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There are no visits when the initial excursion returns to $i$ before hitting $j$, soFor $k\ge 1$, the chain must first hit $j$, return to $j$ before $i$ exactly $k-1$ times, and then hit $i$ before another return to $j$. The Strong Markov property givesThe formula also covers the boundary cases $\alpha =0$ and $\alpha =1$.

Each completed visit to radius $r_2$ begins a new radial excursion. By part (a), the conditional probability that the following excursion reaches radius $R$ before radius $r_1$ isThe Strong Markov property at the successive stopping times ${\tau }_{2k}$ makes these trials independent with the same success probability. Consequently $N(R)$ has a geometric distribution on $\{1,2,\dots \}$ with parameter $q_R$:As $R\to \infty$, $q_R\to 0$ andSlutsky theorem now gives

Let ${\sigma }_x=\inf \{t\ge 0:{\tilde{B}}_t=x\}$. Continuity gives $\{S\ge x\}=\{{\sigma }_x<\infty \}$. On this event, the Strong Markov property says thatis an independent Brownian motion with drift $-\mu$. It reaches level $y$ with probability $\mathbb{P}(S\ge y)$. Therefore

Reflection in $L$ interchanges $x$ and $y$. Relabel the two points if necessary so that $x$ lies on the side of $L$ containing the center $a$; the absolute difference is symmetric in $x$ and $y$. Couple a Brownian motion $B$ started at $x$ to one $B^{\prime }$ started at $y$ by setting $B_t^{\prime }=\varphi (B_t)$ before $T_L$ and $B_t^{\prime }=B_t$ afterwards. The Brownian reflection coupling has the correct marginal laws because reflection is an isometry and the Strong Markov property applies at $T_L$. Once the paths meet on $L$, they agree forever.

If the path from $x$ does not hit the target circle before $T_L$, the coupled paths meet before the relevant uncoupled hitting outcomes can differ. The coupling inequality for total variation therefore gives

By part (a), translated by $a$, starting with $B_0\sim {\nu }_{a,R}$ givesPart (c) says that the same hitting law from ${\nu }_{0,R}$ tends in total variation distance to ${\nu }_{a,r_1}$. On the other hand, every path from the circle of radius $R>r_2$ to the target disc must first hit the circle of radius $r_2$. Part (a) and the Strong Markov property show that its position there has law ${\nu }_{0,r_2}$ and thatindependently of $R$. Letting $R\to \infty$ in part (c) proveswhen $B_0\sim {\nu }_{0,r_2}$.

LetThe clock $A$ is an absolutely continuous function and is strictly increasing: its derivative is positive away from the Brownian zero set, which has zero Lebesgue measure. It also tends to infinity. This is immediate for $\delta =1$; for $\delta >1$, recurrence and the Strong Markov property imply that the Brownian occupation time of, for example, $\{x:1<|x|<2\}$ is unbounded, while the integrand is bounded below there by ${\delta }^2$.

The Strong Markov property and part (b) show that, before $\tau ={\tau }_x\wedge {\tau }_y$,Thus $F(V_{t\wedge \tau })$ is a bounded martingale. From the given stochastic differential equation,The Itô formula says that the drift of $F(V_t)$ isIt must vanish. Dividing by $(1-v{)}^2/2$ and using the algebraic identity supplied in the question gives

The Strong Markov property says that for every almost surely finite stopping time $T$, the process $(B_{T+t}-B_T{)}_{t\ge 0}$ is a standard Brownian motion independent of ${\mathcal{F}}_T$.

Construct the times inductively. Suppose $(B_{T_0},\dots ,B_{T_n})$ has the law of $(S_0,\dots ,S_n)$. Conditional on the past, the martingale increment $S_{n+1}-S_n$ has mean zero and finite second moment. Apply the conditional form of the Skorokhod embedding theorem to this regular conditional law, using the fresh Brownian motion $B_{T_n+t}-B_{T_n}$ supplied by the Strong Markov property. This gives a stopping time increment ${\tau }_{n+1}$ and $T_{n+1}=T_n+{\tau }_{n+1}$ such that the next Brownian increment has the required conditional law. Induction provesfor every $k$.

A stopping time $T$ has $\{T\le n\}$ determined by $X_0,\dots ,X_n$. The Strong Markov property says that, conditional on $T<\infty$ and $X_T=i$, the post-$T$ chain is a fresh chain started at $i$, independent of the pre-$T$ history.

For simple random walk on ${\mathbb{Z}}^d$, $d\ge 3$, the Green-function decay for simple random walk on the integer lattice and the Strong Markov property giveThe series over $k$ converges, so the first of the Borel-Cantelli lemmas says that almost surely only finitely many of the points $x_k$ are ever hit. Moreover, ${\mathbb{Z}}^d$ is a transient graph for $d\ge 3$, so each of those finitely many points is visited only finitely often. Therefore the simple random walk visits $A$ only finitely often almost surely:

The displayed identity is false with the printed non-strict inequality. For example, take $A=B=\{y\}$ and let $r=P_y({\tilde{H}}_y<\infty )\in (0,1)$. The event $L_y\le {\tilde{H}}_y$ says that the walk returns to $y$ at most once, so its left side is ${\mu }_y(1-r^2)$, whereas its right side is $e_{\{y\}}(y)={\mu }_y(1-r)$.

The standard and evidently intended last-exit decomposition for a transient random walk has $L_B<{\tilde{H}}_A$. Decompose that corrected event according to $n=L_B$ and $z=X_n\in B$. The Strong Markov property at time $n$ givesReversibility of the random walk on a graph gives the path-reversal identitySince $e_B(z)={\mu }_z{P}_z({\tilde{H}}_B=\infty )$ is the equilibrium measure of a finite set, summing first over $n$ and then over $z$ yields

The Strong Markov property at the first visit to $0$ gives the singleton equilibrium potential from the Green functionLet $U_n$ be an increasing exhaustion of $G$ by finite sets containing $0$, and let $g_{U_n}(x,0)$ be the Green function of a transient weighted graph stopped on leaving $U_n$. Each $g_{U_n}(\cdot ,0)$ has finite support. The Green identity and the Markov property give, for $m\ge n$,By the monotone convergence theorem, the right side tends to zero as $m,n\to \infty$, while $g_{U_n}(x,0)\uparrow g(x,0)$. Thus $g(\cdot ,0)$ is an $H_0$ limit of finitely supported functions. Dividing by the positive number $g(0,0)$ proves $h\in H_0$.

For planar Brownian motion starting on the positive real axis inside a disk, let $T$ be circular exit and $S$ first contact with the imaginary axis. Reflecting the path after $S$ fixes the disk and swaps positive and negative circular exits. The Strong Markov property makes their probabilities equal on $S<T$. Therefore $\mathbb{P}(T<S)=\mathbb{P}(\mathrm{Re}{B}_T>0)-\mathbb{P}(\mathrm{Re}{B}_T<0)$.