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For a simple symmetric random walk starting at zero, let $T_1$ be its first hitting time of $1$. The infinite mean first passage of a simple symmetric random walk shows that $T_1<\infty$ almost surely but $\mathbb{E}[T_1]=\infty$. Since $S_{T_1}=1$, assuming a finite expected value for $T_1$ would let Wald's equation assert $1=0\mathbb{E}[T_1]=0$, a contradiction. The formal expression $0\cdot \infty$ is not an allowed use of the theorem.^[ref]

Compare $T_{\pm 2}=\inf \{n\ge 0:|S_n|=2\}$. From every surviving position in $\{-1,0,1\}$, three consecutive positive increments force absorption within three steps. Each fresh block has probability $1/8$ of this event, so $\mathbb{P}(T_{\pm 2}>3k)\le (7/8{)}^k$ and $\mathbb{E}[T_{\pm 2}]\le 24$. This establishes finiteness before invoking the theorem. If $e_j$ is the average remaining time from position $j$, the law of total expectation givesso $\mathbb{E}[T_{\pm 2}]=4$. Symmetry gives equal exit probabilities at $-2$ and $2$, and $\mathbb{E}[S_{T_{\pm 2}}]=0$, as Wald's equation predicts. This is a small gambler's ruin problem with two absorbing barriers.

Even for $T_1$, each bounded stopping time $T_1\wedge n$ satisfies $\mathbb{E}[S_{T_1\wedge n}]=0$. The stopped process tends to $1$ almost surely, but its expected value does not tend to $1$. Rare long negative excursions retain the balancing contribution. The existing stopped-walk martingale with almost sure but not L1 convergence explains the related failure of uniform integrability; almost sure convergence alone does not permit exchanging a limit and an expected value.

Take a complete, right-continuous filtration $({\mathcal{F}}_t)$, an increasing family of sigma-algebras. Adapted means that $X_t$ is ${\mathcal{F}}_t$-measurable at each $t$; càdlàg means right-continuous with left limits. A local martingale is an adapted càdlàg process $X$, with integrable initial value, for which there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $X_t^{{\tau }_n}=X_{t\wedge {\tau }_n}$ is a true martingale. Such a sequence is a localizing sequence. A true martingale is integrable at each time and satisfies $\mathbb{E}(X_t\mid {\mathcal{F}}_s)=X_s$ for $s\le t$.

The necessary and sufficient condition is that $X$ is a Class DL process:A family $\mathcal{Z}$ is uniformly integrable if its members are integrable and ${\sup }_{Z\in \mathcal{Z}}\mathbb{E}(|Z|1_{\{|Z|>K\}})\to 0$ as $K\to \infty$. A stopping time $\tau$ satisfies $\{\tau \le t\}\in {\mathcal{F}}_t$ for all $t$.

For necessity, on $[0,T]$ the optional stopping theorem gives $X_{\tau }=\mathbb{E}(X_T\mid {\mathcal{F}}_{\tau })$; conditional expectations of one integrable variable form a uniformly integrable family. For sufficiency, $X_{t\wedge {\tau }_n}\to X_t$ almost surely, and Class DL upgrades this convergence to $L^1$ on each finite horizon. Pass to the limit in $\mathbb{E}(X_{t\wedge {\tau }_n}\mid {\mathcal{F}}_s)=X_{s\wedge {\tau }_n}$ to obtain the martingale identity. Requiring uniform integrability of all times on the entire half-line would be stronger than necessary; Brownian motion is already a counterexample to that stronger requirement.

