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Fix $t>0$ and partition $[0,t]$ into $m$ equal intervals, with $r_k=kt/m$. Round $t\wedge T$ upward within this interval. The corresponding sampled value isEvery time in this sum is at most $t$. The stopping time property makes each indicator function ${\mathcal{F}}_t$-measurable, while the adapted process property makes each sampled value ${\mathcal{F}}_t$-measurable. Consequently $Y_m$ is ${\mathcal{F}}_t$-measurable.

The rounded times approach $t\wedge T$ from the right, so right continuity gives $Y_m\to X_{t\wedge T}$ for the chosen pathwise càdlàg version. If path regularity is instead stated only almost surely, the usual complete filtration handles the exceptional null set; on an incomplete filtration one should formulate that case as existence of an adapted version. At $t=0$, the value is simply $X_0$. Thus $X^T$ is an adapted process, by adaptedness of a stopped right-continuous process. The argument actually needs neither the martingale property nor boundedness of $T$.

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.

The construction is well defined. Since $D$ is contained in a ball, its Brownian exit time is finite almost surely: at successive integer times there is a fixed positive probability that the next independent unit-time Brownian motion increment has length greater than the ball's diameter, forcing an exit. The survival probability is therefore bounded by a geometric sequence. Path continuity gives $B_{{\tau }_D}\in \partial D$. Thus the bounded Dirichlet boundary data give a bounded Borel function $u$ on $D$.

For interior harmonicity, fix a ball with closure in $D$, centred at $x$, and let $\sigma$ be its Brownian exit time. The Strong Markov property, followed by the tower property of conditional expectation, givesThe orthogonal invariance of Brownian motion makes $B_{\sigma }$ uniform on the boundary sphere. Hence $u$ has the spherical mean value property for harmonic functions for every such ball. A locally bounded Borel function with this property is smooth and harmonic: integrating the spherical averages against any smooth radial mollifier gives $u=u\ast \rho$ locally, which first proves smoothness; the mean value property for harmonic functions then implies $\Delta u=0$.

For the boundary limit at $\xi$, choose the harmonic barrier for the Dirichlet problem $h_{\xi }$ from (i). The Itô formula, localized inside $D$, makes $h_{\xi }(B_{t\wedge {\tau }_D})$ a bounded martingale. Compact localization and the bounded convergence theorem, first to the exit and then as $t\to \infty$, yieldFor $\epsilon >0$, if $\partial D\setminus B(\xi ,\epsilon )$ is nonempty, compactness and barrier positivity giveIf that boundary subset is empty, the probability is already zero. ThereforeFirst let $x\to \xi$ and then $\epsilon \downarrow 0$. The continuity of the Dirichlet boundary data proves $u(x)\to f(\xi )$.

Finally, the difference of two continuous solutions is a harmonic function vanishing on the boundary. The maximum principle for harmonic functions on the bounded domain gives that it is zero. Thus the Kakutani solution of the Dirichlet problem exists, attains every prescribed boundary value, and is unique.

Let independent Bernoulli random variables $Y_n$ have success probabilities $2^{-n}$, for $n\ge 1$, and use their natural filtration. DefineEach increment is integrable, finite, and independent of the previous filtration, with expectationThus $X$ is a martingale. The increments have no common finite bound, since a success produces a jump of size $2^n-1$.

Since ${\sum }_n\mathbb{P}(Y_n=1)<\infty$, the First Borel-Cantelli lemma says that only finitely many successes occur almost surely. Consequentlyfor all sufficiently large $n$, with the random sum finite almost surely. Therefore $X_n\to +\infty$ almost surely, so neither finite convergence nor two-sided oscillation occurs. This rare-jump martingale divergence example gives probability zero, rather than one, for the union in (i).

For $A>0$, the first time $T_A=\inf \{n\ge 0:X_n<-A\}$ is a stopping time. On $\{T_A\ge 1\}$, the bounded increments assumption controls the overshoot:Before that time, $X_n\ge -A$. Allowing the possibility $T_A=0$, the processis a nonnegative martingale. Indeed, the stopped martingale in discrete time is a martingale, and the integrable ${\mathcal{F}}_0$-measurable variable $X_0^{-}$ is a constant-in-time martingale. If $T_A=0$, nonnegativity follows from $X_0+X_0^{-}=X_0^{+}$. Its expectations are constant and finite, so the Martingale convergence theorem makes $Y_n$ converge to a finite value almost surely.

