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Fix $x\in (-1,1)$ and start at $(x,y)$ with $y>1$. Write the independent Brownian coordinates as $x+W_t^1$ and $y+W_t^2$. LetOn $\{H<L_y\}$ the path has stayed in the upper strip and first exits the cross-shaped domain at its right or left side. The boundary value is $a$ or $b$, respectively. Let $V=a$ if $x+W_H^1=1$ and $V=b$ if $x+W_H^1=-1$. The Brownian exit from an interval probabilities giveCouple all heights $y$ using the same two Brownian paths. By part (a), $H<\infty$ almost surely, and continuity makes ${\min }_{0\le t\le H}{W}_t^2$ finite. ThusLet $K=\max \{|a|,|b|,|c|,|d|\}$. On the complement of this exceptional event, $f(B_T)=V$; on the event itself, $|f(B_T)-V|\le 2K$. The representation in part (c) therefore yieldsThis proves the harmonic limit along an infinite strip arm:Far up the arm, the bottom is reached with vanishing probability before the horizontal coordinate exits. The limit is consequently determined by the two side values alone.

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,

The random-walk form of the Skorokhod embedding theorem says the following. If $S_n=Y_1+\cdots +Y_n$, where the $Y_i$ are independent and identically distributed withthen on a space carrying a Brownian motion $B$ there are stopping timessuch that $(B_{T_n}{)}_{n\ge 0}$ has the same law as $(S_n{)}_{n\ge 0}$, and the increments $T_n-T_{n-1}$ are independent and identically distributed with mean ${\sigma }^2$.

To prove the one-step statement, first note that every centered distribution is a mixture of centered two-point distributions. Indeed, match the equal-mass size-biased measures $x\mathbb{P}(Y\in dx)$ on $(0,\infty )$ and $|x|\mathbb{P}(Y\in dx)$ on $(-\infty ,0)$. This produces a random pair $(L,R)$ of positive numbers such that, conditionally on $(L,R)$, $Y$ has values $-L,R$ with probabilitiesand $\mathbb{E}[LR]=\mathbb{E}{Y}^2={\sigma }^2$. Include the atom at zero by taking the stopping time zero.

Choose $(L,R)$ independently of $B_t$ and stop Brownian motion on first leaving $(-L,R)$. The Brownian exit from an interval formulas give the displayed two-point probabilities and conditional mean stopping time $LR$. Thus $B_T$ has the law of $Y$ and $\mathbb{E}T={\sigma }^2$.

Starting from $T_0=0$, repeat this construction after each $T_{n-1}$. The Strong Markov property makes the new Brownian increments independent copies of the first embedding, proving the Skorokhod embedding of a centered random walk.