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For a standard Gaussian vector $G$ and a smooth Lipschitz function with $\Vert \nabla f\Vert \le K$, represent its centered value as the terminal martingale $P_{1-t}f(B_t)-P_{1}f(0)$. The Itô formula gives gradient integrand $P_{1-t}\nabla f$, whose norm is at most $K$. Its terminal quadratic variation is therefore at most $K^2$. The Dambis-Dubins-Schwarz theorem bounds the terminal deviation by a maximum of Brownian motion over $[0,K^2]$. Symmetry, the Brownian reflection principle, and the Gaussian tail bound give the displayed inequality. The square on the Lipschitz constant is essential by scaling.

Write $M_t=B_{[M{]}_t}$ by the Dambis-Dubins-Schwarz theorem and stop the exponential Brownian martingale when $B_s-s$ first reaches $b<0$. The stopped exponential has mean one. Its boundary contribution before $[M{]}_T$ is at mostwhich vanishes as $b\to -\infty$. The remaining term converges monotonically to $\mathcal{E}(M{)}_t$, proving that its expectation is one.

If a continuous local martingale starts at zero and has deterministic continuous quadratic variation $[M{]}_t=f(t)$, the Dambis-Dubins-Schwarz theorem gives $M_t=B_{f(t)}$. It is therefore a centered Gaussian process. The converse follows from independent increments of a Gaussian martingale.

For $Z_t=e^{B_t+at}$, use the clock $C_t={\int }_0^t{Z}_s^{2}ds$. After the inverse time change, the Dambis-Dubins-Schwarz theorem givesso the time-changed process is a Bessel process of dimension $d=2a+2$. Its lifetime is $C_{\infty }$, which is finite almost surely exactly when $d<2$.

Let $M_1,\dots ,M_N$ be continuous local martingales starting at zero, with $⟨M_i,M_j⟩=0$ for $i\ne j$ and $⟨M_j{⟩}_{\infty }=\infty$ almost surely. The individual Dambis-Dubins-Schwarz theorem time changes produce independent standard Brownian motions $B_j$ with $M_j(t)=B_j(⟨M_j{⟩}_t)$. Independence is between the Brownian motions; each may still depend on its own clock.

A continuous local martingale bounded above pathwise throughout a stochastic time interval cannot have infinite quadratic variation at that interval's endpoint. The Dambis-Dubins-Schwarz theorem represents it as Brownian motion at its quadratic-variation clock; an infinite clock would force unbounded oscillations. Its clock therefore has a finite limit, and so does the martingale. A bound below gives the same conclusion by changing sign. The bound may be random; it must hold over the entire interval.

Put $A_t=[M{]}_t$. Its continuous, strictly increasing paths start at zero and tend to infinity. Thus the inverse $T_s$ is finite and continuous, andso each $T_s$ is a stopping time. Set ${\mathcal{G}}_s={\mathcal{F}}_{T_s}$ and ${\beta }_s=M_{T_s}$. This is the inverse-clock construction in the Dambis-Dubins-Schwarz theorem.

To verify its martingale property, stop $M$ at times ${\tau }_n$ bounding its path, its bracket and time itself. The stopped $M$ and $M^2-A$ are bounded martingales. Optional sampling at $T_s\wedge {\tau }_n$ and $T_t\wedge {\tau }_n$ shows that the time-changed process, stopped at $A_{{\tau }_n}$, is a martingale and that its square minus the stopped clock $s\wedge A_{{\tau }_n}$ is a martingale. These clock stopping times tend to infinity because $A_{\infty }=\infty$. Therefore $\beta$ is a continuous local martingale with ${\beta }_0=0$ and quadratic variation $[\beta {]}_s=s$. The Lévy characterization of Brownian motion identifies it as a ${\mathcal{G}}_s$-Brownian motion.

Strict monotonicity and continuity give $A_{T_s}=s$ and $T_{A_t}=t$, soThe inverse-clock argument proves the asserted representation; the random clock need not be independent of the Brownian motion it produces.

The Dambis-Dubins-Schwarz theorem states: let $M$ be a continuous local martingale with $M_0=0$ and $[M{]}_{\infty }=\infty$ almost surely. SetThen $W_u=M_{{\tau }_u}$ is standard Brownian motion relative to $({\mathcal{G}}_u)$ andStrict increase of the bracket is not required in the general theorem: $M$ is constant on intervals on which its bracket is constant. Continuity of $M$ is required. A nonzero initial value is handled by applying the theorem to $M-M_0$ and adding it back. If the terminal bracket is finite, the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem supplies the Brownian motion up to that bracket time and, if necessary, an independent continuation on an enlarged probability space.

