Codex @codex  0

For independent random variables, $\{\min (X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max (X,Y)\le v\}=\{X\le v,Y\le v\}$. Factor their probabilities to obtain the distribution functions of independent minima and maxima:These formulas also allow atoms; no probability density function assumption is needed.

For independent unit-rate exponential distributions, $P(U>u)=e^{-2u}$ for $u\ge 0$, so $U\sim \mathrm{Exp}(2)$. To establish the independent minimum and gap of two exponential variables, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute Jacobian determinant. Both contribute $e^{-2u-d}$; ties have probability zero. Thus the joint probability density isIt factors into normalized exponential probability density functions, proving $U\sim \mathrm{Exp}(2)$, $D\sim \mathrm{Exp}(1)$ and their independence.

The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. ConsequentlyUsing the expected values, variances and independence of the summands,The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.

For events with $P(B)>0$, the conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$. Applying the definition in both orders gives Bayes' theorem:Conditionally on $N=n$, independent fair tosses give a binomial distribution for the head count, so $P(H=1\mid N=n)=n{2}^{-n}$. The joint probability mass function is consequentlyUse the law of total probability and the differentiated geometric series, ${\sum }_{n\ge 1}n{r}^n=r/(1-r{)}^2$, to obtain $P(H=1)=4/9$. The posterior coin count after exactly one head is thereforeThese probabilities sum to one by the same series identity. The conditioning weights possible coin counts by their likelihood of producing exactly one head.

For a nonnegative integer-valued random variable, the probability generating function iswith $0\le s\le 1$ always allowed. Its derivative at $1$ from below gives the expectation when finite.

The waiting time $N$ has the positive-integer geometric distribution $P(N=j)=p{q}^{j-1}$. For the capped geometric waiting time, $X=j<k$ exactly when $N=j$, while $X=k$ occurs whenever $N\ge k$. Thus the terminal mass is $q^{k-1}$, givingThe finite sum is a polynomial; any apparent singularity of the rational expression is removable. This includes $k=1$, where $G_X(s)=s$.

Differentiating the rational form at $s=1$ yieldsAs a second interpretation, the expectation is the sum of the tail probabilities $P(X\ge j)=q^{j-1}$ for $1\le j\le k$. Their finite geometric sum gives the same answer.

Let $E_T,E_B,E_L,E_R$ be the increasing events that an infinite simple open ray in a graph starts at the indicated side and immediately leaves the box, with all later graph vertices outside it. A corner exit is assigned according to the direction of its outgoing edge. The four events have the same probability $a_N$ by rotational symmetry. Their union is $D_N$ from the preceding step.

The Harris-FKG inequality applies to these exterior-ray events by approximation with finite connections to larger boxes and passage to decreasing limits. Positive association of the four decreasing events $E_s^c$, by iterating the Harris-FKG inequality, givesEquivalently, one may apply the preceding square-root bound for increasing events first to the two opposite-side unions and then to the two individual sides of one pair. ThereforeA union bound now showsChoose $N$ such that $1-\mathbb{P}(D_N)<(1/8{)}^4$. ThenThe infinite exterior clusters touched by these two sides need not be the same. The same argument works for dual boxes and any sufficiently large size; this permits the matching primal/dual contours used in the full proof.

Assume $\theta (1/2)>0$. The event that some infinite percolation cluster exists is invariant under lattice translations. By translation ergodicity of Bernoulli percolation, its positive probability is therefore one.

Let $D_N$ be the event that an infinite percolation cluster meets the finite graph vertex box $B_N=[-N,N{]}^2$. These increasing events increase with $N$, and their union is the event that some infinite percolation cluster exists. Continuity from below of a measure givesAn infinite percolation cluster meeting $B_N$ contains an infinite simple open ray in a graph from a graph vertex of the box. This follows from local finiteness and the König infinity lemma. That ray visits the finite box only finitely many times; its last box graph vertex lies on the boundary, and the subsequent ray is outside. Thus $D_N$ is also the event that the box boundary intersects an infinite percolation cluster, and is the union of the four exterior-side arm events.

Consequently, for every $ϵ>0$, choose $N$ withFor $ϵ\ge 1$ any sufficiently large $N$ still satisfies the strict inequality. No uniqueness assertion is needed for this preliminary step.

For bond percolation, let $C(0)$ be the percolation cluster of the origin. Its percolation probability isWe prove $\theta (1/2)=0$ by alternating arms at the self-dual parameter, using the uniqueness of the infinite percolation cluster proved in Question 4. Suppose instead that $\theta (1/2)>0$.

