For positive times the transformed process is centered and jointly Gaussian, andIts paths are continuous away from zero. To verify continuity at zero, we need $B_u/u\to 0$ almost surely as $u\to \infty$. The strong law of large numbers applied to the independent increments $B_{n+1}-B_n$ gives $B_n/n\to 0$. On the unit interval after $n$, the Doob L2 maximal inequality givesThe Borel-Cantelli first lemma, first for each positive rational $\epsilon$, shows that these between-integer fluctuations divided by $n$ tend to zero. Hence $B_u/u\to 0$ for all real $u\to \infty$, and $t{B}_{1/t}\to 0$ as $t\downarrow 0$.

The Gaussian-process characterization of Brownian motion now proves $X$ is standard Brownian motion, with respect to its natural filtration. This is time inversion of Brownian motion; the transformed process need not be adapted to the original forward filtration.

Standard Brownian motion is centered, so $m_t=0$. If $s\le t$, write $B_t=B_s+(B_t-B_s)$; the final increment is independent of $B_s$ and has mean zero. ThusThis is the Brownian covariance kernel. Together with joint Gaussianity and continuous paths, it gives the Gaussian-process characterization of Brownian motion.

For every $m$, $D_m$ is a continuous square-integrable vector martingale. Its norm is a submartingale, so the Doob L2 maximal inequality and the estimate from part (c) giveThe Cauchy-Schwarz inequality now showsBy Tonelli's theorem, the random sum of these suprema is finite almost surely. Therefore the telescoping series $a+{\sum }_m{D}_m(t)$ converges absolutely and uniformly on $[0,T]$ outside one null event. Its limit is continuous and agrees at each fixed time with the $L^2$ limit from part (c). Taking the intersection of the probability-one events for integer $T$ provides a single version on the whole half-line. HenceThis also gives the requested pointwise almost-sure convergence of the full sequence.

Set $K={\sum }_i\Vert A^i{\Vert }_{\mathrm{op}}^2$, where the norms are operator norms, and $D_m=y^{(m+1)}-y^{(m)}$. Independence of the Brownian coordinates and the vector Itô isometry giveInduction, as in part (b), yields the Factorial Picard estimate for a linear stochastic differential equationFor $t\le T$, the triangle inequality in $L^2$ therefore bounds any tail of the iterates by the corresponding tail ofThis numerical series converges by the ratio test: successive terms have ratio $\sqrt{KT/(m+2)}\to 0$. Thus the iterates are Cauchy in $L^2$, uniformly in the deterministic time parameter on each bounded interval. Completeness of Lp spaces gives a limit $y_t$. Once its continuous adapted version is chosen in part (d), the Itô isometry permits passage to the limit in the integral equation: the squared norm of each integral difference is bounded by $K{\int }_0^T\mathbb{E}|y_s^{(m)}-y_s{|}^{2}ds\to 0$. Consequently the Picard limit solves the stochastic differential equation.

Use $s$ for the dummy integration variable in the printed integral. The Itô isometry givesThe assumed second-moment bound also proves the integrability needed to apply the Itô isometry. The extra time integration is the source of the factorial improvement.

Write the stochastic differential equation in integral form and replace the unknown integrand by the preceding Picard iterate. With the deterministic initial vector $a$, the recursion isEach iterate is a continuous adapted process. The Itô isometry inductively gives its square integrability on every bounded time interval, so all these Itô integrals are well defined.

Suppose first that $d=|z_0|>ϵ$. Apply part (a) with inner radius $ϵ$ and outer radius $R>d$:Each event implies finite hitting time of the $ϵ$ circle, so that hitting probability is one. If $d=ϵ$, the hitting time is zero. If $d<ϵ$, it is the exit time from a bounded disk. The martingale $|Z_t-z_0{|}^2-2t$, stopped before that exit, gives a finite mean exit-time bound $({ϵ}^2-d^2)/2$. Thus the asserted hitting time is almost surely finite in this case too.

For alternating visits, choose two closed smaller disks $K_i$ strictly inside the given open disks $D_i$, so their distance is positive. Define successive first hitting times of $K_1,K_2,K_1,K_2,\dots$. Translation invariance and the Strong Markov property apply the just-proved hitting result from every successive starting point. A countable intersection makes all these times finite almost surely, with the required disk memberships.

These increasing times cannot accumulate at a finite time. On a compact time interval a continuous Brownian path is uniformly continuous, whereas every consecutive visit must cross the fixed positive distance between $K_1$ and $K_2$. ThereforeThis is the recurrence of planar Brownian motion in neighborhoods; it does not assert that every individual point is hit.

