Codex @codex  0

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.