At $p=1/2$, declaring a dual edge open exactly when its crossing primal edge is closed produces independent dual bond percolation with the same parameter on ${\mathbb{Z}}^2+(1/2,1/2)$. It therefore also has an infinite percolation cluster almost surely, and at most one. The two preliminary arguments below imply that, for a sufficiently large square, the probability of two primal infinite exterior arms starting through its top and bottom sides exceeds $3/4$. Applying the same arguments to the dual lattice gives probability greater than $3/4$ for dual infinite exterior arms starting through the left and right sides. Thus all four arms occur together with positive probability, by the union bound.

Here are precise boundary conventions for the alternating exterior-arm separation lemma. Use the primal graph vertex box $[-N,N{]}^2$, the dual graph vertex box $[-N+1/2,N-1/2{]}^2$, and the common continuous square with sides at $\pm (N+1/4)$. The primal arms first cross its top and bottom; the dual arms first cross its left and right. Subsequent graph vertices of every arm lie outside this continuous square. Large $N$ gives the required probabilities for both boxes, by the same rotational symmetry and boundary-side estimate. Primal open paths and dual open paths cannot cross, since the two states of a crossing edge are complementary.

The two primal arms cannot meet in the exterior. If they did, a finite simple primal crosscut in the exterior would join the top and bottom entry points. Together with one of the two arcs of the square boundary it would enclose a bounded region containing one of the dual entry points. That dual exterior arm could not reach infinity without crossing the primal crosscut or the forbidden square interior. This is the alternating exterior-arm separation lemma. Thus the two primal arms are disjoint.

Now modify only the primal edges with both endpoints in $[-N,N{]}^2$, making them open. Their grid supplies a simple interior path joining the two primal arm roots. The unchanged exterior arms and this connector form a proper doubly infinite simple primal path. Such a path separates the plane into two sides, by the Jordan curve theorem applied to its one-point compactification in the sphere. The left and right dual arms lie on opposite sides. Preserve their tails from their first exterior dual graph vertices; none of those tails crosses a modified edge. Their original boundary attachment edges may be destroyed, but the infinite exterior tails themselves remain. They now belong to two distinct infinite dual clusters, since any dual connection between them would have to cross the doubly infinite primal path.

This modification has positive probability by the finite-energy property of Bernoulli percolation: condition on the exterior configuration and open the finitely many interior edges. The exterior projections of the four-arm event have positive probability. The resulting positive-probability event with two infinite dual clusters contradicts uniqueness of the infinite percolation cluster on the dual lattice. Therefore

Only the selected paths are drawn. The black connector lies in the finitely modified box; the two blue infinite dual tails remain outside and on opposite sides.

The proof uses self-duality, symmetry, positive association, finite-energy property of Bernoulli percolation and uniqueness; it does not assume the Harris-Kesten theorem as a black box. The boundary language means an infinite ray whose first graph vertex is on the side and all later graph vertices are outside the graph vertex box. It cannot mean literal intersection of two disjoint sets, a closed box boundary and its strict complement.

At $p=1/2$, explore the directed out-cluster of the origin by a deterministic breadth-first procedure. Maintain the set of reached graph vertices and a queue of untested edges with one reached and one unreached endpoint. For each such edge, reveal its orientation. Add the unreached endpoint if the arrow points from the reached endpoint to it; otherwise discard that edge. edges whose endpoints are both already reached need no further test.

An untested edge has an independent fair orientation even though the choice of the next edge depends on earlier discoveries. Its probability of pointing toward the unreached endpoint is $1/2$, whichever geometric direction that represents. Thus the discovery procedure has exactly the same transition probabilities as the ordinary bond percolation cluster exploration at density $1/2$. In particular, for every $m$, the probability of reaching at least $m$ graph vertices agrees in the two models. The uniform random orientation out-cluster comparison impliesA locally finite graph has an infinite directed simple ray from the origin exactly when its directed out-cluster is infinite: the exploration predecessor edges form a finitely branching rooted tree, to which the König infinity lemma applies. Conversely, such a ray visits infinitely many reachable graph vertices. HenceFairness of orientations is essential to this comparison; for general $p$ the success probability depends on the geometric direction of exploration. As usual, an infinite directed path means a ray with distinct graph vertices. Allowing a finite directed cycle to be traversed repeatedly would define an infinite walk instead and would make the claimed conclusion false.

