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Let $\tau =T_{-a}\wedge T_b$ for the simple symmetric random walk. From any interior point, $a+b$ successive $+1$ increments force exit through $b$. The conditional probability of this block is $2^{-(a+b)}$, so a blockwise geometric tail bound from a uniform escape probability gives $\tau <\infty$ almost surely.

Apply the bounded optional stopping theorem to $\tau \wedge n$. Since $S_n$ is a martingale, ${\mathbb{E}}_0{S}_{\tau \wedge n}=0$. The stopped positions lie in $[-a,b]$, so the bounded convergence theorem gives ${\mathbb{E}}_0{S}_{\tau }=0$. If $h={\mathbb{P}}_0(T_{-a}<T_b)$, nearest-neighbor jumps ensure $S_{\tau }=-a$ on that event and $S_{\tau }=b$ otherwise. ThusNext apply the bounded optional stopping theorem to the square-minus-time martingale of a simple symmetric random walk:Again the right side converges by the bounded convergence theorem, while the left side increases to ${\mathbb{E}}_0\tau$ by the monotone convergence theorem. ThereforeThis argument justifies passage to the unbounded stopping time rather than applying optional stopping without an integrability check.

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ of the simple symmetric random walk. The event $\{T_1\le n\}={\bigcup }_{k=0}^n\{S_k=1\}$ belongs to ${\mathcal{F}}_n$, so $T_1$ is a stopping time.

For a positive integer $m$, set ${\tau }_m=T_1\wedge T_{-m}$. This exit time from the finite interval is integrable: in each block of $m+1$ steps there is a probability at least $2^{-(m+1)}$ of all steps being positive, which forces exit if it has not already occurred. Consequently the survival probabilities have a geometric upper bound.

We use the bounded-time optional stopping theorem: if $M$ is a martingale and $\sigma$ a bounded stopping time, then $\mathbb{E}{M}_{\sigma }=\mathbb{E}{M}_0$. Apply it at ${\tau }_m\wedge n$ to $S_n$ and to the square-minus-time martingale of a simple symmetric random walk $S_n^2-n$. The stopped positions lie in $[-m,1]$, so dominated convergence and monotone convergence giveWriting $p_m=\mathbb{P}(T_1<T_{-m})$, the first equation becomes $p_m-m(1-p_m)=0$, hence $p_m=m/(m+1)$. The second gives $\mathbb{E}{\tau }_m=p_m+m^2(1-p_m)=m$. Since ${\tau }_m\le T_1$, its expectation satisfies $\mathbb{E}{T}_1\ge m$ for every $m$. ThereforeNevertheless $\mathbb{P}(T_1<\infty )\ge p_m\to 1$. Thus the infinite mean first passage of a simple symmetric random walk occurs despite almost sure finiteness; using unrestricted optional stopping directly at $T_1$ would be unjustified.

For the simple symmetric random walk, $S_n$ is a martingale and the square-minus-time martingale of a simple symmetric random walk isIndeed, conditioning on ${\mathcal{F}}_n$ and using $\mathbb{E}[X_{n+1}]=0$ and $X_{n+1}^2=1$ gives $\mathbb{E}[S_{n+1}^2\mid {\mathcal{F}}_n]=S_n^2+1$.

Apply the optional sampling theorem for a supermartingale to the bounded stopping time $T\wedge n$:Thus the stopped martingale $(S_{T\wedge n})$ is bounded in $L^2$. The L2 martingale convergence theorem gives convergence in $L^2$, and because $T<\infty$ almost surely its limit is $S_T$. Meanwhile the monotone convergence theorem gives $\mathbb{E}[T\wedge n]\to \mathbb{E}[T]$. Therefore

Finite mean is essential. Let $T=\inf \{n\ge 1:S_n=0\}$ be the first return to zero. The one-dimensional simple symmetric random walk is recurrent, so $T<\infty$ almost surely, but its first-return time has infinite mean. Since $S_T=0$,