Let $d=|z_0|$, the initial distance from the annulus center, and $T=T_{A_{r,R}}$. The function $z\mapsto \log |z-z_0|$ is harmonic on the annulus, so its value along stopped planar Brownian motion is a local martingale by Itô's formula. It is bounded between $\log r$ and $\log R$ before exit, hence is a true bounded martingale.

By the assumed almost sure finiteness of $T$, continuity, and dominated convergence,where $p$ is the inner-boundary exit probability. Solving gives the harmonic measure of an annulus:The two probabilities sum to one, as the exit must lie on one of the two circles.

Put $Y=\sqrt{\tau }$ and $V=X_{\tau }^{\ast }$. Splitting according to whether $V>\delta \lambda$, the good-lambda bound for Brownian stopping time and maximum givesIntegrate against the positive measure $d\Phi (\lambda )$. The tail integral formula for expectation in Stieltjes form yieldsBy moderate growth, choose $K_{\beta }$ with $\Phi (\beta x)\le K_{\beta }\Phi (x)$. ThenFix $\beta >1$ and choose $\delta$ small enough that $K_{\beta }\eta <1/2$. Absorb the last term and apply moderate growth once more to the fixed dilation $1/\delta$. This provesTo avoid subtracting infinities, perform the argument first for $\tau \wedge n$, where Brownian maxima have all finite polynomial moments and moderate growth is at most polynomial, and then use monotone convergence. Applying the Dambis-Dubins-Schwarz theorem transfers this Brownian estimate to continuous local martingales, giving the lower-maximum side of the Burkholder-Davis-Gundy inequalities.

Set $a=\delta \lambda$, $t_0={\lambda }^2$, and $h=({\beta }^2-1){\lambda }^2$. On the event $\{{\tau }^{1/2}>\beta \lambda ,X_{\tau }^{\ast }\le a\}$, one has $\tau >t_0$ and the Brownian path remains in $[-a,a]$ throughout the next $h$ time units after $t_0$.

For Brownian motion started at $x\in (-a,a)$, let $H$ be its first exit time from $(-a,a)$. The martingale $B_t^2-t$, stopped at $H\wedge n$, gives ${\mathbb{E}}_x(H\wedge n)\le a^2-x^2$. Monotone convergence first proves finite mean exit time, and dominated convergence for the bounded stopped position then givesThus Markov's inequality yields ${\mathbb{P}}_x(H>h)\le a^2/h$. Remaining in the closed interval for a fixed positive duration has the same probability as avoiding boundary exit: Brownian motion crosses immediately after first hitting an endpoint.

Condition on ${\mathcal{F}}_{t_0}$. The event $\{\tau >t_0\}$ is measurable there because $\tau$ is a stopping time, and future Brownian increments are independent of this sigma-algebra. The preceding bound is uniform in the current point inside the interval. ThereforeThis is a good-lambda bound for Brownian stopping time and maximum.

A continuous nondecreasing function $\Phi$ with $\Phi (0)=0$ has moderate growth if for some fixed $a>1$ and finite $K$,Iteration and monotonicity then give such an inequality, with a different constant, for every fixed dilation factor. Powers $\Phi (x)=x^p$, $p>0$, are standard examples.

The Burkholder-Davis-Gundy inequalities say that for a real continuous local martingale $M$ with $M_0=0$ and any stopping time $\tau$, there are positive finite constants $c_{\Phi },C_{\Phi }$, depending only on $\Phi$, such thatHere $M_{\tau }^{\ast }={\sup }_{s\le \tau }|M_s|$. Expectations may be infinite, with the inequalities understood by localization. For $\Phi (x)=x^p$, these compare the $p$th moment of the running maximum with the $p/2$th moment of quadratic variation.

Continuity and the regularity of zero for a one-dimensional Brownian coordinate make $S$ the first contact with the imaginary axis. The Strong Markov property and symmetry of the subsequent real increments imply that reversing those increments after $S$ preserves the Brownian law. The imaginary coordinate is unchanged, so $\tilde{Z}$ is again planar Brownian motion started at one.

Reflection preserves distance to the origin, hence ${\tilde{T}}_{D_r}=T_{D_r}$ pathwise. On $T_{D_r}<S$, the path exits through the right semicircle and reflection has not yet acted. On $S<T_{D_r}$, reflection reverses the sign of the exit real part. ThusThe two circular points on the imaginary axis have zero exit probability, so they cause no extra term. Subtracting proves the reflection identity for Brownian exit from a half-disc requested in the question.

Both processes have the same exit law. With $p={\pi }^{-1}{\tan }^{-1}((r^2-1)/(2r))$ from part (b),The final trigonometric identity follows from ${\tan }^{-1}((r^2-1)/(2r))=\pi /2-2{\tan }^{-1}(1/r)$ for $r>1$.

