= Solution
Let $V_t=\langle M\rangle_t$, and first suppose $V_\infty=\infty$ almost surely in addition to strict increase. The <Dambis-Dubins-Schwarz theorem> says that
$$
T_u=\inf\{t:V_t>u\},\qquad\mathcal G_u=\mathcal F_{T_u},\qquad B_u=M_{T_u}
$$
define a <Brownian motion> $B$ in the time-changed filtration and give
$$
\boxed{M_t=B_{V_t}.}
$$
Each $T_u$ is a <stopping time> and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.
Here are the <martingale> details behind this <inverse-clock proof of the Dambis-Dubins-Schwarz theorem>. Stopping a <continuous local martingale> when its bracket reaches $n$ makes it an <L2-bounded continuous martingale>. This follows from the stopped <Itô isometry> or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is <uniformly integrable>, and optional sampling is valid even at an unbounded <stopping time> by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}})_{u\geq0}$ is a <martingale>. Thus $B$ is a <continuous local martingale>. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a <local martingale>, so $\langle B\rangle_u=u$.
For completeness, the <Lévy characterization of Brownian motion> follows directly from the <Itô formula>. If a <continuous local martingale> $N$, starting at zero, has bracket $u$, then
$$
\exp\left(i\theta N_u+\frac{\theta^2u}{2}\right)
$$
is a complex <local martingale>. Its modulus is bounded on each deterministic finite horizon, so it is a true <martingale> there. Consequently
$$
\mathbb E\left[e^{i\theta(N_u-N_v)}\mid\mathcal G_v\right]=e^{-\theta^2(u-v)/2}\qquad(v\leq u).
$$
Conditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the <Lévy characterization of multidimensional Brownian motion> when the bracket matrix is $uI$.
The printed strict-increase hypothesis does not imply $V_\infty=\infty$. For example, $M_t=\int_0^te^{-s}\,dW_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_\infty$ can be finite.
On $\{L<\infty\}$ the <martingale> $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an <L2-bounded continuous martingale> which converges; on $\{V_\infty<n\}$ the stopped process is the original one. This proves the <finite-bracket convergence lemma>. Set $T_u=\infty$ when $u\geq L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a <continuous local martingale> with bracket $u\wedge L$. Moreover $L$ is a <stopping time> in $\mathcal G_u=\mathcal F_{T_u}$.
On a product extension add an independent <Brownian motion> $\beta$ in clock time, and put
$$
\widetilde B_u=N_u+\int_0^u\mathbf1_{\{s>L\}}\,d\beta_s.
$$
The two summands have zero <quadratic covariation>, and their brackets are $u\wedge L$ and $(u-L)^+$. Thus $\langle\widetilde B\rangle_u=u$; the proved characterization makes $\widetilde B$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t=\widetilde B_{V_t}$ still holds. This is the <finite-lifetime extension of the Dambis-Dubins-Schwarz theorem>. \b[An infinite clock gives <Brownian motion> on the original space; a finite clock may require the independent extension.]
Back to article page
