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Under the corrected orthogonality hypothesis, the preceding quadratic variation calculation gives $[X{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$. We justify the needed infinite occupation time of one-dimensional Brownian motion, rather than inferring an infinite integral from recurrent visits alone.

Let $T_0=0$, and recursively let $S_n$ be the first exit of $B$ from $(-1,1)$ after $T_n$, and $T_{n+1}$ its first return to zero after $S_n$. Recurrence of one-dimensional Brownian motion makes these stopping times finite almost surely. The Strong Markov property shows that $D_n=S_n-T_n$ are independent and identically distributed random variables, each strictly positive. Choose $\epsilon >0$ with $\mathbb{P}(D_n\ge \epsilon )>0$. The Borel-Cantelli lemma gives infinitely many such durations and hence ${\sum }_n{D}_n=\infty$ almost surely. These intervals are disjoint, and $|B_s|\le 1$ on each, soThis proves the intended conclusion. Without the orthogonality correction, the printed setup gives instead the more general quadratic variation in the previous part; it does not justify identifying that quantity with $A$. The proof here is explicitly for the corrected setup.

A locally defined stochastic process is a pair $(X,T)$, where $T$ is a lifetime stopping time and $X_t(\omega )$ is defined for $0\le t<T(\omega )$, that is, on a stochastic interval. For continuous local problems one uses stopping times $T_n\uparrow T$ with $T_n<T$ on $\{T>0\}$, so the process stopped at each $T_n$ has an ordinary continuous adapted process version. This is an announcing sequence for a stopping time; no value at $T$ or after $T$ is implicit in the pair.

A local solution of a stochastic differential equation on an open domain $U$ takes its values in $U$ before $T$ and, on every such stopped interval, satisfieswith the local integrability needed for the ordinary and Itô integrals. A maximal local solution of a stochastic differential equation has no extension to a strictly later lifetime that agrees with it before $T$. For locally Lipschitz continuous coefficients on $U$, take a nested compact exhaustion $K_n$ of $U$ and the successive exit times: their increasing limit is the maximal lifetime, and on $\{T<\infty \}$ the solution eventually leaves every compact subset of $U$. To obtain the announcing sequence with finite stops even when the path never exits a compact set, cap these exit times by $n$; the capped times still increase to the maximal lifetime. Thus a finite maximal lifetime means exit from the domain or explosion; it need not mean divergence to infinity. For $U=(0,\infty )$, reaching the boundary zero terminates the local solution even if an absorbing extension could be defined for a different domain.

The bounded-time optional stopping theorem states that if $X$ is an integrable discrete-time martingale and $\tau$ is a stopping time bounded by a deterministic integer $N$, then $X_{\tau }$ is integrable and $\mathbb{E}{X}_{\tau }=\mathbb{E}{X}_0$. Indeed $|X_{\tau }|\le {\sum }_{j=0}^N|X_j|$ proves integrability, andThe event $\{\tau \ge j\}=\{\tau >j-1\}$ belongs to ${\mathcal{F}}_{j-1}$. Conditional expectation of each summand given ${\mathcal{F}}_{j-1}$ is zero, soMore generally, if $\sigma \le \tau \le N$ are stopping times, $\mathbb{E}[X_{\tau }\mid {\mathcal{F}}_{\sigma }]=X_{\sigma }$. To prove this, test against $A\in {\mathcal{F}}_{\sigma }$ and apply the same argument to $1_A{1}_{\{\sigma <j\le \tau \}}(X_j-X_{j-1})$; its coefficient is ${\mathcal{F}}_{j-1}$-measurable. Boundedness of the stopping times makes the sums finite, so no limiting optional-stopping hypothesis is being omitted.

The original PDF has $\mathbb{E}|X_T|<\infty$ and $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$. The TeX transcription drops the expected value and absolute value in the first condition, and the absolute value in the second. The proof uses the PDF's conditions.

Because the stopping time $T$ is finite almost surely, $X_{n\wedge T}\to X_T$ almost surely. More strongly,The first term tends to zero by hypothesis; the second does so by the dominated convergence theorem. Thus there is convergence in L1.

Here is the needed L1 convergence implies uniform integrability argument. For any integrable random variables $Z,Y$ and $K>0$, splitting according to $|Y|>K/2$ givesTake $Z=X_{n\wedge T}$ and $Y=X_T$. For large $n$, the first term is uniformly small by convergence in L1, and the second is small for large $K$ by integrability. The finitely many remaining $n$ are handled individually by integrability. Hence the stopped process is uniformly integrable. This proves the stopped-martingale uniform integrability criterion; the martingale assumption is not needed for this particular implication.

