Planar Brownian annular exit probability

Put $a_s=\mathrm{\mid }{Z}_s-y\mathrm{\mid }$. Every path is continuous. For $t\ge 0$, compact attainment and rational approximation give $\{H\le t\}={\bigcap }_{m\ge 1}{\bigcup }_{q\in (\mathbb{Q}\cap [0,t])\cup \{t\}}\{a_q<\epsilon +1\mathrm{/}m\text{ or }{a}_q>R-1\mathrm{/}m\}\mathrm{.}$ The reverse inclusion follows since the continuous nonnegative distance of $a_s$ to the closed set $(-\mathrm{\infty },\epsilon ]\cup [R,\mathrm{\infty })$ then has minimum zero on $[0,t]$. Similarly, for $c=\epsilon$ or $R$, its circle hitting time has event ${\bigcap }_{m\ge 1}{\bigcup }_{q\in (\mathbb{Q}\cap [0,t])\cup \{t\}}\{\mathrm{\mid }{a}_q-c\mathrm{\mid }<1\mathrm{/}m\}\mathrm{.}$ These countable events are in the raw coordinate past, so all three times are stopping times and their comparisons are measurable. On the common probability-one event $Z_0=x$, the initial radius is strictly between the boundaries. Continuity and the intermediate value theorem give $H=S_{\epsilon }\wedge T_R$, the boundary value $\mathrm{\mid }{Z}_H-y\mathrm{\mid }\in \{\epsilon ,R\}$ when $H<\mathrm{\infty }$, and $\mathrm{\mid }{Z}_s-y\mathrm{\mid }\in (\epsilon ,R)$ for $s<H$. These last claims are used only on that event.

Define $\psi (r):=\log r$ for $r\in [\epsilon ,R]$ and extend it to a $C^2$ function on $[0,\mathrm{\infty })$ with $\psi \equiv \log \epsilon$ on $[0,\epsilon \mathrm{/}2]$, $\psi \equiv \log R$ on $[2R,\mathrm{\infty })$, and quintic Hermite splices on $[\epsilon \mathrm{/}2,\epsilon ]$ and $[R,2R]$ that match value, first and second derivative at both joints: on $[\epsilon \mathrm{/}2,\epsilon ]$ use $s\mapsto \log \epsilon +(\log s-\log \epsilon )h(2s\mathrm{/}\epsilon -1)$ and on $[R,2R]$ use $s\mapsto \log R+(\log s-\log R)(1-h(s\mathrm{/}R-1))$, where $h(\theta )=6{\theta }^5-15{\theta }^4+10{\theta }^3$ satisfies $h=h^{\mathrm{\prime }}=h^{\mathrm{\prime }\mathrm{\prime }}=0$ at $\theta =0$ and $h=1$, $h^{\mathrm{\prime }}=h^{\mathrm{\prime }\mathrm{\prime }}=0$ at $\theta =1$. Each splice agrees with the neighbouring branches in value and in its first two derivatives at both endpoints, so the resulting $\psi$ is $C^2$ with bounded first and second derivatives, and $\mathrm{\Phi }(z):=\psi (\mathrm{\mid }z-y\mathrm{\mid })$ is then a bounded $C^2$ function of $z\in {\mathbb{R}}^2$, constant near $y$ and outside the disc of radius $2R$, with $\mathrm{\Phi }(z)=\log \mathrm{\mid }z-y\mathrm{\mid }$ for $\mathrm{\mid }z-y\mathrm{\mid }\in [\epsilon ,R]$; its Laplacian $\mathrm{\Delta }\mathrm{\Phi }(z)={\psi }^{\mathrm{\prime }\mathrm{\prime }}(r)+{\psi }^{\mathrm{\prime }}(r)\mathrm{/}r$ at $r=\mathrm{\mid }z-y\mathrm{\mid }>0$ is continuous and bounded, and it vanishes on $\{\epsilon \le \mathrm{\mid }z-y\mathrm{\mid }\le R\}$ because $\mathrm{\Delta }\log \mathrm{\mid }z-y\mathrm{\mid }=0$ there.

