrow \infty}\over\longrightarrow
1$,
then,
defining
$$h_i=c_i\hbox{,  for }i\in S$$
and
$$
h_{(0,u)}=\frac{1-e^{-{q_0}(1-u)}}{1-e^{-{q_0}}},
$$

\begin{equation}\label{subexp22}
\frac{s_j(t-
v)}{s_i(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow \frac{c_j}{c_i}\hbox{ for all }v\geq 0\hbox{ and for
all }i,j\in\hat S.
\end{equation}
In particular, if the $\tau^{(i)}_0$'s have comparable subexponential tails, with
$$
{\mathbb P}(\tau^{(i)}_0 >t)={\mathbb P}_i(\tau_0>t)\sim a_i{\mathbb P}(\tau^{(1)}_0>t)=a_i{\mathbb P}_1(\tau_0>t)
$$
and
$$
q_{0,i}=0\hbox{ for }i> n,
$$
then,
defining $a_0=0$,
$m=\sum_{i\inC}q_{0,i}a_i/{q_0}$,
$$h_i=1+\frac{a_i}{(e^{{q_0}}-1)m}\hbox{,  for }i\in S$$
and
$$
h_{(0,u)}=\frac{1-e^{-{q_0}(1-u)}}{1-e^{-{q_0}}},
$$
we have that
$$\frac{s_j(t-v)}{s_i(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow \frac{h_j}{h_i}
\hbox{ for all }v\geq 0\hbox{ and for all }i,j\in\hat S.
$$

In general $a$ must be $\tilde P$-superharmonic. If $a$ is $\tilde P$-harmonic then $h$ is $\hat P$-harmonic, so that,
in this case, the restriction of the law of $\tilde X^T$ to
${\mathcal F}_s$ converges weakly to that of $X^\infty$ restricted to ${\mathcal F}_s$, where
$P^\infty$ is the (stochastic) $h$-transform of $\tilde P$.
\end{theorem}
\begin{proof}
The first claim is essentially a restatement of the conditions for
convergence in (\ref{subexp22}).

To prove the second statement, first notice that we may write
$$
\tau^{(i)}=\tau^{(i)}_0+1+\sum_{n=1}^N (\tilde\hold^n+R^n),
$$
where $(\tilde\hold^n)_{n\geq 1}$ are a sequence of iid random variables with
distribution that of the holding time in 0 conditioned on its lying in
(0,1), $N$ is a Geometric($e^{-{q_0}}$) r.v. and the $R^n$'s are as in
section 2 and all are independent.

Now each $R^n$ is a mixture of $\tau^{(i)}_0$s, so, by Lemma \ref{mix},
$$
{\mathbb P}(R^n\geq t)=\sum_{i\in C} \frac{q_{0,i}}{{q_0}}{\mathbb P}(\tau^{(i)}_0\geq t)\sim
\sum_{i\in C} \frac{q_{0,i}}{{q_0}}a_i{\mathbb P}(\tau^{(1)}_0\geq
t)=m{\mathbb P}(\tau^{(1)}_0\geq t).
$$

Now it follows from Lemma \ref{subexp14} that $(\tilde\hold^n+R^n)$
is tail equivalent to $R^n$ and is subexponential and then we deduce,
from Lemmas \ref{subexp15} and \ref{subexp16} that ${\mathbb P}(\tau^{(i)}>t)\sim
(a_i+m(e^{{q_0}}-1)){\mathbb P}(\tau^{(1)}_0\geq t)=m(e^{{q_0}}-1)h_i{\mathbb P}(\tau^{(1)}_0\geq t).$
The last statement follows from the fact that $\tilde X$ is non-explosive and
it is then easy to check (by considering the chain at the epochs when it leaves and returns to 0) that
then $h$ is $\hat P$-harmonic if $a$ is $\tilde P$-harmonic
\end{proof}

\begin{theorem}\label{subexp4}
Suppose that $\tilde X$ is a recurrent birth and death process on ${\mathbb Z}^+$ and, for some
$i$, $\tau^{(i)}_0$ is subexponential, then ${\mathbb P}(\tau^{(j)}_0>t)\sim
\frac{\beta_j}{\beta_i}{\mathbb P}(\tau^{(i)}_0>t)$, where $\beta$ is the unique $\tilde P$
harmonic function on ${\mathbb N}$ with $\beta_1=1$.
\end{theorem}
\begin{proof}
Notice that, since $\tau^{(i)}_0$ is subexponential, it follows that $\mu^{C}=0$
and hence, by Theorem 5.1.1 of \cite{JR3}, there is a unique $\tilde P$-
harmonic $\beta$. It follows that for any $n$, $\sigma_n$, the first exit time of $X$
from the set $\{1,\ldots, n-1\}$ has an exponential tail (i.e its tail decreases to 0 at an exponential rate)
and the exit is
to $n$ with probability $\beta_i/ \beta_n$ if $X$ starts in $i$.

It follows that for each $j\leq i$,
$${\mathbb P}(\tau^{(j)}_0>t)\sim \frac{\beta_j}{\beta_i} {\mathbb P}(\tau^{(i)}_0>t).
$$
Similarly, for $i<n$, $\tau^{(i)}_0=\sigma_n+1_A\tau^{(n)}_0$, where
$A=$($X$ exits $\{1,\ldots,n-1\}$  to $n$), so that
$${\mathbb P}(\tau^{(i)}_0>t)\sim
{\mathbb P}(A){\mathbb P}(\tau^{(n)}_0>t)=\frac{\beta_i}{\beta_n}{\mathbb P}(\tau^{(n)}_0>t).
$$
\end{proof}
The following is an immediate consequence of Theorems \ref{subexp3} and
\ref{subexp4}.
\begin{corollary}\label{bd2}
If $\tilde X$ is a birth and death process on ${\mathbb Z}^+$ and, for some
$i$, $\tau^{(i)}_0$ is subexponential, and for some $n$, $q_{0,j}=0$ for $j>n$ then the conclusion of Theorem
\ref{subexp3} holds.
\end{corollary}
\begin{remark}
If $\mu^{C}=0$ and the process conditioned on not hitting 0 until
time $T$ converges vaguely, then the $\tau^{(i)}_0$'s must have comparable {\em
heavy}
tails. If, in fact the convergence is weak (i.e. to an honest process)
then the vector $a$ must be harmonic for $\tilde P$.

