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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 $\kappa >0$, divide the Boundary-point Bessel flow for SLE by $\sqrt{\kappa }$ and replace $W$ by $-W$. The resulting Bessel process has dimensionsince its drift is $\frac{\delta -1}{2R_t}=2/(\kappa R_t)$. The Hitting-zero classification for a Bessel process suggests the threshold $\delta =2$, or $\kappa =4$. Here is a direct verification including accessibility in finite time.

The infinitesimal generator of $X$ is $\mathcal{L}=\frac{\kappa }{2}{\partial }_x^2+(2/x){\partial }_x$. An increasing scale function of a one-dimensional diffusion isIt satisfies $\mathcal{L}s=0$. For $0<\epsilon <b<R$, the boundary hitting probability from a diffusion scale function, or optional stopping theorem applied to $s(X)$, givesFor $0<\kappa <4$, $s(\epsilon )\to -\infty$, so this probability tends to zero as $\epsilon \downarrow 0$. A finite zero hit has a bounded path before the hit and hence precedes $T_R$ for some integer $R$; taking a countable union proves that zero is never hit.

For $\kappa >4$, write $a=1-4/\kappa >0$. Taking the inner boundary to zero gives $1-(b/R{)}^a$. This limit really concerns a finite zero hit: the Itô formula givesStopped on $[\epsilon ,R]$, it impliesuniformly in $\epsilon$. The increasing limit of these exit times is finite almost surely; the stopped squared process has a continuous extension, so its limiting lower endpoint is zero. ThusLet $R\to \infty$ to obtain ${\mathbb{P}}_b(T_0<\infty )=1$. When $\kappa =0$, the explicit solution is $X_t=\sqrt{b^2+4t}$, which stays positive.

Combining this with the SLE boundary swallowing criterion gives the critical parameter and the two regimes:

At $\kappa =4$, the Boundary-point Bessel flow for SLE has dimension two. Its scale function of a one-dimensional diffusion is $s(x)=\log x$, givingThe same bounded-path and countable-union argument as in part (c) excludes a finite zero hit. Hence the critical case belongs to the boundary-avoiding regime:The countable argument of part (d) gives no visits to any positive real point almost surely. By Reflection invariance of Brownian motion, reflected SLE has the same law, so no negative real point is visited either. Thus the set of real points visited is exactly $\{0\}$. Recurrence of a two-dimensional Bessel process near zero does not mean that it hits zero.

The boundary assertion is interpreted at positive finite times: a chordal SLE already starts on the domain boundary, and its marked target is another boundary point. The precise conclusion in the complex upper half-plane is $\gamma (0,\infty )\subseteq \mathbb{H}$ almost surely.

For $0<\kappa \le 4$ and a fixed real $x>0$, the centered Boundary-point Bessel flow for SLE $V_t=g_t(x)-U_t$ solvesThus $R_t=V_t/\sqrt{\kappa }$ is a Bessel process driven by $-B$ of dimensionThe Hitting-zero classification for a Bessel process says that, started positively, it never hits zero when $\delta \ge 2$. For $x<0$, reflect the equation and apply the same result. Therefore every fixed nonzero boundary point has infinite swallowing time. A countable intersection gives this simultaneously for all nonzero rational boundary points.

To extend this countable conclusion to every real $x>0$, choose rational $q$ with $0<q<x$. The difference of two centered boundary solutions satisfiesso it remains positive while both flows exist. On any finite time interval, $V_t(q)$ is bounded away from zero, hence so is $V_t(x)$. Moreover $V_t(x)=x-U_t+{\int }_0^{t}2/V_s(x)ds$ remains bounded there, so the differential equation continues for the whole interval. Its uniform separation from the Loewner driving function also allows each flow map to continue as a holomorphic function on a complex neighborhood of $x$. Thus $x$ is outside $K_t$ and cannot be visited by the Loewner trace. Reflection gives the negative real axis. This uses countably many Bessel processes followed by deterministic flow comparison, rather than an uncountable union of probability-zero events.

