= Solution
{c}
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> with
$$
\langle U\rangle_t=\langle\Theta\rangle_t=\int_0^t\frac{ds}{|Z_s|^2}=:\tau(t),\qquad
\langle U,\Theta\rangle_t=0.
$$
The 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 gives
$$
\boxed{Z_t=\exp(\zeta_{\tau(t)}),\qquad\zeta_0=0.}
$$
To identify the inverse clock, write $A=\tau^{-1}$. The derivative of an inverse function and the representation just obtained give
$$
A'(s)=|Z_{A(s)}|^2=e^{2\zeta_s^1},\qquad
A(s)=\int_0^s e^{2\zeta_u^1}du.
$$
Consequently
$$
\boxed{\tau(t)=\int_0^t|Z_s|^{-2}ds
=\inf\left\{s\geq0:\int_0^s e^{2\zeta_u^1}du>t\right\}.}
$$
The 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.
Back to article page