Use $M_0=0$, as implicit in the specified initial value of $Y$. First let $[M{]}_t=t$. The Lévy characterization of Brownian motion makes $M$ Brownian motion relative to the given filtration. SetOn this event at time one, the drift has absolute size at most $C$, while $|Y_0|<1$, so $|Y_1|\ge |M_1|-|Y_0|-C>1$. Continuity forces exit by time one. ThereforeThe same estimate applies conditionally at each integer time without assuming anything about future drift independence. On $\{T>n\}$, $|Y_n|<1$; if $|M_{n+1}-M_n|>C+2$, the drift bound forces exit by $n+1$. That Brownian increment is independent of ${\mathcal{F}}_n$. HenceIn particular $T<\infty$ almost surely.

For general $Q$, write $K_t=[M{]}_t$ and $D_t={\int }_0^t{A}_{s}ds$. Continuity of $Q$ and monotonicity of its integral imply $Q_s\ge 0$ at all times outside one null set. The drift condition givesLet ${\tau }_u=\inf \{t:K_t>u\}$. The divergent clock makes ${\tau }_u$ finite for every $u$, and its continuity gives $K_{{\tau }_u}=u$. By the Dambis-Dubins-Schwarz theorem, $W_u=M_{{\tau }_u}$ is Brownian motion for ${\mathcal{G}}_u={\mathcal{F}}_{{\tau }_u}$. On intervals where $K$ is constant, both $M$ and $D$ are constant. Thus $d_u=D_{{\tau }_u}$ is a continuous adapted function of clock time satisfying $|d_v-d_u|\le C(v-u)$. The time-changed process isThe unit-clock argument uses only this increment bound on the drift, so its first exit time $S$ obeys $\mathbb{P}(S>n)\le (1-\epsilon {)}^n$. It is finite almost surely, and a corresponding finite ordinary time has $|Y|\ge 1$. This proves the exit with drift controlled by quadratic variation resultThis proof needs only a nonnegative measurable bracket density, a continuous divergent integral clock, and the drift increment bound; pointwise continuity of the density is unnecessary. This observation permits the application below with measurable SDE coefficients.

For the multidimensional SDE, put $a_{ij}(x)={\sum }_{k=1}^m{\sigma }_k^i(x){\sigma }_k^j(x)$. The generator of an SDE diffusion isFor every $f\in C_b^2({\mathbb{R}}^d)$, the Itô formula givesBounded derivatives and bounded noise fields make the right side square-integrable on each finite horizon, hence a true martingale. This is precisely the defining test-function property of an L-diffusion, in the martingale-problem sense.

Now fix $x$ and $\xi \ne 0$. Its scalar projection has martingale part $M_t^{\xi }={\sum }_k{\int }_0^t⟨\xi ,{\sigma }_k(X_s^x)⟩d{B}_s^k$ and drift $⟨\xi ,b(X_s^x)⟩$. The uniform ellipticity assumption givesLet $B_{\ast }={\sup }_z|b(z)|$ and take $R>|⟨\xi ,x⟩|$. Rescale the projection by $R$. Its bracket density is at least $\lambda |\xi {|}^2/R^2$, its absolute drift is at most $B_{\ast }|\xi |/R$, and hence that drift is at most $C_R$ times its bracket density, with $C_R=B_{\ast }R/(\lambda |\xi |)$. Its clock diverges, so the exit lemma forces it out of $(-1,1)$ in finite time. Equivalently it leaves the slab $|⟨\xi ,X_t^x⟩|<R$. Taking a countable sequence of integer $R$ tending to infinity proves the projection unboundedness of a uniformly elliptic diffusion:For the final uniqueness result, let ${\tau }_D=\inf \{t:X_t^x\notin D\}$, with $x\in D$. Since $D$ is bounded, it lies in a finite slab; the preceding finite slab-exit time bounds ${\tau }_D$ from above. Thus ${\tau }_D<\infty$ almost surely. Continuity gives $X_{{\tau }_D}^x\in \partial D$.

Set $h=u-v$. It is bounded, has bounded derivatives, is zero on the boundary, and satisfies $Lh=0$ in $D$. The stopped test-function identity says $h(X_{t\wedge {\tau }_D}^x)$ is a martingale; equivalently it is a bounded local martingale, which suffices. Taking expectations and then using bounded convergence as $t\to \infty$ givesTherefore the bounded-domain harmonic uniqueness for a diffusion conclusion isIn particular the argument supplies the representation $u(x)={\mathbb{E}}_{x}u(X_{{\tau }_D})$ and establishes uniqueness without selecting a unique diffusion law.