For the standard normal pair $G=(G_1,G_2)$ of [F3], every bounded $C^2$ function $f$ with bounded derivatives satisfies $\mathbb{E}[{\mathrm{\partial }}_{i}f(z+\sqrt{r}G)G_i]=\sqrt{r}\mathbb{E}[{\mathrm{\partial }}_i^{2}f(z+\sqrt{r}G)]$ for $i=1,2$ and $r>0$. Indeed, the law of $G$ is the product of the two standard normal laws by [F3], so Fubini expresses the expectation as an iterated integral. Fix the other coordinate and integrate by parts in the chosen one-dimensional coordinate on $[-L,L]$ with the compact theorem of [F4] using ${\phi }^{\mathrm{\prime }}=-u\phi$ and the bounded factor $u\mapsto {\mathrm{\partial }}_{i}f(z+\sqrt{r}(u{e}_i+v{e}_j))$, where $j\mathrm{\ne }i$ and the other Gaussian coordinate $v$ is fixed; the boundary terms vanish as $L\to \mathrm{\infty }$ because $\phi$ decays rapidly and the derivative factor is bounded, and dominated convergence identifies the limit. On each finite interval the integrands are continuous, so [F8] identifies the compact integration-by-parts identity with its Lebesgue version. The bounds are independent of the fixed other coordinate; Fubini completes that coordinate integration. Summing the two coordinates gives $\mathbb{E}[\mathrm{\nabla }f(z+\sqrt{r}G)\cdot G]=\sqrt{r}\mathbb{E}[\mathrm{\Delta }f(z+\sqrt{r}G)]$.

Let ${\mathcal{G}}_s:=\sigma (Z_u:0\le u\le s)$ be the raw coordinate filtration. For every bounded Borel $g:{\mathbb{R}}^2\to \mathbb{R}$ and $0\le s\le t$ one has $E_x[g(Z_t)\mathrm{\mid }{\mathcal{G}}_s]=Q_{t-s}g(Z_s)$ almost surely: apply the conditioning lemma to the known state $Z_s$ and independent noise $Z_t-Z_s$.

The standard one-dimensional Brownian motion $Z^{(1)}-x_1$ (zero-start almost surely) hits the level $y_1+R-x_1$ almost surely. At that time $\mathrm{\mid }{Z}_t-y\mathrm{\mid }\ge \mathrm{\mid }{Z}_t^{(1)}-y_1\mathrm{\mid }=R$, so $H$ is no larger and is finite $P_x$-almost surely.

Fix a bounded $C^2$ function $f$ with bounded first and second derivatives and put $Q_{r}f(z):=\mathbb{E}[f(z+\sqrt{r}G)]=\int f(z+u){\mu }_r(du)$ with ${\mu }_r$ the law of $\sqrt{r}G$. Then $r\mapsto Q_{r}f(z)$ is differentiable on $(0,\mathrm{\infty })$ with $\frac{d}{dr}{Q}_{r}f(z)=\frac{1}{2}{Q}_r\mathrm{\Delta }f(z)$: differentiating the expectation is licensed by the mean value theorem in [F8] and dominated convergence on a neighborhood bounded away from $r=0$, because $\mathrm{\nabla }f$ is bounded and $\mathrm{\mid }G\mathrm{\mid }$ is integrable, and the resulting expression $\mathbb{E}[\mathrm{\nabla }f(z+\sqrt{r}G)\cdot G\mathrm{/}(2\sqrt{r})]$ is $\frac{1}{2}\mathbb{E}[\mathrm{\Delta }f(z+\sqrt{r}G)]$ by step 1.3.

Let $f$ and $Q$ be as in step 2.2. Continuity of $\mathrm{\Delta }f$ and bounded convergence imply that $r\mapsto Q_r\mathrm{\Delta }f(z)$ is continuous, including at zero. For $0<\delta <h$, the fundamental theorem of calculus applied to the continuous integrand $r\mapsto \frac{1}{2}{Q}_r\mathrm{\Delta }f(z)$ on $[\delta ,h]$ gives $\frac{1}{2}{\int }_{\delta }^h{Q}_r\mathrm{\Delta }f(z)dr=Q_{h}f(z)-Q_{\delta }f(z)$ by step 2.2; [F8] identifies this compact calculus integral with the Lebesgue integral. Letting $\delta \downarrow 0$, dominated convergence gives $Q_{\delta }f(z)\to f(z)$ because $f$ is continuous and bounded, and the integrals converge by monotone convergence on the nonnegative and negative parts; hence $Q_{h}f(z)-f(z)=\frac{1}{2}{\int }_0^h{Q}_r\mathrm{\Delta }f(z)dr$ for every $h>0$.

