 Solution

Let $f(y)=|y|$ away from zero. Its first and second partial derivatives areThus its Laplace operator is $\Delta f(y)=(n-1)/|y|$. The quadratic covariations of the independent Brownian coordinates satisfy $d[B_j,B_k{]}_s={\delta }_{jk}ds$. The stopped Itô formula therefore givesFor $t<\tau$, this is the required integral representation.

To identify the driving Brownian motion, the stopped stochastic term alone is insufficient: its quadratic variation is $t\wedge \tau$. Extend its unit direction after exit. Define a vector predictable process $u(s)$ byPredictability follows from the continuous adapted direction before stopping and the left-continuous adapted indicator $1_{\{s\le \tau \}}$. This $W$ is a zero-starting continuous martingale, andThe Lévy characterization of Brownian motion identifies $W$ as a one-dimensional Brownian motion. Before $\tau$, the preceding representation becomes the stochastic differential equation of the dimension-$n$ Bessel process:The fully stopped equation has both integrals stopped at $\tau$.

For $n=2$, apply the one-dimensional Itô formula to the logarithm. Before $\tau$, $d[X{]}_t=dt$, so the two drift terms cancel:Consequently,Indeed $X_s\ge r$ before stopping, so the Itô isometry bounds the stochastic term's second moment by $t/r^2$, and $\log X_{t\wedge \tau }\in [\log r,\log R]$. This is the martingale assertion under the paper's standing stopping assumption. Without stopping, the same calculation establishes only a local martingale; the stopped qualification is essential for the stated argument.