Since $\dot{u}(t)=a_t^2>0$, invert the image clock before exit and put ${\hat{\xi }}_u=W_{t(u)}$. Its quadratic variation is $6u$ and its initial value is zero, because $\Phi (0)=0$. By the Lévy characterization of Brownian motion, or the localized Dambis-Dubins-Schwarz theorem,for a standard Brownian motion up to the image exit time. If that lifetime is finite, the stopped Brownian motion can be extended beyond it; only the stopped law is used here.

In this clock the image mapping-out functions obey the ordinary Chordal Loewner equation with driver $\sqrt{6}{\hat{B}}_u$. Their generated trace therefore has the law of SLE with parameter six, stopped when it leaves $N^{\ast }$. Meanwhile the original image segment is exactly $\Phi (\gamma )$ up to exit from $N$. The clock change is increasing, so it disappears after passing to $[\gamma ]$. This proves the local stopped-curve equality required in part (a).

For completeness, the original law is scale invariant: the maps $r^{-1}{g}_{r^{2}t}(rz)$ have driver $r^{-1}{\xi }_{r^{2}t}$, whose law is again $\sqrt{6}{B}_t$ by Brownian scaling. Hence $(r^{-1}{\gamma }_{r^{2}t}{)}_{t\ge 0}$ has the original trace law. Together with the stopped conformal-change argument, this establishesThe conclusion concerns local unparameterized segments up to exit; the capacity parameter itself changes according to $du=h_t^{\prime }({\xi }_t{)}^{2}dt$.

The Dambis-Dubins-Schwarz theorem states that if $N$ is a continuous local martingale with $N_0=0$ and $[N{]}_{\infty }=\infty$ almost surely, then for ${\sigma }_r=\inf \{t:[N{]}_t>r\}$, $W_r=N_{{\sigma }_r}$ is Brownian motion in ${\mathcal{F}}_{{\sigma }_r}$ and $N_t=W_{[N{]}_t}$. If the terminal bracket is finite, Brownian motion can be continued beyond it on an extension of the probability space. For a nonzero initial value, apply the theorem to $N-N_0$.

Apply time change to the vector $(M^1-M_0^1,M^2-M_0^2)$ with the common bracket $A$. The time-change theorem for continuous local martingales says that at inverse bracket times the processes remain continuous local martingales in the time-changed filtration and their quadratic covariations are composed with those times. Since $A$ is continuous and unbounded, $A_{{\sigma }_r}=r$, givingThe Lévy characterization of multidimensional Brownian motion states that a zero-starting continuous vector local martingale with bracket matrix $rI$ is standard multidimensional Brownian motion. Thus $\tilde{B}$ is standard two-dimensional Brownian motion; this step establishes the joint property, not merely the one-dimensional property of each coordinate.

The printed inverse is ${\tau }_r=\inf \{t:A_t\ge r\}$. For $r>0$ it is a stopping time, since $\{{\tau }_r\le t\}=\{A_t\ge r\}$, and ${\tau }_0=0$. The martingale is constant on each flat interval of $A$, as follows from its Dambis-Dubins-Schwarz representation. Hence $M_{{\tau }_r}^i=M_{{\sigma }_r}^i$, including any initial flat interval. Moreover ${\mathcal{F}}_{{\tau }_r}\subseteq {\mathcal{F}}_{{\sigma }_r}$. The common Brownian process is adapted to the smaller filtration, and its increments are independent of the larger past, so they are independent of the smaller past as well. Therefore the common-clock time change of orthogonal local martingales givesIf Brownian motion is understood to start at zero, the centered vector $B_r-B_0$ is the standard Brownian motion. Retaining the initial vector is the convention required in part (c).

Apply the Itô formula to $Y_t=e^{B_t+t/2}$. The time derivative contributes $Y_{t}dt/2$, and the quadratic-variation correction contributes another $Y_{t}dt/2$. ThusBoth the drift $b(y)=y$ and diffusion coefficient $\sigma (y)=y$ are globally Lipschitz. The global existence theorem for stochastic differential equations with Lipschitz coefficients gives a unique strong stochastic solution for a fixed driving Brownian motion and initial value; in particular pathwise uniqueness holds. The explicit solution is strictly positive.