Write $\Delta X_{n+1}=X_{n+1}-X_n$ and $Y={\sum }_{n\ge 0}|\Delta X_{n+1}|1_{\{T>n\}}$. Since $T$ is a stopping time, $\{T>n\}\in {\mathcal{F}}_n$. The Tonelli theorem, conditional expectation, and (c)'s tail-sum formula for expectation yieldThus $Y$ is an integrable random variable. Since $X_0=0$, telescoping gives $|X_T|\le Y$ and $|X_n|1_{\{T>n\}}\le Y{1}_{\{T>n\}}$. The dominated convergence theorem gives $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$, so (b) gives uniform integrability of the stopped martingale.

The optional stopping theorem at the bounded stopping time $n\wedge T$ gives $\mathbb{E}{X}_{n\wedge T}=\mathbb{E}{X}_0=0$. Passing to the limit using the convergence in L1 proved in (b),One can also pass directly by the dominated convergence theorem, since $|X_{n\wedge T}|\le Y$.

Put $D_t=B_t-B_t^{\prime }$. By independence of the two Brownian motions, $D_t/\sqrt{2}$ is a standard Brownian motion: its independent increments have the normal distribution with variance equal to elapsed time. The stopping time $\tau$ is the first time $D$ hits $-1$ or $3$.

To justify finiteness without presupposing the desired hitting probability, each unit-time increment of $D$ has the normal distribution $N(0,2)$. An increment of absolute value greater than $4$ forces exit from $[-1,3]$ if the path was inside at its start. These independent increments give $\mathbb{P}(\tau >n)\le (1-p{)}^n$, where $p=\mathbb{P}(|N(0,2)|>4)>0$. Hence $\tau <\infty$ almost surely.

The stopped martingale $D_{t\wedge \tau }$ lies in $[-1,3]$. The optional stopping theorem at $t\wedge \tau$ gives $\mathbb{E}{D}_{t\wedge \tau }=0$, and the bounded convergence theorem gives $\mathbb{E}{D}_{\tau }=0$. Writing $q=\mathbb{P}(D_{\tau }=-1)$, path continuity gives $0=-q+3(1-q)$. Therefore, as in Brownian exit from an interval,

For a stopping time $T$ of a Brownian filtration, define the reflected stochastic processThe Brownian reflection at a stopping time form of the Brownian reflection principle states that $\hat{B}$ is again standard Brownian motion. If $T=\infty$, leave the path unchanged; the second branch is used only when $T<t$.

For an almost surely finite $T$, the Strong Markov property says that $(B_{T+s}-B_T{)}_{s\ge 0}$ is a fresh Brownian motion independent of ${\mathcal{F}}_T$. Its negative has the same probability distribution, so reflecting the future preserves the full path probability distribution. Allowing $T=\infty$ follows by applying this argument to $T\wedge n$ and restricting to each fixed finite time interval.

Let $T_y=\inf \{s\ge 0:B_s=y\}$. Reflect at this stopping time, using the Brownian reflection principle from (a). Path continuity gives $\{S_t\ge y\}=\{T_y\le t\}$, and on this event the reflected endpoint is ${\hat{B}}_t=2y-B_t$. ConsequentlySince $x\le y$, an endpoint ${\hat{B}}_t\ge 2y-x\ge y$ forces the reflected path to have hit $y$ by time $t$. Reflection preserves the first hit of $y$, so the condition $T_y\le t$ is redundant on the right. As $\hat{B}$ has the same probability distribution as $B$, the joint distribution of Brownian motion and its running maximum satisfiesThis argument includes $x=y$; at positive $t$ the boundary endpoint has zero probability under the normal distribution. At $t=0$ both sides vanish because $y>0$.

Taking $x=y=a$ in (c) and using that $B_t$ has no atom at $a$ givesThis is the distribution function of the Brownian first-passage time; (b) ensures that it is a proper probability distribution. Comparing these distribution functions gives the Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ for $c>0$.

For an integer $n\ge 1$, set $T_0=0$ and successively hit the levels $a,2a,\dots ,na$. The Strong Markov property at these finite stopping times and spatial translation imply thatare independent random variables, each with the probability distribution of $T_a$. Thus, for independent copies $T_a^{(j)}$,ThereforeThe paper's definition is the strictly stable distribution convention, with no centering term, a special case of a stable distribution.