First justify that the logarithm is defined at all times from one onward. For planar Brownian motion started at $z\ne 0$, stop on exiting the annulus $\epsilon <|x|<R$, where $\epsilon <|z|<R$. This exit time $\rho$ is finite almost surely: the stopped identity $|B_t{|}^2-2t$ gives $\mathbb{E}(t\wedge \rho )\le R^2/2$. The function $f(x)=\log |x|$ is harmonic on the annulus, so its stopped process is a bounded martingale. Optional stopping givesHitting zero before the outer circle would require hitting every inner circle first. It therefore has probability zero. Taking a countable sequence of outer radii tending to infinity proves that planar Brownian motion avoids a fixed point. Since $B_1$ has a density and is nonzero almost surely, condition on ${\mathcal{F}}_1$ to apply this result to the future path.

For $x\ne 0$, $\nabla f(x)=x/|x{|}^2$ and $\Delta f(x)=0$ in two dimensions. The Itô formula on annuli therefore givesTo localize explicitly, let ${\tau }_n=\inf \{t\ge 1:|B_t|\notin (1/n,n)\}$. These stopping times increase to infinity almost surely: on each compact time interval the continuous path has a finite maximum and, by point avoidance, a positive minimum radius. Up to ${\tau }_n$, the stochastic integrand has norm at most $n$. Its stopped integral is square-integrable on every finite horizon. The initial value $\log |B_1|$ is integrable, as follows from the Rayleigh-density calculation in part d. Hence $X^{{\tau }_n}$ is a true martingale from time one and $X$ is a local martingale. This also gives its quadratic variation from time one as ${\int }_1^t|B_s{|}^{-2}ds$.

Use the tree's natural filtration. The stopping times take values in the process horizon $\{0,1,2\}$. Construct a dominating supermartingale $Y$ by backward comparison, the finite-horizon Snell envelope. At time two put $Y_2={\xi }_2$. At time one the continuation values areCompare these with immediate rewards $11$ and $7$ to obtain $Y_1=12$ on the upper branch and $Y_1=7$ on the lower branch. Finally setThese definitions explicitly give $Y_t\ge {\xi }_t$, $Y_0=\mathbb{E}{Y}_1$, and $Y_1\ge \mathbb{E}[Y_2\mid {\mathcal{F}}_1]$. Thus $Y$ is a supermartingale. By part (ii), its stopped process is a supermartingale, so for every allowed $\tau$,The reward process itself need not be a supermartingale; the constructed majorant is what proves the bound.

Let $Q$ be the given risk-neutral measure, and write $B_t=B_0{e}^{rt}$. Take $V$ to be classical on $t<T$ and continuous at its prescribed terminal value. By Itô formula and the pricing equation for a local volatility model,soThe discounted added asset is a local martingale, as is $S_t/B_t$, while $B_t/B_t=1$. Consequently $Q$ is an equivalent local martingale measure for all three assets.

Here is the arbitrage argument, including the usual admissible trading strategy condition. If $X$ is the discounted wealth of a self-financing strategy, it is a stochastic integral against the discounted asset prices, plus its initial value, and hence a local martingale. Assume $X_t\ge -c$ for a fixed $c$, as required for an admissible trading strategy. For a localizing sequence ${\tau }_n$, the stopped process $X^{{\tau }_n}$ is a martingale. Apply the conditional Fatou lemma to $X_{t\wedge {\tau }_n}+c\ge 0$: for $s\le t$, continuity and the stopped martingale identity give ${\mathbb{E}}_Q[X_t+c\mid {\mathcal{F}}_s]\le X_s+c$. Thus $X$ is a supermartingale. An arbitrage would have $X_0=0$, $X_T\ge 0$ and positive probability of $X_T>0$. Equivalence of $P,Q$ makes that probability positive under $Q$, contradicting ${\mathbb{E}}_Q[X_T]\le 0$. Therefore the augmented market has no admissible arbitrage.

Nonnegativity of $V$ gives the needed price process, but does not by itself make its discounted local martingale a true martingale. If $V$ is bounded, the latter follows by bounded localization, and its price is also $B_t{\mathbb{E}}_Q[g(S_T)/B_T\mid {\mathcal{F}}_t]$. The arbitrage proof above applies to the stated nonnegative classical solution without assuming that additional bound.

