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For independent one-dimensional Brownian motions $\beta ,\theta$, the real and imaginary parts of $Z$ are continuous local martingales with equal bracket $A$ and zero cross bracket. The Divergence of a driftless Brownian exponential clock makes $A_{\infty }=\infty$, and the common-clock time change of orthogonal local martingales makes $Z_{A^{-1}(r)}$ planar Brownian motion started at $1$. The exponential never vanishes, proving that this Brownian motion never hits zero. Recurrent visits of $\beta$ to arbitrarily negative levels prove visits arbitrarily near zero at arbitrarily large Brownian times.

For independent standard Brownian motions $B,W$, Itô formula gives $dY=e^W\cos BdW-e^W\sin BdB$: the two second-order drifts cancel. Independence gives zero cross quadratic covariation, so the displayed bracket follows. The process itself is not Brownian because its bracket is random. Its inverse-bracket time change is Brownian motion starting at one, in the time-changed filtration; subtracting one gives standard Brownian motion. The Divergence of a driftless Brownian exponential clock guarantees that the inverse is finite at every new time.

Put $C_u={\int }_0^u{e}^{2W_s}ds$. Its paths are continuous and strictly increasing. We also need $C_{\infty }=\infty$ to ensure that the printed inverse is finite for every $t$.

Here is a duration argument for Divergence of a driftless Brownian exponential clock. Start at zero and choose successive return times $T_j$ to zero, each at least one time unit after the preceding return. Recurrence of one-dimensional Brownian motion makes all these times finite. By the Strong Markov property, each unit interval after $T_j$ has the same fixed positive probability that the Brownian path stays above $-1$. The intervals are disjoint, and repeated strong Markov conditioning makes these trials independent. Infinitely many succeed almost surely, by the Borel-Cantelli lemma. Each successful interval contributes at least $e^{-2}$ to $C_{\infty }$, proving divergence.

Therefore ${\tau }_t$ is the unique finite inverse of $C$, with $C_{{\tau }_t}=t$. It is an increasing family of stopping times, since $\{{\tau }_t\le u\}=\{C_u\ge t\}$. The continuous-local-martingale time-change theorem states that inverse-bracket time change preserves the local-martingale property in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}$ and transforms the bracket by composition. Applied to $Y$, this givesThe Lévy characterization of Brownian motion now proves that $Z$ is Brownian motion starting at one; $Z-1$ is standard Brownian motion, in the time-changed filtration. If the term Brownian motion is reserved for a process starting at zero, $Z$ itself fails only that normalization. The substantive Brownian property, including independent Gaussian increments in the new time, holds.

For $X=e^{\beta }\cos \theta$ and $Y=e^{\beta }\sin \theta$, apply the two-variable Itô formula. Independence gives $[\beta ,\theta ]=0$, while each individual bracket is $t$. The second derivatives in the two variables cancel, yieldingTheir integrands are continuous adapted and locally bounded, so both coordinates are continuous local martingales. The quadratic variation of a stochastic integral gives

Here is a full proof that the clock diverges. Let $S_0=0$, and successively defineBy recurrence of one-dimensional Brownian motion, all these times are finite almost surely. By the Strong Markov property at the returns $S_j$, the variables $D_j=(U_j-S_j)\wedge 1$ are independent identically distributed, lie in $(0,1]$ almost surely, and have a strictly positive mean. The strong law of large numbers gives ${\sum }_j{D}_j=\infty$ almost surely. On $[S_j,U_j]$, ${\beta }_s\ge -1$, soThe segments are disjoint, and their divergent duration also ensures they cannot accumulate at a finite time. This proves Divergence of a driftless Brownian exponential clock and