On $\{T_A=\infty \}$ this proves finite convergence of $X_n$. Taking the countable union over positive integer $A$, we conclude that $X_n$ converges finitely on the event $\{{\inf }_n{X}_n>-\infty \}$. Apply the same reasoning to the martingale $-X$ to obtain finite convergence on $\{{\sup }_n{X}_n<\infty \}$.

Outside the event of finite convergence, both the infimum and the supremum of the sequence must therefore be infinite in the respective directions. Removing any finite initial segment cannot change this, because each of its values is finite. HenceThis is the bounded-increment martingale convergence-or-oscillation dichotomy. The $X_0^{-}$ term ensures the proof also covers an unbounded integrable initial value.

Let $K={\sup }_T\mathbb{E}[e^{M_T/2}]<\infty$, with $T$ ranging over finite stopping times. To locate the half-threshold for the exponential-martingale Hölder bound, put $u=\sqrt{pq}>\sqrt{q}$. Since $u(u-1)$ is increasing for $u>1$,On the other hand, taking $q=1+\epsilon$ and $p=1+{\epsilon }^2$ gives $C(p,q)\to 1/2$ as $\epsilon \downarrow 0$. Hence

Fix $0<a<1$. Choose $p,q>1$ with $aC(p,q)<1/2$, and put $\lambda =2aC(p,q)\in (0,1)$. Part (a) and the Jensen inequality give, for every finite stopping time,Thus the entire stopped family has a uniform $L^p$ bound for some $p>1$. By uniform integrability from an Lp bound, it is uniformly integrable.

To check that this local martingale is a true martingale, stop it by a localizing sequence. At each fixed time the stopped variables are uniformly integrable by the same bound; taking limits in their conditional martingale identities proves the unstopped identity. The uniform bound over all finite stopping times then makes it a uniformly integrable martingale, with terminal expectation one. Therefore

The scalar Lévy characterization of Brownian motion states that an adapted process $X$ starting at zero is a Brownian motion if and only if it is a continuous local martingale with

For necessity, the centered independent increments make a Brownian motion a martingale. Their conditional second moments show that $X_t^2-t$ is also a martingale. The defining uniqueness of the quadratic variation compensator gives $⟨X{⟩}_t=t$.

For sufficiency, fix $\theta \in \mathbb{R}$ and apply the Itô formula toThe time drift cancels the second-order Itô term, leaving $d{Z}_t=i\theta Z_{t}d{X}_t$. Its real and imaginary parts are local martingales. On any deterministic interval $[0,T]$, $|Z_t|=e^{{\theta }^{2}t/2}\le e^{{\theta }^{2}T/2}$, so the bounded local martingale criterion makes them true martingales. ThereforePart (b) proves the Brownian property. This also supplies a proof of Lévy's characterization of Brownian motion through conditional Fourier transforms.

Write ${\delta }_n=2^{-n}$ and let $Q_n(U)$ denote the printed squared-increment sum for a process $U$, on grid times $k{\delta }_n$. We construct its limit first for bounded martingales, then use localizing sequences and the finite variation part of a semimartingale.

Continuity and monotonicity of the bounded-martingale limit. Let $U$ be a uniformly bounded continuous martingale. The given result supplies a limit $R^U$ in uniform convergence on compacts in probability. Each $Q_n(U)$ is continuous and adapted. A subsequence converges almost surely uniformly on each compact time interval, by choosing summable error probabilities and diagonalizing. Its limit therefore has a continuous version; in the usual completed filtration this version is adapted.

The sums $Q_n(U)$ themselves can decrease between grid points, because the last partial increment is being squared. To establish monotonicity, use instead the quadratic variation from completed grid incrementsThese step processes are nondecreasing, and pathwise uniform continuity giveswhere ${\omega }_U(\delta ;T)=\sup \{|U_t-U_s|:s,t\le T,|t-s|\le \delta \}$. Thus the same almost surely uniform subsequence of $S_n$ converges to $R^U$, proving that $R^U$ is nondecreasing. Also $R_0^U=0$.