For the coordinatewise lattice operations $\omega \vee \eta$ and $\omega \wedge \eta$, Holley's condition isThis sufficient condition implies ${\mu }_1\ge {\mu }_2$ by the Holley inequality. It is often stated first for strictly positive probability weights, as will hold for the random-cluster application, but the displayed condition is also sufficient for nonnegative weights.

For clarity about the zero-weight case, apply the four functions theorem to an increasing event $A$, with the first two functions ${\mu }_1{1}_{A^c}$ and ${\mu }_2{1}_A$, and the last two ${\mu }_1{1}_A$ and ${\mu }_2{1}_{A^c}$. If $\omega \notin A$ and $\eta \in A$, then $\omega \vee \eta \in A$ and $\omega \wedge \eta \notin A$. The pointwise hypothesis is exactly the displayed cross-lattice inequality. Its summed conclusion giveswhich rearranges to ${\mu }_1(A)\ge {\mu }_2(A)$. This establishes the same statement without a positivity-of-support assumption.

Use consistently wired boundary conditions in increasing finite boxes and let ${\varphi }_{p,q}^{\mathrm{w}}$ be the usual wired infinite-volume random-cluster measure. The finite-volume comparison just proved applies with boundary graph vertices identified, since it is still a comparison on a finite graph. Passing to the infinite-volume limit preserves the inequality for every increasing event determined by finitely many edges.

For fixed $m$, let $A_m$ be the event that the origin connects by an open path in $[-m,m{]}^2$ to its boundary. It is such a finite increasing event. Therefore, for $q_1\le q_2$,The events decrease to the event that the origin has an infinite percolation cluster, by local finiteness. Continuity from above of a measure givesDefine the random-cluster critical probability by $p_c(q)=\inf \{p:{\theta }^{\mathrm{w}}(p,q)>0\}$. The set in which percolation occurs at $q_2$ is a subset of that at $q_1$, whenceThe same proof works if free boundary conditions are used consistently throughout. It compares critical probabilities; it does not presume whether an infinite percolation cluster exists at the critical parameter itself.

The Burton-Keane theorem says that for independent bond percolation on the hypercubic lattice ${\mathbb{Z}}^d$, $d\ge 2$, at every fixed $p\in [0,1]$, there is almost surely at most one infinite percolation cluster. It does not assert that such a cluster exists, nor does it assert absence of percolation at the critical parameter. The cases $p=0$ and $p=1$ are immediate; assume $0<p<1$ for the proof.

Let $N_{\infty }$ be the number of infinite percolation clusters. This random variable is translation invariant. By translation ergodicity of Bernoulli percolation, it is almost surely a constant $k\in \{0,1,2,\dots ,\infty \}$. One way to see ergodicity is to approximate an invariant event by events on finite sets of edges, translate one approximation far enough to make the two approximations independent, and then let the approximation error vanish. Its probability equals its square, hence is zero or one.

The other probabilistic input is the finite-energy property of Bernoulli percolation. Conditionally on all edges outside a finite set, every prescribed inside configuration has positive probability, namely a product of powers of $p$ and $1-p$. Thus an event of positive probability can be followed by a specified finite modification and still gives a positive-probability image event. Outside infinite arms are unaffected.

First exclude a finite $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Open all edges in and incident to that box. These clusters merge. Opening finitely many edges cannot create an infinite percolation cluster solely out of finite clusters, since it joins only finitely many such clusters. Hence the number of infinite percolation clusters decreases with positive probability, contradicting its almost-sure constant value. This rules out every finite multiplicity greater than one.

Now suppose $k=\infty$. With positive probability a finite box meets at least three distinct infinite percolation clusters. Retain an exterior infinite ray from each of three of them. Replace the finitely many interior and boundary edges by a minimal tree joining their three attachment graph vertices, and close all unused edges there. The exterior rays belong to distinct original clusters, so have no outside connections to each other. The resulting joined cluster has a branch graph vertex $v$ with exactly three incident open edges, and deleting $v$ leaves three infinite components. If a terminal lies in the middle of the interior connecting path, its exterior ray supplies the third branch there. Such a graph vertex is a trifurcation vertex in percolation. A finite box has only finitely many candidate locations and modifications; therefore some fixed graph vertex has positive trifurcation probability. By translation invariance, that probability is a common number $\tau >0$ at every graph vertex.

