 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 givesSince $Y_t=|B_t{|}^{\delta }$, this becomeswhere ${\hat{B}}_t={\int }_0^t\mathrm{sign}(B_s)d{B}_s$ 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 $\hat{B}$ as $B$.