# The probability that a killed random walk on $[-N,N]$ escapes before dying

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 $$N \to \infty.$$

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 $$$$p_0 = \frac 13 p_1 + \frac 13 p_{-1}.$$$$ More generally, for $$-N + 1 \leq i \leq N-1$$ we have $$$$p_i = \frac 13 p_{i+1} + \frac 13 p_{i-1}$$$$ with boundary conditions $$p_N = p_{-N} = 1$$. Solving this difference equation we get $$$$p_n = \frac{\psi^n + \psi^{-n}}{\psi^N + \psi^{-N}} = \frac{\cosh(n \log \psi)}{\cosh(N \log \psi)}.$$$$ where $$\psi = \frac{3 + \sqrt{5}}{2}$$. In particular, for $$n=0$$ we obtain $$$$p_0 = \frac{1}{\cosh(N \log \psi)}.$$$$