To exclude a return to the starting point, use the boundary extension of the inverse map for a continuous Loewner trace: $g_s^{-1}$ is continuous on the closed complex upper half-plane and maps $U_s$ to $\gamma (s)$. This follows from the continuous trace theorem and the Caratheodory boundary extension theorem; it does not assume simplicity. At each deterministic rational time $s$, the Conformal Markov property of SLE makes the mapped future a fresh SLE, so its trace lies in $\mathbb{H}\cup \{0\}$ by the preceding argument. Mapping back with $g_s^{-1}$ therefore showsIf a point $q=\gamma (r)$ were revisited at $t>r$, choose rational $s\in (r,t)$ with $\gamma (s)\ne q$. Such an $s$ exists because continuity and strictly increasing half-plane capacity forbid a constant trace on an interval. But $q\in K_s\cup \mathbb{R}$, contradicting the displayed inclusion. This excludes all self-contacts, including returns to zero, and proves the simple curve property needed when discussing later swallowing times.

For $\kappa =0$, the deterministic Loewner trace is $\gamma (t)=2i\sqrt{t}$, so the conclusion follows directly. Hence the boundary-intersection threshold for SLE givesMapping to another marked simply connected domain carries positive-time points into its interior. The literal statement including the starting point is false, since $\gamma (0)$ is prescribed to be on the domain boundary.

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.

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.

Start the Bessel process at $x>0$ and let ${\tau }_0$ be its first hit of zero. Away from zero its stochastic differential equation isApply the Itô formula to $f(x)=x^{2-\delta }$, localizing first to a compact interval in $(0,\infty )$. Its drift cancels:Thus the Bessel power local martingale satisfiesFor $\delta =2$, this power is simply the constant one. For $\delta >2$, it is a continuous local martingale for all time, as zero is inaccessible.

Here is a proof of that inaccessibility and of the stronger minimum assertion. For $0<r<x<R$, stop the power at ${\tau }_r\wedge {\tau }_R$. The stopped process is bounded, so the optional stopping theorem givesExit from the bounded interval occurs almost surely. Taking $r\downarrow 0$ with $R$ fixed shows that zero cannot be hit before $R$, hence cannot be hit in finite time. Taking $R\to \infty$ instead givesIf the all-time infimum were zero, every level $r>0$ below $x$ would be hit, by continuity. Letting $r\downarrow 0$ proves the all-time minimum of a transient Bessel process assertion:The first claim in the printed question needs a qualification when $\delta <2$: it holds before the first hit of zero, or for the process stopped there, rather than for the usual reflecting continuation. For example, at $\delta =1$ the proposed power is $X$ itself, a Reflected Brownian motion; the Tanaka formula contains a nonzero local time of a semimartingale term, so it is not a local martingale after reflection. A positive starting point is also needed for the negative power when $\delta >2$; an entrance process started at zero would give an infinite value at time zero.

For $\kappa >0$ and $x>0$, before the boundary point is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $d{U}_t=\sqrt{\kappa }d{B}_t$ giveSince $-B$ is again a standard Brownian motion, this is the Boundary-point Bessel flow for SLE withFor $x<0$, it is $-X_t$ that is the nonnegative Bessel process, with the same dimension; the formula without this sign convention is a signed Bessel flow.

When $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X_t$ never reaches zero. The borderline $\delta =2$ uses the scale function of a one-dimensional diffusion $\log x$: for $0<r<x<R$,Thus the critical case also cannot hit zero in finite time, though its all-time infimum is zero.

Apply the non-swallowing assertion simultaneously to all nonzero rational boundary points. The order-preserving real Loewner flow then keeps every compact real interval away from the origin in the surviving boundary. A boundary contact away from the starting point would cut off a nonempty real interval, swallowing a rational point, so the trace avoids $\mathbb{R}\setminus \{0\}$.

