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The generating function is $F_0(s) = \sum_{n=1}^{\infty}f_0(n)s^n$.

$f_0(n)$ is the probability of "first" return after n steps.

If the particle is certain to return to origin, i.e.

$F_0(1) = \sum_{n=1}^{\infty}f_0(n) = 1$

So the expected time is infinity.

$F_0^{'}(1) = \sum_{n=1}^{\infty}nf_0(n) = \infty$

However, I lack the intuition why it is infinity but not finite number. I understand for nonsymmetric it is infinity, but wonder what is the intuition that for symmetric it is also infinity.

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    A related question is here (just use $p_0=1/2$ for symmetric case): http://math.stackexchange.com/questions/2087996/bounded-random-walk-on-one-side-only-are-you-guaranteed-to-hit-the-bound/2088236#2088236 – 2017-01-23
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    Oh, for expectations, the random time $T$ needed to go one step left satisfies $$E[T] = \underbrace{E[T|\mbox{first go left}]}_{1}(1/2) + \underbrace{E[T|\mbox{first go right}]}_{1 + 2E[T]}(1/2)$$ and the only solution is $E[T]=\infty$. – 2017-01-23

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