Localization of a continuous local martingale. Subtract the initial value, which does not affect increments, so that $M_0=0$. SetContinuity gives ${\tau }_m\uparrow \infty$ almost surely and makes $M^{{\tau }_m}$ bounded. The bounded local martingale criterion makes it a true martingale. Let $R^m$ be its continuous, adapted, nondecreasing limit.

The discrete sums commute exactly with stopping:For $\ell \le m$, uniqueness of the probability limit and stability of uniform convergence on compacts in probability under stopping giveup to indistinguishability. Taking a common null set for the countably many pairs, patch these processes into $R^M$ by setting $R_t^M=R_t^m$ when $t\le {\tau }_m$. Compatibility makes this definition independent of $m$. It is continuous, adapted, and nondecreasing. For each $T,\epsilon >0$,Let $n\to \infty$ and then $m\to \infty$. This proves the localization and patching of quadratic variation.

Adding finite variation. Decompose the continuous semimartingale as $X=X_0+M+A$, where $M$ is a continuous local martingale starting at zero and $A$ is continuous, adapted, and locally of finite variation. For every compact interval, the sum of the absolute increments of $A$ is at most its total variation of a function. Hence, pathwise,The Cauchy-Schwarz inequality bounds the mixed increment sum byThe first factor is bounded in probability, since $Q_n(M)\to R^M$ uniformly on compacts in probability; the second tends to zero almost surely. Thus the mixed term tends to zero in probability. Expanding the square provesThe constructed $R=R^M$ is continuous, nondecreasing, and adapted. This is the quadratic variation of $X$, and expresses the fact that finite-variation terms do not change quadratic variation.

The strong law for Brownian motion says $W_s/s\to 0$ almost surely. For sufficiently large $s$, therefore $2W_s-s\le -s/2$, givingThus $J_t={\int }_0^t{e}^{W_s-s/2}d{B}_s$ has finite terminal quadratic variation. Stopping when its bracket reaches each integer makes it an $L^2$-bounded martingale, which converges by the Martingale convergence theorem. Patching these limits shows thatis a well-defined finite almost sure limit. This construction uses local square integrability; a global $L^2$ bound is not required.

Conditional on the entire $W$ path, the independent Brownian motion $B$ still supplies a centered Gaussian stochastic integral, with variance $V_{\infty }\in (0,\infty )$. The conditionally Gaussian stochastic integral with an independent integrator therefore has no atoms, even after averaging over $W$.

Also $e^{-W_t+t/2}\to \infty$ almost surely. For each fixed $x$, the relation $X_t=e^{-W_t+t/2}(x-J_t)$ then gives, outside the null event $J_{\infty }=x$,Part (b) identifies the cumulative distribution:Hence $J_{\infty }$ has the Standard Cauchy distribution. This establishes the Cauchy law of an infinite-horizon Brownian exponential integral.

For $t>0$, write $A_t=t^{-1}{\int }_0^t{W}_{u}du$. Deterministic integrals of the Gaussian process $W$ are Gaussian, as follows by taking $L^2$ limits of Riemann sums, so $(W,A)$ is jointly Gaussian. For $0<s\le t$, integrating the Brownian covariance kernel givesFor the last equality, integrate ${\int }_0^t\min (u,v)dv=ut-u^2/2$ over $0\le u\le s$ and divide by $st$. ConsequentlyThe only nonzero choice giving Brownian covariance is $\boxed{c=3}$; even the variance condition at $s=t$ requires $c(c-3)=0$.

Define ${\hat{W}}_0=0$. Since $|A_t|\le {\sup }_{u\le t}|W_u|\to 0$ almost surely as $t\downarrow 0$, the transformed process is continuous at zero, as well as at positive times. It is centered Gaussian and has covariance $s\wedge t$, so part (c) provesThe transformed process is Brownian in its own natural filtration. This is the Brownian motion transform by three times its running average. The filtration qualification matters: for $t>s>0$,which is not identically zero. The Gaussian covariance calculation identifies the Brownian filtration generated by $\hat{W}$; it does not make the transformed process a martingale in the larger original Brownian filtration.

