= Solution
Use $A_t=\int_0^tH_s^2ds$ and its inverse $\tau_u$ as above. Strict increase and continuity make $u\mapsto\tau_u$ continuous, with $A_{\tau_u}=u$ and $\tau_{A_t}=t$. Thus $W_u=M_{\tau_u}$ is continuous and <adapted> to $\mathcal G_u=\mathcal F_{\tau_u}$, with $W_0=0$.
The essential remaining point is independence of increments, which does not follow merely from the <Gaussian> marginal in part (a). Fix $0\le u<v$ and $\lambda\in\mathbb R$. The complex exponential
$$
\exp\left(i\lambda M_{t\wedge\tau_v}
+\frac{\lambda^2}{2}A_{t\wedge\tau_v}\right)
$$
is bounded by $e^{\lambda^2v/2}$ in modulus, hence is a <uniformly integrable> <martingale>. The <optional sampling theorem> at the finite, possibly unbounded, times $\tau_u\le\tau_v$ gives
$$
\mathbb E\left[e^{i\lambda M_{\tau_v}+\lambda^2v/2}\mid\mathcal F_{\tau_u}\right]
=e^{i\lambda M_{\tau_u}+\lambda^2u/2}.
$$
Divide by the nonzero right-hand exponential. Then
$$
\boxed{\mathbb E[e^{i\lambda(W_v-W_u)}\mid\mathcal G_u]
=e^{-\lambda^2(v-u)/2}.}
$$
This deterministic <conditional characteristic function> says that $W_v-W_u$ is $N(0,v-u)$ and independent of $\mathcal G_u$. Successively conditioning proves independence of every finite family of increments. Together with continuity and the zero initial value, this proves that $W$ is <Brownian motion>. The inverse-clock identity gives $W_{A_t}=M_{\tau_{A_t}}=M_t$, completing the requested special-case proof of the <Dubins-Schwarz theorem>.
Back to article page
