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A bounded stopping time is a stopping time bounded almost surely by a deterministic finite constant. In discrete time, a bound $T\le N$ makes $M_T$ integrable whenever $M_0,\dots ,M_N$ are integrable random variables, and the optional stopping theorem applies to a martingale without additional limiting hypotheses. Almost-sure finiteness alone is weaker than boundedness.

Let $X$ be an integrable adapted càdlàg process under the usual conditions for a filtration. Then $X$ is a martingale if and only if $X_T$ is integrable and $\mathbb{E}{X}_T=\mathbb{E}{X}_0$ for every bounded stopping time $T$.

The forward implication is the optional stopping theorem. Conversely, fix deterministic $s<t$ and $A\in {\mathcal{F}}_s$, and use the stopping time $U=s{1}_A+t{1}_{A^c}$. Comparing $\mathbb{E}{X}_U$ with $\mathbb{E}{X}_t$ gives $\mathbb{E}[(X_t-X_s)1_A]=0$. Since this holds for every such $A$, it is the defining conditional expectation identity of a martingale. The converse uses only integrable adapted coordinates; the stated path assumptions justify the forward assertion at all bounded random times.

Let $a<i<b$ and $a<j<b$ be integers. A simple symmetric random walk started at $i$ exits the integer interval at $\tau =\inf \{n\ge 0:S_n\in \{a,b\}\}$. Its expected occupation of $j$ before exit, including its initial state, isA block of $b-a$ consecutive positive steps forces exit, so the geometric tail bound from a uniform escape probability proves $\tau <\infty$ almost surely. Optional stopping at the bounded stopping time $n\wedge \tau$, followed by the bounded convergence theorem for $S_{n\wedge \tau }\in [a,b]$, gives the gambler's ruin probabilitiesTranslate the absolute value martingale of a simple symmetric random walk from zero to $j$. Bounded-time optional stopping yieldsUse monotone convergence on the increasing visit count and the bounded convergence theorem on the right. Substituting the exit probabilities giveswhich simplifies to the displayed expression by considering $i\le j$ and $i\ge j$. In particular, $G_{a,b}(0,0)=-2ab/(b-a)$ when $a<0<b$.

The proof establishes finite expected value of the visit count. Therefore each translated stopped absolute-value martingale is dominated in absolute value by a constant plus that total count, an integrable random variable; its family of values is uniformly integrable. This supplies the limit justification without assuming uniform integrability in advance.

For a random walk with independent and identically distributed random variables as increments and finite moment-generating function at $\theta$, put $\kappa (\theta )=\log \mathbb{E}{e}^{\theta X_1}$. Then $e^{\theta S_n-n\kappa (\theta )}$ is a martingale by conditioning on the next independent increment. In particular, if $r>0$ and $\mathbb{E}{r}^{X_1}=1$, then $r^{S_n}$ is a martingale. Stopping at a finite interval exit can make this process bounded, allowing the dominated convergence theorem to justify passage from bounded stopping times to the exit time.

In discrete time, let $V$ be a supermartingale and let $\sigma \le \tau$ be bounded stopping times. Then the stopped values are integrable andFor a martingale, equality holds. The same bounded-time statement holds for a càdlàg supermartingale under the usual conditions for a filtration. In discrete time, expand stopped increments and use that $1_{\{\tau \ge k\}}$ is known at time $k-1$; continuous time follows by approximation and the path regularity.

Almost-sure finiteness of $\tau$ alone is insufficient. A simple symmetric random walk started at zero hits one almost surely, but its value there is one rather than its initial expectation zero. Conditions such as uniform integrability of a stopped uniformly integrable martingale, or bounded-increment optional stopping with integrable time, justify appropriate unbounded-time versions.

The random walk $S_n$ is a martingale, since its integrable increments are independent of the past and have mean zero. For the bounded stopping time $n\wedge \eta$, the bounded optional stopping theorem, proved in the next question, gives $\mathbb{E}{S}_{n\wedge \eta }=x$.

