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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.

Put $\tau =T_r\wedge T_R$. First establish almost sure finiteness. The Itô formula gives the martingale $|B_t{|}^2-2t$ for planar Brownian motion. Stop it at $t\wedge T_R$. The stopped path has norm at most $R$, soBy monotone convergence, $\mathbb{E}{T}_R<\infty$, and hence $\tau <\infty$ almost surely.

The function $h(y)=\log |y|$ is a harmonic function away from $0$: in polar coordinates its Laplacian is $h^{\prime \prime }(\rho )+h^{\prime }(\rho )/\rho =-{\rho }^{-2}+{\rho }^{-2}=0$. The Itô formula shows $h(B_{t\wedge \tau })$ is a local martingale. It is bounded between $\log r$ and $\log R$, so it is a true martingale and passage to $\tau$ is justified by dominated convergence. Writing $p={\mathbb{P}}_x(T_r<T_R)$ and using continuity at exit givesSolving yields the planar Brownian annulus hitting probabilityThe stopping argument stays away from the singularity of the logarithm until the calculation is complete.

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 $L$ be the fixed line and choose $y\in L$. Take orthonormal vectors $e_1,e_2$ spanning the plane perpendicular to $L$. The processis standard planar Brownian motion starting from $Y_0\ne 0$. Indeed the projected increments are centered Gaussian random vectors with covariance $(t-s)I_2$, are independent on disjoint intervals, and have continuous paths.

The event $B_t\in L$ is exactly $Y_t=0$. The polar point for planar Brownian motion result in (b) therefore givesThis is a projection criterion for Brownian avoidance of affine subspaces: avoiding a fixed point in a two-dimensional projection excludes hitting the original line.

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.

Let $X_i$ be independent and identically distributed random variables with mean $0$ and variance ${\sigma }^2\in (0,\infty )$, and put $S_0=0$, $S_k={\sum }_{i=1}^k{X}_i$. Define the polygonal interpolationwith the endpoint $W_n(1)=S_n/(\sigma \sqrt{n})$. The Donsker invariance principle stateswhere $B$ is standard one-dimensional Brownian motion and the arrow denotes weak convergence of probability measures on this path space. Equivalently, the step interpolation converges in the Skorokhod space with its $J_1$ topology. The continuous interpolation version is sufficient for the maximum in part (c).

One continuous Girsanov theorem formulation is as follows. Let $L$ be a zero-starting continuous local martingale such that $Z=\mathcal{E}(L)$ is a strictly positive uniformly integrable martingale, and define $d\mathbb{Q}=Z_{\infty }d\mathbb{P}$. For every continuous $\mathbb{P}$-local martingale $N$,On a finite horizon the same assertion holds with terminal density $Z_T$, provided $Z$ is a true martingale through $T$. In particular, if $L_t={\int }_0^t{\theta }_{s}d{B}_s$, thenis a Brownian motion under $\mathbb{Q}$, on the specified horizon. Its quadratic variation remains $t$ under this absolutely continuous measure (equivalent on every finite-horizon sigma-algebra), and the Lévy characterization of Brownian motion identifies it as Brownian. The density sign and drift subtraction are paired: choosing the exponential of $-\int \theta dB$ instead produces $B+\int \theta ds$.

Use the predictable processThe value at zero is irrelevant to the Brownian integral. The numerator is bounded and ${\sigma }_s\ge \delta >0$, so $\theta$ is bounded and $[L{]}_{\infty }={\int }_0^T{\theta }_s^{2}ds$ has a deterministic bound. Part (a) supplies a uniformly integrable density martingale. Define a measure on the entire given sigma-algebra byIts density is positive and has expectation one, so $\mathbb{Q}$ is a probability measure equivalent to $\mathbb{P}$. The Girsanov theorem makes ${\tilde{B}}_t=B_t-{\int }_0^{t\wedge T}({\nu }_s-{\mu }_s)/{\sigma }_{s}ds$ a Brownian motion under $\mathbb{Q}$. For $t\le T$, substitute $dB=d\tilde{B}+\theta dt$:Continuity of $\sigma$ gives pathwise local square integrability, even though it is not assumed bounded. Equivalence preserves this property. This is finite-horizon drift replacement by a change of measure.

The sampling rule in the PDF is with replacement. Thus conditional on the current proportion $y=j/n$, the next count is $W\sim \mathrm{Bin}(n,y)$. Put $\Delta =W/n-y$ and $q=y(1-y)$. Direct binomial moments giveThe fourth centered moment of a binomial distribution implies $\mathbb{E}|\Delta {|}^3\le (\mathbb{E}{\Delta }^2\mathbb{E}{\Delta }^4{)}^{1/2}\le C{n}^{-3/2}$. The accelerated generator of the Wright–Fisher binomial sampling chain isFor $f\in C^3([0,1])$, Taylor's formula uniformly yieldsThe initial condition is $Z_0^n=\lfloor xn\rfloor$, so $X_0^n\to x$. The endpoints are absorbing.

