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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 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.

Put $A_t={\int }_0^t{H}_s^{2}ds$. It is continuous and tends to infinity almost surely, so $A_{T_{\sigma }}={\sigma }^2$. Define the right-continuous inverse $T(u)=\inf \{t:A_t>u\}$ andThe time-change theorem for local martingales shows that $W$ is a continuous local martingale in the time-changed filtration, andFor completeness, this proves the required case of the Dambis-Dubins-Schwarz theorem: for every $\theta \in \mathbb{R}$, Itô formula makes $\exp (i\theta W_u+{\theta }^{2}u/2)$ a local martingale; stopping and conditioning show that $W_v-W_u$ has conditional characteristic function $e^{-{\theta }^2(v-u)/2}$. Hence the increments are independent centered normal variables with the Brownian variances, and continuity makes $W$ a Brownian motion. ConsequentlyThus $X_{\sigma }$ is centered Gaussian with variance ${\sigma }^2$.

The Dambis-Dubins-Schwarz theorem says that there is Brownian motion $W$ such thatWhen $[M]$ is strictly increasing, define its inverseand set $W_s=M_{T_s}$. Optional sampling shows that $W$ is a continuous local martingale, while time change gives $[W{]}_s=s$. The Lévy characterization of Brownian motion makes $W$ Brownian, and inverse time change gives the displayed representation.

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.

Put $A_t=[M{]}_t$ and use the Dambis-Dubins-Schwarz theorem to write $M_t=B_{A_t}$, enlarging the space if necessary after the terminal clock value. In the time-changed filtration, $A_t$ is a stopping time. For $b<0$, let $S_b=\inf \{s:B_s-s=b\}$. Applying part c with $R=A_t$ givesOn $\{S_b\le A_t\}$, the first integrand is at most $e^b{e}^{A_t/2}$. The assumed Novikov condition therefore impliesas $b\to -\infty$, uniformly for $t\le T$. Meanwhile $mathbf{1}_{\{A_t<S_b\}}\uparrow 1$, so the monotone convergence theorem gives $\mathbb{E}\mathcal{E}(M{)}_t=1$. A nonnegative local martingale with constant expectation is a martingale. Thus $\mathcal{E}(M^T)$ is a martingale, proving the Novikov condition.

Let $C_t={\int }_0^t{Z}_s^{2}ds$ and ${\tau }_u=\inf \{t:C_t>u\}$. By the Dambis-Dubins-Schwarz theorem, $W_u=M_{{\tau }_u}$ is Brownian motion. Setting $Y_u=Z_{{\tau }_u}$ and using $du=Z_t^{2}dt$ transforms the decomposition in part b intoComparison with the Bessel process equation $d{Y}_u=d{W}_u+(d-1)(2Y_u{)}^{-1}du$ givesThus an exponential Brownian motion with drift becomes a Bessel process under its quadratic-variation time change; this is an Exponential Brownian-to-Bessel time change.

For $Y=X^{\alpha }$, the Itô formula and the Bessel equation giveUse the clockand its inverse. The Dambis-Dubins-Schwarz theorem turns the first term into Brownian motion, while division of the drift by the clock rate givesHence $\tilde{Y}$ is a Bessel process of dimensionThis is the Power time change of a Bessel process.

Because $v\in L^1(0,1)$, its indefinite integral $V$ is bounded, so $e^{-V}$ is bounded above and away from zero. Hence $\psi$ extends continuously and strictly increasingly to $[0,1]$. The quadratic-variation clock of $\psi (X)$ iswhose rate is bounded above and away from zero before exit. The Dambis-Dubins-Schwarz theorem therefore identifies $\psi (X)$, up to an equivalent time change, with Brownian motion in the bounded interval $(0,\psi (1))$; in particular, $\mathcal{T}<\infty$ almost surely.

The bounded stopped local martingale $\psi (X_{t\wedge \mathcal{T}})$ is a martingale. If $q={\mathbb{P}}_x(X_{\mathcal{T}}=0)$, the optional sampling theorem givesThereforeThis is the boundary hitting probability from a diffusion scale function.

For $x>0$, the Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ givePut $D_t=V_t^r-V_t^1$. The common Brownian term cancels, soThe Itô formula applied to $Z_t=\log D_t-\log V_t^1$ givesSincethe clock $q(u)={\int }_0^u(V_s^1{)}^{-2}ds$ and its inverse $\sigma$ turn the local-martingale term into Brownian motion by the Dambis-Dubins-Schwarz theorem. ThereforeThis is the SLE boundary-point logarithmic separation diffusion.

Set $M_t={\int }_0^t{H}_{s}d{B}_s$. Its quadratic variation is $[M{]}_t={\int }_0^t{H}_s^{2}ds$, which is continuous and tends to infinity almost surely by assumption. The stated stopping time is the inverse clock at level one, so $[M{]}_{\tau }=1$. The Dambis-Dubins-Schwarz theorem gives

The Dambis-Dubins-Schwarz theorem states that if $M$ is a continuous local martingale with $M_0=0$ and $[M{]}_{\infty }=\infty$, then, forthe process $W_s=M_{{\tau }_s}$ is a standard Brownian motion and $M_t=W_{[M{]}_t}$. If $[M{]}_{\infty }<\infty$, one obtains the same representation after enlarging the probability space and continuing $W$ independently beyond $[M{]}_{\infty }$.

True. The Dambis-Dubins-Schwarz theorem, with an independent continuation of the Brownian motion if $f$ is bounded, representsBecause $f$ is deterministic, every finite vector $(M_{t_1},\dots ,M_{t_n})$ is a finite vector of a Brownian motion at deterministic times and therefore has a multivariate normal distribution. Hence $M$ is a Gaussian process. This is the deterministic quadratic variation characterizes a Gaussian continuous local martingale result.

For a fixed interior point and parameter four, the SLE angle process has zero drift and satisfies ${\theta }_t=\arg z+2{\int }_0^t{Y}_s|Z_s{|}^{-2}d{B}_s$. Its values in $(0,\pi )$ make this continuous local martingale a true bounded martingale. The identity $\log ({{\rm Y}}_t/{{\rm Y}}_0)=-[\theta {]}_t$, together with the Dambis-Dubins-Schwarz theorem, prevents its Loewner conformal radius from vanishing at a finite lifetime. Since the parameter-four trace is simple and does not meet the real boundary at positive times, finite swallowing would require a visit to the point and vanishing radius. Thus its Loewner swallowing time is infinite almost surely. This gives a global continuous bounded martingale, not just one defined before swallowing.

For two boundary points $1<r$, let $V_t^x=g_t(x)-U_t$ and set $Z_t=\log (V_t^r-V_t^1)-\log V_t^1$. After the clock $q(u)={\int }_0^u(V_s^1{)}^{-2}ds$, the Dambis-Dubins-Schwarz theorem givesThis one-dimensional diffusion process compares the swallowing times of nearby boundary points.

Changing the target boundary point by a conformal automorphism of the upper half-plane preserves the local chordal SLE law at parameter six, up to a capacity time change and until the relevant boundary point is swallowed. The transformed SLE driving function has zero drift at six; the Dambis-Dubins-Schwarz theorem makes it Brownian in image capacity time.