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Each marginal is a Bessel process of dimension $\delta =1+2a\in (1,2)$, but finite lifetime can also be proved directly. Apply the Itô formula before $X$ reaches zero:With clock $u(t)={\int }_0^t{X}_s^{-2}ds$, the Itô integral is a Brownian motion $W_u$. The inverse clock satisfiesThe clock runs through all $u\ge 0$ up to the maximal lifetime: a finite terminal clock would leave $X$ at a finite positive limit, allowing continuation, or, at an infinite physical lifetime, would make its clock grow at a positive rate, a contradiction. There is no finite-time explosion to infinity; for example $d{X}_t^2=2X_{t}d{B}_t+(1+2a)dt$ and localization bounds the probability of reaching arbitrarily large levels on a fixed time interval. Since $W_u/u\to 0$ almost surely and $1-2a>0$, the last exponential integral has finite limit, while $X_{t(u)}\to 0$. ThereforeThe same argument applies to $Y$, whose noise $-B$ is also Brownian motion, so its lifetime is finite too. No independence between the two lifetimes is being claimed.

Set $S=X+Y$ and $U=Y/S$ before $T=\zeta \wedge \tau$. Brownian motion terms cancel in the sum:Thus simultaneous extinction is impossible. At the finite first lifetime, precisely one coordinate is zero and the other remains positive. Because $S$ has finite variation, differentiating the ratio givesAfter clock change $v={\int }_0^t{S}_s^{-2}ds$, its generator isA scale function of a one-dimensional diffusion $H$ solves $\mathcal{L}H=0$, soBoth endpoint singularities are integrable since $a<1/2$. Normalize $H(0)=0,H(1)=1$ to obtainThe Itô formula makes $H(U_{t\wedge T})$ a bounded martingale, first by localization away from the endpoints and then by bounded convergence theorem. Its terminal value is one on $\{\zeta <\tau \}$, when $U_T=1$, and zero on the opposite event, when $U_T=0$. optional stopping theorem therefore proves the oppositely driven Bessel exit probability

Here $\mathrm{B}$ is the beta function and $\Gamma$ is the Gamma function.

The integral printed in the question is incorrect for general $a$. Its first exponent is $2-4a$, whereas the two coupled equations give $2a$. At $a=1/4$, which lies in the allowed interval, its integral from zero diverges for every positive upper limit; no finite normalizing constant gives a probability. Moreover exchanging $X$ and $Y$ while changing $B$ to $-B$ shows that at $x=y$ the probability must be $1/2$. The corrected symmetric integral has this value for every $a\in (0,1/2)$, whereas the printed asymmetric normalized integral generally does not. For $a>1/4$, even when the printed integral converges, its derivative density $H_{\mathrm{print}}^{\prime }(u)\propto u^{-(2-4a)}(1-u{)}^{-2a}$ satisfiesso it is harmonic for the required exit problem only at $a=1/3$. The two exponents agree exactly at that value.

To relate that value to percolation, write the two SLE boundary gaps as $g_t(x)-{\xi }_t$ and ${\xi }_t-g_t(-y)$. In Brownian-time units $s=\kappa t$, they have the displayed coupled equations with $a=2/\kappa$, after reversing the Brownian motion sign. Hence $a=1/3$ corresponds to $\kappa =6$. The event is that the positive marked point $x$ is swallowed before the negative marked point $-y$.

For a quadrilateral with boundary points in the order $(-y,0,x,\infty )$, its conformal cross-ratio coordinate is $u=y/(x+y)$. In the critical percolation exploration, wire the arcs $(-y,0)$ and $(x,\infty )$ blue and the other two arcs yellow. Exploration from zero, stopped when one of the two marked sides is swallowed, detects a blue crossing between those two blue arcs exactly in the event just computed. The continuum exploration has the SLE6 law; its Locality property of SLE lets one stop the ordinary chordal exploration before the distant boundary colors affect it. Thus the crossing probability isthe Cardy boundary crossing formula. This is the rigorous scaling-limit formula for critical site percolation on the triangular lattice; applying it to other critical planar models requires the corresponding universality assumption. At this percolation value of $a$, the printed integral is already the correct one.

