Show that hitting time of random walk has finite expectation Let $(S_n)_{n\geq0}$ be a random walk in $\mathbb{R}$, starting from $0$, with steps of mean $\mu$ and variance $\sigma^2\in(0,\infty)$. Fix $a,b\in\mathbb{R}$ with $a<0<b$ and set $$T=\inf\{t\geq0:S_n\leq a \textrm{ or } S_n\geq b\}.$$
I am tasked with showing that $\mathbb{E}(T)<\infty$, but don't know where to start. Advice would be greatly appreciated!
 A: I'll look at the simple random walk case first: steps are $\pm1$ each with probability $\tfrac12$.
Wheenver the walk is inside $(a, b)$, if it takes $L := b-a$ steps to the right, then it certainly exits $(a, b)$ via the $b$-endpoint. What is the probability of this? Well, it's just $2^{-L}$. Thus, $T$ is dominated by $L \cdot \operatorname{Geom}(2^{-L})$:
$$
\Pr( T \ge k*L ) \le \Pr( \operatorname{Geom}(2^{-L}) \ge k ).
$$
In particular, $E(T) \le L \cdot E( \operatorname{Geom}(2^{-L}) ) < \infty$.
For the general case, there must be some $\epsilon > 0$ such that a step of distance at least $\epsilon$ is made to either the right or the left with probability at least $\epsilon$; this holds because the variance is positive, ie some distance is moved with positive probability. Wlog assume that this step is to the right. If we take $L/\epsilon$ of these steps, then we get at least to $b$. So now, $T \lesssim \operatorname{Geom}(\epsilon^{L/\epsilon}) \cdot L/\epsilon$. This bound is much cruder, but you still have finite expectation!
