showing that ito's integral on brownian motion normalized by bessel process is a martingale

Question

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I couldn't prove (ii), I've tried to use levy's theorem together with (i) (and also without (i)) but it doesn't work out.

Answer 1

I think Levy's theorem should work. We have that $X$ is a continuous local martingale because it's a stochastic integral, so we just need to check $\langle X, X \rangle_t = t$. We compute

\begin{align*} \langle X, X \rangle_t &= \sum_{i=1}^n \int_0^t \left( \frac 1R W^i \right)^2 ds \\ &= \int_0^t \frac 1{R^2} \sum_{i=1}^n \left(W^i \right)^2 ds \\ &= \int_0^t \frac 1{R^2} |W|^2 ds \\ &= \int_0^t ds = t \end{align*}

so by Levy's characterization, $X$ is a Brownian motion.