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For a compact H-hull $A$ whose closure avoids a real boundary point $x$, the Brownian excursion in the upper half-plane from $x$ to infinity avoids $A$ with the displayed probability. From an interior point $z$, the corresponding Doob h-transform avoidance probability is $\mathrm{Im}{g}_A(z)/\mathrm{Im}z$. Letting $z=x+i\epsilon$ approach the boundary and using the Schwarz reflection principle gives the derivative. In particular $0<g_A^{\prime }(x)\le 1$ at every regular real boundary point away from the hull.

The excursion from $0$ to infinity is the Doob h-transform of killed planar Brownian motion with $h(z)=\mathrm{Im}z$, using the entrance law at zero. Equivalently, its real part is a standard Brownian motion and its independent imaginary part is a Bessel process of dimension three started at zero. This planar, infinite-duration excursion differs from the one-dimensional excursion on a finite time interval.

Killed planar Brownian motion conditioned to exit at a smooth boundary point $a$ is the Doob h-transform by its positive Poisson kernel, normalized at the starting point. Its generator is $\frac{1}{2}\Delta +\nabla \log h\cdot \nabla$. Conditioning on shrinking boundary arcs yields the same localized laws and specifies a canonical version of an otherwise probability-zero conditioning event.

The scalar stochastic differential equation $d{X}_t=\tanh X_{t}dt+d{W}_t$ has a pathwise unique global strong solution, since its coefficients are globally Lipschitz. The positive martingale $e^{t/2}/\cosh X_t$ gives a Girsanov theorem change of measure under which $X_t-X_0$ is Brownian motion. For $X_0=x$, its transition density isThis is a Doob h-transform with $h=\cosh$, and a mixture of $N(x+t,t)$ and $N(x-t,t)$ with weights $e^x/(2\cosh x)$ and $e^{-x}/(2\cosh x)$.

For a simple symmetric random walk absorbed at $0,r$, the probability of absorption at $0$ is $h_i=(r-i)/r$. Conditioning on that absorption gives the Doob h-transform with transition probabilities $(r-i+1)/[2(r-i)]$ downwards and $(r-i-1)/[2(r-i)]$ upwards. If $g_i={\mathbb{E}}_i[\tau 1_{\{{\tau }_0<{\tau }_r\}}]$, first-step analysis gives $g_i=h_i+(g_{i-1}+g_{i+1})/2$, $g_0=g_r=0$. Its solution is $g_i=i(r-i)(2r-i)/(3r)$, and dividing by $h_i$ gives the conditional duration.

Put $h_0=1$ and $h_r=0$. The Markov property and first-step conditioning give $h_i={\sum }_j{P}_{ij}{h}_j$. Consider a finite history ending at state $i$ and having positive conditional probability given $A$. The chance of eventual absorption at zero after that history is $h_i$: the future depends only on the present state, and if zero has already been reached, absorption keeps the chain there. If the next state is $j$, this chance is $h_j$. Bayes' rule therefore givesThis depends on the history only through $i$, proving the conditional Markov property. The conditional transition matrix isIts row sums are one by harmonicity of $h$, and $Q_{ir}=0$. State $r$ is never visited under the conditioning and is naturally removed; any row assigned there on an enlarged formal state space is immaterial, since conditioning starting at $r$ is undefined. State zero stays absorbing. For a random initial law $\nu$, its conditional law is ${\nu }_i{h}_i/{\sum }_j{\nu }_j{h}_j$. This is the Doob h-transform with the absorption probability as harmonic function.

For the simple symmetric walk, $h_i=(r-i)/r$ solves the linear harmonic recurrence and its boundary values. ThusIn particular, the conditioned chain moves downward with probability one from $r-1$.

