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If $u$ is a bounded harmonic function on a domain $D$, continuous on its closure, and the Brownian exit time $T_D$ is almost surely finite from $z$, then $u(z)={\mathbb{E}}_z[u(B_{T_D})]$. The Itô formula and compact localization make $u(B_{t\wedge T_D})$ a bounded martingale. bounded convergence theorem at the exit time proves the representation even if the domain is unbounded. Removing boundedness of $u$ can invalidate this conclusion through loss of uniform integrability.

For a finite Markov chain, let $T$ be the first hitting time of $D$, assume $T<\infty$ almost surely, $f=0$ on $D$ and $Pf=f-\delta$ on its complement, with $\delta >0$. The stopped martingale $f(X_{n\wedge T})+\delta (n\wedge T)$ has mean $f(x)$. Boundedness of $f$ and bounded convergence theorem give $\mathbb{E}f(X_{n\wedge T})\to 0$; monotone convergence then gives the stated finite expected hitting time. No drift condition is required after entry into $D$.

Take a birth-death chain on $\{0,1,2,3,4\}$. At each interior state, move up one state with probability $p=1/2$ and down one state with probability $q=1/2$. The endpoints are absorbing states: $P_{00}=P_{44}=1$. This is a small gambler's ruin model in which the state can represent capital, and reaching $4$ is success.

Let $T_a=\inf \{n\ge 0:X_n=a\}$ be the hitting time of endpoint $a$, and let $T=\min (T_0,T_4)$ be the first endpoint hitting time. The desired hitting probability is $h_i={\mathbb{P}}_i(T_4<T_0)$. First-step analysis and the Markov property giveThe interior equation says adjacent differences are equal. Four equal increments must bridge the boundary values $0$ and $1$, soFor example, from state $2$ the hitting probability is $1/2$, and its one-step average is $\frac{1}{2}(1/4)+\frac{1}{2}(3/4)=1/2$. Individual sample paths can wander back and forth, but the same averaging rule already accounts for all of them.

Now ask how long it takes to reach either endpoint. The expected hitting time $t_i={\mathbb{E}}_{i}T$ counts steps, including the transition that first reaches an endpoint. First-step analysis givesThis is a discrete Poisson equation, $(I-P)t=1$ in the interior. The extra $1$ counts the step just taken. The quadratic valuessatisfy the equations; from state $2$, $t_2=1+\frac{1}{2}(3)+\frac{1}{2}(3)=4$ steps. The hitting probability is harmonic, whereas the expected hitting time has a nonzero source term. These are special cases of expected duration of symmetric gambler's ruin; see Hao Wu's MIT lecture 16, gambler's ruin, slide 9.

For a biased random walk with $p=0.6$ upward and $q=0.4$ downward, keep the same absorbing states. The interior harmonic equation becomes $h_i=p{h}_{i+1}+q{h}_{i-1}$. Using biased gambler's ruin probability with these boundary values givesThe three interior hitting probabilities are approximately $0.415385$, $0.692308$, and $0.876923$. Thus starting at $2$ now succeeds about $69.23\%$ of the time. A direct check is $0.6(57/65)+0.4(27/65)=9/13$. The curved profile reflects the unequal transition weights: its successive increments shrink by the factor $q/p=2/3$. The formula, with the endpoint denoted $T$ there, is verified in Karger and Lynch's MIT lecture 26, sections 2.3-2.4; the five-state numbers here are direct substitutions.

Why does harmonic averaging produce the correct hitting probability? With the natural filtration ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$, the Markov property gives the conditional expectation identityFor a function harmonic everywhere, this makes $h(X_n)$ a martingale whenever it is integrable. For a function harmonic only in the interior, $M_n=h(X_{n\wedge T})$ is a stopped martingale. In either five-state example, $0\le M_n\le 1$. Also, from any interior state at most three consecutive upward steps reach $4$, with probability at least $p^3>0$. Repeated three-step blocks give ${\mathbb{P}}_i(T>3k)\le (1-p^3{)}^k$, so $T$ is almost surely finite and has finite expected value; compare finite-interval exit times have geometric tails. Applying optional stopping at $n\wedge T$ and then the bounded convergence theorem givesThe boundedness and finite hitting time justify this passage to the limit. See Steven Lalley's notes, harmonic functions and stopped expectations, theorem 1, and stopped martingale in discrete time for this bounded version of optional stopping.