Define $M_t:=\mathrm{\Phi }(Z_t)-\mathrm{\Phi }(Z_0)-\frac{1}{2}{\int }_0^t\mathrm{\Delta }\mathrm{\Phi }(Z_r)dr$. The map $(r,\omega )\mapsto Z_r(\omega )$ restricted to $[0,t]$ is $\mathcal{B}([0,t])\otimes {\mathcal{G}}_t$-measurable: finite deterministic grid approximations to the continuous paths, using only coordinates at times at most $t$, converge pointwise. Parameter integration therefore makes the drift integral ${\mathcal{G}}_t$-measurable. Thus $M$ is adapted, continuous on every path and integrable on each finite horizon, with $\mathrm{\mid }{M}_t\mathrm{\mid }\le 2\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{\mathrm{\infty }}+(t\mathrm{/}2)\mathrm{\parallel }\mathrm{\Delta }\mathrm{\Phi }{\mathrm{\parallel }}_{\mathrm{\infty }}$. For $s\le t$ and $A\in {\mathcal{G}}_s$, Fubini on the bounded finite-time integrands and step 1.4 give ${\int }_A{\int }_s^t\mathrm{\Delta }\mathrm{\Phi }(Z_r)drd{P}_x={\int }_A{\int }_0^{t-s}{Q}_u\mathrm{\Delta }\mathrm{\Phi }(Z_s)dud{P}_x\mathrm{.}$ By step 3.1 the inner integral is $2(Q_{t-s}\mathrm{\Phi }(Z_s)-\mathrm{\Phi }(Z_s))$, while step 1.4 gives ${\int }_A\mathrm{\Phi }(Z_t)d{P}_x={\int }_A{Q}_{t-s}\mathrm{\Phi }(Z_s)d{P}_x$. Thus ${\int }_A(M_t-M_s)d{P}_x=0$, and $M$ is a martingale.

Fix $n\ge 1$ and let $H_n:=H\wedge n$. For each $m\ge 1$ put $\delta :=2^{-m}$, define the discrete filtration ${\mathcal{D}}_k:={\mathcal{G}}_{k\delta }$ and the discrete martingale $Y_k:=M_{k\delta }$, which satisfies $\mathbb{E}[Y_{k+1}\mathrm{\mid }{\mathcal{D}}_k]=Y_k$ by [F5] and step 4.1. The integer-valued ceiling ${\rho }_m:=\lceil 2^m{H}_n\rceil$ is a stopping time for $({\mathcal{D}}_k)$, since $\{{\rho }_m\le k\}=\{H_n\le k\delta \}\in {\mathcal{G}}_{k\delta }={\mathcal{D}}_k$ by step 1.1, and it is bounded by $\lceil 2^{m}n\rceil$. Applying [F5] with $\sigma =0$ and $\tau ={\rho }_m$ gives $\mathbb{E}[M_{{\rho }_m\delta }]=0$.

Expanding step 5.1 and letting $m\to \mathrm{\infty }$, ${\rho }_m\delta \downarrow H_n$ and $Z_{{\rho }_m\delta }\to Z_{H_n}$ by continuity. Dominated convergence on $[0,n+1]$ gives $E_x[\mathrm{\Phi }(Z_{H_n})]=\mathrm{\Phi }(x)+\frac{1}{2}{E}_x{\int }_0^{H_n}\mathrm{\Delta }\mathrm{\Phi }(Z_r)dr\mathrm{.}$ Since $\mathrm{\mid }{Z}_r-y\mathrm{\mid }\in (\epsilon ,R)$ for $r<H$, the integral vanishes, and $E_x[\mathrm{\Phi }(Z_{H_n})]=\log \mathrm{\mid }x-y\mathrm{\mid }$.

Define $Z_H$ by literal evaluation when $H<\mathrm{\infty }$ and as $y$ otherwise. This is measurable by finite-grid approximation to $Z_{H\wedge n}$ and passage to the limit on $\{H<\mathrm{\infty }\}$. Letting $n\to \mathrm{\infty }$, continuity and bounded convergence give $E_x[\mathrm{\Phi }(Z_H)]=\log \mathrm{\mid }x-y\mathrm{\mid }$. Moreover $H=S_{\epsilon }\wedge T_R$ and $\mathrm{\mid }{Z}_H-y\mathrm{\mid }\in \{\epsilon ,R\}$ almost surely, so $\mathrm{\Phi }(Z_H)=\log \epsilon$ on $\{S_{\epsilon }<T_R\}$ and $\log R$ on $\{T_R<S_{\epsilon }\}$. The tie event can include paths with both times infinite, but it has $P_x$-probability zero because $H<\mathrm{\infty }$ almost surely; a finite tie is impossible when $\epsilon <R$.

Therefore $\log \mathrm{\mid }x-y\mathrm{\mid }=\log \epsilon P_x(S_{\epsilon }<T_R)+\log R{P}_x(T_R<S_{\epsilon })$, and the two probabilities sum to one by step 7.1. Solving gives the stated formula.

The hypotheses are exactly those used: $0<\epsilon <\mathrm{\mid }x-y\mathrm{\mid }<R$ makes $\log$ finite and the end annulus nondegenerate, the case $x=y$ is excluded, the degenerate case $\epsilon =R$ is excluded because the formula's denominator vanishes there, and the truncated times $H_n$ are bounded so that the discrete optional sampling theorem applies. AC covers [F7] and the Countable Choice bridge in [F8], and no countable or dependent choice beyond AC is spent: the integration by parts, the fundamental theorem of calculus and the optional sampling theorem used here are the compact and discrete statements cited in [F4]-[F5].