Why the Brownian motion stopped at the local time is a martingale?

335 Views Asked by At

It's exercise 1 in Chapter 7 from Kai Lai Chung's book: Introduction to Stochastic Integration.

If I have 2 independent Brownian motion $B$ and $\tilde{B}$ that both start from $0$, and $L$ is the local time corresponding to $B$ from position $0$, that is $$L(t)=|B_t|-\int_0^t sgn(B_s)dB_s$$ Also our filtration is $\mathcal{F}_t$ which is generated by $B$ and $\sigma\{ \tilde{B}_s, 0\leq s\leq t\}$

We are asked to show that the triple $\{\tilde{B}_{L(t)}, \mathcal{F}_{L(t)}, t\geq 0 \}$ is a continuous $L^2$-martingale with quadratic variation $\{L(t), t\geq 0 \}$.

Simply I got stuck at the very beginning since I don't think I could even use Doob's optional stopping theorem, which would require the stopping time to be bounded or the stochastic process to be uniformly integrable, which both fail here.

Therefore I don't even know how to prove that triple is a martingale, let alone the quadratic variation, though it looks plausible since for a Brownian motion $B_t$, it's corresponding quadratic variation process is $t$.