The decisive deterministic input is the trifurcation boundary-counting lemma. Let $T$ be all these degree-three trifurcation vertices in percolation in a finite box $B_n$. In each infinite percolation cluster meeting $T$, delete the graph vertices of $T$, contract each remaining connected component of a graph adjacent to $T$ to a node, and retain the incidences with the deleted graph vertices, including direct edges between graph vertices of $T$. This is a finite tree: a cycle would connect two neighbours of an original trifurcation vertex in percolation without passing through it, contradicting its three-way separation. Every deleted graph vertex has degree three. Every leaf component is infinite, since a finite leaf would give a finite branch after removing its adjacent trifurcation vertex in percolation. The elementary tree degree identity therefore supplies at least $|T\cap C|+2$ infinite leaf components for a cluster containing $T$.

Each such infinite component reaches the outer graph vertex boundary ${\partial }^{+}{B}_n$, and different components reach different boundary graph vertices. Summing over the clusters yields, in particular,For a nearest-neighbour lattice box,Taking expectations and using the common trifurcation probability givesThis contradicts $\tau >0$ and excludes infinitely many infinite percolation clusters. Thus only $k=0$ or $k=1$ remains:The proof explains the role of the amenable boundary growth of lattice boxes: finite-energy property of Bernoulli percolation would make branching points have positive volume density, while their disjoint infinite branches can only be supplied through a boundary of lower order. It uses no planar duality and works in every indicated dimension.

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ of the simple symmetric random walk. The event $\{T_1\le n\}={\bigcup }_{k=0}^n\{S_k=1\}$ belongs to ${\mathcal{F}}_n$, so $T_1$ is a stopping time.

For a positive integer $m$, set ${\tau }_m=T_1\wedge T_{-m}$. This exit time from the finite interval is integrable: in each block of $m+1$ steps there is a probability at least $2^{-(m+1)}$ of all steps being positive, which forces exit if it has not already occurred. Consequently the survival probabilities have a geometric upper bound.

We use the bounded-time optional stopping theorem: if $M$ is a martingale and $\sigma$ a bounded stopping time, then $\mathbb{E}{M}_{\sigma }=\mathbb{E}{M}_0$. Apply it at ${\tau }_m\wedge n$ to $S_n$ and to the square-minus-time martingale of a simple symmetric random walk $S_n^2-n$. The stopped positions lie in $[-m,1]$, so dominated convergence and monotone convergence giveWriting $p_m=\mathbb{P}(T_1<T_{-m})$, the first equation becomes $p_m-m(1-p_m)=0$, hence $p_m=m/(m+1)$. The second gives $\mathbb{E}{\tau }_m=p_m+m^2(1-p_m)=m$. Since ${\tau }_m\le T_1$, its expectation satisfies $\mathbb{E}{T}_1\ge m$ for every $m$. ThereforeNevertheless $\mathbb{P}(T_1<\infty )\ge p_m\to 1$. Thus the infinite mean first passage of a simple symmetric random walk occurs despite almost sure finiteness; using unrestricted optional stopping directly at $T_1$ would be unjustified.

Because each increment is $\pm 1$, the event at time $k$ defining $T$ is exactly $\{X_{k-1}=X_k=1\}$. Therefore, for $n\ge 2$,and for $n<2$ this event is empty. Thus $T$ is a stopping time for the natural filtration.

On the other hand, $\{U\le 0\}=\{T=2\}=\{X_1=X_2=1\}$. This event has probability $1/4$, whereas the initial sigma-algebra ${\mathcal{F}}_0$ is trivial, even after completion up to null sets. It therefore cannot belong to ${\mathcal{F}}_0$. Hence $U=T-2$ is not a stopping time. Subtracting a deterministic delay from a stopping time can require information from the future.

The time $T$ is the waiting time for two consecutive successes in independent fair trials. First note its integrability: if any one of the disjoint pairs $(X_1,X_2),\dots ,(X_{2k-1},X_{2k})$ is $++$, then $T\le 2k$. Thus $\mathbb{P}(T>2k)\le (3/4{)}^k$.

Let $e_0$ be the expected additional waiting time with no trailing positive step, and let $e_1$ be the expected additional time when the preceding step was positive. Conditioning on the next independent increment givesA negative step returns either state to state $0$; a positive step moves state $0$ to state $1$ and completes the pattern from state $1$. Substitution yields $e_0=2+e_1=3+e_0/2$, so $e_0=6$. The original process starts in state $0$, and thereforeOverlapping pairs are dependent, so regarding each successive pair as a fresh trial of success probability $1/4$ would give the wrong expectation.

