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Because $f$ is a continuous probability density, there are a point $z$, a radius $r>0$, and $ϵ>0$ such thatBy the recurrence of planar Brownian motion, the smaller disc $B(z,r/2)$ is visited at arbitrarily large times. Starting anywhere in that smaller disc, Brownian continuity and compactness give a uniform probability $\delta >0$ of staying in $B(z,r)$ for a fixed time $u>0$.

Apply the Strong Markov property at successive visits separated by at least $u$. The conditional probability of each stay event is at least $\delta$, so the conditional Borel-Cantelli lemma gives infinitely many successful stays almost surely. Every success adds at least $ϵu$ to $A_t$. Since $A_t$ is nondecreasing,

Letand choose $x$ attaining the maximum. For every $t$, the Strong Markov property at time $t$ givesThusIf $t\ge t_{\mathrm{mix}}(1/4)$, thenSince $\{X_t\in A\}\subseteq \{T_A\le t\}$, this is impossible when $t<m/4$. Allowing for integer times, one may take any smaller absolute constant, for example

After leaving zero, the jump chain avoids every future reset with probabilityThis probability is positive exactly when ${\prod }_{i\ge 1}(1+{\rho }_i)<\infty$. If $q>0$, the probability of ever returning to zero is less than one, so the irreducible chain is transient. If $q=0$, a reset to zero occurs almost surely after each departure; by the Strong Markov property this happens repeatedly, so the chain is recurrent. Therefore

A random time $\tau$ is a stopping time for the natural filtration $({\mathcal{F}}_n)$ of a Markov chain when $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$. The Strong Markov property says that, conditionally on $\tau <\infty$ and $X_{\tau }=i$, the process after $\tau$ is a fresh copy of the chain started from $i$, independent of the pre-$\tau$ history.

Before capture define the half-separationFor $D_n=k\ge 2$, independence of the two moves gives the transition probabilitiesIgnore the holding steps. The resulting embedded chain is a biased random walk that moves left with probability $q$ and right with probability $1-q$. Its expected number of moves needed to descend one level is $1/(2q-1)$; this follows either from first-step analysis or from its drift $1-2q$. A non-holding move occurs with probability $1/2$ at each time, so its mean waiting time is two. Hence, for every $m\ge 2$, the expected time to go from separation $2m$ to $2(m-1)$ is

Let $T_k$ be the expected capture time starting from $D_0=k$, so $T_0=0$. The Strong Markov property at successive first passages to lower levels and part (a) giveAt $D_n=1$, the next state is zero with probability $r/2$, one with probability $1/2$, and two with probability $(1-r)/2$. First-step analysis therefore givesorSolving,Substituting $\mu =2/(2q-1)$ now yieldswhich is the required expected capture time.

A random time $T$ is a stopping time when the event $\{T\le n\}$ is determined by $X_0,\dots ,X_n$. The Strong Markov property says that, conditionally on $T<\infty$ and $X_T=x$, the process $(X_{T+k}{)}_{k\ge 0}$ is a fresh Markov chain started at $x$, independent of the history before $T$.

Use the state space $\{1,2,4\}\times {\mathbb{Z}}_{\ge 0}$, with $(2,0)$ absorbing. Observe the chain only when it is at square 2. From $(2,k)$, the change $Y$ in wealth by the next return to square 2 has distributionIndeed, heads lands on square 3 and returns to square 2 after losing £1. After tails reaches square 4, the remaining two or three moves give the other cases.

For $r=2/3$,Thus $r^{M_j}$ is a martingale for the embedded wealth random walk $M_j$. Stopping when it first reaches $m-1$ or a large upper level and then letting that level tend to infinity givesThe upper-bound contribution vanishes because $0<r<1$; equivalently, this is the smaller probability solution of the first-step equation.

By part (a) and repeated use of the Strong Markov property, the probability of descending from $(2,k)$ to $(2,0)$ isStarting from square 1 with £$m$, heads moves to $(2,m)$, whereas tails lands on square 3, loses £1, and moves to $(2,m-1)$. Therefore the loss probability is

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.

Consider an occurrence counted by $G_m$. Before the next counted occurrence, the exploration must first take a downward step of the simple symmetric random walk, which has probability $1/2$, and then choose the decrement $-1$ among the three equally likely values of $\xi$, which has probability $1/3$. Thus it terminates the visits to the current record value with probabilityThe Strong Markov property at successive counted occurrences makes these trials independent. Therefore $G_m$ has the geometric distribution on $\{1,2,\dots \}$ with parameter $1/6$, andfor every $j\ge 1$.