Let $\tau$ be the first hit of $\{0,r\}$ in the original walk and let $g_i={\mathbb{E}}_i[\tau 1_A]$. Its absorption time has finite expectation: from any transient state a run of at most $r$ downward steps hits zero with probability at least $2^{-r}$, giving a geometric tail bound by blocks. First-step conditioning then givesEquivalently, $g_{i+1}-2g_i+g_{i-1}=-2(r-i)/r$. The cubic polynomialhas exactly these second differences and boundary values. It is the unique solution, since the difference of any two solutions is a linear sequence vanishing at both endpoints. Conditional on $A$, $\tau$ is the first time in zero, so division by $h_i$ provesThis is the duration of gambler's ruin conditioned on a chosen boundary.

At a real point $x$ outside the hull's closure, the Schwarz reflection principle makes the mapping-out function analytic across the boundary and $g_K^{\prime }(x)>0$. Use the course's Brownian excursion restriction identity: for an excursion from $x$ to infinity in the complex upper half-plane,One way to obtain this identity is to start its Doob h-transform at $x+i\epsilon$: its avoidance probability is $\mathrm{Im}{g}_K(x+i\epsilon )/\epsilon$, which tends to $g_K^{\prime }(x)$ by reflection. Thus the boundary derivative is an excursion avoidance probability, and in particular it is at most one.

Let $D$ be the filled unit half-disc. Since $K\subseteq D$, avoiding $D$ implies avoiding $K$. The monotonicity of boundary derivatives of mapping-out functions and part (b) give, for $|x|>1$,The upper bound is attained by the empty hull and the lower by the filled half-disc, so the comparison is sharp on either real tail.

Define $k(x,t)=h_x(x,t)/h(x,t)={\partial }_x\log h(x,t)$. Positivity and smoothness of $h$ make $k$ continuous and bounded on compact space-time sets. Brownian paths remain in compact spatial sets on each finite horizon, so ${\int }_0^{T}k(X_s,s{)}^{2}ds<\infty$ almost surely for every $T$. Thusis a well-defined continuous local martingale with $M_0=0$. The backward heat equation from part (b) removes the drift in Itô's formula and yieldsBy the allowed fact $Z_0=1$ and the uniqueness in part (a),

The finite-horizon form of the Girsanov theorem used here says: if $M_0=0$ is a continuous local martingale and its stochastic exponential $Z$ is a strictly positive true martingale of mean one on $[0,T]$, then under $d\mathbb{Q}{|}_{{\mathcal{F}}_T}=Z_{T}d\mathbb{P}{|}_{{\mathcal{F}}_T}$ every continuous $\mathbb{P}$-local martingale $N$ becomes the $\mathbb{Q}$-local martingale $N-[N,M]$. Quadratic variation is unchanged by this locally equivalent change of measure.

All the assumptions hold. The integrand defining $M$ is locally square integrable as checked above; $Z$ is strictly positive; and it is already a true martingale because it is the density process. On each finite horizon $Z_t=\mathbb{E}[Z_T\mid {\mathcal{F}}_t]$, so it is uniformly integrable on that horizon. There is no need to establish a separate Novikov condition, nor to assume equivalence on the infinite-time sigma-algebra.

Take $N=X$. Its covariation with $M$ is $[X,M{]}_t={\int }_0^{t}k(X_s,s)ds$. Hence $B_t=X_t-{\int }_0^{t}k(X_s,s)ds$ is a continuous $\mathbb{Q}$-local martingale starting at zero with $[B{]}_t=t$. The Lévy characterization of Brownian motion makes it Brownian under $\mathbb{Q}$. Since this argument holds on every finite horizon for the same process $B$,This is the Brownian change of measure by a positive space-time harmonic function, a Doob h-transform with drift ${\partial }_x\log h$.