We give the path-space argument, rather than only a formal generator calculation. Regard $X^n$ as a càdlàg process in $D([0,1],[0,1])$ with the Skorokhod $J_1$ topology. The chain proportion is a bounded martingale, with predictable square compensator $V_t^n=n^{-1}{\sum }_{k<\lfloor nt\rfloor }{X}_{k/n}^n(1-X_{k/n}^n)$. For bounded stopping times ${\tau }_n\le {\sigma }_n\le {\tau }_n+{\delta }_n$, optional sampling gives $\mathbb{E}|X_{{\sigma }_n}^n-X_{{\tau }_n}^n{|}^2=\mathbb{E}(V_{{\sigma }_n}^n-V_{{\tau }_n}^n)\le ({\delta }_n+n^{-1})/4$. This bound uses the deterministic number of grid jumps in an interval of length at most ${\delta }_n$. Compact containment is automatic. The Aldous tightness criterion therefore proves tightness.

Furthermore, the fourth-moment bound and a union bound giveEvery subsequential limit is continuous. For each smooth $f$, the discrete compensated processis a martingale. Generator convergence and continuity of any limiting path turn the sum into ${\int }_0^t\frac{1}{2}{X}_s(1-X_s)f^{\prime \prime }(X_s)ds$. These compensated processes are uniformly bounded on $[0,1]$, so their martingale identities pass to the limit against bounded continuous functions of past coordinates. A Monotone class theorem argument extends the identities to the natural filtration. Smooth approximation in the $C^2$ norm extends the test class. Thus every limit solves the martingale problem for $Lf(y)=\frac{1}{2}y(1-y)f^{\prime \prime }(y)$ on $[0,1]$.

To identify this problem with the stated SDE, tests equal to $y$ and $y^2$ on $[0,1]$ (using bounded smooth extensions) give a continuous martingale with bracket ${\int }_0^t{X}_s(1-X_s)ds$. On an enlargement carrying independent Brownian motion $W$, setIts bracket is $t$, and the Lévy characterization of Brownian motion makes it Brownian. The zero-coefficient portion of $dX$ has zero bracket, so $d{X}_t=\sqrt{X_t(1-X_t)}d{B}_t$. Conversely Itô's formula shows that any such SDE solution solves the limiting martingale problem. The allowed uniqueness of the SDE, understood at least as uniqueness in law, therefore makes every subsequential limit have the same law. ConsequentlyThe limit is the Wright–Fisher diffusion. Linear interpolation has the same weak limit in the uniform topology because the maximum jump tends to zero. The results applied are Aldous tightness, the vanishing-jump continuous-limit criterion, passage of uniformly bounded martingale identities under weak convergence, Lévy characterization, and uniqueness in law; the estimates above verify their hypotheses.

Before $T(x)$, the centered image satisfies the stochastic differential equationChanging the sign of the Brownian motion shows that $X/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa$. Here is a direct proof of the relevant hitting property, including finite lifetime.

Put $p=1-4/\kappa >0$. By the Itô formula, the drift of $X^p$ isThus $X^p$ is a local martingale and a scale function of a one-dimensional diffusion. Fix $R>x$. The nonnegative functionvanishes at both endpoints and satisfies $\mathcal{L}{u}_R=-1$, for $\mathcal{L}=(\kappa /2){\partial }_{vv}+(2/v){\partial }_v$. Stop first on $[ϵ,R]$, and also at a deterministic time. The Itô formula gives $\mathbb{E}(t\wedge {\tau }_{ϵ,R})\le u_R(x)$. Let $t\to \infty$ and then $ϵ\downarrow 0$. It follows that the first exit ${\tau }_{0,R}$ is finite almost surely, not merely that the path approaches a boundary at infinite time.

Optional stopping applied to $X^p$ on this bounded interval givesOn $\{{\tau }_0=\infty \}$ the process must therefore hit every $R>x$ before zero. Letting $R\to \infty$ makes that probability zero. ConsequentlyAt $\kappa =4$ the dimension is two; zero is not hit from a positive starting point. This endpoint must not be included in the finite-swallowing assertion.

Put $X_t=g_t(x)-W_t$, $Y_t=g_t(y)-W_t$ and $D_t=Y_t-X_t$. Before $T(x)$, order preservation gives $0<X_t<Y_t$ and $D_t>0$. The common noise cancels from their difference:Since $D$ has finite variation, applying the Itô formula to $Z=Y/D$ gives the SLE two-boundary-point ratio diffusionWith the strictly increasing clock $u(t)={\int }_0^t{D}_s^{-2}ds$, the Dambis-Dubins-Schwarz theorem yields a Brownian motion $B$ for which the time-changed process obeysThe sign of the new Brownian motion has absorbed the minus sign above.

The domain Markov property of a chordal Loewner chain says that, conditionally on the past before $T(x)$, the mapped future is a fresh SLE. Its boundary marked points are $X_t,Y_t$. HenceLocalize, for example, where $D_t\ge 1/n$, $Z_t\in [1+1/n,n]$ and $t\le n$. On each such stopped interval this is a bounded conditional-expectation martingale, and therefore a genuine martingale. No smoothness assumption on $f$ is needed for this argument. The exit calculation below identifies it explicitly.