\end{remark}
\begin{remark}
Suppose that $X$ is a birth and death process, with birth rates $b_i$
equal to the corresponding death rates. If the rates are
decreasing in $i$, then $\tau^{(1)}_0$ is subexponential.

To see this, first observe that, by
conditioning on the first jump, we obtain that
$${\mathbb P}(\tau^{(1)}_0>t)=\frac{1}{2}{\mathbb P}(E_1>t)+\frac{1}{2}{\mathbb P}(E_1+\tau^{(2)}_0>t),
$$
where $E_1$ is the first waiting time in state $1$.
Now, since
$$\tau^{(2)}_0=\tau^{(2)}_1+\tau^{(1)}_0
$$
and
since $\tau^{(2)}_1$ stochastically dominates $\tau^{(1)}_0$,
we obtain the desired result that
$$
\limsup_{t\rightarrow \infty}\frac{\overline{F^{(2)}}(t)}{\overline F(t)}\leq 2,
$$
where $F$ is the distribution function of $\tau^{(1)}_0$. The result now
follows by (\ref{subdef2}).
\end{remark}
\section{Some concluding remarks}
Sigman \cite{Sigman} gives some conditions which ensure that a random
variable has a subexponential tail.

Many obvious examples exist of the $\alpha$-recurrent case. We have
exhibited a few examples in the $\alpha$-transient case always assuming that $C$
is irreducible. If it is not, then in principle we can divide $C$
into communicating classes $\{C_l:\; l\in L\}$, where $L$ is some countable or finite index set.
It is easy to show
that
$$\phi\leq \inf_{l\in L} \mu^{C_l}.
$$
By adapting the proof of Theorem \ref{subexp1}, it is easy to see that
if $\tau$ is subexponential but $\mu^{C_l}>0$ for some $l\in L$,
then $\frac{s_i(t-v)}{s_j(t)}\buildrel {t\rightarrow \infty}\over\longrightarrow 1$ for $i,j\in
C_l\cup\{\{0\}\times [0,1)$ and so, as in Theorem \ref{subexp2},
weak convergence of the conditioned chains is not
possible if each $\mu^{C_l}>0$.
Conversely, if $\min\limits_{l\in L} \mu^{C_l}=\mu^{C_{l^*}}$ and $X$ restricted to
$C_{l^*}$ is $\alpha$-recurrent then $\phi=\mu^{C_{l^*}}$
and a suitably adapted version of Theorem \ref{rec1} and Corollary
\ref{cor1} will apply.


\section{Introduction}

The motivation of the following work has its origin in experimental physics. Some long molecules are solvable in a liquid (for instance imogolite in water or DNA in lithium) and the molecules forming the liquid show an intermittent dynamics, alternating diffusion in the bulb and adsorption on the long molecules. For the physicist's point of view, it is very important to have as precise as possible knowledge of the statistics of these brownian flights.

In \cite{GKLSZ} a connection is established between the statistics of the long flight lengths and the geometry of the long molecules (more precisely their Minkowski dimension).  This connection has been made rigorous in \cite{BLZ}. These two papers concern almost exclusively lengths. How does one check experimentally the results? A very powerful tool for that is relaxation methods in nuclear magnetic resonance (see \cite{CDLPZ}): but this method only allows to compute (the statistics of the) duration of long flights. Some heuristic link between time and length was derived  in \cite{GKLSZ}, \cite{CDLPZ}. The aim of this paper is to make this heuristics rigorous.

 \section{Geometric Backgound}
 In the sequel, $\Omega$ will always denote a domain in $\R^d$ with compact boundary.  The crucial tool we need to use is the notion of Whintey cubes. We thus recall the
 \begin{Proposition}\label{Whitney}  (cf. \cite{Grafakos}, p. 463)
Given any non-empty open proper subset $\Omega$ of $\R^d$, there exists a family of closed dyadic cubes $\{Q_j\}_j$ such that
\begin{itemize}
\item $\bigcup_j Q_j=\Omega$ and the cubes $Q_j$'s have disjoint interiors
\item $\sqrt{d}\ell(Q_j)\le\dist(Q_j,\partial\Omega)\le 4\sqrt{d}\ell(Q_j)$
\item if $Q_j$ and $Q_k$ touch then $\ell(Q_j)\le 4\ell(Q_k)$
\item for a given Whitney cube $Q_j$ there are at most $12^d$ Whitney cubes $Q_k$'s that touch $Q_j$.
\end{itemize}
\end{Proposition}

In this statement, $\ell(Q)$ stands for the side-length of the cube $Q$ and, for $\lambda>0$,  $\lambda Q$ is the cube of the same center and of sidelength $\lambda\ell(Q)$.
For $k\in\Z$, we denote by ${\mathcal Q}_k$, the collection of Whitney cubes $Q_j$ with $\ell(Q_j)=2^k$.
We also recall the definition of the Minkowski sausage: for $r>0$, $$M_r =\{x\in\Omega\;;\; \dist(x,\partial\Omega)\le r\}$$ and
 $$\Gamma_r=\{x\in\Omega\; ;\;\dist(x,\partial\Omega)=r\}$$

We then define  ${\mathcal S}_r$ as the collection of Whitney cubes intersecting $\Gamma_r$. Notice that ${\mathcal S}_r$ is a finite set.

\begin{Definition}
Let $\varepsilon>0$. We will call Brownian flight the random process $F_t, t\ge 0$ consisting in picking at random with equiprobability one of the dyadic Whitney cubes of  ${\mathcal S}_{\varepsilon}$  and starting from the center of the cube a Brownian motion $B_t$ killed once it reaches $\partial\Omega$. We denote by $\tau_{\Omega}= \inf\{t\; ;\; F_t\notin\Omega\}$ the lifetime of this process.
\end{Definition}

We are interested in the asymptotics of $\p(\tau_{\Omega}>t)$ as $t$ grows, but this needs some explanation:

It is well known that, if $\Omega$ is bounded, this quantity decreases exponentially, as $t\rightarrow\infty$, as $e^{\lambda t}$, where $\lambda$ is the first eigenvalue of the Laplacian. We define
$$R_{\Omega}=\min\left\{1,\sup_{x\in\Omega} \dist(x,\partial\Omega)\right\}.$$
 Our aim is to evaluate $\p(\tau_{\Omega}>t)$ in the interval $\epsilon^2\le t\le R_{\Omega}^2$, and this independently of $\varepsilon$. In fact, the estimate we are looking for is thus an estimate with respect to $\varepsilon$ rather than for "pure" $t$.