Uniqueness has a short maximum principle explanation. On a finite state space, the difference of two harmonic extensions with the same boundary conditions is zero on the boundary. If its positive maximum occurs inside, the weighted average forces every successor with positive transition probability to have the same maximum. A path to the boundary propagates that maximum to zero, a contradiction. Applying the argument to the negative of the difference proves uniqueness. Reachability of the boundary is essential: an interior closed communicating class can retain an arbitrary constant.

// Target: markov-process.bigb

Write $f=u+iv$. The Cauchy-Riemann equations give $v_x=-u_y$, $v_y=u_x$ and $\Delta u=\Delta v=0$. Since the two Brownian coordinates are independent, $[X{]}_t=[Y{]}_t=t$ and $[X,Y{]}_t=0$. The Itô formula therefore becomesStop in balls on which the derivatives are bounded. The stopped integrals are martingales, proving that $U$ and $V$ are local martingales. Their coefficient rows have equal squared length and are orthogonal. ThusThese are quadratic covariations of an analytic Brownian image; orthogonality does not in general imply independence of the two transformed coordinates.

For the exit law, put $c=2a^2$. Squaring sends the positive quadrant to the upper half-plane, sends $a+ia$ to $ic$, and the Möbius transformation $w\mapsto (w-ic)/(w+ic)$ sends the upper half-plane to the unit disk with the starting point sent to zero. Its composition is the displayed conformal map. The exit time is finite: it is the minimum of the two almost surely finite one-dimensional hitting times of zero. There is no hit at the quadrant's corner, since a planar Brownian motion started away from zero avoids that point.

We can justify the uniform image exit law without assuming a conformal-time-change theorem. For each integer $n\ge 1$, the real and imaginary parts of $f(Z_{t\wedge T}{)}^n$ are bounded martingales obtained from harmonic functions. At the start their values are zero. The bounded convergence theorem at the finite exit time gives $\mathbb{E}f(Z_T{)}^n=0$. The boundary image has modulus one, and conjugation gives the vanishing negative Fourier moments as well. Trigonometric polynomials are dense among continuous functions on the unit circle, so these moments characterize the uniform distribution there.

Consequently the real variable $W=Z_T^2$ is the inverse image of a uniform point $e^{i\Theta }$ under that Möbius transformation:The Jacobian of this change of variable gives the centered Cauchy distribution of scale $c$:Positive $w$ corresponds to exit at $\sqrt{w}$ on the real axis; negative $w$ corresponds to exit at $i\sqrt{-w}$ on the imaginary axis. Pulling back by $w=\pm r^2$ therefore gives the complete boundary lawHere $dr$ on each axis denotes length along that axis, not two-dimensional area measure. Each density integrates to $1/2$, with no remaining atoms. Equivalently the chosen axis is a fair Bernoulli choice independent of the radius, andThis is Brownian exit law from a quadrant, obtained as a harmonic measure and not as an absolutely continuous planar density.

For probability measures on a metric space, weak convergence of probability measures ${\mu }_n\Rightarrow \mu$ means $\int fd{\mu }_n\to \int fd\mu$ for every bounded continuous real function $f$. On the real line it is equivalently convergence of the cumulative distribution functions at every continuity point of the limiting cumulative distribution function. Bounded continuous upper and lower approximations to interval indicators give one direction; a partition at continuity points, uniform approximation on a compact interval, and control of the tails give the reverse direction.

For a probability measure $\mu$ on $\mathbb{R}$, its characteristic function is ${\phi }_{\mu }(t)=\int e^{itx}\mu (dx)$. The Lévy continuity theorem states: ${\mu }_n\Rightarrow \mu$ implies ${\phi }_{{\mu }_n}(t)\to {\phi }_{\mu }(t)$ for every real $t$; conversely, if these characteristic functions have a pointwise limit $\phi$ which is continuous at zero, there is a unique probability measure $\mu$ with characteristic function $\phi$, and ${\mu }_n\Rightarrow \mu$. In particular, for a specified limiting probability measure, weak convergence is equivalent to pointwise convergence of its characteristic functions.

