Continuous local martingale

A continuous local martingale is a local martingale whose sample paths are continuous almost surely. Its quadratic variation determines much of its pathwise behaviour.

If $M$ is a continuous local martingale and $f\in C^2(\mathbb{R})$, then finite variation of $f(M)$ forces $f(M_t)=f(M_0)$. The Itô formula and the quadratic variation of a stochastic integral give ${\int }_0^t{f}^{\prime }(M_s{)}^{2}d[M{]}_s=0$. The occupation-times formula implies that $L_t^a(M)=0$ almost everywhere where $f^{\prime }(a)\ne 0$. On the zero set of $f^{\prime }$, its derivative $f^{\prime \prime }$ vanishes almost everywhere: zeros at which $f^{\prime \prime }\ne 0$ are isolated and hence countable. Thus ${\int }_0^t|f^{\prime \prime }(M_s)|d[M{]}_s=\int |f^{\prime \prime }(a)|L_t^a(M)da=0$. Both terms in the Itô formula vanish.

Suppose $X,A$ are continuous, start at zero, and $A$ is increasing. If $e^{X-A/2}$ and $e^{-X-A/2}$ are local martingales, their logarithms first recover adaptation of $X,A$ and the semimartingale property of $X$. In the decomposition $X=N+V$, the two Itô drift measures give $dV+(d[N]-dA)/2=0$ and $-dV+(d[N]-dA)/2=0$. Thus $V=0$ and $[X]=A$. All real exponential parameters may be assumed, but these two suffice.

A process has conditionally symmetric increments if, for all deterministic $s\le t$ and bounded measurable $g$,Equivalently, the conditional characteristic function of every increment is invariant under changing the sign of its argument. This property is stronger than unconditional symmetry and is different from independent increments. It supplies identities between positive and negative exponential terminal martingales.

Suppose $X$ is a continuous local martingale starting at zero, with conditionally symmetric increments, and terminal conditional expectations have continuous martingale versions. ThenTo prove this, set $M_t=\mathbb{E}[e^{i\theta X_T}\mid {\mathcal{F}}_t]$. Symmetry makes $e^{-2i\theta X_t}{M}_t$ a martingale, and the Itô product rule gives $d⟨M,X⟩=i\theta Md⟨X⟩$. Consequently $e^{-i\theta X_t-{\theta }^2⟨X{⟩}_t/2}{M}_t$ is a bounded local martingale, hence a martingale. Evaluating at the endpoints proves the formula. If also $X_T\sim N(0,T)$ for all $T$, the values of the bracket Laplace transform at $1$ and $2$ force $⟨X{⟩}_T=T$ almost surely. Continuity and the Lévy characterization of Brownian motion then identify $X$ as Brownian motion.

A continuous local martingale converges to a finite limit on the event that its terminal bracket is finite. Stopping at each integer bracket level gives an L2-bounded martingale and its almost sure limit. On the finite-bracket event, some such stopped process coincides with the original process forever. This supplies terminal values for a finite DDS lifetime.

Let $X$ be a continuous local martingale with $X_0=0$ and bracket $A=[X]$. On each horizon under consideration, assume $\mathbb{E}{\sup }_{s\le t}|X_s{|}^4<\infty$ and $\mathbb{E}{A}_t^2<\infty$. The Itô formula makes $X^4-6X^{2}A+3A^2$ a local martingale; these moment assumptions give an integrable running supremum and hence a true martingale. In particular,If additionally $\mathbb{E}{X}_t^2=t$ and $\mathrm{Cov}(X_t^2,A_t)=0$, the Itô isometry gives $\mathbb{E}{A}_t=t$ and thereforeIf the fourth moment equals $3t^2$ at every time under these assumptions, then $A_t=t$ on all rational times almost surely, hence at all times by continuity. The Lévy characterization of Brownian motion identifies $X$ as Brownian motion. Equality of fourth moments without the covariance hypothesis does not give this argument.

If a local martingale has an integrable running absolute supremum on each fixed finite horizon, localization and conditional dominated convergence make it a true martingale. This upgrades polynomial Itô local martingales when maximal moments and bracket moments control all terms.

A continuous local martingale $M$ with $M_0=0$ and deterministically bounded quadratic variation on every finite horizon is a square-integrable martingale, with $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t$. Localization, the Doob L2 maximal inequality, and the resulting integrable bound on the squared supremum justify removing the localizing sequence.

A continuous local martingale with finite variation is almost surely constant. After localization, its quadratic variation is both zero, because it has finite variation, and the quantity controlling its $L^2$ increments, because it is a martingale.

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.

For every $p>0$ there are constants $c_p,C_p>0$ such that a continuous local martingale $M$ with $M_0=0$ satisfiesfor every stopping time $T$ for which the quantities are finite.

Real and imaginary parts of an analytic Brownian image, after subtracting their starting values, have equal quadratic variation. The Burkholder-Davis-Gundy inequalities compare their centered maximal moments with constants depending only on $p$. A boundary-point normalization $P(0)=0$ does not remove the starting imaginary value. For $P(z)=iz$ on a wedge of opening $\alpha \pi <\pi /2$, $\sup |\mathrm{Im}z|\le \tan (\alpha \pi )\sup |\mathrm{Re}z|$. If $\alpha p<1$ both relevant moments are finite, so an uncentered comparison requires a constant at least ${\cot }^p(\alpha \pi )$; no bound depending only on $p$ can hold over all such wedges. For fixed $\alpha >0$, polynomials with finite real-part maximal moment have degree $d<1/(\alpha p)$ by the wedge radial-maximum tail. They form a finite-dimensional vector space, and equivalence of norms in finite dimensions gives a valid constant depending on $p,\alpha$.

For Brownian motion started at zero, condition at time ${\lambda }^2$. If the stopping time exceeds ${\beta }^2{\lambda }^2$ while the running absolute maximum stays below $\delta \lambda$, the path must remain in a fixed interval for another $({\beta }^2-1){\lambda }^2$ time units. Its expected exit time from this interval is at most ${\delta }^2{\lambda }^2$. Markov's inequality yields the displayed bound. Integrating it against $d\Phi$ for a function of moderate growth, then absorbing the small tail term, proves $\mathbb{E}\Phi (\sqrt{\tau })\le C_{\Phi }\mathbb{E}\Phi (B_{\tau }^{\ast })$.

A continuous nondecreasing function $\Phi :[0,\infty )\to [0,\infty )$ with $\Phi (0)=0$ has moderate growth if the displayed inequality holds for some $a>1$ and finite $K$, uniformly in $x\ge 0$. Iteration gives a bound for any fixed positive dilation, and at most polynomial growth at infinity. The powers $x^p$, $p>0$, satisfy the condition. It allows a good-lambda tail comparison to be integrated into a moment comparison.

For $p\ge 2$ and a continuous local martingale starting at zero, the Itô formula for $|X{|}^p$, the Doob Lp maximal inequality and Hölder's inequality yield $\mathbb{E}{\sup }_{s\le t}|X_s{|}^p\le C_p\mathbb{E}⟨X{⟩}_t^{p/2}$. One possible constant is $C_p=[p(p-1)(p/(p-1){)}^p/2{]}^{p/2}$. Stop at increasing absolute-value levels to remove an initial boundedness assumption. This proves the upper half of the Burkholder-Davis-Gundy inequalities in this range.