The same argument can be applied to the future after every rational time, using the Conformal Markov property of SLE. If the trace revisited an earlier point, choose a rational time strictly between the two visits. In the domain with that initial segment removed, the later visit would be a contact with an old boundary point away from the current growing tip. Under the mapping-out map, such a contact cuts off a real interval and contradicts the preceding non-swallowing result. This is the usual crosscut argument converting boundary non-swallowing into absence of self-intersections. Taking the countable intersection over rational restart times provesFor $\kappa =0$, the equation is deterministic and gives the simple slit $(0,2i\sqrt{t}]$. The division by $\sqrt{\kappa }$ is unnecessary in that case.

A Brownian excursion in the upper half-plane from $0$ to infinity is killed planar Brownian motion conditioned to stay in $\mathbb{H}$, with its entrance law at $0$. Precisely, it is the Doob h-transform for $h(z)=\mathrm{Im}z$, obtained by starting at $i\epsilon$ and letting $\epsilon \downarrow 0$. Equivalently,where $W$ is standard Brownian motion and the independent process $R$ is a dimension-three Bessel process started at zero. The transformed generator is $\frac{1}{2}\Delta +y^{-1}{\partial }_y$, which explains this representation.

First compute avoidance from an interior point $z$. Let $\tau$ be the exit time of ordinary planar Brownian motion from $\mathbb{H}\setminus A$. The harmonic correctionhas boundary values $\mathrm{Im}z$ on $A$ and zero on the real axis. The hydrodynamic normalization at infinity gives $g_A(z)=z+\mathrm{hcap}(A)/z+O(|z{|}^{-2})$. Together with the maximum principle for harmonic functions this gives the useful bound $0\le v\le {\sup }_{a\in A}\mathrm{Im}a$. Since ordinary Brownian motion hits the real axis almost surely, the optional stopping theorem applied to this bounded harmonic function givesThe Doob h-transform weights a stopped path by its final height divided by its initial height. Applying this at the first hit of $A$, with bounded-time localization and then a limit, therefore yieldsFor the entrance law the same result follows by conditioning at a small positive time and letting that time decrease to zero: the avoidance function has the boundary limit computed next, and the excursion is initially in a neighbourhood disjoint from $A$.

Because the closure of the hull avoids $0$, the Schwarz reflection principle makes $g_A$ analytic in a neighbourhood of zero, with real $g_A(0)$ and positive real $g_A^{\prime }(0)$. ThusIn fact, the reflected Taylor expansion gives $\mathrm{Im}{g}_A(z)/\mathrm{Im}z=g_A^{\prime }(0)+O(|z|)$ uniformly as $z\to 0$ inside the half-plane, so the same limit applies to the entrance process. The map in the question is ${\psi }_A=g_A-g_A(0)$, so its derivative agrees with $g_A^{\prime }$. We obtain the restriction probability of a Brownian half-plane excursion:

For a real boundary point $x\ne 0$, setAfter a deterministic rescaling of time, the Boundary-point Bessel flow for SLE says that $X$ is a Bessel process of dimensionWhen $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X$ never reaches zero. Thus no nonzero real boundary point is swallowed. The standard Loewner trace criterion then implies that each new tip is attached only to the preceding tip and the trace never intersects its past, so it is simple. For $\kappa =0$, the equation is driven by zero and generates a vertical slit. Hence ${\mathrm{SLE}}_{\kappa }$ is simple for $0\le \kappa \le 4$.

Fix $x\in \mathbb{R}\setminus \{0\}$ and, before $x$ is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ show, after changing the sign of the Brownian motion, thatThus $X$ is a Bessel process of dimensionA Bessel process hits zero exactly when $\delta <2$, which here is equivalent to $\kappa >4$. Hitting zero is precisely the swallowing of a nonzero boundary point by the SLE hull. For $\kappa \le 4$ the trace is simple and swallows no such point; for $\kappa >4$ swallowing occurs through a boundary contact. Hence the trace intersects $\partial \mathbb{H}\setminus \{0\}$ exactly when $\kappa >4$.

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.