Let's say each $X_i$ is a simple random variable taking on values 1 or -1 with probability $1/2$ each. Then $S_n = \sum_{i=1}^{n} X_i$ is a random walk. Set $T = \min \{n\in\mathbb{N} \, : \, S_n = 1\}$. One can show that $E[T] = \infty$.
Can we choose the $X_i$ instead in such a way that $E[T] < \infty$? They don't necessarily have to be identically distributed, but I would be interested to see if we can do it with the restriction that each $X_i$ takes on two possible values, each with probability $1/2$.
Let $X_n$ be a simple random variable that takes on values $2^{n-1}$ or $-2^{n-1}$, each with probability $1/2$.
For each possible random walk, there is some smallest $n$ such that $X_n$ is a positive step (the sole exception, where all $X_i$ are negative, occurs with probability zero anyway). Then for all $X_k$ where $k<n$, $X_k = -2^{k-1}$. So $$S_n = \sum_{i=1}^{n} X_i = 2^{n-1} - (2^{n-2} + 2^{n-3} + \cdots + 1) = 1$$ Since $S_k$ is negative for $k<n$, in this case, $T = n$.
In general, $P(T = n) = P(X_n > 0 \text{ and } X_k < 0 \text{ for all } k<n) = 1/2^{n}$. We then compute $$E[T] = \sum_{k=1}^{\infty} \frac{k}{2^k} = 2 < \infty$$