Martingale

An integrable adapted process $(M_n)$ is a martingale when $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$.

A martingale is square integrable when its value at every time belongs to L2 space. This implies ordinary integrability by Cauchy-Schwarz inequality. The condition is separate from a uniform bound on these second moments over an infinite horizon. An Itô integral is square integrable on a finite interval when its predictable integrand has finite expected integrated square, by Itô isometry.

Let $0<\alpha <1$, $X_0=x\in [0,1]$, and let $B_{n+1}$ be a binary random variable satisfying $\mathbb{E}[B_{n+1}\mid {\mathcal{F}}_n]=X_n$. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable and $B_{n+1}$ is ${\mathcal{F}}_{n+1}$-measurable. The updateremains in $[0,1]$ and defines a bounded martingale. For example, take independent uniform random variables $U_{n+1}$ independent of the current filtration and set $B_{n+1}=1_{\{U_{n+1}\le X_n\}}$.

The binary identity $B_{n+1}^2=B_{n+1}$ givesand henceThe L2-bounded martingale convergence theorem gives a terminal random variable $X_{\infty }$ with almost sure convergence and convergence in L2. Iteration followed by the bounded convergence theorem yieldsThus $X_{\infty }$ takes only $0$ and $1$, and preservation of its expected value gives a Bernoulli distribution with parameter $x$. The terminal conditional expectation representation gives the exact mean squared errorThe finite-time paths can stay strictly inside $(0,1)$ while their limit lies entirely at the endpoints. The positive conditional variance in the interior makes permanent convergence to an interior value impossible.

Three players with positive integer holdings exchange a token by choosing an unordered active pair uniformly, then choosing the transfer direction fairly. Let $N$ be the conserved total, and $\tau$ the stopping time of the first elimination. Given holdings $(x,y,z)$ before $\tau$, the conditional expectation of the change of their product is $-(x+y+z)/3=-N/3$: transfers in each pair decrease the mean product by the third holding. Thus the displayed stopped process is a martingale. Its nonnegative product term gives $\mathbb{E}[n\wedge \tau ]\le 3abc/N$, so the monotone convergence theorem first proves integrability of $\tau$. The bounded product then vanishes by the dominated convergence theorem, yielding $\mathbb{E}[\tau ]=3abc/N$. Continuing the deterministic time term beyond elimination instead produces a submartingale.

Let $\xi$ be a random walk, independent of centered, independent time-space signs $h$, and let $0<\epsilon <1$. Averaging over the walk gives a positive martingale in the environment filtration: each new time layer has conditional mean-one weight. Hence $\mathbb{E}{Z}_t=1$ and the martingale convergence theorem gives a finite nonnegative almost-sure limit. Preserving its mean requires an additional condition such as bounded second moments; positivity of finite-time weights alone does not suffice.

Delete the first $m$ environment factors from the path-weight expectation and call the result $Z_T^{(m)}$. Pointwise, $(1-\epsilon {)}^m{Z}_T^{(m)}\le Z_T\le (1+\epsilon {)}^m{Z}_T^{(m)}$. Thus $\{{\mathrm{liminf}}_T{Z}_T=0\}$ equals $\{{\mathrm{liminf}}_T{Z}_T^{(m)}=0\}$ and belongs to every future-time environment sigma-algebra. The Kolmogorov zero-one law makes its probability zero or one. If the nonnegative martingale limit has mean one, the probability cannot be one, so its limit is strictly positive almost surely. The pointwise comparison establishes a genuine tail event, rather than merely invariance outside an unspecified exceptional set.

Here $I_t={\sum }_{j=1}^t{1}_{\{{\xi }_j^{(1)}={\xi }_j^{(2)}\}}$ counts simultaneous collisions of two independent walks. A shared environment sign contributes $\mathbb{E}(1+\epsilon h{)}^2=1+{\epsilon }^2$; distinct signs contribute one. If the difference walk returns to zero with probability $\rho <1$, its total number of positive-time returns has law $\mathbb{P}(I_{\infty }=k)=(1-\rho ){\rho }^k$. Therefore the second moments are bounded by $(1-\rho )/(1-\rho (1+{\epsilon }^2))$ whenever $\rho (1+{\epsilon }^2)<1$. Convergence in L2 then preserves the limit's mean one.