Let $D$ be the unit disc and normalize its positive Poisson kernel at the boundary point $1$ byIt is harmonic in $D$. Let $p_D(t,z,w)$ be the killed Brownian transition density, for planar Brownian motion with generator $\frac{1}{2}\Delta$. The Doob h-transform has sub-Markov transition densityIts missing mass represents paths already absorbed at the distinguished boundary point. Equivalently, for a stopping time $\sigma$ before exit from a compact subdomain, its law is weighted relative to ordinary killed Brownian motion by $h(B_{\sigma })/h(B_0)$. These stopped laws are consistent and define a diffusion up to its lifetime. Its generator isThis is a rigorous Brownian motion conditioned to exit at a boundary point, not conditioning on an event of positive probability. Exhaustion of $D$ shows that its terminal boundary limit is $1$; its lifetime is finite. For example its expected lifetime from $0$ is ${\int }_D{G}_D^{\mathrm{BM}}(0,w)h(w)dw$, which is finite: the Green function vanishes linearly near the smooth boundary and cancels the Poisson-kernel singularity there, while its logarithmic singularity at $0$ is integrable.

One can also condition ordinary Brownian motion on exiting through an arc $I_{ϵ}$ about $1$, and then let its length decrease to zero. The conditional harmonic functions ${\omega }_D(z,I_{ϵ})/{\omega }_D(0,I_{ϵ})$ converge locally uniformly to $h(z)$. Thus their stopped conditional laws converge to this same diffusion. Both constructions specify the conditioning unambiguously.

Under $\mathbb{Q}$, $X_T=x+W_T^{\mathbb{Q}}$ has density ${\phi }_T(z-x)=(2\pi T{)}^{-1/2}\exp (-(z-x{)}^2/(2T))$. The reciprocal Radon-Nikodym derivative depends only on $X_T$:Consequently the probability density function under the original measure isIt integrates to one because ${\mathbb{E}}_{\mathbb{Q}}\cosh X_T=e^{T/2}\cosh x$. Completing the square also gives the useful mixture distribution formThus the terminal law is a mixture of $N(x+T,T)$ and $N(x-T,T)$ with the displayed positive weights. The diffusion with hyperbolic tangent drift density also exhibits the Doob h-transform with $h(z)=\cosh z$.

A Brownian excursion in the upper half-plane from $0$ to infinity is killed planar Brownian motion conditioned to stay in $\mathbb{H}$, with its entrance law at $0$. Precisely, it is the Doob h-transform for $h(z)=\mathrm{Im}z$, obtained by starting at $i\epsilon$ and letting $\epsilon \downarrow 0$. Equivalently,where $W$ is standard Brownian motion and the independent process $R$ is a dimension-three Bessel process started at zero. The transformed generator is $\frac{1}{2}\Delta +y^{-1}{\partial }_y$, which explains this representation.

First compute avoidance from an interior point $z$. Let $\tau$ be the exit time of ordinary planar Brownian motion from $\mathbb{H}\setminus A$. The harmonic correctionhas boundary values $\mathrm{Im}z$ on $A$ and zero on the real axis. The hydrodynamic normalization at infinity gives $g_A(z)=z+\mathrm{hcap}(A)/z+O(|z{|}^{-2})$. Together with the maximum principle for harmonic functions this gives the useful bound $0\le v\le {\sup }_{a\in A}\mathrm{Im}a$. Since ordinary Brownian motion hits the real axis almost surely, the optional stopping theorem applied to this bounded harmonic function givesThe Doob h-transform weights a stopped path by its final height divided by its initial height. Applying this at the first hit of $A$, with bounded-time localization and then a limit, therefore yieldsFor the entrance law the same result follows by conditioning at a small positive time and letting that time decrease to zero: the avoidance function has the boundary limit computed next, and the excursion is initially in a neighbourhood disjoint from $A$.

Because the closure of the hull avoids $0$, the Schwarz reflection principle makes $g_A$ analytic in a neighbourhood of zero, with real $g_A(0)$ and positive real $g_A^{\prime }(0)$. ThusIn fact, the reflected Taylor expansion gives $\mathrm{Im}{g}_A(z)/\mathrm{Im}z=g_A^{\prime }(0)+O(|z|)$ uniformly as $z\to 0$ inside the half-plane, so the same limit applies to the entrance process. The map in the question is ${\psi }_A=g_A-g_A(0)$, so its derivative agrees with $g_A^{\prime }$. We obtain the restriction probability of a Brownian half-plane excursion: