Dambis-Dubins-Schwarz theorem

Let $M$ be a continuous local martingale with $M_0=0$ and $[M{]}_{\infty }=\infty$. Forthe process $B_s=M_{{\tau }_s}$ is Brownian motion andWhen $[M{]}_{\infty }<\infty$, one can enlarge the probability space and continue $B$ independently beyond that time.

For a continuous local martingale with infinite terminal quadratic variation, the Dambis-Dubins-Schwarz theorem represents $M_t-M_0=W_{[M{]}_t}$. Its continuous clock traverses all nonnegative times. Standard one-dimensional Brownian motion has unbounded positive and negative oscillations, by the Brownian reflection principle or recurrence of one-dimensional Brownian motion. Continuity bounds every path on compact time intervals, so these large values occur at arbitrarily late times. Pulling them back through the clock proves the displayed conclusion almost surely, and in particular ${\mathrm{limsup}}_{t\to \infty }|M_t|=\infty$.

For independent standard Brownian motions $B,W$, Itô formula gives $dY=e^W\cos BdW-e^W\sin BdB$: the two second-order drifts cancel. Independence gives zero cross quadratic covariation, so the displayed bracket follows. The process itself is not Brownian because its bracket is random. Its inverse-bracket time change is Brownian motion starting at one, in the time-changed filtration; subtracting one gives standard Brownian motion. The Divergence of a driftless Brownian exponential clock guarantees that the inverse is finite at every new time.

Let $M$ be a zero-starting continuous local martingale and let $T<\infty$ be a stopping time with $[M{]}_{t\wedge T}\le a$ and $[M{]}_T=a$ for a deterministic $a$. The complex exponential $\exp (iu{M}_{t\wedge T}+u^2[M{]}_{t\wedge T}/2)$ is a bounded local martingale, hence a true martingale. Its terminal expectation is one, giving the characteristic function $\mathbb{E}{e}^{iu{M}_T}=e^{-u^{2}a/2}$. Conditional optional sampling between two bracket levels also gives independent Gaussian increments in inverse-bracket time; Gaussian one-time marginals alone do not establish that independence.

For a continuous local martingale $M$ with divergent bracket $K$ and a continuous drift $D$ satisfying $|D_t-D_s|\le C(K_t-K_s)$, the process $y+M+D$, $|y|<1$, exits $(-1,1)$ almost surely. In unit bracket time a Brownian increment of absolute size greater than $C+2$ forces exit within one unit, giving a geometric survival bound. The inverse bracket clock transfers this finite exit to ordinary time.

// Target: probability-and-statistics.bigb

If a continuous vector local martingale has bracket matrix $A_{t}I$ with $A_{\infty }=\infty$, inverse-clock time change gives bracket matrix $rI$. The Lévy characterization of multidimensional Brownian motion then gives Brownian motion in the common time-changed filtration. A continuous local martingale is constant on flat intervals of its bracket, so inverses using $A_t\ge r$ and $A_t>r$ give the same martingale values. The filtration at the earlier inverse is smaller; adaptation and independence from the larger past transfer the Brownian property to it. Nonzero initial values are retained by adding the initial vector back.

A continuous local martingale bounded above pathwise throughout a stochastic time interval cannot have infinite quadratic variation at that interval's endpoint. The Dambis-Dubins-Schwarz theorem represents it as Brownian motion at its quadratic-variation clock; an infinite clock would force unbounded oscillations. Its clock therefore has a finite limit, and so does the martingale. A bound below gives the same conclusion by changing sign. The bound may be random; it must hold over the entire interval.

Strict increase of a bracket does not force an infinite terminal value. Up to $L=⟨M{⟩}_{\infty }$, the inverse-clock martingale continues at a finite lifetime by its terminal limit, with bracket $u\wedge L$. On an independent product extension, add ${\int }_0^u{1}_{\{s>L\}}d{\beta }_s$. Its bracket fills $(u-L{)}^{+}$ and its cross variation with the original part vanishes. The result is Brownian for all clock times and still represents $M_t$.

For a continuous strictly increasing unbounded bracket $V$, the inverse times $T_u=\inf \{t:V_t>u\}$ satisfy $V_{T_u}=u$ and $T_{V_t}=t$. Optional sampling of the original martingale stopped at bounded bracket levels makes $M_{T_u}$ a continuous local martingale in ${\mathcal{F}}_{T_u}$. Time-changing the square-minus-bracket martingale gives its bracket $u$; the Lévy characterization of Brownian motion identifies it as Brownian.

If a continuous local martingale satisfiesfor a nonnegative predictable process $A$, then, after enlarging the probability space if necessary, there is a Brownian motion $B$ such thatAn independent Brownian motion supplies noise on the set where $A=0$.

The Brownian motion $B$ in the Dambis-Dubins-Schwarz theorem representation $M_t=B_{[M{]}_t}$ need not be independent of the bracket clock. Let $W$ be standard Brownian motion and takeThe Gaussian fourth moment and the Brownian covariance kernel give $\mathbb{E}[(W_t^2-t)W_s^2]=2s^2$ for $s\le t$, and henceIf the entire driving Brownian motion $B$ were independent of the clock $A$, conditioning on $A_t$ would give $\mathbb{E}[B_{A_t}{A}_t]=0$, contradicting this identity. Gaussian increments in bracket time therefore do not justify an independent random-clock model.