Fix $t$ and put $\sigma =t\wedge \tau$. The stopping time property implies that $\sigma$ is an ${\mathcal{F}}_t$-measurable random variable with values in $[0,t]$: for $s<t$,and for $s\ge t$ the event is the whole space. By part (c), the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable.

The evaluation map $\omega \mapsto (\sigma (\omega ),\omega )$ is measurable from $(\Omega ,{\mathcal{F}}_t)$ to this product space: the inverse image of a measurable rectangle $A\times C$ is $\{\sigma \in A\}\cap C$. Composing it with the jointly measurable stochastic process gives an ${\mathcal{F}}_t$-measurable random variableSince this holds for every fixed $t$, the stopped process is adapted. If path regularity holds only almost surely, first apply the proof to its pathwise regular representative; with a completed filtration, the original stopped variable differs only on a null event and is also ${\mathcal{F}}_t$-measurable. The adaptedness of a stopped right-continuous process requires neither boundedness of $\tau$ nor a martingale assumption; $\tau =\infty$ is harmless because $t\wedge \tau \le t$.

The first entrance time ${\tau }_K$ is a stopping time, because $\{{\tau }_K\le n\}={\bigcup }_{j=0}^n\{M_j\le -K\}\in {\mathcal{F}}_n$. Before this time $M_n>-K$; at entrance, the bounded increments give $M_{{\tau }_K}>-K-L$. Thus the stopped martingaleis a nonnegative martingale. Here the stopped process is a martingale since its increment is $1_{\{{\tau }_K>n\}}(M_{n+1}-M_n)$, whose conditional expectation given ${\mathcal{F}}_n$ is zero.

The martingale convergence theorem states that a nonnegative martingale converges almost surely to a finite integrable random variable; indeed its expectations are constant, giving the required bound in the L1 norm. Apply this theorem to $(N_n)$. On $\{{\tau }_K=\infty \}$ the stopped process equals the original one at every time. HenceThe overshoot bound is why the assumption of bounded increments matters.

Put $P_n=X_n{Y}_n{Z}_n$. Before $\tau$, all three players are active. Each unordered pair is selected with probability $1/3$, and each direction of transfer then has conditional probability $1/2$. Given the current holdings $(x,y,z)$, a transfer within the first pair has mean productFor transfers within the other two pairs the analogous values are $xyz-y$ and $xyz-x$. Averaging over pairs givesAt and after elimination the product is zero, since an eliminated player never re-enters. In particular the stopped process has incrementThe indicator function is ${\mathcal{F}}_n$-measurable, and the expression in parentheses has zero conditional expectation on its support. The stopped process is adapted and integrable: at fixed $n$, its absolute value is bounded by $N^3+nN/3$. HenceThis is the product martingale for three-player token exchange.

Three players with positive integer holdings exchange a token by choosing an unordered active pair uniformly, then choosing the transfer direction fairly. Let $N$ be the conserved total, and $\tau$ the stopping time of the first elimination. Given holdings $(x,y,z)$ before $\tau$, the conditional expectation of the change of their product is $-(x+y+z)/3=-N/3$: transfers in each pair decrease the mean product by the third holding. Thus the displayed stopped process is a martingale. Its nonnegative product term gives $\mathbb{E}[n\wedge \tau ]\le 3abc/N$, so the monotone convergence theorem first proves integrability of $\tau$. The bounded product then vanishes by the dominated convergence theorem, yielding $\mathbb{E}[\tau ]=3abc/N$. Continuing the deterministic time term beyond elimination instead produces a submartingale.

The stopped stochastic process follows $X$ until the stopping time $\tau$ and thereafter keeps the value $X_{\tau }$ when $\tau$ is finite. For a pathwise right-continuous adapted process, the stopped process is adapted, by progressive measurability and measurable evaluation at $t\wedge \tau$. When $X$ is a martingale, further stopping and integrability conditions determine whether its stopped process is a martingale.