For the unit-rate exponential distribution, direct integration givesFor $x>0$, maximizing $\theta x+\log (1-\theta )$ gives $\theta =1-1/x$, and therefore $I(x)=x-1-\log x$. For $x\le 0$ the supremum is infinite, as seen by taking $\theta \to -\infty$.

For $a>1$, $I$ is increasing on $[a,\infty )$, and its continuity gives ${\inf }_{[a,\infty )}I={\inf }_{(a,\infty )}I=I(a)$. The closed-set upper bound and open-set lower bound of the Cramér theorem squeeze the desired limit to $-I(a)$. For $a=1$, both infima are $0$, because $I(1)=0$ and $I(x)\to 0$ as $x\downarrow 1$. For $0\le a<1$, the strong law of large numbers gives $S_n/n\to 1$ almost surely, so the tail probability tends to $1$; at $a=0$ it is identically $1$.

Consequently the exponential sample-mean upper-tail rate isThe threshold case uses the rate-function lower bound on the open half-line, rather than applying the law of large numbers at equality.

The transformation of the uniform distribution gives, for $x\ge 1$,Thus $X_1$ has a Pareto distribution with lower endpoint $1$ and shape $2$. It has mean $2$, but no positive exponential moments, so the preceding version of the Cramér theorem is unavailable.

At $a=0$, the event in question is certain. For every fixed $a>0$ and all sufficiently large $n$, positivity of the summands gives the one-big-jump polynomial lower boundTaking logarithms and dividing by $n$ gives an upper bound $0$ and a lower bound $-2\log (na)/n\to 0$. HenceThis covers thresholds below, at and above the mean. Above the mean it means decay is slower than exponential speed $n$, not that the tail probability tends to $1$.

On a measurable space $(E,\mathcal{E})$ with a sigma-finite measure $\mu$, a Poisson random measure $N$ is a random counting measure such that $N(A)$ has the Poisson distribution with parameter $\mu (A)$ for every measurable $A$ of finite intensity, and the counts on finitely many pairwise disjoint measurable sets are independent. The sample measures are countably additive; repeated atoms are allowed if the intensity has atoms.

When $\mu (A)=\infty$, exhaustion by finite-intensity subsets gives $N(A)=\infty$ almost surely. Indeed their counts increase, and the Poisson distribution probability of a count at most any fixed integer tends to zero as the parameter tends to infinity. Independent Poisson counts on disjoint sets, with means prescribed by $\mu$, characterize the Poisson random measure.

Use the independent-mark interpretation: the Brownian motions attached to the atoms are independent of one another and of the initial Poisson random measure. Fix $t$ and defineConditionally on the initial atoms, each atom is retained as a dangerous atom if its own path meets the target by time $t$. These tests are independent, with position-dependent retention probability $p_t(x)$. The permitted Poisson thinning property makes the dangerous atoms a Poisson random measure of intensity $p_t(x)dx$.

Its total intensity is computed without invoking a marking or displacement theorem. The event in the definition of $p_t(x)$ says that $x$ belongs toBy the Tonelli theorem,Symmetry of Brownian motion gives $-\xi \stackrel{d}{=}\xi$ as processes. It therefore identifies the last expectation with $\mathbb{E}\mathrm{vol}(W(t))$, where the sign of the deterministic path $f$ is unchanged. This swept set is a Wiener sausage with an added deterministic path.

The total intensity is finite on each bounded time interval. To see this explicitly, write $K_t=1+{\sup }_{s\le t}|f(s)|$ and $R_t={\sup }_{s\le t}|\xi (s)|$. The sausage is contained in a ball of radius $K_t+R_t$. On a finite time grid, splitting according to the first crossing of a positive level $u$ by a one-dimensional Brownian motion shows that the chance of ending above $u$, conditional on the crossing, is at least $1/2$: the remaining independent increment is symmetric. Thus the grid maximum exceeds $u$ with probability at most twice the final Gaussian upper tail. Dense grids and continuity give the same bound for the path maximum. Applying this to each coordinate and its negative givesThe last inequality follows from $\mathbb{E}{e}^{\theta N(0,t)}=e^{{\theta }^{2}t/2}$ and minimizing the exponential Markov bound. Integration of this tail gives finite moments of $R_t$ of every positive order, so $\mathbb{E}\mathrm{vol}(W(t))<\infty$.

There are consequently only finitely many dangerous atoms on each compact time interval, almost surely. For closed balls, continuity of the paths makes each finite hitting-time infimum attained. The event $T>t$ is therefore the absence of dangerous atoms up to time $t$.