We study first a simple example, since we will need the partial result anyhow.
Let $[0,a]$ be a real segment and take $x\in(0,a)$. The probability that brownian motion started at $x$ has not exit the interval $(0,a)$ by time $t$, $\p(\tau_x>t)$, is given by the following equivalent formulas (see \cite{Feller}, pg. 342)
$$\p(\tau_x>t)=\frac{1}{\sqrt{2\pi}}\int_{-x/\sqrt{t}}^{-x/\sqrt{t}}\exp\left({-{\frac12}y^2}\right)dy+ {\sqrt{\frac{2}{\pi}}}\sum_{k=1}^{\infty}(-1)^k\int_{(ka-x)\sqrt{t}}^{(ka+x)\sqrt{t}}\exp\left({-{\frac12}y^2}\right)dy $$
and
$$\p(\tau_x>t)=\frac{4}{\pi} \sum_{k=1}^{\infty}\exp\left({-\frac{(2n+1)^2\pi^2}{2a^2}t}\right)\sin\frac{(2n+1)\pi x}{a}.$$
By symmetry we can assume $x\le \frac{a}{2}$. If $\frac{a}{\sqrt t}$ is not too big, say $\frac{a}{\sqrt t}<1/2$, we have an easy estimate of $\p(\tau_x>t)$ using the first formula :
$\p(\tau_x>t)\sim \frac{x}{\sqrt t}$.
On the other hand, for $\frac{a}{\sqrt t}$ not too small we get, by the second formula, that
$\p(\tau_x>t)\le \exp\left({-\frac{\pi^2}{2a^2}t}\right)$. Hence, that there is a change of regime, the decay of  $\p(\tau_x>t)$ with $t$ going from polynomial to exponential and the quantity $\frac{x}{\sqrt t}$ is relevant for small times.

In a higher dimensional context, let $Q$ be the cube centered at $0$ and of side $r$. By the preceeding remark it follows that, $T$ denoting the  exit time from $Q$ of the Brownian motion starting at $0$, we have
\begin{equation}\label{cubeestimates}
\p(T>t)\le c\left(\frac{r}{\sqrt{t}}\right)^d
\end{equation}
where $c$ depends only on $d$. Here, $\varepsilon=r\simeq R_{\Omega}$ and the opposite inequality thus holds for  $\varepsilon^2\le t\le R_{\Omega}^2$.

If the Brownian motion is started at distance $\varepsilon$ from the boundary then the exit time is essentially the exit time from a half space and we thus get
\begin{equation}\label{cubetimetable}
\p(T>t)\le c\left(\frac{\varepsilon}{\sqrt{t}}\right)= c\left(\frac{\varepsilon}{\sqrt{t}}\right)^{d-1+2-d},
\end{equation}
and, as we will see, the opposite inequality is valid for $t$ not too big.

Our goal is to extend (\ref{cubetimetable}) to general domains with rough boundary. In order to describe the domain of validity of our result let us recall a few definitions. Let $K$ be a compact subset of $\R^d$. For $j\ge 0$ let $N_j$ be the number of dyadic cubes of the $j$-th generation (i.e of size $2^{-j}$) that intersect $K$.

\begin{Definition}
The Minkowski dimension of $K$ is
$$d_M(K)=\limsup_{j\to\infty}\frac{\log_2(N_j)}{j}$$
\end{Definition}

Returning to our situation, we can define similarly the Whitney dimension of $\partial \Omega$ as
\begin{equation}\label{Whitneydimension}
d_W=d_W(\partial \Omega)=\limsup_{j\to\infty}\frac{\log_2(W_j)}{j},
\end{equation}
where $W_j$ is the number of elements of ${\mathcal Q}_j$.

Under very mild conditions (see   \cite{Bishop3}, \cite{JK}, \cite{BLZ}) these two dimensions coincide.
If the boundary of $\Omega$ has some self similarity we can moreover say that there is a constant $c>0$ such that
\begin{equation}\label{hypothesis}
\frac1c{\varepsilon}^{d_M}\le\#{\mathcal S}_{\varepsilon}\le c {\varepsilon}^{d_M},
\end{equation}
for all $\varepsilon\le R_{\Omega}$, where $d_M=d_M(\partial \Omega)$.

In our main theorem we will assume that our domain $\Omega$ satisfies (\ref{hypothesis}).

We also suppose that the domain $\Omega$ satisfies so-called $\Delta$-regularity condition (see also \cite {JW}, \cite{Ancona5}, \cite{hkm}): there exists $L>0$ such that for all $x\in\Omega$, if $d_x=\dist(x,\partial\Omega)<R_{\Omega}$ then
\begin{equation}\label{capacitycondition}
\ds\omega_{\B(x,2d_x)\cap\Omega}^x\left(\partial\Omega\right)\ge L,
\end{equation}
where $\ds\omega_{\B(x,2d_x)\cap\Omega}^x$ is the distribution law of the hitting point of Brownian motion starting at x and killed when reaching the boundary of  $\B(x,2d_x)\cap\Omega)$.
This is a very mild condition (satisfied, for instance, by all domains in $\R^2$ with non-trivial connected boundary) that appears frequently in related literature in various forms (for instance ``uniform capacity condition'' or Hardy inequality).

