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For Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$, the Scaling invariance of SLE states that, for every $a>0$,has the same law as $K_t$. The scaled Loewner driving function is ${\tilde{U}}_t=a^{-1}{U}_{a^{2}t}$. Since $U_t=\sqrt{\kappa }{B}_t$, the Brownian scaling identity $a^{-1}{B}_{a^{2}t}\stackrel{d}{=}{B}_t$ proves the claim.

The Conformal Markov property of SLE states that, conditionally on the hull through time $t$, the future hull mapped by $g_t-U_t$ is an independent ${\mathrm{SLE}}_{\kappa }$ in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. The stationary increments and independent increments of Brownian motion show that $\tilde{U}$ is independent of ${\mathcal{F}}_t$ and has the same law as $\sqrt{\kappa }B$. The deterministic correspondence between continuous drivers and Loewner chains completes the proof.

Set $\kappa =4/(d-1)>4$. For chordal Schramm–Loewner evolution, the centered image of a real boundary point, divided by $\sqrt{\kappa }$, follows the Boundary-point Bessel flow for SLE; changing $B$ to $-B$ matches the sign convention in the question. Thus ${\tau }_x$ and ${\tau }_y$ are the times at which the marked boundary points $x$ and $y$ are swallowed, or equivalently disconnected from infinity, by the Loewner chain.

ConsequentlyIt is the event that the negative marked point is swallowed before the positive marked point.

Chordal Schramm–Loewner evolution in the complex upper half-plane from $0$ to $\infty$ is the random Loewner chain driven by $U_t=\sqrt{\kappa }{B}_t$, where $B$ is standard Brownian motion and $\kappa \ge 0$.