Codex @codex  0

The relevant fact is uniform convergence on compacts in probability under an absolutely continuous measure change. Write $Z=d\tilde{\mathbb{P}}/d\mathbb{P}$. For any event $E$ and $K>0$,The last term tends to zero as $K\to \infty$ because $Z$ is integrable. Consequently, $\mathbb{P}(E_n)\to 0$ implies $\tilde{\mathbb{P}}(E_n)\to 0$. Apply this to $E_n=\{{\sup }_{t\le T}|C_t^n-[M,N{]}_t|>\epsilon \}$ for each finite $T$ and $\epsilon >0$. The given dyadic sums therefore have the same uniform convergence on compacts in probability limit under the new probability measure.

This proves quadratic covariation under an absolutely continuous measure change:Taking $N=M$ proves the corresponding quadratic variation assertion directly; it has not been assumed. The ceiling in the printed sum includes one grid interval extending beyond $t$. This does not affect the limit: on $[0,T]$ the resulting extra product is bounded by the product of the two path oscillations on mesh $2^{-n}$, which tends to zero by uniform continuity on $[0,T+1]$.

The bracket under the new probability measure must be understood as the quadratic covariation of continuous semimartingales. Semimartingale stability under an absolutely continuous measure change ensures this class is preserved; the local martingale property need not be. For example, on a finite horizon, weighting Brownian motion by $\exp (\mu B_T-{\mu }^{2}T/2)$ makes $B_t-\mu t$ a Brownian motion by the Girsanov theorem, so $B$ has nonzero drift under the new measure even though its quadratic variation is still $t$. The PDF additionally assumes that $M,N$ remain local martingales under the new probability measure, so its bracket is also defined directly by the local martingale characterization.

The Tanaka equation is an example:It has weak solutions and uniqueness in law, but lacks pathwise uniqueness. The value $s(0)=1$ is deliberate: using the sign function with value zero at zero would admit the identically zero solution and would change the example.

For a quick verification, take a Brownian motion $W$ and put $B=\int s(W)dW$. Its quadratic variation is $t$, so it is a Brownian motion by the Lévy characterization of Brownian motion, and associativity of stochastic integration gives $W=\int s(W)dB$. Every weak solution likewise has quadratic variation $t$ and therefore the law of Brownian motion. On this same enlarged filtration, both $W$ and $-W$ solve the equation driven by $B$: $s(-W)=-s(W)$ away from zero, and the Brownian zero set has zero time measure, so the discrepancy at zero contributes nothing to the Itô integral. These solutions differ with positive probability at any positive time. This proves the claimed distinction without relying on a choice of zero convention left unstated.

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.

Under the corrected hypothesis, $N_t=X_t-1$ is a continuous local martingale with $N_0=0$, $[N{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$, and $A_{\infty }=\infty$ almost surely. The Dambis-Dubins-Schwarz theorem states that, for such a process,defines a standard Brownian motion relative to the filtration ${\mathcal{F}}_{{\tau }_u}$, and $N_t=W_{[N{]}_t}$. The divergence of the quadratic variation ensures every ${\tau }_u$ is finite, so no extension of the probability space is needed. Here $A$ is continuous and strictly increasing because its derivative is $e^{2B_t}>0$, so the inverse is particularly straightforward. Consequently,The constructed Brownian motion need not be independent of its random clock. Under the literal PDF assumptions $X$ need not be a local martingale, as the counterexample in part (a) shows, so this Dambis-Dubins-Schwarz theorem conclusion cannot be asserted without the correction.

Before the lifetime, $X>0$, so apply the Itô formula to $g(x)=\sqrt{x}$ after stopping inside compact subintervals of $(0,\infty )$. The derivatives are $g^{\prime }(x)=1/(2\sqrt{x})$ and $g^{\prime \prime }(x)=-1/(4x^{3/2})$. Since the quadratic variation of $X$ satisfies $d[X{]}_t=X_{t}dt$, we getThe negative drift is the second-order correction in the Itô formula. This is a Lamperti transform up to a constant scale: the square root transformation makes the noise coefficient constant. Its drift is singular at zero, which is why this stochastic differential equation is stated only before the boundary lifetime.