\section{Time and length estimates for Brownian flights}
We can now state the main result. Let $\Omega$ be a bounded domain in $\R^d$ satisfying (\ref{hypothesis}) and (\ref{capacitycondition}). If $\tau_{\Omega}$ denotes the life-time of a Brownian flight $F_t$ with parameter $\varepsilon$ we have

\begin{Theorem}\label{Th1}
There exists $c>0$ depending only on constants in  (\ref{hypothesis}), (\ref{capacitycondition}) (and in particular not on $\varepsilon$) such that  \begin{equation}\label{Th1.1}
\frac1c\left(\frac{\varepsilon}{\sqrt t}\right)^{d_{M}+2-d}\le\p(\tau_{\Omega}>t)
\end{equation}
and
\begin{equation}\label{Th1.2}
\p(\tau_{\Omega}>t)\le c  \left(\frac{\varepsilon}{\sqrt t}\right)^{d_{M}+2-d}\left|\log\left(\frac{\varepsilon}{\sqrt t}\right)\right|^{2d},
\end{equation}
for all $\varepsilon^2<t<R_{\Omega}^2$.
\end{Theorem}

This theorem has a ``cousin'' theorem, which was proved in \cite{BLZ}.

\begin{Theorem}\label{main2}
Let $\varepsilon<r<R_{\Omega}$. The probability that the hitting point of $F$ is at distance greater than $r$ from the starting point $x$ is comparable to
\begin{equation}\label{main2.1}
\left(\frac{\#{\mathcal S}_r}{\#{\mathcal S}_{\varepsilon}}\right)^{d_M}\left(\frac{r}{\varepsilon}\right)^{d-2}
\end{equation}
\end{Theorem}

Notice that we do not assume ($\ref{hypothesis}$) for this theorem. If we do, we have
\begin{equation}\label{main2.2}
\left(\frac{\#{\mathcal S}_r}{\#{\mathcal S}_{\varepsilon}}\right)^{d_M}\left(\frac{r}{\varepsilon}\right)^{d-2}\sim \left(\frac{r}{\varepsilon}\right)^{d_M-(d-2)}
\end{equation}

Notice that the quantity on the left of (\ref{main2.2}) is the same as the one in (\ref{Th1.1}) where we have replaced $r$ by $\sqrt{t}$ which is coherent with standard behaviour of Brownian motion.

\section{Proof of the theorem \ref{Th1}}

For $s>0$ we denote by $\beta_s$ the total time  spent by Brownian flight $F_t$  in the Minkowski sausage $\{x\in\Omega\;;\; \dist(x,\partial\Omega)\le s\}$ and
$\delta\beta_s=\beta_s-\beta_{s/2}$.

We define analogue quantities more adapted to the Whitney decomposition ; namely, $\delta \tilde\beta_{2^k}$ will denote the time spent by $F_t$ inside ${\tilde M}_k=\bigcup\{Q\;;\;Q\in{\mathcal Q}_k\}$.

If $Q$ is a Whintney cube we define the "vicinity" of $Q$ as $$\tilde Q=Q\cup\bigcup\{Q'\;;\;Q'\in{\mathcal E}\},$$
where $${\mathcal E}=\{Q'\in\bigcup_k{\mathcal Q}_k\;;\; \lambda Q\cap Q'\not= \emptyset \mbox{ and } \lambda Q'\cap Q\not=\emptyset\},$$
and $\lambda= 8\sqrt{d}$ satisfy that for all Whitney cubes
$$\lambda Q\supset \B(x_Q,2\dist(x_Q,\partial\Omega)),$$ $x_Q$ being the center of $Q$.

We may now start the proof and we begin with

\subsection{The upper bound}

We separate the event $\{\tau_{\Omega}>t\}$ by the partition $\{\tau_{\Omega}=\beta_{\sqrt{t}}\}$ and $\{\tau_{\Omega}>\beta_{\sqrt{t}}\}$, that is whether the process goes or does not go at distance $\sqrt{t}$ from the boundary. From theorem \ref{main2} we get
$$\p(\tau_{\Omega}>t\mbox{ and }\tau_{\Omega}>\beta_{\sqrt{t}})\le \p(\tau_{\Omega}>\beta_{\sqrt{t}})\le c \left(\frac{\varepsilon}{\sqrt{t}}\right)^{d_M+2-d}.$$

In order to estimate the term $\p(\tau_{\Omega}=\beta_{\sqrt{t}}>t)$ we begin by writing $$\ds\tau_{\Omega}=\sum_{k=-\infty}^{\log_2{\sqrt{t}}}\delta\beta_{2^k},$$
and thus, putting $k_0=\log_2{\epsilon}$ and $k_1=\log_2{\sqrt{t}}$,
\begin{equation}\label{sum}
\p\left( \sum_{k=-\infty}^{\log_2{\sqrt{t}}}\delta\beta_{2^k} >t \right)\le \sum_{k=k_0+1}^{k_1}\p\left(\delta\beta_{2^k}>\frac{t}{2(k_1-k_0)} \right)+\p\left(\beta_{2^{k_0}}>\frac{t}{2}\right)
\end{equation}

We now invoke the following lemma whose proof is post poned to the next section.

\begin{Lemma}\label{Technical1} There exists a number $k^*$ depending only on $d$ and constants $C$ , $0<p<1$ depending only on $d$ and $L$ such that for all $t>0$, $k\in\Z$ and $N\in\N$  we have
\begin{eqnarray*}
&&\p(\delta\beta_{2^k}>t)\le Cp^N
+\\
&&\p\left(\exists Q \in{\mathcal Q}_k\cup...\cup {\mathcal Q}_{k-k^*}; \exists\; 0<s_1<s_2<t \mbox{ with }F_{[s_1,s_2]}\subset\tilde Q \mbox{ and } s_2-s_1>t/N\right)
\end{eqnarray*}
\end{Lemma}

Using this lemma, we get
 \begin{eqnarray*}
&&\p\left(\delta\beta_{2^j}>\frac{t}{2(k_1-k_0)}\right)\le cp^N+\\
&&\p\left(\exists Q\in{\mathcal Q}_{j}\cup...\cup{\mathcal Q}_{j-k^*}\; \exists s_1<s_2<\tau_{\Omega}\;;\;F_{[s_1,s_2]}\subset\tilde Q\;,\; s_2-s_1>\frac{t}{2(k_1-k_0)N}\right)
\end{eqnarray*}

By (\ref{cubeestimates}), (\ref{main2.2}) and the strong Markov property of Brownian motion we then have :

 \begin{eqnarray*}
&&\p\left(\delta\beta_{2^j}>\frac{t}{2(k_1-k_0)}\right)\le cp^N+\\
&&\p\left(\exists Q\in{\mathcal Q}_{j}\cup...\cup{\mathcal Q}_{j-k^*}\; \exists s_1\tau_{\Omega}\;;\;F_{s_1}\in\tilde Q\right)\times \\
&&\p\left(\exists s_2>s_1\;,\; F_{[s_1,s_2]}\subset\tilde Q\mbox{ and }s_2-s_1>\frac{t}{2(k_1-k_0)N}\right)\le \\
&& cp^N+c\left(\frac{2^j}{\sqrt{\frac{t}{2(k_1-k_0)N}}}\right)^d\left(\frac{\varepsilon}{2^j}\right)^{d_M+2-d}
\end{eqnarray*}

Suming up the first term in (\ref{sum}) we get
 \begin{eqnarray*}
&&\sum_{k=k_0+1}^{k_1}\p\left(\delta\beta_{2^k}>\frac{t}{2(k_1-k_0)} \right)\le c(k_1-k_0)p^N+
c\sum_{k=k_0+1}^{k_1}\left(\frac{2^k}{\sqrt{\frac{t}{2(k_1-k_0)N}}}\right)^d \left(\frac{\varepsilon}{2^k}\right)^{d_M+2-d}\\
&&\le c(k_1-k_0)p^N+c(2(k_1-k_0)N)^d\varepsilon^{d_M+d-2}\left(\frac{1}{\sqrt{t}}\right)^d\sum_{k=k_0+1}^{k_1} 2^{2d-d_M+2}\\
&&\le  c(k_1-k_0)p^N+c(2(k_1-k_0)N)^d\varepsilon^{d_M+d-2}\left(\frac{1}{\sqrt{t}}\right)^d 2^{k_1(2d-d_M+2)}\\
&&\le  cp^N\log_2\left(\frac{\sqrt{t}}{\varepsilon}\right)+c\left(\log_2\left(\frac{\sqrt{t}}{\varepsilon}\right)N\right)^d\left(\frac{\varepsilon}{\sqrt{t}}\right)^{d_M+2-d}
\end{eqnarray*}

Take $N\simeq (d_M+2-d) \log_p\left(\frac{\varepsilon}{\sqrt{t}}\right)$ to obtain
\begin{equation}\label{firstterm}
\sum_{k=k_0+1}^{k_1}\p\left(\delta\beta_{2^k}>\frac{t}{2(k_1-k_0)} \right)\le c \left(\log\left(\frac{\varepsilon}{\sqrt{t}}\right)\right)^{2d} \left(\frac{\varepsilon}{\sqrt{t}}\right)^{d_M+2-d}
\end{equation}

To bound the second term $\p\left(\beta_{2^{k_0}}>\frac{t}{2}\right)$ of the sum (\ref{sum}) we need a lemma of the same nature as lemma \ref{Technical1}. The proof of this lemma is also post poned to the next section.
\begin{Lemma}\label{Technical4}
Let ${\mathcal R}_{k_0}$ be the collection of all dyadic cubes of sidelength $2^{k_0}$ intersecting $\partial\Omega$. There exist constants  $C$ , $0<p<1$ depending only on $d$ and $L$ such that for all $t>0$, $k_0\in\Z$ and $N\in\N$
\begin{eqnarray*}
&&\p(\delta\beta_{2^{k_0}}>t)\le Cp^N
+\\
&&\p\left(\exists Q \in{\mathcal R}_{k_0}; \exists\; 0<s_1<s_2<t \mbox{ with }F_{[s_1,s_2]}\subset NQ \mbox{ and } s_2-s_1>t/N\right)
\end{eqnarray*}
\end{Lemma}

We therefore deduce
$$\p(\delta\beta_{2^{k_0}}>\frac{t}{2})\le cp^N+\left(\frac{2^{k_0}N}{\sqrt{t}}\right)^d.$$
We minimize on $N\simeq\log_p(\varepsilon)$ to get
$$\p(\delta\beta_{2^{k_0}}>\frac{t}{2})\le c\left(\log(\varepsilon)\frac{\varepsilon}{\sqrt{t}}\right)^d.$$

Combining this last inequality and (\ref{firstterm}) we get the upper bound, since $d_M+d-2\le d$.

\subsection{The Lower Bound}\label{lb}
Following the same reasoning for ${k_1}=[\log_2\sqrt{t}]+1$ we get
$$\p(\tau_{\Omega}>t)\ge \p(\exists s_1>0 \mbox{ s.t. }F_{s_1}\in\bigcup_{Q\in{\mathcal Q}_{k_1}}Q\mbox{ and }\exists {s_2>s_1+t} \mbox{ s.t. } F_{[s_1,s_2]}\in\bigcup_{Q\in{\mathcal Q}_{k_1}}2 Q)$$
Using strong Markov property the this probability can be written as the product of
$\p(\exists s_1>0 \mbox{ s.t. }F_{s_1}\in{\mathcal Q}_{k_1} \mbox)$ with $\p(\exists {s_2>s_1+t} \mbox{ s.t. } F_{[s_1,s_2]}\in\bigcup_{Q\in{\mathcal Q}_{k_1}}2 Q)$. The second term of the product  is greater than the probability that Brownian motion exits a cube of size $2^{k_1+1}\simeq\sqrt{t}$ at time greater that $t$ which is bounded below by a positive constant depending only on $d$. The first one is simply the probability that Brownian flight gets to ${\mathcal Q}_{k_1}$ which is equivalent to $\left(\frac{\epsilon}{\sqrt{t}}\right)^{d_{M}+2-d}$ and the proof is complete.

\section{Proofs of lemmas}
Let un first deal with lemma  \ref{Technical1}. The proof of  \ref{Technical4} is quite similar and will hence be abridged.
\subsection{Proof of lemma \ref{Technical1}}
We need the following
\begin{Lemma}\label{key} Under the $\Delta$-regularity hypothesis
the probability that BM touches more than $N$ Whitney cubes of a given size decreases as $Cp^N$, with $0<p<1$, $C$ a positive constant.
\end{Lemma}

 The proof of the lemma relies on an annuli reasoning.
 \begin{Proof}
 Le $(B_t)_{t>0}$ be Brownian motion started at any point $x\in\Omega$ and choose $k\in\Z$. Choose any $Q\in{\mathcal Q}_k$ and let $\lambda Q$ be the cube of the same center but $\lambda$ times the side-length $\ell(Q)$  of $Q$. By the definition of Whitney cubes, there is a $\lambda= 8\sqrt{d}$ depending only on $d$  such that $$\frac{\lambda}{2} \ell(Q)\le\dist(Q,\partial\Omega)\le 2\lambda \ell(Q).$$

Suppose that there exists $t_0>0$ such that $B_{t_0}\in Q$.
By the $\Delta$-regularity condition (\ref{capacitycondition}), the probability that there exists $t_1>t_0$ with $B_{[t_0,t_1]}\subset\Omega$ and $B_{t_1}\notin\lambda Q$ is bounded above by $p<1$ depending only on $L,\lambda$:
\begin{equation}\label{one}
\p\left(\exists t_1>t_0\; ; \; B_{[t_0,t_1]}\subset\Omega \mbox{ and }B_{t_1}\notin\lambda Q\Big|\exists t_0>0\;;\; B_{t_0}\in Q\right)<p
\end{equation}

On the other hand, the number of Whitney cubes of ${\mathcal Q}_k$ lying inside $\lambda Q$ is bounded by a constant $c_1=c_1(d)$. The probability that there exists a Whitney cube $Q_1\in {\mathcal Q}_k$ outside $\lambda Q$ that is visited by Brownian motion is hence bounded above by $p<1$.

We study probability that there exist Whitney cubes $Q_1,...,Q_m\in{\mathcal Q}_k$ such that $Q_1\cap\lambda Q=Q_2\cap\lambda Q_1=...=Q_m\cap\lambda Q_{m-1}=\emptyset$ all visited by Brownian motion.
It is sufficient to prove that this probability decays exponentially with $m$.

By the strong Markov property the probability that there exists $t_m>t_{m-1}>...>t_0$ such that $B_{t_0}\in Q$ , $B_{t_1}\in Q_1$ ... , $B_{t_m}\in Q_m$ is given by
\begin{eqnarray*}
&&\p\left(\exists\; t_m>t_{m-1}>...>t_0\; ;\mbox{ and }Q,...Q_m \mbox{ as above such that } B_{t_m}\in Q_m,...,B_{t_0}\in Q\right)\\
&&=\p\left(\exists\; t_m>t_{m-1}\; ;\;B_{t_m}\in Q_m|\exists\; t_{m-1}>...>t_0\; B_{t_{m-1}}\in Q_{m-1},...,B_{t_0}\in Q\right)\cdot\\
&&\p\left( \exists\; t_{m-1}>...>t_0\; B_{t_{m-1}}\in Q_{m-1},...,B_{t_0}\in Q\right)\\
&&=\p\left(\exists\; t_m>t_{m-1}\; ;\;B_{t_m}\in Q_m \mbox{ with }Q_m\cap\lambda Q_{m-1}=\emptyset|\exists\; t_{m-1}\; ;B_{t_{m-1}}\in Q_{m-1}\right)\cdot\\
&&\p\left( \exists\; t_{m-1}>...>t_0\; B_{t_{m-1}}\in Q_{m-1},...,B_{t_0}\in Q\right)
\end{eqnarray*}
Now, by (\ref{one}),
$$\p\left(\exists\; t_m>t_{m-1}\; ;\;B_{t_m}\in Q_m \mbox{ with }Q_m\cap\lambda Q_{m-1}=\emptyset|\exists\; t_{m-1}\; ;B_{t_{m-1}}\in Q_{m-1}\right)<p.$$
By induction we get that
$$\p\left(\exists\; t_m>t_{m-1}>...>t_0\; ;\mbox{ and }Q,...Q_m \mbox{ as above such that } B_{t_m}\in Q_m,...,B_{t_0}\in Q\right)<p^m$$ and hence the lemma.
\end{Proof}

Recall that for a given dyadic Whitney cube $Q $ we have defined the vicinity $\tilde Q$ of $Q$ as the union of all Whitney cubes $Q'$ verifying
$$Q'\cap\lambda Q\not=\emptyset\mbox{ and }Q\cap\lambda Q'\not=\emptyset.$$
We can easily check that there are less than $(100\sqrt{d})^d$ such cubes $Q'$ of size at most $\ell(Q)/12$  (the constants are not optimal).
We say that the $k$-level layers are visited at least $n$ times if there exist $t_0<s_1<t_1<...<s_n<t_n$ satisfying
$$B_{t_j}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q\mbox{ and }B_{s_j}\notin\bigcup_{Q\in{\mathcal S}_{{2^k}}}\tilde Q,$$ for all $j=1,...,n$. For any $k\in\Z$ note
$$\nu_k=\sup\{n\in\N\; ;\; \mbox{the $k$-level layers are visited at least $n$ times}\}$$
\begin{Lemma}\label{technical0}
There exists $0<p<1$ and a positive constant $C$ such that, given $k\in\Z$, for all $n\in\N$
$$\p(\nu_k>n)\le p^n \p\left(\exists t_0>0\mbox{ and } Q\in{\mathcal S}_{2^k}\; ; \; B_{t_0}\in Q\right).$$
\end{Lemma}
\begin{Proof}
The arguments as similar as in lemma \ref{key}. We only need to prove that $\p(\nu_k>1)<p$ and apply strong Markov property.
We have
\begin{eqnarray*}
\p(\nu_k>1)& \le &\p\left(\exists 0<t_0<s_1<t_1\; ,\; Q\in{\mathcal S}_{2^k} \; ; \; B_{t_0}\in Q\; , \; B_{s_1}\notin \bigcup_{Q\in{\mathcal S}_{{2^k}}}\tilde Q\; , \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q\right)\\
&=& \p\left(\exists t_1>s_1>t_0\; ; \;  B_{s_1}\notin \bigcup_{Q\in{\mathcal S}_{{2^k}}}\tilde Q\; , \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q\Big| \exists t_0>0 \; ; \; B_{t_0}\in Q\in{\mathcal S}_{2^k}\right)\\
&\times &\p\left(\exists t_0>0\mbox{ and } Q\in{\mathcal S}_{2^k}\; ; \; B_{t_0}\in Q\right)
\end{eqnarray*}
To abbreviate formulas we note $ \p_c(.)=\p\left(.| \exists t_0>0 \; ; \; B_{t_0}\in Q\in{\mathcal S}_{2^k}\right)$. With this notation,
\begin{eqnarray*}
&&\p_c\left(\exists t_1>s_1>t_0\; ; \;  B_{s_1}\notin \bigcup_{Q\in{\mathcal S}_{{2^k}}}\tilde Q\; , \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q)\right)=\\
&&\p_c\left(\exists t_1>s_1\; ; \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q\; \Big| A\right)\p_c(A)+\p_c\left(\exists t_1>s_1\; ; \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q\; \Big| B\right)\p_c(B)
\end{eqnarray*}
where $$A=\left\{\exists s_1>t_0\; ; \;  B_{s_1}\notin \bigcup_{Q\in{\mathcal S}_{{2^k}}}\lambda Q\right\}\mbox{ and }$$
$$B=\left\{\exists s_1>t_0\; ; \;  B_{s_1}\in Q'\ , \,\lambda Q'\cap \bigcup_{Q\in{\mathcal S}_{{2^k}}}Q=\emptyset\;, \;B_{s_1}\in \bigcup_{Q\in{\mathcal S}_{{2^k}}}\lambda Q\right\}$$
form a partition of the event $\left\{\exists s_1>t_0\; ; \;  B_{s_1}\notin \bigcup_{Q\in{\mathcal S}_{{2^k}}}\tilde Q\right\}$.

By (\ref{one}), $\p_c(A)\le p$. Similarly, by the strong Markov property of Brownian motion and by (\ref{one}),
$$\p_c\left(\exists t_1>s_1\; ; \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q\; \Big| B\right)= \p\left(\exists t_1>s_1\; ; \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q\; \Big| B_{s_1}\in Q' \right)\le p.$$
We deduce that
$$\p_c\left(\exists t_1>s_1>t_0\; ; \;  B_{s_1}\notin \bigcup_{Q\in{\mathcal S}_{{2^k}}}\tilde Q\; , \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q)\right)\le \p_c(A)+p(1-\p_c(A))$$

The function $t\mapsto t+p(1-t)$ being increasing on $[0,p]$  we get
$$\p_c\left(\exists t_1>s_1>t_0\; ; \;  B_{s_1}\notin \bigcup_{Q\in{\mathcal S}_{{2^k}}}\tilde Q\; , \; B_{t_1}\in\bigcup_{Q\in{\mathcal S}_{2^k}}Q)\right)\le 2p-p^2<1$$
and the lemma is proven.
\end{Proof}

Remark that by definition of dyadic Whitney cubes there exist $k^*$ depending only on the dimension of the space ($k^*=[\log_2(8\sqrt{d})]+3$ will do) such that for all $k\in\Z$,
\begin{equation}\label{k*}
\{x\in\Omega\;;\;2^{k-1}\le \dist(x,\partial\Omega)\le 2^k\}\subset\bigcup_{j=k-k^*}^k\bigcup_{Q\in{\mathcal Q}_j}Q.
\end{equation}

\begin{Proof}{\bf of lemma \ref{Technical1}}
We clearly have
\begin{equation}\label{partition}
\p(\delta\beta_{2^k}>t)\le \p(\delta\beta_{2^k}>t\; ,\;\nu_k+...+\nu_{k-k^*}>N)+\p(\delta\beta_{2^k}>t\; ,\;\nu_k+ ...+\nu_{k-k^*}\le N)
\end{equation}
Given $t>0$, $k\in\Z$,  by lemma \ref{technical0} we have, for all $N$,
$$ \p(\delta\beta_{2^k}>t\; ,\;\nu_k+...+\nu_{k-k^*}>N)< k^*p^{\frac{1}{k^*}N} \p\left(\exists t_0>0\mbox{ and }Q\in{\mathcal S}_{2^{k-k^*}}\; ; \; B_{t_0}\in Q\right)<c\tilde p^N$$
Let us estimate the second term of the sum (\ref{partition}).
By (\ref{k*}) and the definition of $\nu_k$ we get
 \begin{eqnarray*}
 &&\p(\delta\beta_{2^k}>t\; ,\;\nu_k+...+\nu_{k-k^*}\le N )\\
&&\le\p(\exists l\le N\; , \;Q_1,...,Q_l\in\bigcup_{j={k-k^*}}^k\mathcal{Q}_j\;;\;Q_s\cap\tilde Q_{s-1}=\emptyset\;,\;\forall s=2,...,l\mbox { and } \\
&&\exists  t_1<s_1\le t_2<s_2<...\le t_l<s_l\; ; \;B_{[t_i,s_i]}\subset \tilde Q_s\;\forall s=1,...,l\mbox{ and }\sum_{i=1}^ls_i-t_i>t).
\end{eqnarray*}
Since $l\le N$ we get that
$\p(\delta\beta_{2^k}>t\; ,\;\nu_k+...+\nu_{k-k^*}\le N ) $ is bounded above by
$$\p\left(\exists Q \in{\mathcal Q}_k\cup...\cup {\mathcal Q}_{k-k^*}; \exists\; 0<t_{i_0}<s_{i_0}\mbox{ with }B_{[t_{i_0},s_{i_0}]}\subset\tilde Q \mbox{ and } s_{i_0}-t_{i_0}>t/N\right),$$
which completes the proof.
\end{Proof}
\subsection{Proof of lemma \ref{Technical4}}
The ideas are the same but we will work with cubes touching the boundary instead of Whitney cubes.
\begin{Lemma}\label{key1} Under the $\Delta$-regularity hypothesis, the probability that BM started at distance $\varepsilon<r$ from the boundary gets at distance greater than $R$ from the starting point without leaving the Minkowski sausage $M_r=\{x\in\Omega\;;\; \dist(x,\partial\Omega)\le r\}$ is bounded above by $cp^{R/r}$ where $c>0$ and $0<p<1$ are constants (depending only on $L$ of the $\Delta$-regularity hypothesis and on $d$).
\end{Lemma}
The proof, similar to the one of lemma \ref{key}, is therefore abridged.
\begin{Proof}
Let $x\in M_r$ and consider the annuli centered at $x$ of inner radii $4\ell r$ and outer radii $4(\ell+1)r$ where $\ell=0,...,\left[\frac{R}{4r}\right]$. Brownian motion started at $x$ and moving at distance $R$ from $x$ before exiting $\Omega$ must go through all these annuli.
The probability of going through such an annulus while staying at distance at most $r$ from the boundary is bounded by a $p_0\in (0,1)$ by the $\Delta$-regularity hypothesis. To see this take any point $y$ in the middle of the annulus (i.e. at distance $\frac{6\ell+6}r$ from $x$) and consider the ball of center $y$ and radius $2r$. If $d_y<r$, the probability to exit the ball without  touching $\partial\Omega$ is uniformly bounded away from $1$ by the same hypothesis. This probability being greater than the probability of going through the annulus  we have the statement.
By the independence of the ``crossing annuli" events we get that the probability that Brownian motion  goes through all the annulli is smaller that $cp_0^{\left[\frac{R}{4r}\right]}\sim c\tilde p^{\left[\frac{R}{r}\right]}$.
\end{Proof}