Write $\alpha =2r/{\sigma }^2\in (0,1)$. In a continuation interval, the Euler differential equation has power solutions $s$ and $s^{-\alpha }$: substitution of $s^{\beta }$ gives $(\beta -1)(\beta +\alpha )=0$. A decreasing bounded value uses the second solution.

Let $b$ be the exercise boundary. Value matching and smooth fit require $A{b}^{-\alpha }=(1+b{)}^{-1}$ and $-\alpha A{b}^{-\alpha -1}=-(1+b{)}^{-2}$. Dividing gives $b/(1+b)=\alpha$. ConsequentlyThis is a perpetual reciprocal-payoff American option.

To verify the obstacle inequality, observe that $s^{\alpha }/(1+s)$ increases up to $b$ and decreases afterwards, because its logarithmic derivative is $\alpha /s-1/(1+s)$. Therefore $V(s)\ge (1+s{)}^{-1}$ for $s>b$. In the continuation region $({\mathcal{L}}_r-r)V=0$. In the exercise region,The quadratic is convex. At the two endpoints of $[0,b]$ it equals $-\alpha /2$ and $-\alpha /[2(1-\alpha )]$, respectively, so it is negative throughout that interval. Thus the obstacle problem is satisfied on both regions, with value matching and smooth fit at $b$. The second derivative has a jump at $b$; the equation is understood piecewise and in the generalized Itô sense described in part (a).

For optimality, use the Risk-neutral measure for the Black-Scholes model, under which $d{S}_t=r{S}_{t}dt+\sigma S_{t}d{W}_t^{\mathbb{Q}}$. The Itô formula makes $e^{-rt}V(S_t)$ a nonnegative supermartingale, so every exercise time $\tau$ satisfiesBefore hitting the exercise region, its drift vanishes. Since $0\le V\le 1$, the stopped process is a bounded martingale. Apply the optional sampling theorem at ${\tau }_{\ast }\wedge T$ and let $T\to \infty$. The contribution from $\{{\tau }_{\ast }>T\}$ is at most $e^{-rT}$, while the boundary value equals the payoff. Hence equality holds forIf $S_0\le b$, exercise immediately; otherwise wait for the first down-crossing. If that time is infinite, the discounted payout is zero. This proves both the value and the optimal stopping policy, and part (a) supplies its superhedge.

Use the conditional second-moment matrix $V_t=\mathbb{E}[P_t{P}_t^{\top }\mid {\mathcal{F}}_{t-1}]$ from the PDF. By part (b), the filtration is finite-state at every finite time. On a time-$(t-1)$ atom let $p_1,\dots ,p_m$ be the successor price vectors and $q_1,\dots ,q_m>0$ their conditional probabilities. Then $m\le n$ andPositive definiteness gives rank $n$, so $m\ge n$. Hence $m=n$, and the $n$ successor vectors have linear independence.

We use the finite-horizon fundamental theorem of asset pricing in deflator form: an arbitrage-free finite market admits a strictly positive adapted process $D$, with $D_0=1$, such that $DP$ is a martingale. Equivalently, on every step,Fix a finite horizon containing the step in question. On the parent atom, write $d_i=D_t/D_{t-1}>0$ on successor $i$ and $p_0=P_{t-1}$. ThenThe proposed values $z_i=p_i^{\top }{V}^{-1}{p}_0$ satisfy precisely the same equation:Because the $p_i$ have linear independence and the $q_i$ are positive, this linear system has a unique solution. Thus $z_i=d_i>0$, provingThis is the positive regression deflator in a complete finite market. It also proves the suggested conclusion: with $Y_0=1$ and $Y_t={\prod }_{u=1}^t{Z}_u$,Hence $Y$ is a strictly positive martingale deflator. Finite-state structure makes these expectations integrable on each finite horizon. Positive definiteness alone would not ensure positivity of the regression factor; market completeness and the positive deflator are essential.