Expectation of hitting time of random walk with I.I.D. Geometric increments? Let $X_k$ be i.i.d. Geometric distributed with success probability $p$(i.e. $\mathbb{E}X_k=1/p$), and $S_n=\sum_{k=1}^{n}X_k$.
I want to compute the expectation of the hitting time of $(N,\infty)$, i.e. $\mathbb{E}\tau$ with $\tau:=\inf\{n\ge 1:S_n> N\}$.
Since the "overshot" $S_\tau$ might be far larger than $n$, the simple martingale approach using $M_n:=S_n-\frac np$ does not seem useful...
Also, how to compute $\mathbb{E}S_\tau$?
 A: The fact that the increments are Geometric is really important to this problem.
In particular, since the increments are Geometric, they have the memoryless property, and so the "overhang" $S_\tau - N$ has a $\mathrm{Geometric}(p)$ distribution (ie. $S_\tau - N\sim \mathrm{Geometric}(p)$ so that $\mathbf{E} S_\tau = N + 1/p$).
Specifically, since the $X_i$ are Geometric, they can be thought of as increments of a Bernoulli process.
That is, let $Y_1, Y_2, \dotsc$ be iid $\mathrm{Bernoulli}(p)$ random variables.
Then $S_n$ is number of the trial in which we see the $n$th success and $X_i$ is the number of trials between the $(i-1)$th and $i$th success.
Formally,
$$
S_n
= \min\Bigl\{k : \sum_{i=1}^k Y_i \geq n\Bigr\},
\quad
X_i = S_i - S_{i-1}.
$$
Now, the overhang random variable $S_\tau - N$ being greater than some $x$ means exactly that there are no successes in $\{N+1, \dotsc, N+x\}$.
That is,
\begin{align*}
\Pr(S_\tau - N > x)
= \Pr(Y_{N+1} = \dotsc = Y_{N+x} = 0)
= (1 - p)^x,
\end{align*}
and so $S_\tau - N \sim \mathrm{Geometric}(p)$.
As @aaron-montgomery mentions in the comments, the overhang usually doesn't have the same distribution as an increment.
However, in the special case of the memoryless distributions (Geometric and Exponential), it is the same.
Thus, we can use the usual optional stopping technique (since $\tau$ is bounded by $N$) to find that
\begin{align*}
0
&= \mathbf{E}(S_\tau - \tau/p) \\
&= N + \frac{1}{p} - \frac{\mathbf{E} \tau}{p}, \\
\mathbf{E}\tau
&= pN + 1.
\end{align*}

Actually, we can use the Bernoulli process to make a simpler computation.
In particular, since $\tau$ is the first $n$ such that $S_n$ exceeds $N$, we have that $\tau-1$ is the number of successes out of the first $N$ Bernoulli trials.
That is
$$
\tau - 1
\sim \mathrm{Binomial}(N, p)
$$
and so
$$
\mathbf{E} \tau
= pN + 1,
$$
agreeing with our previous result!

(note: In a previous version of this answer, I misread your definition of $\tau$ as $\inf\{n : S_n \geq N\}$, which led to a different answer)