The forward implication follows by testing cosine and sine. The dominated convergence theorem shows that every characteristic function is continuous at zero and has value one there. For the converse, first establish tightness of probability measures. By Fubini's theorem, for $h>0$,The integrand on the right is at least $1/2$ when $|x|\ge 2/h$. This proves the characteristic-function tightness boundThe integrals converge for each $h$ by dominated convergence, and continuity of $\phi$ at zero makes the limit as small as desired by choosing $h$ small. Thus all sufficiently late measures have a common tail bound; the finitely many earlier measures also have small tails once the radius is increased. This is uniform tightness of probability measures.

We give the required compactness argument explicitly. For any subsequence, choose a further diagonal subsequence whose cumulative distribution functions $F_n$ converge at every rational $q$, to $f(q)$. DefineThis function is nondecreasing and right-continuous: given a rational $q>x$ with $f(q)$ close to $F(x)$, the same bound holds for $F(y)$ when $x<y<q$. Tightness gives $F(-\infty )=0$ and $F(+\infty )=1$, so $F$ is a cumulative distribution function of a probability measure $\nu$. For rationals $r<x<s$, $F_n(r)\le F_n(x)\le F_n(s)$. The limits bracket $F(x-)$ and $F(x)$; at a continuity point of $F$ they coincide. Thus $F_n(x)\to F(x)$ at all such points. To obtain weak convergence, choose a compact interval with small tails and partition it using continuity points of $F$. A bounded continuous function is uniformly continuous there, so step-function approximations and convergence of interval probabilities prove convergence of its integral. This proves Helly selection for distribution functions and supplies a weak subsequential limit $\nu$.

Its characteristic function is $\phi$ by the already proved forward implication. We also prove uniqueness rather than assume it. If two probability measures $\nu ,\rho$ have the same characteristic function, convolve each with a centered normal distribution of variance ${\epsilon }^2>0$. Their densities are, by the Gaussian integral and Fubini's theorem,and likewise for $\rho$. The integrable Gaussian factor justifies exchanging the integrals, so these densities agree. For every bounded continuous $f$, letting $\epsilon \downarrow 0$ in $\int \mathbb{E}[f(x+\epsilon Z)]\nu (dx)$, with $Z$ a standard normal random variable, gives $\int fd\nu$ by bounded convergence theorem. Hence the two measures have equal integrals against all these test functions, and approximating interval indicators shows $\nu =\rho$. This proves the uniqueness theorem for characteristic functions.

Every subsequence therefore has a further subsequence weakly converging to the same unique $\mu$. If the original sequence failed weak convergence, some bounded continuous test function would have a subsequence of integrals staying a fixed distance from its integral under $\mu$, contradicting that further convergence. This completes both directions of the theorem. Continuity at zero is essential: the laws $N(0,n)$ have pointwise limiting characteristic function zero away from zero and one at zero, and are not tight.

For the finite probability distribution, put ${\xi }_j=(X_j+1)/2$. These are independent fair zero-one variables, andThe binary integer ${\sum }_{j=1}^n{2}^{n-j}{\xi }_j$ is uniform on $\{0,\dots ,2^n-1\}$. HenceThese are the midpoints of $2^n$ equal subintervals of $[-1,1]$. For every bounded continuous $f$, the Riemann sums giveThus $\boxed{Y_n\Rightarrow \mathrm{Unif}[-1,1]}$ in the sense of weak convergence of probability measures.

For the almost-sure and $L^2$ assertions, the Dyadic Rademacher series $Y={\sum }_{j\ge 1}{2}^{-j}{X}_j$ converges absolutely on every sign sequence, since $|X_j|=1$. Its tail satisfies $|Y-Y_n|\le 2^{-n}$, proving almost sure convergence and convergence in L2. More precisely, zero means and independence eliminate the cross terms:This identity follows first for finite tails and then by dominated convergence. Bounded continuous test functions and bounded convergence theorem identify the probability distribution of $Y$ with the weak limit above, namely the uniform distribution on $[-1,1]$.

For the requested product identity, each Rademacher random variable has characteristic function $\mathbb{E}{e}^{it{X}_j}=\cos t$. The characteristic function of a sum of independent variables therefore yieldsBy part (a), or directly by bounded convergence theorem, this converges to the characteristic function of a uniform distribution, $\frac{1}{2}{\int }_{-1}^1{e}^{itx}dx=\sin t/t$. Consequently the dyadic cosine product isBoth sides are one at zero. At zeros of the sinc function the identity is understood as the limit of the finite products, so no logarithm or division by a vanishing factor is needed.

Let $B$ be a standard planar Brownian motion started at $z$. Neighbourhood recurrence means that, with probability one, every nonempty open subset of ${\mathbb{R}}^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, ${\mathbb{P}}_z(\exists t\ge 0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.

We prove both claims from the planar Brownian annulus hitting probability. Fix a center $w$ and radii $0<r<\rho =|z-w|<R$, and define the stopping times ${\tau }_r=\inf \{t:|B_t-w|=r\}$ and ${\tau }_R=\inf \{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite almost surely: a coordinate of Brownian motion cannot stay forever in a bounded interval, by the Brownian reflection principle, so exit from the outer disk is finite. On the annulus,The Itô formula therefore makes $\log |B_{t\wedge \sigma }-w|$ a local martingale. It is bounded between $\log r$ and $\log R$, hence is a genuine martingale. The optional stopping theorem at $t\wedge \sigma$, followed by the bounded convergence theorem, givesThusThe exit can only be at one of the two distinct boundaries by path continuity.

For a fixed $r$, let $R\to \infty$. These events increase to $\{{\tau }_r<\infty \}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit almost surely from any starting point outside it. An initial point inside it already counts as a hit.

To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the Markov property and the preceding disk-hitting result imply that some time $t\ge n$ visits this disk with probability one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A countable basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is recurrence of planar Brownian motion, or equivalently neighbourhood recurrence.

For point avoidance, keep $R>\rho$ fixed and let $r\downarrow 0$. Hitting $w$ before ${\tau }_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has probability zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A countable union now gives $\boxed{{\mathbb{P}}_z(T_w<\infty )=0(z\ne w)}$, the polar point for planar Brownian motion assertion.

Finally, even if the process starts at $w$, for each deterministic $s>0$ its normal distribution has a density, so $\mathbb{P}(B_s=w)=0$. The Markov property at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time.

First derive the Laplace functional of a Poisson random measure. For a simple nonnegative $g={\sum }_j{t}_j{1}_{A_j}$ on disjoint finite-intensity sets, independent Poisson random variables giveIncreasing simple approximations, monotone convergence for the integrals, and bounded convergence theorem for the random exponentials prove the same identity for every nonnegative measurable $g$.

Here the intensity is Lebesgue measure. Put $A=B(0,r)$ andThe integral is finite because $f$ has compact support. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$, $t\ge 0$:Differentiate from the right at $t=0$. The difference quotient on the random side is dominated by $N_r$, which is integrable with mean $v_d{r}^d$; the intensity integral is over a finite-volume ball. We obtain the count-weighted Laplace functional of a Poisson random measure:At $r=0$, the ball is empty and both sides are zero.

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.

The global drift hypothesis as printed is impossible on a finite state space. Indeed, let $x_{\ast }$ minimize $f$. Every possible value of $f(X_{n+1})$ is at least $f(x_{\ast })$, socontradicting a decrease by $\delta >0$. Equivalently, its formal iteration would give ${\mathbb{E}}_{x}f(X_n)=f(x)-n\delta$, whereas $f$ is bounded on the finite state space. This is the constant nonzero drift is impossible on a finite Markov chain obstruction. Thus there is no genuine chain satisfying all the literal premises.

The meaningful intended version imposes the drift only on $S\setminus D$, and stops at $D$. In the natural filtration ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$, the Markov property then gives, on $\{T>n\}$,The correct stopped martingale isIt is adapted and integrable, since $f$ is bounded and $n\wedge T\le n$. Its increment equalswhose conditional expectation is zero. Taking expectations yieldsSince $T<\infty$ almost surely and $f(X_T)=0$, the bounded convergence theorem gives ${\mathbb{E}}_{x}f(X_{n\wedge T})\to 0$. The monotone convergence theorem gives ${\mathbb{E}}_x(n\wedge T)\uparrow {\mathbb{E}}_{x}T$, without presupposing its finiteness. Consequently the constant-drift hitting-time identity isFor $x\in D$, both sides are zero. Under the repaired hypothesis the unstopped $f(X_n)+n\delta$ need not be a martingale after hitting $D$; it is the displayed stopped martingale that proves the result.

Let $\tau =T_{-a}\wedge T_b$ for the simple symmetric random walk. From any interior point, $a+b$ successive $+1$ increments force exit through $b$. The conditional probability of this block is $2^{-(a+b)}$, so a blockwise geometric tail bound from a uniform escape probability gives $\tau <\infty$ almost surely.

Apply the bounded optional stopping theorem to $\tau \wedge n$. Since $S_n$ is a martingale, ${\mathbb{E}}_0{S}_{\tau \wedge n}=0$. The stopped positions lie in $[-a,b]$, so the bounded convergence theorem gives ${\mathbb{E}}_0{S}_{\tau }=0$. If $h={\mathbb{P}}_0(T_{-a}<T_b)$, nearest-neighbor jumps ensure $S_{\tau }=-a$ on that event and $S_{\tau }=b$ otherwise. ThusNext apply the bounded optional stopping theorem to the square-minus-time martingale of a simple symmetric random walk:Again the right side converges by the bounded convergence theorem, while the left side increases to ${\mathbb{E}}_0\tau$ by the monotone convergence theorem. ThereforeThis argument justifies passage to the unbounded stopping time rather than applying optional stopping without an integrability check.

The construction is well defined. Since $D$ is contained in a ball, its Brownian exit time is finite almost surely: at successive integer times there is a fixed positive probability that the next independent unit-time Brownian motion increment has length greater than the ball's diameter, forcing an exit. The survival probability is therefore bounded by a geometric sequence. Path continuity gives $B_{{\tau }_D}\in \partial D$. Thus the bounded Dirichlet boundary data give a bounded Borel function $u$ on $D$.

For interior harmonicity, fix a ball with closure in $D$, centred at $x$, and let $\sigma$ be its Brownian exit time. The Strong Markov property, followed by the tower property of conditional expectation, givesThe orthogonal invariance of Brownian motion makes $B_{\sigma }$ uniform on the boundary sphere. Hence $u$ has the spherical mean value property for harmonic functions for every such ball. A locally bounded Borel function with this property is smooth and harmonic: integrating the spherical averages against any smooth radial mollifier gives $u=u\ast \rho$ locally, which first proves smoothness; the mean value property for harmonic functions then implies $\Delta u=0$.

For the boundary limit at $\xi$, choose the harmonic barrier for the Dirichlet problem $h_{\xi }$ from (i). The Itô formula, localized inside $D$, makes $h_{\xi }(B_{t\wedge {\tau }_D})$ a bounded martingale. Compact localization and the bounded convergence theorem, first to the exit and then as $t\to \infty$, yieldFor $\epsilon >0$, if $\partial D\setminus B(\xi ,\epsilon )$ is nonempty, compactness and barrier positivity giveIf that boundary subset is empty, the probability is already zero. ThereforeFirst let $x\to \xi$ and then $\epsilon \downarrow 0$. The continuity of the Dirichlet boundary data proves $u(x)\to f(\xi )$.

Finally, the difference of two continuous solutions is a harmonic function vanishing on the boundary. The maximum principle for harmonic functions on the bounded domain gives that it is zero. Thus the Kakutani solution of the Dirichlet problem exists, attains every prescribed boundary value, and is unique.

First take finite endpoints $\ell <r$ and a starting point $z\in (\ell ,r)$. For the Brownian exit time $H=\inf \{t:B_t\notin (\ell ,r)\}$, the stopped path stays in $[\ell ,r]$. The process $(B_t-z{)}^2-t$ is a martingale. Bounded-time optional stopping theorem givesBy the monotone convergence theorem, $\mathbb{E}H<\infty$, so exit from every bounded interval is almost sure:For completeness, a proper unbounded interval also has almost sure exit. Applying optional stopping theorem to $B_{H\wedge t}$ and then bounded convergence theorem gives the endpoint probabilitiesFor a fixed lower level $\ell$, let $r\to \infty$. Hitting $\ell$ before $r$ is a subset of ever hitting $\ell$, and its probability tends to one. Similarly every fixed higher level is hit almost surely. After a level is reached, the same argument from that level shows that a level strictly outside a given proper interval is eventually reached. This proves exit even when endpoints are included. A starting point already outside needs no argument. The whole real line has no exit, so the phrase “every interval” is understood to exclude that trivial exception. Countably many rational levels make the exit conclusion simultaneous for all proper intervals. The bounded-interval assertion is the one used below.

The domain is the union of the horizontal and vertical strips. The four connected components of its boundary carry the constants $a,b,c,d$ in the northeast, northwest, southwest and southeast, respectively.

First justify finiteness of $T$. A small open ball around $(2,2)$ lies outside the domain. By part (b), planar Brownian motion visits that ball almost surely in finite time. A continuous path from $z\in D$ to that ball must meet $\partial D$ first. Hence $T<\infty$ almost surely, and $B_T\in \partial D$ by continuity.

The harmonic function $u$ is smooth inside $D$. The Itô formula, localized on relatively compact subdomains, makes $u(B_{t\wedge T})$ a local martingale. Its boundary definition uses the continuous extension of $u$. Since $u$ is bounded, this stopped process is a true martingale with uniform integrability: pass through the compact exhaustion using bounded convergence theorem. In particular ${\mathbb{E}}_{z}u(B_{t\wedge T})=u(z)$.

Now $B_{t\wedge T}\to B_T$ almost surely as $t\to \infty$. The dominated convergence theorem, boundedness of $u$, and its continuous boundary values give the bounded harmonic representation with almost sure Brownian exit:Almost sure exit and boundedness of the harmonic function are both used. The domain's unboundedness causes no loss of expectation here because the stopped function values are uniformly bounded.

Put $D_t=B_t-B_t^{\prime }$. By independence of the two Brownian motions, $D_t/\sqrt{2}$ is a standard Brownian motion: its independent increments have the normal distribution with variance equal to elapsed time. The stopping time $\tau$ is the first time $D$ hits $-1$ or $3$.

To justify finiteness without presupposing the desired hitting probability, each unit-time increment of $D$ has the normal distribution $N(0,2)$. An increment of absolute value greater than $4$ forces exit from $[-1,3]$ if the path was inside at its start. These independent increments give $\mathbb{P}(\tau >n)\le (1-p{)}^n$, where $p=\mathbb{P}(|N(0,2)|>4)>0$. Hence $\tau <\infty$ almost surely.

The stopped martingale $D_{t\wedge \tau }$ lies in $[-1,3]$. The optional stopping theorem at $t\wedge \tau$ gives $\mathbb{E}{D}_{t\wedge \tau }=0$, and the bounded convergence theorem gives $\mathbb{E}{D}_{\tau }=0$. Writing $q=\mathbb{P}(D_{\tau }=-1)$, path continuity gives $0=-q+3(1-q)$. Therefore, as in Brownian exit from an interval,

Let $\tau =T_a\wedge T_b$. The walk exits the finite interval $[a,b]$ almost surely, and the stopped martingale $\varphi (S_{n\wedge \tau })$ is bounded between ${\lambda }^b$ and ${\lambda }^a$. The optional stopping theorem and bounded convergence theorem giveSolving for the first probability gives the biased gambler's ruin probability

Because $\tilde{F}$ solves the differential equation from part (c), the Itô formula makes $\tilde{F}(V_t)$ a local martingale before $\tau$. The defining improper integral converges at both endpoints: its integrand is asymptotic to $(-u{)}^{1-d}$ near zero and to $|u{|}^{d-3}$ at minus infinity. Since $1<d<2$, both exponents are integrable. Henceso the stopped local martingale is a bounded martingale.

If ${\tau }_x<{\tau }_y$, then $V_t\to 0$ and $\tilde{F}(V_t)\to 0$. If ${\tau }_y<{\tau }_x$, then $V_t\to -\infty$ and $\tilde{F}(V_t)\to I$. The hitting times cannot coincide because $X_t-Y_t$ stays positive. The optional sampling theorem for a supermartingale and bounded convergence theorem therefore giveConsequentlywhich is the Two-sided SLE boundary swallowing probability.

Let $T=\inf \{t\ge 0:X_t\notin (-a,b)\}$, and suppose the jumps of $X$ have absolute value at most $c$. A nonconstant centered finite-variance Lévy process oscillates, so $T<\infty$ almost surely. Before $T$ the process lies in $(-a,b)$, and at $T$ its bounded overshoot gives $X_T\in [-a-c,b+c]$. Thus the variables $X_{T\wedge n}^2$ are uniformly bounded.

Apply the optional sampling theorem for a supermartingale to the martingale from part (c):Bounded convergence theorem on the left and monotone convergence theorem on the right yield