I'm trying to solve this problem and don't know where to start. If someone could prove it or tell me how or point me to any relevant information I'd very much appreciate it.
Let $(s_n)_{n\geq0}$ be a 1-dimensional, unbiased random walk. For $a\in\mathbb Z^{*}$, let $T_a=\inf\left\{n\geq0:s_n=a\right\}$. Prove that $\mathbb E(T_a)=\infty$.
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By symmetry, it suffices to show that $T_1$ is not integrable. But $T_1=1$ with probability $\frac12$ and, also with probability $\frac12$, $T_1=1+T'+T''$, where $T'$ and $T''$ are i.i.d. copies of $T_1$. To wit, if $S_1=1$, $T_1=1$ but, if $S_1=-1$, one already spent one step to go to $-1$ and now one must come back to $0$, which takes $T'$ time steps, and again go to $+1$, which takes $T''$ time steps. Thus, $$ \mathbb E(T_1)=\tfrac12\cdot1+\tfrac12\cdot(1+\mathbb E(T')+\mathbb E(T'')) $$ Note that the identity above is entirely rigorous as long as one only adds nonnegative terms, finite or infinite. Now, use $\mathbb E(T')=\mathbb E(T'')=\mathbb E(T_1)$ to get $$\mathbb E(T_1)=1+\mathbb E(T_1)$$ and note that this last identity has a unique solution in $\overline{\mathbb R_+}=[0,+\infty]$, which is $$\mathbb E(T_1)=+\infty$$ Perhaps more constructively, the identity above also completely determines the distribution of $T_1$ since, for every $s$ in $(0,1]$, it yields $$E(s^{T_1})=\tfrac12\cdot(s+s\cdot E(s^{T_1})^2)$$ which shows that $$E(s^{T_1})=\frac{1-\sqrt{1-s^2}}s$$
If we restrict to $a\in\mathbb Z^+$, the expected first hitting time $t_a=\mathbb E(T_a)$ has to satisfy the recurrence
$$ t_a=1+\frac12\left(t_{a-1}+t_{a+1}\right)\;. $$
Since we can solve this for $t_{a+1}$, there can't be a least $a$ for which $t_a$ is infinite, so $t_a$ is either finite for all $a$ or infinite for all $a$. If it's finite for all $a$, it has to solve the above recurrence. The solutions are $t_a=-a^2+ma+n$, which would imply negative first hitting times for sufficiently large $a$. It follows that $t_a$ is infinite for all $a$.