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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.

Define the stopped martingaleIt is nonnegative: before the first visit to $1$, the nearest-neighbour random walk is at most $0$, and afterwards the stopped position is $1$. Each $M_n$ is integrable, with $|M_n|\le n+1$. MoreoverThe indicator is ${\mathcal{F}}_n$-measurable and the increment has conditional mean zero, so $M_n$ is a martingale with $\mathbb{E}{M}_n=M_0=1$.

Part (a) shows $T_1<\infty$ almost surely, so every such path eventually has $M_n=0$. Thus $M_n\to 0$ almost surely. But $\mathbb{E}|M_n-0|=1$ for every $n$, so there is no convergence in L1. Any possible L1 convergence limit would agree with the almost sure limit. This is a nonnegative martingale with almost sure convergence but no convergence in L1, and hence it is not uniformly integrable.

For a nonnegative submartingale $(M_k{)}_{0\le k\le n}$ and $M_n^{\ast }={\max }_{0\le k\le n}{M}_k$, the weak form of the Doob maximal inequality isTo prove it, let $A_k$ be the event that $k$ is the first index with $M_k\ge u$. These events are disjoint, $A_k\in {\mathcal{F}}_k$, and their union is $A=\{M_n^{\ast }\ge u\}$. The submartingale property gives $\mathbb{E}[M_n\mid {\mathcal{F}}_k]\ge M_k$. Multiplying by $1_{A_k}$, taking expectations and summing yields $\mathbb{E}[M_n{1}_A]\ge {\sum }_k\mathbb{E}[M_k{1}_{A_k}]\ge u\mathbb{P}(A)$. Nonnegativity gives the second inequality.

The associated Doob Lp maximal inequality, for $p>1$ and $M_n\in L^p$, isIndeed $M_k\le \mathbb{E}[M_n\mid {\mathcal{F}}_k]$, so conditional Jensen inequality first ensures finite $p$th moments for all $M_k$ and hence for the finite maximum. Integrating the weak inequality against $p{u}^{p-2}du$ and using the Tonelli theorem givesThe last step is Holder inequality. Dividing, unless the maximum is zero, proves the stated norm bound. For an arbitrary martingale, its absolute value is a nonnegative submartingale by conditional Jensen inequality, so both corresponding absolute-maximum bounds follow.

The moment-generating function of a normal distribution givesIndependence of $X_{n+1}$ from ${\mathcal{F}}_n$ impliesEvery exponential here is integrable. Thus the process is a martingale exactly when $-\lambda +{\lambda }^2/2=0$. Its two roots are $0$ and $2$, and the unique positive root isThe resulting exponential martingale of a random walk is nonnegative and has expectation $1$ at every finite time.

Put $\tau =T_r\wedge T_R$. First establish almost sure finiteness. The Itô formula gives the martingale $|B_t{|}^2-2t$ for planar Brownian motion. Stop it at $t\wedge T_R$. The stopped path has norm at most $R$, soBy monotone convergence, $\mathbb{E}{T}_R<\infty$, and hence $\tau <\infty$ almost surely.

The function $h(y)=\log |y|$ is a harmonic function away from $0$: in polar coordinates its Laplacian is $h^{\prime \prime }(\rho )+h^{\prime }(\rho )/\rho =-{\rho }^{-2}+{\rho }^{-2}=0$. The Itô formula shows $h(B_{t\wedge \tau })$ is a local martingale. It is bounded between $\log r$ and $\log R$, so it is a true martingale and passage to $\tau$ is justified by dominated convergence. Writing $p={\mathbb{P}}_x(T_r<T_R)$ and using continuity at exit givesSolving yields the planar Brownian annulus hitting probabilityThe stopping argument stays away from the singularity of the logarithm until the calculation is complete.

An elementary predictable process with stopping-time intervals has the form $H={\sum }_{j=0}^{r-1}{h}_j{1}_{({\tau }_j,{\tau }_{j+1}]}$, where the stopping times are ordered and $h_j$ is bounded and ${\mathcal{F}}_{{\tau }_j}$-measurable. Deterministic endpoints give the usual simple predictable process. Define the zero-starting stochastic integral byCommon refinement and telescoping show that this definition is independent of the representation. Use $M_{\infty }$ for an infinite endpoint. The L2-bounded martingale convergence theorem gives convergence of $M_t$ in $L^2$ to $M_{\infty }$ and $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$.

Put $D_j=M_{{\tau }_{j+1}}-M_{{\tau }_j}$. The optional sampling theorem gives $\mathbb{E}[D_j\mid {\mathcal{F}}_{{\tau }_j}]=0$. For $j<k$, the variable $h_j{D}_j{h}_k$ is ${\mathcal{F}}_{{\tau }_k}$-measurable; hence $\mathbb{E}[h_j{D}_j{h}_k{D}_k]=0$. Applying the same theorem to the assumed martingale $M^2-[M]$ givesConsequently the Itô isometry follows:The unbounded stopping times cause no optional-sampling gap. The $L^2$ terminal representation makes $M$ uniformly integrable and makes the family of stopped $M^2$ uniformly integrable, by conditional Jensen applied to $M_{\infty }^2$. Also $\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2<\infty$, from monotone convergence theorem and the assumed square-compensator identity. Thus $M^2-[M]$ has the needed uniform integrability, and truncating stopping times and passing in $L^1$ is legitimate. Bounded coefficients are part of the simple-integrand convention; more general coefficients require the displayed square-integrability condition.