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Fix an outer radius $R>\rho =|z_0|$. By continuity, reaching zero before radius $R$ requires first reaching every radius $\epsilon \in (0,\rho )$. The planar Brownian annulus hitting probability from (i) therefore impliesThus $\mathbb{P}({\tau }_0<{\tau }_R)=0$ for every fixed $R>\rho$.

On any continuous sample path that reaches zero in finite time, the radius has a finite maximum before that time. There is therefore some integer $m>\rho$ for which zero is reached before radius $m$. ConsequentlyEvery event on the right has probability zero, soThis proves that planar Brownian motion avoids a fixed point even though it can reach arbitrarily small positive radii.

For $g(x)=\log |x|$ on ${\mathbb{R}}^2\setminus \{0\}$,The Itô formula, stopped in a compact annulus, therefore givesThis proves the local martingale property up to a possible hitting of the origin. For the annulus exit time $T$, $M_{t\wedge T}$ is bounded between $\log r$ and $\log R$, so it is a true uniformly integrable martingale.

The exit is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, so stopping at $T\wedge t$ gives $2\mathbb{E}(T\wedge t)=\mathbb{E}|B_{T\wedge t}{|}^2-1\le R^2-1$. Letting $t\to \infty$ proves $\mathbb{E}T<\infty$. Bounded convergence in the logarithmic martingale now yieldsIf $p$ is the probability of reaching radius $r$ before radius $R$, continuity givesHitting zero before reaching radius $R$ would require reaching every positive radius $r$ first. Its probability is bounded by $p$, which tends to zero as $r\downarrow 0$. Any finite-time path hitting zero is bounded before that time and hence hits zero before exiting some integer-radius disk. A countable union over those disks proves $B_t\ne 0$ for every $t\ge 0$, almost surely. This is planar Brownian motion avoids a fixed point; it also makes $M$ a globally defined continuous local martingale.

To decide whether it is a true martingale, compute its expectation directly. Condition on $B_0$ if necessary and use rotation to take $B_0=(1,0)$ and write $B_t=B_0+W_t$. Conditional on $\rho =|W_t|$, the angle of $W_t$ is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by taking the real part of the convergent power series for $\log (1+\rho e^{i\theta })$: each nonconstant Fourier term integrates to zero. For $\rho >1$ factor out $\rho$ and apply the same argument to $1/\rho$. The equality at one follows by the integrable limiting logarithmic singularity. All interchanges are valid: the shifted Gaussian density is bounded near the origin, where ${\int }_0^{1}r|\log r|dr<\infty$, and its Gaussian tail controls logarithmic growth.

The radius $\rho$ has density $(\rho /t)e^{-{\rho }^2/(2t)}$. Consequently, for $t>0$,Since $M_0=0$, $M$ is not a martingale, despite integrability at each deterministic time. This is the logarithmic radius of planar Brownian motion example.

The optional hint is consistent with this conclusion. The upper-radius exit $T(2,0)$ is finite and never occurs at zero. After that exit the stopped square is $(\log 2{)}^2$; before it, a radius in $[1/2,2]$ also gives a logarithmic square no larger than $(\log 2{)}^2$, while a radius below $1/2$ contributes precisely the extra indicated term. This proves the hinted inequality without assuming that the unstopped process is a true martingale.

For the first equation use an integrating factor. The Itô product rule gives $d(e^{\lambda t}{X}_t)=\sigma e^{\lambda t}d{B}_t$, soThe Gaussianity of deterministic Brownian stochastic integrals follows by approximating its deterministic integrand by step functions and using the Itô isometry. HenceThe variance is positive also when $\lambda <0$, since numerator and denominator then both have negative sign. This is the explicit Ornstein-Uhlenbeck solution with initial value one.

For the second equation, take independent standard Brownian motions $U,V$ and define $R_t=\sqrt{(1+U_t{)}^2+V_t^2}$. The associated planar Brownian motion does not hit the origin. To check this rather than silently differentiate at zero, stop its radius upon hitting $\epsilon$ or $L$, with $0<\epsilon <1<L$. The function $\log r$ is harmonic on that annulus, so the optional stopping theorem givesLet $\epsilon \downarrow 0$: hitting zero before $T_L$ has probability zero. Taking integer $L\to \infty$ covers every possible finite hitting time, since continuous paths are bounded on compact intervals. Thus planar Brownian motion avoids a fixed point and $R$ stays positive.