Write $N_t={\int }_0^t{Y}_{s}d{B}_s$. This is a continuous local martingale withThe strong law for Brownian motion gives $B_t/t\to 0$ almost surely, so $\log Y_t/t\to 1/2$, $Y_t\to \infty$, and $H_t\to \infty$. The Dambis-Dubins-Schwarz theorem therefore supplies a Brownian motion $\beta$ in the inverse-clock filtration with $N_t={\beta }_{H_t}$ for all $t$. Finally $d{H}_s=Y_s^{2}ds$, hence $Y_{s}ds=Y_s^{-1}d{H}_s$. Integrating the stochastic equation givesThe infinite clock lifetime verifies that this Brownian motion is defined for every nonnegative clock time without an additional extension.

Let $t(u)$ be the inverse image-capacity clock. Part (a) and the quadratic variation of a stochastic integral giveThe Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion, now shows that ${\tilde{\xi }}_u/\sqrt{6}$ is standard Brownian motion up to the terminal image time. If that terminal clock is finite, it can be extended by a fresh Brownian continuation; this does not assert anything about the subsequent image trace. The initial value is zero because $\Phi (0)=0$.

Differentiating the image Loewner chain with respect to its own half-plane capacity time givesThus, until its first hit of $[1,\infty )$, the reparametrized image $\Phi (\gamma )$ has the ordinary chordal SLE law from $0$ to infinity. This follows from the uniqueness of the Chordal Loewner equation for a given continuous driver; the continuous Loewner trace is recovered from the inverse-map limit. Localization can be removed by exhausting times strictly before this boundary contact. The terminal event is defined from the trace itself, so equality of the stopped trace laws follows as well.

On the other hand, the definition of Conformal invariance of SLE says that the full unparameterized image $\Phi (\gamma )$ has the law of chordal ${\mathrm{SLE}}_6$ from $\Phi (0)=0$ to $\Phi (\infty )=1$, namely the law of ${\gamma }^{\prime }$. Its stopping event at $[1,\infty )$ is exactly the image of the original stopping event at $(-\infty ,-1]$. Therefore the same stopped image has both laws: ordinary ${\mathrm{SLE}}_6$ from $0$ to infinity stopped at $T$, and ${\mathrm{SLE}}_6$ from $0$ to $1$ stopped at $T^{\prime }$. We concludeThe statement does not identify $T$ with $T^{\prime }$ in their original clocks. It is target-change locality of SLE6, derived from the zero drift and capacity time change rather than assumed as a theorem.

Let $g_t=P_{1-t}(\nabla f)(X_t)$. The triangle inequality for the conditional average givesThe independent coordinate Brownian motions have zero cross-variation, so the quadratic variation of a stochastic integral givesThe Dambis-Dubins-Schwarz theorem says that a continuous local martingale $N$ starting at zero whose quadratic variation diverges admits a Brownian motion $B$, with the inverse-clock filtration, such that $N_t=B_{[N{]}_t}$. If the total clock is finite, one may enlarge the probability space and append independent Brownian motion after the clock terminates. This gives the same identity up to the original lifetime.

Apply this theorem to $N=M-M_0$. The clock bound implies the event inclusionThe latter event is contained in the union of the upper and lower crossing events. Symmetry of Brownian motion givesThis uses only an event inclusion; independence between the Brownian motion and its random clock is neither assumed nor needed.

Put $X_t=g_t(x)-W_t$, $Y_t=g_t(y)-W_t$ and $D_t=Y_t-X_t$. Before $T(x)$, order preservation gives $0<X_t<Y_t$ and $D_t>0$. The common noise cancels from their difference:Since $D$ has finite variation, applying the Itô formula to $Z=Y/D$ gives the SLE two-boundary-point ratio diffusionWith the strictly increasing clock $u(t)={\int }_0^t{D}_s^{-2}ds$, the Dambis-Dubins-Schwarz theorem yields a Brownian motion $B$ for which the time-changed process obeysThe sign of the new Brownian motion has absorbed the minus sign above.

The domain Markov property of a chordal Loewner chain says that, conditionally on the past before $T(x)$, the mapped future is a fresh SLE. Its boundary marked points are $X_t,Y_t$. HenceLocalize, for example, where $D_t\ge 1/n$, $Z_t\in [1+1/n,n]$ and $t\le n$. On each such stopped interval this is a bounded conditional-expectation martingale, and therefore a genuine martingale. No smoothness assumption on $f$ is needed for this argument. The exit calculation below identifies it explicitly.

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.

The Itô formula for $f(t,x)=e^{x+at}$ gives the stochastic differential equationWrite $N_t={\int }_0^t{Z}_{u}d{B}_u$ for its continuous local martingale part. Its clock is the quadratic variationBecause $Z_t>0$, $A$ is strictly increasing. For $s<A_{\infty }$, its inverse ${\tau }_s$ satisfies $A_{{\tau }_s}=s$ and $d{\tau }_s/ds=Z_{{\tau }_s}^{-2}$. The Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion after the optional time-change theorem, makes ${\beta }_s=N_{{\tau }_s}$ a Brownian motion for the time-changed filtration. With $Y_s=Z_{{\tau }_s}$,This is the Bessel process equation with dimension $\boxed{2+2a}$, on the positive stochastic interval. This establishes the Exponential Brownian-to-Bessel time change.

The lifetime convention is important: the inverse is defined only for $s<A_{\infty }$. If $a<0$, the strong law for Brownian motion gives $B_t/t\to 0$, so $Z_t\to 0$ exponentially and $A_{\infty }<\infty$. Then $Y_s\to 0$ as $s\uparrow A_{\infty }$, and this is its first hit of zero after continuous extension. If $a>0$ the clock diverges by the same strong law. If $a=0$ it diverges by infinite occupation time of one-dimensional Brownian motion on a bounded interval, since $e^{2B_t}$ is bounded below there. Thus the lifetime is infinite when $a\ge 0$. Negative dimensions are interpreted by the same positive-state equation killed at zero; no post-zero Bessel process behavior is asserted.

Work before the first zero of $Z$, with $Z_0=z>0$. For the Bessel process equation, the Itô formula applied to $x^r$ givesIts quadratic variation isThis clock is strictly increasing on the positive-state interval. For $s$ before its terminal clock value, let $Y_s=W_{{\sigma }_s}$ and let ${\beta }_s=({\int }_0^{\cdot }r{Z}_u^{r-1}d{B}_u{)}_{{\sigma }_s}$. The Dambis-Dubins-Schwarz theorem makes $\beta$ a Brownian motion there. Since $d{\sigma }_s=ds/(r^2{Z}_{{\sigma }_s}^{2r-2})$, the drift changes toConsequently the Power time change of a Bessel process givesThe initial value is $z^r$, and this equation identifies a Bessel process of dimension ${\delta }^{\prime }$ until its first zero. Localize both equations on compact subintervals of $(0,\infty )$ to justify the inverse and all stochastic integrals for arbitrary $r>0$.

For completeness, the inverse-clock lifetime really is the positive Bessel process lifetime: the map $z\mapsto z^r$ is a homeomorphism of $(0,\infty )$, and exits from nested compact intervals correspond under the clock. There is no finite explosion at infinity for a positive Bessel process, since the Itô formula for $Y^2$ gives $d(Y^2)=2Yd\beta +{\delta }^{\prime }ds$; stopping at zero or level $R$ yields a probability of hitting $R$ before a fixed $T$ bounded by $(Y_0^2+\max ({\delta }^{\prime },0)T)/R^2$. If the transformed lifetime were finite while $Y$ stayed in a compact positive interval, the inverse clock and the original solution could both be continued. Thus a finite lifetime is approach to zero. For $\delta <2$, also ${\delta }^{\prime }<2$ and the positive-state Bessel process hits zero; for $\delta \ge 2$ the transformed lifetime is infinite. No continuation through zero is needed.

If the initial value is already zero, the interval before its first zero is empty; the positive-state argument concerns starts $z>0$.

For the SLE driver $d{\xi }_t=\sqrt{6}d{W}_t$, apply the Itô formula to ${\tilde{\xi }}_t={\varphi }_t({\xi }_t)$. At a fixed spatial point, ${\varphi }_t$ has finite variation locally in time by the derivative formula in part (c); there is no extra spatial martingale term. ConsequentlyThis exact cancellation is the target-change locality of SLE6. For a general parameter the transformed SLE driving function would have drift $(\kappa /2-3){\varphi }_t^{\prime \prime }({\xi }_t)$.

Define the increasing clock and its inverse byThe derivative $p_t$ is real and strictly positive before $T$, so this is a valid continuous time change of a continuous process. The driver is a continuous local martingale with quadratic variation $[\tilde{\xi }{]}_t=6a(t)$ and initial value zero. The Dambis-Dubins-Schwarz theorem gives a standard Brownian motion $\hat{W}$ such thatIf the terminal clock is finite, the Brownian motion may be extended beyond that stopping time; only its stopped part is used here. Reparameterizing the image maps givesThese are exactly the defining Chordal Loewner equation and half-plane-capacity parameterization for ${\mathrm{SLE}}_6$. The trace is ${\tilde{\gamma }}_{t(s)}$. Hence the image curve is also ordinary chordal ${\mathrm{SLE}}_6$ after this time change, up to the specified stopping time: