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Let $X_t$ be a continuous time simple random walk in $\mathbb{Z}$ starting at $0$, let $\tau^*$ be an exponential r.v of parameter $1$. What is the probability $$ \mathbb{P}(\tau ^* \ge \tau_{N})? $$ Where $\tau_N = \inf \{t \in \mathbb{R}: X_t \in \{-N,N\} \}$.

My first attempt was to use the optional stopping theorem, but it didn't seem to be enough to extract the probability I am interested in.

EDIT: If an analytical representation is not possible, is there asymptotics as $N \to \infty.$

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Let $p_n$ denote the probability of escape before being killed when started from $n \in \mathbb{Z}$. We are interested in the value of $p_0$. By the memoryless property of the exponential distribution, conditioning on the first step we get \begin{equation} p_0 = \frac 13 p_1 + \frac 13 p_{-1}. \end{equation} More generally, for $-N + 1 \leq i \leq N-1$ we have \begin{equation} p_i = \frac 13 p_{i+1} + \frac 13 p_{i-1} \end{equation} with boundary conditions $p_N = p_{-N} = 1$. Solving this difference equation we get \begin{equation} p_n = \frac{\psi^n + \psi^{-n}}{\psi^N + \psi^{-N}} = \frac{\cosh(n \log \psi)}{\cosh(N \log \psi)}. \end{equation} where $\psi = \frac{3 + \sqrt{5}}{2}$. In particular, for $n=0$ we obtain \begin{equation} p_0 = \frac{1}{\cosh(N \log \psi)}. \end{equation}

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