If $0<x<r$, the value immediately before exit lies in $(0,r)$, soBefore exit the stopped value has absolute value less than $r$, and after exit it equals $S_{\eta }$. Consequently $|S_{n\wedge \eta }|\le r+|X_{\eta }|$ for every $n$. This is an integrable dominating random variable by the preceding part. Since $\eta <\infty$ with probability one, the dominated convergence theorem givesFor $r\le x$, $S_{\eta }=S_0=x$ directly, without any convention about $X_0$. Thus the expected stopped position is well-defined and has the stated value for every $r>0$.

Let $T\le N$ for a deterministic integer $N$. The stopped martingale has the finite-sum representationEach summand is integrable, and $\{T\ge k\}=\{T>k-1\}\in {\mathcal{F}}_{k-1}$ by the stopping time property. Therefore the conditional expectation identity for a martingale givesTaking expectations in the finite sum proves the bounded optional stopping theorem:No limiting argument or uniform-integrability assumption is needed for a bounded stopping time.

Put $Z_n=X_{n\wedge T}$. The stopped martingale in discrete time identityshows that $Z$ is a martingale: the indicator function is ${\mathcal{F}}_n$-measurable and the increment has zero conditional expectation. Integrability follows because $Z_n$ is selected from the finitely many integrable values $X_0,\dots ,X_n$.

For the bounded stopping time $\tau =n\wedge T$, the optional stopping theorem in its conditional form givesHere ${\mathcal{F}}_{\tau }$ is the stopping-time sigma-algebra. Indeed, for $A\in {\mathcal{F}}_{\tau }$, partition $A$ into $A\cap \{\tau =k\}\in {\mathcal{F}}_k$ and apply the martingale identity $\mathbb{E}[X_n\mid {\mathcal{F}}_k]=X_k$ on each piece. This proves the displayed conditional expectation identity directly.

Let $C={\sup }_n\mathbb{E}|X_n|<\infty$ and $A_n=\{|Z_n|>K\}\in {\mathcal{F}}_{n\wedge T}$. Conditional absolute-value domination and the Markov inequality give $\mathbb{P}(A_n)\le C/K$ and, for any $R>0$,First choose $R$ using the uniform integrability of $X$, and then choose $K$. The estimate is uniform in $n$, proving uniform integrability of a stopped uniformly integrable martingale. No finiteness assumption on $T$ is needed.

Recall the stopping-time sigma-algebra:For the proposed random time,The first set belongs to ${\mathcal{F}}_t$. Since $S\le T$, the second can be writtenwhich also belongs to ${\mathcal{F}}_t$. Therefore $U$ is a stopping time, as in pasting ordered stopping times. Moreover $S\le U\le T$, so $U$ is a bounded stopping time.

Fix deterministic $0\le s<t$ and $A\in {\mathcal{F}}_s$. The pasting ordered stopping times argument shows that $U=s{1}_A+t{1}_{A^c}$ is a bounded stopping time. The given stopped-expectation property, applied to $U$ and to the deterministic stopping time $t$, yieldsThis holds for every $A\in {\mathcal{F}}_s$. Since $X_s$ is ${\mathcal{F}}_s$-measurable and both time values are integrable, it is exactly the defining test for conditional expectation:Hence $X$ is a martingale. This proof of the characterization of a martingale by bounded continuous-time stopped expectations uses only deterministic two-time pastings; the càdlàg assumption and the usual conditions for a filtration are stronger than needed for this implication.

For the converse in the characterization of a martingale by stopped expectations, fix $n\ge 0$ and $A\in {\mathcal{F}}_n$. Define the bounded stopping timeIndeed, its only nontrivial sublevel event is $\{T_A\le n\}=A^c\in {\mathcal{F}}_n$. Applying the assumed stopped-expectation equality to $T_A$ and to the deterministic stopping time $n$ givesThis holds for every $A\in {\mathcal{F}}_n$. Since $M_n$ is adapted and both variables are integrable, the defining test-event property of conditional expectation yieldsTogether with the given adaptation and integrability, this is exactly the martingale property.