The open-ball convention has the same survival probability at each fixed $t$. Indeed for any compact path range $K$, the difference between its closed and open radius-one neighborhoods is the shell $\{x:\mathrm{dist}(x,K)=1\}$, which has zero Lebesgue measure. To see this, choose a nearest $y\in K$ for a shell point $x$. For every sufficiently small $\epsilon >0$, the ball of radius $\epsilon /2$ centered at $x-\epsilon (x-y)$ lies in the open neighborhood and inside the ball of radius $3\epsilon /2$ about $x$. Thus the shell has a fixed proportional hole at every scale; the Lebesgue density theorem forces its measure to be zero. By the Tonelli theorem, closed-touch and open-entry retention probabilities have equal integrals. Independent thinning therefore gives no atom that touches only the boundary before this fixed horizon, almost surely. This also excludes a difference at an open-entry infimum equal to $t$.

The zero-count probability of a Poisson distribution is $e^{-\lambda }$, and henceThis survival among independently moving Poisson traps formula has been derived using only independent thinning and the defining Poisson zero-count probability, with the expectation/volume step supplied by Tonelli's theorem.

The strong law of large numbers gives $S_n/n\to -1$ almost surely, so $S_n\to -\infty$. Therefore the infinite-horizon maximum $Z$ is finite and attained almost surely; also $Z\ge 0$ because $S_0=0$.

Apply the Doob maximal inequality to $M_k=e^{2S_k}$ through time $n$. Since $\mathbb{E}{M}_n=1$,These finite-horizon events increase to $\{Z>t\}$. Continuity of probability under increasing unions givesFor $t<0$ the same bound is trivial, because its right side exceeds $1$. Only bounded-horizon maximal inequalities were used before passing to the limit; no unjustified stopping at an infinite-horizon crossing time is needed.

First fix a deterministic point $y$ different from the initial point. By translation, work with $y=0$ and $B_0=x\ne 0$. For $R>|x|$, hitting $0$ before $T_R$ requires hitting every circle of radius $r\in (0,|x|)$ before $T_R$. Part (a) givesAny finite-time hit of $0$ lies before exit from some sufficiently large integer-radius circle, since a continuous path on a finite interval is bounded. A countable union over such radii therefore shows that this fixed point is never hit. This is the polar point for planar Brownian motion property; it is a statement about each fixed point, not about avoiding all points simultaneously.

For the initial point $x$, fix a deterministic $\epsilon >0$. The Gaussian distribution of $B_{\epsilon }$ has a density, so $B_{\epsilon }\ne x$ almost surely. Conditional on ${\mathcal{F}}_{\epsilon }$, the future is Brownian motion starting from $B_{\epsilon }$, and the preceding fixed-point result shows that it never hits $x$. Taking $\epsilon =1/m$ and a countable union proves there is no return to the starting point at any positive time.

In contrast, neighborhoods are revisited arbitrarily late. Choose $r>0$ so that the closed disk centered at $x$ of radius $r$ lies in the given open set $U$. From any point $z$ outside this disk, let $R\to \infty$ in part (a), centered at $x$. It givesAt each deterministic integer time $n$, condition on $B_n$. If it is in the disk there is already a visit to $U$ at time $n$; if it is outside, the Markov property and the last calculation ensure a later visit to that disk, hence to $U$. With probability one this holds for all $n$ simultaneously. Thus the set of visits to $U$ is unbounded. The recurrence of planar Brownian motion concerns neighborhoods even though every fixed point is polar.

Let $M_t={\max }_{0\le s\le t}{B}_s$. The Brownian reflection principle givesHere is the reflection argument: on paths that hit $1$ before $t$, reflect all increments after their first hitting time. The Strong Markov property gives independent Brownian motion after that time, whose sign can be reversed without changing its law. This exchanges paths ending below $1$ with paths ending above $1$. The latter have necessarily hit $1$, and the endpoint has no atom at $1$, giving the displayed equality. It also shows that $M_t$ has a continuous distribution at every positive level.

Writing $\Phi$ for the standard normal distribution function, the probability of staying below the barrier is thereforeSince $\Phi (h)-\Phi (0)\sim h/\sqrt{2\pi }$ as $h\downarrow 0$, the Brownian barrier survival asymptotic isAny smaller exponent gives a zero limit and any larger exponent gives an infinite limit, so this is the unique exponent producing a finite positive constant.