The radial Itô formula is now legitimate and givesThe sum of the two martingale terms has bracket $t$ and starts at zero, so it is a standard Brownian motion $\beta$ by Lévy characterization of Brownian motion. Thus $dR=d\beta +(2R{)}^{-1}dt$ with $R_0=1$. The law-determination assumption permitted in the question identifies the law of $Y$ with that of this dimension-two Bessel process. For independent standard normal random variables $G_1,G_2$,This is the two-dimensional Bessel transition law and answers the requested distribution without needing a special-function density.

First justify that the logarithm is defined at all times from one onward. For planar Brownian motion started at $z\ne 0$, stop on exiting the annulus $\epsilon <|x|<R$, where $\epsilon <|z|<R$. This exit time $\rho$ is finite almost surely: the stopped identity $|B_t{|}^2-2t$ gives $\mathbb{E}(t\wedge \rho )\le R^2/2$. The function $f(x)=\log |x|$ is harmonic on the annulus, so its stopped process is a bounded martingale. Optional stopping givesHitting zero before the outer circle would require hitting every inner circle first. It therefore has probability zero. Taking a countable sequence of outer radii tending to infinity proves that planar Brownian motion avoids a fixed point. Since $B_1$ has a density and is nonzero almost surely, condition on ${\mathcal{F}}_1$ to apply this result to the future path.

For $x\ne 0$, $\nabla f(x)=x/|x{|}^2$ and $\Delta f(x)=0$ in two dimensions. The Itô formula on annuli therefore givesTo localize explicitly, let ${\tau }_n=\inf \{t\ge 1:|B_t|\notin (1/n,n)\}$. These stopping times increase to infinity almost surely: on each compact time interval the continuous path has a finite maximum and, by point avoidance, a positive minimum radius. Up to ${\tau }_n$, the stochastic integrand has norm at most $n$. Its stopped integral is square-integrable on every finite horizon. The initial value $\log |B_1|$ is integrable, as follows from the Rayleigh-density calculation in part d. Hence $X^{{\tau }_n}$ is a true martingale from time one and $X$ is a local martingale. This also gives its quadratic variation from time one as ${\int }_1^t|B_s{|}^{-2}ds$.

Use first that planar Brownian motion avoids a fixed point, so the path starting at $1$ never reaches zero. Its continuous logarithmic lift, initialized at zero, has real part $U_t=\log |Z_t|$ and imaginary part the continuous unwrapped angle ${\Theta }_t$. Local branches of the complex logarithm, patched along the path, and the Itô formula show that both coordinates are continuous local martingales withThe clock is finite on every compact time interval, since a continuous nonvanishing path stays a positive distance from zero there. It is continuous and strictly increasing. Moreover $\tau (\infty )=\infty$: by recurrence, planar Brownian motion spends infinite total time in a fixed disk, and inside the unit disk the clock integrand is at least one. This is the divergence of a positive planar Brownian occupation integral.

The common-clock time change of orthogonal local martingales, using the Lévy characterization of multidimensional Brownian motion, supplies independent Brownian coordinates ${\zeta }^1,{\zeta }^2$ such that $U_t+i{\Theta }_t={\zeta }_{\tau (t)}$. Exponentiating givesTo identify the inverse clock, write $A={\tau }^{-1}$. The derivative of an inverse function and the representation just obtained giveConsequentlyThe strict inequality in this generalized inverse has the same infimum as the unique inverse value, since $A$ is continuous and strictly increasing onto the half-line.

A dimension-two Bessel process starting at $r>0$ has the displayed marginal law, where $G_1,G_2$ are independent standard normal random variables. Take the norm of a planar Brownian motion starting at $(r,0)$. Planar Brownian motion avoids a fixed point, so the norm stays positive; Itô formula gives drift $1/(2R)$ and a martingale term of bracket $t$. The Lévy characterization of Brownian motion identifies that term as its driving Brownian motion. The squared radius divided by $t$ has a noncentral chi-squared distribution with two degrees of freedom and noncentrality $r^2/t$.