If $A\subseteq C$, Brownian motion exits $\mathbb{H}\setminus C$ no later than it exits $\mathbb{H}\setminus A$. The Strong Markov property and the nonnegative harmonic function $h_A$ show that the expected exit height for $C$ is at least that for $A$. Taking the limits in the Brownian representation of half-plane capacity proves the monotonicity of half-plane capacity

For $0<\kappa \le 4$, the trace meets the real boundary only at its starting point, so the intersection has zero Lebesgue measure. Suppose $4<\kappa <8$. A boundary point swallowed by an interval need not itself lie on the trace. The stated fact that $\mathbb{P}({\tau }_x={\tau }_y)>0$ for $0<x<y$, combined with the Strong Markov property and Scaling invariance of SLE at successively nested swallowed intervals, implies that a fixed deterministic $x\ne 0$ is swallowed in an interval rather than hit by the trace with probability one. Equivalently,Applying the Tonelli theorem to the random indicator of the boundary trace givesThe nonnegative random measure is therefore zero almost surely.

View each of the three level-one triangles in $G_2$ as a copy of $G_1$. Observe the walk only when it moves from one corner of such a copy to a different corner. By symmetry, the next of the two other corners is equally likely, so this embedded chain is the simple random walk on the coarse graph $G_1$. Part (a) says it makes an expected five coarse transitions before reaching $B$ or $C$.

Within each level-one copy, the mean time for one coarse transition is again five by part (a). The Strong Markov property at successive coarse-corner hitting times therefore gives

Let $q={\mathbb{P}}_i(T_i<\infty )$. By the Strong Markov property, after every return to $i$ the chance of another return is again $q$. HenceThus $V_i$ is geometric with success parameter $1-q$; when $q=1$, it is infinite almost surely.

Yes. Quotient only the first coordinate modulo $\mathbb{Z}$. The resulting process lives on $\mathbb{T}\times {\mathbb{R}}^2$ and the image of $S_1$ is the radius-$ϵ$ ball about $(0,0,0)$. The two noncompact coordinates form planar Brownian motion, which is recurrent; during its infinitely many returns to a smaller disc, the independent circle coordinate has a fixed positive chance of lying in the required interval. The Strong Markov property then shows that the target ball is visited infinitely often. Lifting back proves that $S_1$ is hit at unbounded times.

Observe the lazy walk on ${\mathbb{Z}}_{2^k}$ only after every second genuine jump. Two independent nearest-neighbor signs have sum $-2,0,2$ with probabilities $1/4,1/2,1/4$. Division by two therefore produces one step of the lazy simple random walk on ${\mathbb{Z}}_{2^{k-1}}$. The Strong Markov property at successive jump times proves

For $x\ne 0$, the Strong Markov property at the first step gives the discrete mean-value identityAt $x=0$, $v(0)=1$ while the same average is at most one. Thus $v$ is a bounded superharmonic function on ${\mathbb{Z}}^3$. Conditioning on the natural filtration and using the one-step Markov property givesTherefore $(v(X_n))$ is a nonnegative supermartingale.

The pair $(B,\tilde{B})$ is a Brownian motion in ${\mathbb{R}}^{2d}$. Assume inductively that $T_{i-1}<\infty$. By the Strong Markov property at $T_{i-1}$, the differenceis a one-dimensional Brownian motion with variance rate two, started from its current value. By recurrence of one-dimensional Brownian motion, it hits zero in finite time almost surely. Hence $T_i<\infty$. Induction through $i=1,\dots ,d$ proves $T_d<\infty$ almost surely.

Start Brownian motions at arbitrary $x,y\in {\mathbb{R}}^d$. Use the successive meeting times from part b, but after coordinate $i$ meets, drive that coordinate of the second process with the first process's increments forever. The Strong Markov property shows that each marginal remains a $d$-dimensional Brownian motion. By part b every coordinate is eventually locked, so the resulting coordinatewise coalescing coupling of Brownian motions has an almost surely finite coalescence time $T$.

Because $f$ is bounded and harmonic, Dynkin formula for Brownian motion shows that $f(B_t^x)$ and $f(B_t^y)$ are bounded martingales. ThereforeSince $T<\infty$ almost surely, the right side tends to zero. Thus $f(x)=f(y)$ for all $x,y$, proving the Brownian coupling proof of the harmonic Liouville theorem.

Let $\sigma$ be the first time Brownian motion started outside the unit half-disc reaches its semicircular boundary. A path that reaches $A$ must first cross that semicircle. The Strong Markov property at $\sigma$ givesTake $z=iy$, multiply by $y$, and let $y\to \infty$. The Brownian representation of half-plane capacity identifies the left side with $\mathrm{hcap}(A)$, while part b givesSince the exit height is between zero and one, the dominated convergence theorem applies and yields

Starting from $i$, the probability of reaching $j$ before returning to $i$ is $\alpha$. Conditional on reaching $j$, each visit to $j$ is followed by a hit on $i$ before the next return to $j$ with probability $\alpha$, by symmetry and the Strong Markov property. For $\alpha >0$, the number of visits after entry is therefore geometric on $\{1,2,\dots \}$ with mean $1/\alpha$. Hence the two-state excursion visit law givesIn the degenerate case $\alpha =0$, the chain never reaches $j$ before its return to $i$, so $N=0$ almost surely and ${\mathbb{E}}_{i}N=0$.