Regard a chord as an oriented curve modulo increasing reparameterization, and write $[\gamma ]$ for that class. Scale invariance of $\mu$ means that $[r\gamma ]$ has the same law for each $r>0$. The locality of a chord law under conformal changes of neighborhoods is the following stopped-curve property: if $N,N^{\ast }$ are neighborhoods of the starting boundary point and $\Phi :N\to N^{\ast }$ is a boundary-preserving conformal isomorphism with $\Phi (0)=0$, thenThe two stopped curves are sampled from the same original law $\mu$; equality is for their unparameterized initial segments. A normalization of ${\Phi }^{\prime }(0)$ can be imposed in an equivalent formulation, since scale invariance removes the positive dilation. In a hull-removal formulation, the Locality property of SLE says that a modification away from the starting point leaves the initial law unchanged until the curve reaches the modified part of the domain.

By the tip correspondence in the conjugacy,The time-dependent Itô formula is valid after localization away from the pole and the terminal time $S$. The map's time variation is of finite variation, so there is no extra stochastic cross-variation term. Using the supplied conjugacy derivative,ConsequentlyThe cancellation is special to $\kappa =6$: for a general SLE parameter the drift would be $(\kappa /2-3){\Phi }_t^{\prime \prime }({\xi }_t)dt$. This is the transformed SLE driving function, and explains the special Locality property of SLE at six.

For the Locality property of SLE, take a compact H-hull $A$ away from $0$, and map $\mathbb{H}\setminus A$ conformally to $\mathbb{H}$ by ${\psi }_A$, fixing $0$ and infinity and with derivative one at infinity. Up to the first contact with $A$, the image under ${\psi }_A$ of an ${\mathrm{SLE}}_6$ grown in $\mathbb{H}$ has the same law as ${\mathrm{SLE}}_6$ grown in $\mathbb{H}$, after the change to its own half-plane-capacity parameterization. Equivalently, ${\mathrm{SLE}}_6$ in the smaller domain has the same initial law as the curve in the larger domain until the latter encounters the removed hull. There is no conditioning on eventual avoidance.

The chordal restriction property for ${\mathrm{SLE}}_{8/3}$ concerns the whole curve. Conditional on avoiding $A$, applying ${\psi }_A$ gives the original ${\mathrm{SLE}}_{8/3}$ law, modulo increasing reparameterization:Its associated avoidance probability, from the SLE eight-thirds restriction martingale, isThus locality identifies a stopped, unconditioned law, whereas restriction identifies a conditioned law of the entire curve.

The Locality property of SLE says the following. Let $D\subset \mathbb{H}$ be simply connected and agree with $\mathbb{H}$ in a neighborhood of $0$, and let $\psi :D\to \mathbb{H}$ be conformal with $\psi (0)=0$. The image under $\psi$ of an ${\mathrm{SLE}}_6$ in $\mathbb{H}$, stopped when it first leaves $D$, has the law of an ${\mathrm{SLE}}_6$ in $\mathbb{H}$, up to the corresponding stopping time and a change of half-plane-capacity parameterization.

Explore the two stopped curves in the opposite order. Begin with an ${\mathrm{SLE}}_6$ from $1$ to $\infty$, stopped on leaving $B(1,1)$, and then, in its unbounded complementary component, draw an ${\mathrm{SLE}}_6$ from $-1$ to $\infty$, stopped on leaving $B(-1,1)$. Before their respective stopping times, each curve is separated from the neighborhood in which the other hull changes the domain. The Locality property of SLE therefore says that mapping out the other stopped hull does not change either stopped marginal law.

The two exploration orders consequently define the same joint law for the pair of stopped hulls. Disintegrating this joint law with respect to the second curve shows that, conditional on ${\gamma }_2{|}_{[0,{\tau }_2]}$, the first curve is an ${\mathrm{SLE}}_6$ in the unbounded component offrom $-1$ to $\infty$, stopped when it leaves $B(-1,1)$, as required.

Choose a Möbius transformation of $\mathbb{H}$ that fixes $0$ and exchanges $x$ with $\infty$. By Conformal invariance of SLE, it transforms an ${\mathrm{SLE}}_6$ from $0$ to $x$ into one from $0$ to $\infty$. Until $x$ is disconnected from infinity, the discrepancy between the two target domains lies beyond the component visible from the growing tip. The Locality property of SLE therefore makes the two initial curve laws identical up to that disconnection time. Hence an ${\mathrm{SLE}}_6$ from $0$ to $x$, stopped at ${\tau }_x$, has the law of an ${\mathrm{SLE}}_6$ from $0$ to $\infty$ stopped when it disconnects $x$ from infinity.