Variance of the exponential hitting time of a random walk

Question

Let $\left(S_n\right)_{n\geq 0}$ be a symmetric random walk on integers starting at 0. Fix a positive integer $a$, and define $\tau=\min\{n\geq 0:|S_{n}|=a\}$. How could we compute the variance $Var\left(e^{-\tau}\right)$? I thought that the hitting time $|S_n|=a$ suggests using the quadratic martingale $S_{n}^2-n$, but trying to construct a 'product martingale' by normalizing terms $e^{-S_n^2+S_{n-1}^2}$ wasn't particularly fruitful. Also, I am aware of the way to compute $\mathbb{P}\left(\tau=2k-1\right)$ for each $k$ but I couldn't find an obvious modification that gives the variance (and directly calculating the variance using the probabilities would be too complicated). Any hints or insights are appreciated. Many thanks in advance!

Answer 1

There is an exponential, so it would be nice to use the exponential martingale! Note that if you write $S_n = X_1 + \cdots + X_n$, then for any $t \in \mathbb{R}$, $$ \mathbb{E} \left ( e^{t S_n} \right ) = \mathbb{E} \left ( e^{t X_1} \right )^n = \cosh^n t, $$ and thus, by independence, $M_n = e^{t S_n} (\cosh t)^{-n}$ is a martingale. This martingale stopped at $\tau$ is bounded, so the optional stopping theorem applies without issue, and therefore $$ \mathbb{E} \left ( e^{t S_{\tau}} (\cosh t)^{- \tau} \right ) = 1. $$ By symmetry, $S_{\tau}$ is independent of $\tau$ and is $\pm a$ with probability $1/2$, so you get that $$ \mathbb{E}((\cosh t)^{-\tau}) = \frac{1}{\cosh (t a)}. $$ Now, if you take $t$ such that $\cosh t = e$, you get $\mathbb{E}(e^{-\tau})$, and similarly to get the second moment and the variance.