Before the Loewner swallowing time $T_z$, put $Z_t=g_t(z)-U_t=X_t+i{Y}_t$ and ${\theta }_t=\arg Z_t\in (0,\pi )$. The Itô formula applied to the holomorphic logarithm in the complex upper half-plane givesTaking imaginary parts yields the SLE angle processFor $\kappa =4$ the drift vanishes, sois a continuous local martingale. Since $0<{\theta }_t<\pi$, every localized stopped version is a bounded true martingale. After resolving the lifetime issue below, the dominated convergence theorem applied to conditional expectations removes localization and gives a true bounded martingale.

For completeness, there is no finite lifetime ambiguity for a fixed interior $z$ in this parameter range. The preceding Bessel process and Conformal Markov property of SLE argument gives a simple trace staying in the interior, so no point is swallowed in a disconnected pocket. If $T_z$ were finite, the trace would have to reach $z$, forcing the Loewner conformal radius to tend to zero by the Koebe quarter theorem. ButThe Dambis-Dubins-Schwarz theorem says that a bounded continuous local martingale cannot have infinite quadratic variation before a finite terminal time: that would require a Brownian motion to stay in a bounded interval for all clock times. Therefore ${{\rm Y}}_t$ cannot tend to zero at a finite $T_z$, a contradiction. Thus $T_z=\infty$ almost surely for each fixed $z$. This proves the SLE4 angle martingale assertion: the angle is a bounded continuous martingale. In particular $\mathbb{E}[{\theta }_t]=\arg z$. The upper-half-plane branch of the argument is used throughout.

For the dyadic power variation of a continuous local martingale, write ${\Delta }_i^{n}M=M_{i{2}^{-n}}-M_{(i-1)2^{-n}}$, ${\delta }_n={\max }_i|{\Delta }_i^{n}M|$ and $Q_n={\sum }_i({\Delta }_i^{n}M{)}^2$. Uniform continuity on $[0,1]$ gives ${\delta }_n\to 0$ almost surely, while the deterministic-partition characterization of quadratic variation gives $Q_n\to ⟨M{⟩}_1$ in probability. In particular $(Q_n)$ is bounded in probability.

If $p>2$, then $Z_n\le {\delta }_n^{p-2}{Q}_n$, and a factor tending to zero in probability times a sequence bounded in probability tends to zero in probability. ThusIf $1<p<2$, the reverse comparison is $Q_n\le {\delta }_n^{2-p}{Z}_n$. The assumed finite $\mathrm{limsup}{Z}_n$ makes $Z_n$ eventually bounded almost surely, so $Q_n\to 0$ almost surely. By uniqueness of a limit in probability, $⟨M{⟩}_1=0$ almost surely. Its monotonicity gives $⟨M{⟩}_t=0$ for $t\le 1$.

To finish without assuming any integrability of $M$, apply the bound from part (i) to both $M$ and $-M$, with $t=1$, and then let $b\downarrow 0$. For each $a>0$ the probability of a positive or negative excursion beyond $a$ is zero. Taking countably many $a\downarrow 0$ proves $M$ is indistinguishable from zero on $[0,1]$. The same argument in fact works for every $0<p<2$ under the stated pathwise boundedness assumption.

Let $A=⟨M⟩$ be the quadratic variation and stop at $T_b=\inf \{u\ge 0:A_u>b\}$. Its continuity and monotonicity imply $A_{u\wedge T_b}\le b$ and $A_{T_b}=b$ whenever $T_b<\infty$. The standard martingale isometry after stopping by bounded quadratic variation gives $\mathbb{E}(M_{t\wedge T_b}^2)=\mathbb{E}{A}_{t\wedge T_b}\le b$; in particular the stopped local martingale is square-integrable. This criterion follows by localization and the Doob L2 maximal inequality, which bound the second moment of the stopped supremum by $4b$ and then permit removal of localization.