We say that the Minkowki sausage $M_r$ is visited by the Brownian motion at least $k$ times if there exist $t_0<s_1<t_1<...<s_n<t_n<\tau_{\Omega}$ satisfying $B_{t_i}\in M_r$ for all $i=0,...,n$ and $B_{s_i}\notin M_{4r}$.

In a similar way with $\nu_k$ we define $\xi_k$ as the
$$\xi_{r}=\sup\{n\in\N\;;\ M_r \mbox{ has been visited at least } k \mbox{ times}\}.$$
\begin{Lemma}\label{technical1}
There exists $0<p<1$ depening only on the $\Delta$-regularity's $L$ such that, given $r>0$, for all $n\in\N$
$$\p(\xi_{r}>n)\le p^n.$$
\end{Lemma}
\begin{Proof}
The proof of this lemma is a straightforward  application of the $\Delta$-regularity condition.
It suffices to show that there exists $0<p<1$ such that
$$\p(\left(\exists t_1>s_1> t_0\; ; \; B_{t_0}\in M_r\, ,\,B_{s_1}\notin M_{4r}\,,\, B_{t_1}\in M_r \right)<p,$$
and then apply  the Markov property.
Remark that, the probability
$$\p(\left(\exists t_1>s_1> t_0\; ; \; B_{t_0}\in M_r\, ,\,B_{s_1}\notin M_{4r}\right)$$ is smaller than the probability that brownian motion started at $t_0$ exits a ball of radius $2r>2\dist(B_{t_0},\partial\Omega)$ without hitting $\partial\Omega$. By the $\Delta$-regularity this last probability is bounded by a constant $p<1$.
\end{Proof}
\begin{Proof}{\bf of lemma \ref{Technical4}}
As before we get :
\begin{equation}\label{partitionbis}
\p(\delta\beta_{2^{k_0}}>t)=\p(\delta\beta_{2^{k_0}}>t\; ,\;\xi_{2^{k_0}}>N)+\p(\delta\beta_{2^{k_0}}>t\; ,\;\xi_{2^{k_0}}\le N)
\end{equation}
By lemma \ref{technical1} we get $\p(\delta\beta_{2^{k_0}}>t\; ,\;\xi_{2^{k_0}}>N)\le p^N$, where $0<p<1$, for all $N\in\N$.
Let us now deal with the second term of the sum.
$$\p(\delta\beta_{2^{k_0}}>t\; ,\;\xi_{2^{k_0}}\le N)\le \p(\exists s_1<s_2<\tau_{\Omega}\;;\; s_2-s_1>t/N\,,\,B_{[s_1,s_2]}\subset M_{4r})$$
Using lemma \ref{key1} we get that, for $R>4r$,  this probability is bounded by
$$cp^\frac{R}{r}+  \p\left(\exists s_1<s_2<\tau_{\Omega}\;;\; s_2-s_1>t/N\,,\,B_{[s_1,s_2]}\subset M_{4r}\cap\B(B_{s_1},R)\right),$$ and the statement of the lemma  \ref{Technical4} follows on taking $N=\left[\frac{R}{r}\right]$
\end{Proof}
\section{Further Comments}
We should point out that hypothesis  (\ref{hypothesis})  in theorem \ref{Th1} can be dropped; in this case, the same reasoning as in subsection \ref{lb} gives the lower bound $$\left(\frac{\#{\mathcal S}_{\sqrt t}}{\#{\mathcal S}_{\varepsilon}}\right)^{d_M}\left(\frac{{\sqrt t}}{\varepsilon}\right)^{d-2}.$$ The best upper bound is less evident; nevertheless a slight improvement of the above proofs gives
$$\p(\tau_{\Omega}>t)\le c\left(\log\frac{\sqrt t}{\epsilon}\right)^{cd}\sup_N\left(2^{-Nd}+\sum_{k=0}^N\left(\frac{\#{\mathcal S}_{\sqrt t/2^k}}{\#{\mathcal S}_{\varepsilon}}\right)^{d_M}\left(\frac{2^{-k}{\sqrt t}}{\varepsilon}\right)^{d-2}\right).$$

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