Let $X,Y$ be two continuous solutions on the same filtered probability space, driven by the same Brownian motion and with $X_0=Y_0$ almost surely. Write $D=X-Y$ and choose a common Lipschitz constant $L$ for $b,\sigma$. Introduce the stopping time ${\tau }_n=\inf \{t\ge 0:|X_t|+|Y_t|\ge n\}$. On $\{{\tau }_n=0\}$ the stopped difference is zero; before ${\tau }_n$, both coefficients and the difference are bounded.

The Itô formula and the zero expectation of the stopped square-integrable Itô integral yieldThe Gronwall inequality implies $\mathbb{E}{D}_{t\wedge {\tau }_n}^2=0$. Continuous solution paths are bounded on each compact time interval, so ${\tau }_n\uparrow \infty$ almost surely. Letting $n\to \infty$ proves $X_t=Y_t$ almost surely at each fixed time, and a countable dense set of times together with continuity gives indistinguishability of stochastic processes. Therefore the stochastic differential equation has pathwise uniqueness.

Let $U_t=(1,0,0)+(B_t^1,B_t^2,B_t^3)$, with three independent standard Brownian motions, and set $R_t=|U_t|$ and $M_t=R_t^{-1}$. The radial stochastic process $R$ is a three-dimensional Bessel process, which never hits zero by the Hitting-zero classification for a Bessel process. In three dimensions $u\mapsto |u{|}^{-1}$ is a harmonic function away from zero: direct differentiation gives $\Delta (|u{|}^{-1})=0$.

For the stopping times $T_n=\inf \{t:R_t\notin (1/n,n)\}$, $n\ge 2$, the Itô formula givesThe coefficients are bounded by $n^2$, so each stopped Itô integral is a true martingale on every finite horizon. The stopping times form a localizing sequence; as permitted, its divergence need not be proved here. Thus $M$ is a nonnegative continuous local martingale.

To see that $M$ is not a true martingale, integrate the three-dimensional multivariate normal density of $U_t$ in spherical coordinates. The angular integral yieldswhere $\Phi$ is the standard normal distribution function. Therefore the reciprocal of a three-dimensional Bessel process is a strict local martingale.

Take the stopping timesContinuity gives $|M_{t\wedge S_n}|\le n$, with no overshoot at the first crossing. Stopping preserves the local martingale property, and the bounded case proved in question 4(i) makes each $M^{S_n}$ a continuous true martingale.

Every continuous path is bounded on each compact time interval, so $S_n\uparrow \infty$ almost surely. Hence this is also a localizing sequence consisting of bounded stopped martingales.

Work with a locally growing Loewner chain started at $0$, with $A_0=\varnothing$ and the half-plane-capacity parameterization $\mathrm{hcap}(A_t)=2t$. The Conformal Markov property of SLE combines invariance under conformal changes of the marked domain with restarting after an initial hull has been removed.

In the half-plane this means, first, that $(r^{-1}{A}_{r^{2}t}{)}_{t\ge 0}$ has the same law as $(A_t{)}_{t\ge 0}$ for every $r>0$. Second, conditionally on ${\mathcal{F}}_t$, the future hulls mapped by $g_t-U_t$,with the usual hull closure, have the original law and are independent of ${\mathcal{F}}_t$. Their capacity is $2s$ by the half-plane-capacity composition rule. One may require the same restarting property at finite stopping times.

Equivalently, the driver has the scaling law and conditional increment lawThe conformal invariance clause matters: the restarting property alone would also allow a Brownian driver with deterministic drift.

Take an increasing localizing sequence ${\tau }_n$ for $X$ and define the stopping timesThe predictable process condition $K_{u+1}\in {\mathcal{F}}_u$ makes $R_n$ a stopping time. For each fixed finite time range, the finitely many coefficients $K_1,\dots ,K_t$ are finite almost surely; hence $T_n\uparrow \infty$ almost surely.

The stopped martingale transform can be writtenIts coefficient is ${\mathcal{F}}_{s-1}$-measurable and bounded in absolute value by $n$. Each summand is consequently integrable with zero conditional expectation given ${\mathcal{F}}_{s-1}$, since $X^{{\tau }_n}$ is a true martingale. Thus $M^{T_n}$ is a true martingale andThe original PDF fixes $M_0=0$; the TeX extraction omits this equality.