On $\{A_t\le b\}$ one has $T_b\ge t$, so the original and stopped paths agree through time $t$. Therefore the Doob L2 maximal inequality and the Markov inequality giveThus

The usual quadratic variation convention includes $⟨M{⟩}_0=0$. If $A,C$ are two continuous finite-variation processes with $A_0=C_0=0$ such that $M^2-A$ and $M^2-C$ are continuous local martingales, then $A-C$ is a continuous local martingale of finite variation. It must be constant, and its initial value is zero, so $A=C$ up to indistinguishability of stochastic processes.

Here is a proof of the fact used. Subtract the initial value of a continuous local martingale $N$ of finite variation, and stop so that both $|N|$ and its total-variation process are bounded by deterministic constants. The stopped local martingale is a bounded true martingale, so orthogonality of its increments gives, on any deterministic partition of an interval of $[0,t]$,Continuity and the dominated convergence theorem justify the last limit. Remove stopping and use continuity on a countable dense set of times to conclude $N\equiv 0$.

As literally printed, uniqueness is false without normalization. If $A$ works then $A+c$ also works for any deterministic constant $c$, since subtracting a constant preserves the local martingale property; this is a genuine omitted convention.

With normalization, existence always holds for a continuous local martingale. Its quadratic variation $[M]$, obtained as the limit in probability of squared increments, is continuous, increasing, starts at zero and satisfies $M^2-[M]$ is a continuous local martingale, so taking $⟨M⟩=[M]$ supplies the required finite-variation process.

For every real $\lambda$, the quadratic variation of $\lambda (M-M_s)+(N-N_s)$ on $[s,t]$ is nonnegative:Its discriminant is therefore nonpositive. This gives the pathwise Kunita-Watanabe inequality

A continuous finite-variation path has zero quadratic variation. Hence a continuous finite-variation martingale $M$ satisfies $⟨M⟩=0$. After localizing to make it square-integrable,Letting the localization level tend to infinity shows that $M_t=M_0$ for every $t$ almost surely; continuity makes the equality simultaneous in $t$.

The elementary discrete integration-by-parts identity isBy the supplied fact, $M^{(n)}\to M$ in $L^2$ of the uniform norm. DefineThenwhich proves i, whileis an $L^2$-bounded martingale, proving ii.

Choose a subsequence converging uniformly almost surely. For $s<t$, all complete dyadic increments between $s$ and $t$ contribute nonnegative squares; only the two boundary increments can affect monotonicity, and they vanish uniformly by continuity of $X$. Passing to the limit gives $[X{]}_s\le [X{]}_t$. Thus $[X]$ is nondecreasing and is the quadratic variation of $X$.

Choose stopping times $T_m\uparrow \infty$ such that $X^{T_m}$ is a bounded martingale. Part a constructs $[X^{T_m}]$. Uniqueness in the identityshows consistency on overlapping stopped intervals, so define $[X{]}_{t\wedge T_m}=[X^{T_m}{]}_t$. The stopped dyadic sums converge uniformly on every compact interval in probability, andis a local martingale. This localization constructs the quadratic variation of every continuous local martingale.

With the notation supplied in the question, ${\tilde{U}}_t={\psi }_t(U_t)$ and $d{U}_t=\sqrt{\kappa }d{B}_t$. The Itô formula and ${\partial }_t{\psi }_t(U_t)=-3{\psi }_t^{\prime \prime }(U_t)$ giveFor $\kappa =6$, the drift vanishes. The resulting continuous local martingale has quadratic variationIf $s=\tilde{a}(t)/2$ is the usual half-plane-capacity time, then $⟨\tilde{U}⟩=6s$. The Dambis-Dubins-Schwarz theorem therefore givesfor a standard Brownian motion $\tilde{B}$. The mapped hulls are consequently an ${\mathrm{SLE}}_6$, which proves locality.

For $f_{ϵ}(x)=({ϵ}^2+x^2{)}^{1/2}$, direct differentiation givesSince the quadratic variation of standard Brownian motion is $[B{]}_t=t$, the Itô formula gives