= Solution
Work before the first zero of $Z$, with $Z_0=z>0$. For the <Bessel process> equation, the <Itô formula> applied to $x^r$ gives
$$
\boxed{dW_t=rZ_t^{r-1}dB_t+\frac{r(\delta+r-2)}2Z_t^{r-2}dt,\qquad W_0=z^r.}
$$
Its <quadratic variation> is
$$
C_t=[W]_t=r^2\int_0^tZ_u^{2r-2}du.
$$
This clock is strictly increasing on the positive-state interval. For $s$ before its terminal clock value, let $Y_s=W_{\sigma_s}$ and let $\beta_s=(\int_0^\cdot rZ_u^{r-1}dB_u)_{\sigma_s}$. The <Dambis-Dubins-Schwarz theorem> makes $\beta$ a <Brownian motion> there. Since $d\sigma_s=ds/(r^2Z_{\sigma_s}^{2r-2})$, the drift changes to
$$
\frac{r(\delta+r-2)}2\frac{Z_{\sigma_s}^{r-2}}{r^2Z_{\sigma_s}^{2r-2}}=\frac{\delta+r-2}{2rY_s}.
$$
Consequently the \b[<Power time change of a Bessel process>] gives
$$
\boxed{dY_s=d\beta_s+\frac{\delta'-1}{2Y_s}ds,\qquad\delta'=2+\frac{\delta-2}{r}.}
$$
The initial value is $z^r$, and this equation identifies a <Bessel process> of dimension $\delta'$ until its first zero. Localize both equations on compact subintervals of $(0,\infty)$ to justify the inverse and all <stochastic integrals> for arbitrary $r>0$.
For completeness, the inverse-clock lifetime really is the positive <Bessel process> lifetime: the map $z\mapsto z^r$ is a homeomorphism of $(0,\infty)$, and exits from nested compact intervals correspond under the clock. There is no finite explosion at infinity for a positive <Bessel process>, since the <Itô formula> for $Y^2$ gives $d(Y^2)=2Y\,d\beta+\delta'ds$; stopping at zero or level $R$ yields a probability of hitting $R$ before a fixed $T$ bounded by $(Y_0^2+\max(\delta',0)T)/R^2$. If the transformed lifetime were finite while $Y$ stayed in a compact positive interval, the inverse clock and the original solution could both be continued. Thus a finite lifetime is approach to zero. For $\delta<2$, also $\delta'<2$ and the positive-state <Bessel process> hits zero; for $\delta\geq2$ the transformed lifetime is infinite. No continuation through zero is needed.
If the initial value is already zero, the interval before its first zero is empty; the positive-state argument concerns starts $z>0$.
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