$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ > [!theorem] Dambis-Dubins-Schwarz > Let $M$ be a CLM such that $\braket{M,M}_\infty = \infty$ a.s.. Then, there exists a BM $\left(\beta_s\right)_{s\geq 0}$ such that a.s., for all $t\geq 0$,$M_t = \beta_{\braket{M,M}_t}.$ - We write $\beta$ and not $B$ to remind ourselves that $\beta$ will not be adapted with respect to $\left(\FFF_t\right)$, but rather a time-changed filtration. - $\braket{M,M}_\infty = \infty$ can be dropped! >[!proof]- Proof sketch > The details of the proof seem unenlightening. Here's the breakdown. I ignore all almost-sure details (and more). > - The intuition is that we want to use the increasing process $\braket{M,M}_t$ to measure time. To this end: > - The time change is exactly $\tau_r = \inf\{t\geq 0: \braket{M,M}_t \geq r\}$ > - The BM is exactly $\beta_r = M_{\tau_r}$. (Thus, "$\beta_{\braket{M,M}_t} = M_{\tau_{\braket{M,M}_t}} = M_tquot;). > - The filtration is $\GGG_r = \FFF_{\tau_r}$. > - Now, we simply show $\beta_s$ and $\beta_s^2 - s$ are martingales. > - $\beta_s$ is obviously a martingale:$\EE\left[\beta_s | \GGG_t\right] = \EE\left[M_{\tau_s} | \FFF_{\tau_t}\right] = \beta_t$ > - $\beta^2_s - s$ is also obviously a martingale:$\EE[\beta_s^2 - s | \GGG_t] = \EE[M_{\tau_s}^2 - \braket{M,M}_{\tau_s} | \FFF_{\tau_t}] = M^2_{\tau_t} - \braket{M,M}_{\tau_t} = \beta^2_t - t.$ >[!claim] Random technical Proposition >Let $M, N$ be two CLMs with $M_0 = N_0 = 0$. Suppose $\braket{M,M}_t = \braket{N,N}_t$ for all $t\geq 0$, $\braket{M,N}_t = 0$ for every $t\geq 0$, a.s., and $\braket{M,M}_\infty = \braket{N,N}_\infty = \infty$, a.s.. > >Then, the corresponding BMs $\beta$ and $\gamma$ are independent. Proof omitted; you want to show that $\braket{\beta, \gamma} = 0$ and hit with [[Levy's Characterization of Brownian Motion]].