= Solution
On $[\tau,\sigma)$ the Brownian path does not meet zero, so the <power function> $x\mapsto|x|^\delta$ is twice continuously differentiable along the path. The <Itô formula> gives
$$
dY_t=\delta|B_t|^{\delta-1}\operatorname{sign}(B_t)dB_t+\frac{\delta(\delta-1)}2|B_t|^{\delta-2}dt.
$$
Since $Y_t=|B_t|^\delta$, this becomes
$$
dY_t=\delta Y_t^{(\delta-1)/\delta}d\widehat B_t+\frac{\delta(\delta-1)}2Y_t^{(\delta-2)/\delta}dt,
$$
where $\widehat B_t=\int_0^t\operatorname{sign}(B_s)dB_s$ is a standard <Brownian motion> by the <Lévy characterization of Brownian motion>. Thus the displayed equation in the paper is valid after the customary renaming of $\widehat B$ as $B$.
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