Use the nonnegative martingale $(Y_n)$ and filtration $({\mathcal{H}}_n)$ established above. Fix a finite $n$ and letFor $j<n$, $\{T\le j\}$ is the union of the threshold-crossing events up to $j$, so $T$ is a bounded stopping time. The optional stopping theorem for a martingale at a bounded stopping time gives $\mathbb{E}{Y}_T=\mathbb{E}{Y}_0=1$. On $H$ one has $Y_T\ge \lambda$; on its complement one still has $Y_T\ge 0$. ThereforeOnly a bounded stopping time has been used, so no extra uniform integrability assumption at an unbounded hitting time is implicit. This is also the Doob maximal inequality for a nonnegative submartingale specialized to a mean-one martingale.

The optional stopping theorem at a bounded stopping time states that an integrable martingale $Z_n$ satisfies $\mathbb{E}{Z}_T=\mathbb{E}{Z}_0$ whenever $T\le n$ for a fixed finite $n$. Apply this first to $S$ at $T=n\wedge \tau$. The simple symmetric random walk cannot cross an integer endpoint without hitting it, soThe preceding part gives $\tau <\infty$ almost surely, hence $S_{n\wedge \tau }\to S_{\tau }$ almost surely. The constant bound $C=\max \{-a,b\}$ permits the dominated convergence theorem, yielding $\mathbb{E}{S}_{\tau }=0$. Since $S_{\tau }$ is either $a$ or $b$, solving $ap+b(1-p)=0$ gives its full probability distribution:Now letThis counts the visit at time $0$, and excludes the exit time. Apply the same bounded-time optional stopping theorem to $M_n=|S_n|-L_n$. For each $n$ it givesThe visit counts increase to $L_{\tau }$, so monotone convergence applies to the left. Dominated convergence applies to the right using $|S_{n\wedge \tau }|\le C$. Consequently $L_{\tau }$ is an integrable random variable andThis is the origin case of expected occupation before interval exit for a simple symmetric random walk. The argument obtains integrability of the total occupation count before using any unbounded-time optional stopping for $M$, so it does not presuppose the uniform integrability asked about next. The endpoint probabilities also agree with the standard gambler's ruin calculation.^[ref]

The bounded-time optional sampling theorem says that for an integrable discrete-time martingale $(U_n)$ and a bounded stopping time $S$, $\mathbb{E}[U_S]=\mathbb{E}[U_0]$. It requires no assumption about an eventual unbounded stopping time.

Apply this theorem with $S=n\wedge \tau$, or simply take the expectation of the stopped martingale in part (ii). For $N>0$ we obtainSince $P_{n\wedge \tau }\ge 0$,The monotone convergence theorem now yields $\mathbb{E}[\tau ]\le 3abc/N<\infty$. In particular $\tau <\infty$ almost surely, which has been established without assuming that optional stopping is valid at $\tau$.

For each sample path with finite $\tau$, $P_{n\wedge \tau }$ eventually equals $P_{\tau }=0$. Also $0\le P_{n\wedge \tau }\le N^3$. The dominated convergence theorem gives $\mathbb{E}[P_{n\wedge \tau }]\to 0$. Taking the limit of a sequence in the earlier identity therefore provesFor example, initial holdings $(1,1,1)$ give expectation one, as the first transfer must eliminate its giver. This computes the first elimination time, rather than the later time when one player owns all tokens.

If $S\le T$ are stopping times and $A\in {\mathcal{F}}_S$, then $U=S{1}_A+T{1}_{A^c}$ is a stopping time. The key identity isThis uses the ordering $S\le T$: without it, $A$ need not be known when $T$ occurs. Such pastings test the martingale identity through expectations at bounded stopping times.

The information available at a stopping time $T$ is the sigma-algebraIn discrete time, $A\cap \{T=k\}\in {\mathcal{F}}_k$ for $A\in {\mathcal{F}}_T$. A bounded stopping time permits conditional forms of the optional stopping theorem with respect to this sigma-algebra.

Let an integrable adapted reward process $Z$ be dominated by a supermartingale $V$ over a finite discrete horizon. For every bounded stopping time $\tau$ within that horizon, the optional stopping theorem givesWhen $V_0$ is a deterministic constant, the latter bound is $\mathbb{E}{Z}_{\tau }\le V_0$. An expectation is required on the right when $V_0$ is random. The Snell envelope is the least such supermartingale majorant and turns this upper bound into the optimal expected reward. For continuous time, use a càdlàg supermartingale under the usual conditions for a filtration and ensure integrability of the stopped reward.