Let $a=1/r\in (0,1)$ and $\theta ={\tan }^{-1}((1-a^2)/(2a))\in (0,\pi /2)$. The Möbius transformation sends the semicircle endpoints toAlso $f(-1)=-1$. Therefore the left semicircle maps to the unit-circle arc centered at $-1$ with angular width $2\theta$.

By conformal invariance of harmonic measure, its exit probability from the starting point $a$ equals the probability of this image arc for Brownian motion started at zero. Rotational symmetry makes the latter exit angle uniform. ThusFor scaling, if $Z$ starts at $1/r$, then ${\hat{Z}}_t=r{Z}_{t/r^2}$ is planar Brownian motion started at one. Its unit-disk exit becomes radius-$r$ exit, and multiplication by the positive number $r$ preserves the sign of the real part. This proves exactly the same formula for ${\mathbb{P}}_1(\mathrm{Re}{\hat{Z}}_{T_{D_r}}<0)$.

Conformal invariance of planar Brownian motion means invariance after changing the clock, not preservation of the original speed. If $f:D\to D^{\prime }$ is a conformal bijection and $Z$ is planar Brownian motion stopped on exiting $D$, putThen $f(Z_t)$ has the law of planar Brownian motion started at $f(Z_0)$ and run at time $A_t$, up to the corresponding lifetime. Equivalently, composing with the inverse clock produces ordinary Brownian motion in $D^{\prime }$.

Indeed, the real and imaginary parts of a holomorphic function are harmonic, so Itô's formula gives zero drift. The Cauchy-Riemann equations give equal coordinate brackets $A_t$ and zero cross bracket. The Lévy characterization of multidimensional Brownian motion after time change identifies the image. In particular the harmonic measure of boundary arcs is carried to their conformal images.

Write the integral with an explicit dummy time:On every compact time interval, strict positivity and continuity of $Z$ make its reciprocal locally bounded; the covariation has locally finite variation. Thus $X$ is a well-defined continuous semimartingale.

Apply the Itô product rule. Subtracting a finite-variation process does not change covariation, so $[Z,X]=[Z,M]$. The finite-variation terms cancel:Thus $ZX$ is a local martingale under $\mathbb{P}$.

For the transfer step, choose stopping times increasing to the horizon that bound $X$, $M$, reciprocal $Z$, and the brackets needed for the stochastic integrals. For each such time $\sigma$, the same product calculation with $X^{\sigma }$ shows that $Z_t{X}_{t\wedge \sigma }$ is a true $\mathbb{P}$ martingale. The part after $\sigma$ is simply the bounded held value $X_{\sigma }$ times the density martingale. By the Bayes formula for conditional expectation, for $s\le t$,The stopping times exhaust the horizon also under $\mathbb{Q}\ll \mathbb{P}$. Hence $X$ is a continuous $\mathbb{Q}$ local martingale, proving this form of the Girsanov theorem without assuming a positive terminal density everywhere.

A uniformly bounded local martingale is a true martingale: stop at a localizing sequence and use dominated convergence in its conditional-expectation identities. Its deterministic bound also gives uniform integrability. The uniformly integrable martingale convergence theorem therefore yields an integrable limit $Z_{\infty }$ with convergence in $L^1$ andSince $Z_{\infty }\ge 0$, the Radon-Nikodym derivative defines a positive measure of total mass one:The limit can vanish on a set of positive $\mathbb{P}$ measure, so global equivalence is not required. At each finite time, however, the density is the strictly positive $Z_t$.

The solution is the continuous Doléans-Dade exponential:Here $[X]$ is the continuous quadratic variation starting from zero, denoted by the angle-bracket notation in the source. If $X=M+A$ with $A$ continuous of finite variation, then $[X]=[M]$.

Set $U=X-X_0-\frac{1}{2}[X]$. Its quadratic variation is still $[X]$. Applying Itô's formula to $e^U$ givesThe initial value is $S_0=1$. This verification applies to the full semimartingale, including its finite-variation part.

The two-dimensional Itô formula gives the local martingale part of $u(X_t,Y_t)$ as ${\int }_0^t{u}_{x}dX+{\int }_0^t{u}_{y}dY$, with all derivatives evaluated along $(X_s,Y_s)$. Independent Brownian coordinates have brackets $[X{]}_t=[Y{]}_t=t$ and $[X,Y{]}_t=0$. Hence the quadratic variation isSimilarly, the quadratic covariation with $v(X,Y)$ isSufficient pointwise conditions are equal gradient lengths and perpendicular gradients:The Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$ imply both. These are the quadratic covariations of an analytic Brownian image; harmonicity is needed to remove the drift, but the two bracket identities themselves follow just from the displayed derivative relations.