For a compensated jump martingale starting at zero, with jumps bounded in absolute value by $c$ and predictable quadratic variation at most $v$ on the interval, the displayed maximal bound holds. For $0<\theta c<3$, the elementary exponential boundmakes the exponential of $\theta M$ minus that multiple of its predictable bracket a nonnegative supermartingale. Stop at the first crossing of $\eta$ and use its mean bound. Taking $\theta =\eta /(v+c\eta /3)$ yields the one-sided bound; apply the same argument to $-M$ and use the union bound for the displayed two-sided version. The elementary estimate follows from the exponential series and $k!\ge 23^{k-2}$ for $k\ge 2$. If $v=0$, the compensated martingale is constant and the crossing probability is zero; the parameter choice above is needed only for $v>0$.

For a nonnegative martingale, the displayed maximal probability bound follows by stopping at the first crossing of $a$. The stopped expectation is at most its initial expectation, and on the crossing event its value is at least $a$. Bound the stopping time first and pass to the limit using nonnegativity. This form is useful for the Poisson maximal concentration bound.

In a finite population that repeatedly lets one uniformly chosen individual copy a uniformly chosen distinct individual, the fraction of either type is a bounded martingale. Conditional probabilities of an increase and decrease coincide. Absorption at one of the two unanimous states occurs almost surely, since from every state there is a uniformly positive probability of reaching unanimity in a fixed finite block of updates. Bounded convergence implies that the probability of eventual unanimity of one type equals its initial fraction.

In a uniformly shuffled deck with $r$ red and $b$ black cards remaining, the conditional probability that the next two are red is $q(r,b)$. If at least three cards remain, $q(r,b)$ is the weighted average of $q(r-1,b)$ and $q(r,b-1)$ after one reveal. Hence these probabilities form a bounded martingale until only two cards remain. Any adapted choice of exactly one betting time before that deadline has the same winning probability by bounded optional stopping. Waiting creates no advantage when a bet must eventually be made.

A nonnegative martingale takes nonnegative values almost surely at each time. In discrete time its expectations are constant, so it is bounded in $L^1$ and the martingale convergence theorem gives a finite integrable almost sure limit. Its expectations need not converge to the expectation of that limit; uniform integrability is needed for convergence in L1.

Expose independent coordinates successively and take the conditional mean of a function after each exposure. These conditional means form a martingale from $\mathbb{E}f$ to $f$. A coordinate oscillation bound $c_i$ gives a conditional interval of length $c_i$ for the corresponding increment, by coupling the unexposed coordinates. Applying the Hoeffding lemma to these conditional ranges proves the McDiarmid inequality.

A martingale can diverge to $+\infty$ almost surely when its increments are finite but not uniformly bounded. For independent Bernoulli random variables with $\mathbb{P}(Y_n=1)=2^{-n}$,has centered integrable increments. The First Borel-Cantelli lemma makes the rare negative jumps occur only finitely often, after which every increment equals one. The expectation remains zero at every finite time despite this pathwise divergence.

An integrable adapted process is a martingale if all its values at bounded stopping times are integrable with the same expectation as $X_0$. For the converse, compare deterministic $t$ with $U=s{1}_A+t{1}_{A^c}$, where $s<t$ and $A\in {\mathcal{F}}_s$. The resulting identity $\mathbb{E}[(X_t-X_s)1_A]=0$ is precisely the defining test for conditional expectation. In the forward direction, use the optional stopping theorem for a càdlàg martingale under the usual conditions for a filtration.

If a martingale has increments with $|M_i-M_{i-1}|\le c_i$ almost surely for deterministic $c_i$, then each one-sided deviation of magnitude $t>0$ from $M_0$ has probability at most $\exp (-t^2/(2{\sum }_i{c}_i^2))$. If all $c_i$ vanish, the martingale is constant. The inequality gives concentration from bounded increments without requiring independent increments.

For deterministic bounds $|M_k-M_{k-1}|\le c_k$, the conditional mean-zero increment has exponential moment at most $e^{{\theta }^2{c}_k^2/2}$. Iteration, the Markov inequality and optimization at $\theta =x/\sum c_k^2$ give each one-sided tail, and the union bound gives the displayed inequality. The factor two in the exponent denominator cannot be omitted: a single symmetric increment of size one has absolute deviation one with probability one.

If a real sequence converges finitely, its successive increments tend to zero. A martingale whose increments stay bounded away from zero on every sample path cannot have a finite almost-sure limit, regardless of its zero conditional mean increments.

Once a nonnegative martingale hits zero, it stays zero almost surely. Conditional expectation on $\{M_n=0\}$ gives $\mathbb{E}[M_{n+1}{1}_{\{M_n=0\}}]=0$; nonnegativity forces the next value to vanish there. A countable intersection makes these assertions simultaneous for all times, proving absorption at the first zero.

Let a nonnegative martingale start at one, let $T$ be its first zero, and let $T_R$ be its first hit of $[R,\infty )$. Suppose $\mathbb{E}[(M_{n+1}-M_n{)}^2\mid {\mathcal{F}}_n]\ge {\sigma }^2{1}_{\{T>n\}}$ and $M_{T_R}\le \tau R$ on finite hits, for thresholds $R\ge 1$. Then $M_{T\wedge T_R\wedge n}^2-{\sigma }^2(T\wedge T_R\wedge n)$ is a submartingale. The bounded stopped values have mean one, so their second moment is at most $\tau R$. Consequently $\mathbb{E}(T\wedge T_R)\le (\tau R-1)/{\sigma }^2$. For $\tau \ge 1$ this implies the weaker bound ${\tau }^{2}R/{\sigma }^2$. The threshold restriction prevents an impossible overshoot requirement below the starting value.

Under the conditional-variance and overshoot assumptions of the absorption-time bound from conditional variance and overshoot, $\mathbb{P}(T>n)\le 2\tau /(\sigma \sqrt{n})$. Bound the probability by $\mathbb{P}(T_R<\infty )+\mathbb{E}(T\wedge T_R)/n$ and optimize $1/R+{\tau }^{2}R/({\sigma }^{2}n)$. If the resulting bound is below one, its minimizing threshold is greater than one; otherwise the probability bound is trivial.

A martingale difference sequence is an adapted sequence of integrable random variables $D_n$, $n\ge 1$, with $\mathbb{E}[D_n\mid {\mathcal{F}}_{n-1}]=0$. Its partial sums are a martingale, and martingale increments give such sequences. If the differences are square-integrable, they are uncorrelated random variables: for $j<k$, conditioning on ${\mathcal{F}}_{k-1}$ gives $\mathbb{E}{D}_j{D}_k=0$.

A time-indexed increment $D_j$ with $\mathbb{E}(D_j\mid {\mathcal{F}}_{j-1})=0$ is a martingale difference relative to its history. Square-integrable differences at different indices are orthogonal, allowing their conditional variances to accumulate. They need not be independent.

If a martingale has increments bounded by a single deterministic constant, then $M_n/n\to 0$ almost surely. Its increments form a uniformly $L^2$-bounded martingale difference sequence; apply the strong law for uniformly L2-bounded uncorrelated random variables to their partial sums and note $M_0/n\to 0$. No square-integrability of the initial value is needed beyond the usual martingale integrability.

For a real continuous function $f$ on $[0,1]$, let $G_n$ be its secant slope on each length-$2^{-n}$ dyadic cell. Regard $([0,1],\mathcal{B},dt)$ as a probability space and use the filtration of half-open dyadic cells with the endpoint $\{1\}$ as a separate null cell. The mean of the two child slopes equals the parent slope, so $(G_n)$ is a martingale. Its integral gives the linear interpolation of $f$ on that grid. If $f$ is Lipschitz continuous, these slopes are bounded by its Lipschitz constant; the Lp martingale convergence theorem then supplies a bounded integral density for $f$.

For an $L$-Lipschitz function $f$ on $[0,1]$, its slopes on the dyadic partition, evaluated at a uniform distribution input, form a bounded martingale for the dyadic filtration. The uniformly integrable martingale convergence theorem yields a bounded terminal random variable $g(X)$, with $g$ chosen as a Borel measurable function by the Doob-Dynkin lemma. The terminal conditional expectation identity says that each slope is the average of $g$ over the corresponding dyadic interval. Integrating over adjacent dyadic intervals and taking a telescoping sum proves the displayed representation first at dyadic endpoints. The Lipschitz condition for $f$ and the bound $|g|\le L$ extend it to every endpoint by continuity. This proves absolute continuity of a function without presupposing a derivative.

For fair independent and identically distributed random variables ${\xi }_j$ with the Bernoulli distribution, $M_n=2^n{1}_{\{{\xi }_1=\cdots ={\xi }_n=1\}}$ is a nonnegative martingale. It has almost sure convergence to zero but $\mathbb{E}{M}_n=1$, so it lacks convergence in L1.

If $(M_n)$ is a square-integrable martingale, then its increments are orthogonal in $L^2$: for $i<j$,More generally, multiplying the later increment by any square-integrable quantity measurable before it still gives expectation zero whenever the product is integrable.

An integrable adapted process $(X_n)$ is a submartingale when

For a submartingale with ${\sup }_n\mathbb{E}|X_n|=C<\infty$, its positive part is a submartingale by the conditional Jensen inequality. Thus the displayed conditional expectations increase in $p$. Their limits are finite almost surely and have expectations at most $C$, by the monotone convergence theorem. Conditional monotone convergence and the tower property of conditional expectation give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Moreover $M_n\ge X_n^{+}$. Consequently $Y_n=M_n-X_n$ is a nonnegative supermartingale, ${\sup }_n\mathbb{E}{Y}_n\le 2C$, and $X=M-Y$. The resulting martingale need not have uniform integrability.

Under the usual filtration conditions, a càdlàg submartingale of Class D on a finite horizon has a unique decomposition into a uniformly integrable martingale and a predictable integrable nondecreasing process starting at zero. A general continuous semimartingale decomposition instead allows an arbitrary continuous finite-variation drift. A convex function of a semimartingale need not be a submartingale when that drift has unrestricted sign.

If $(X_m{)}_{m\le n}$ is a nonnegative submartingale, then for every $a>0$,Stop at the first crossing of $a$ and use the submartingale property on that event.

For a nonnegative submartingale with maximum $M_t$, the first-moment maximal estimate is $\lambda \mathbb{P}(M_t\ge \lambda )\le \mathbb{E}[X_t{1}_{\{M_t\ge \lambda \}}]$. Integrate $2\lambda$ times the tail probability only up to $R$, obtaining $\mathbb{E}(M_t\wedge R{)}^2\le 2\mathbb{E}[X_t(M_t\wedge R)]$. The Cauchy-Schwarz inequality gives $\Vert M_t\wedge R{\Vert }_2\le 2\Vert X_t{\Vert }_2$, and monotone convergence gives the result as $R\to \infty$. Truncation avoids assuming beforehand that the maximum has a finite second moment.

// Target: brownian-motion.bigb

For a square-integrable martingale with $M_0=0$ and $V=\mathbb{E}{M}_n^2$, apply the Doob maximal inequality to $(M_k+c{)}^2$ for $c\ge 0$. An upper crossing of $x$ implies a square crossing of $(x+c{)}^2$, giving $(V+c^2)/(x+c{)}^2$. Minimization at $c=V/x$ gives the displayed bound. A one-step martingale taking $x$ with probability $V/(V+x^2)$ and $-V/x$ otherwise attains equality when $V>0$.

For independent zero-mean square-integrable increments, their partial sums form a martingale, and their squares form a nonnegative submartingale. Apply the Doob maximal inequality to the squared process at threshold $x^2$. Independence and zero means make its terminal second moment the sum of the increment variances. Identical distributions are not required.

For a centered square-integrable martingale with $v=\mathbb{E}{M}_n^2$ and $\lambda >0$, apply the Doob maximal inequality for a nonnegative submartingale to $(M_k+c{)}^2$ for $c\ge 0$. Conditional Jensen inequality gives the submartingale property, and crossing $\lambda$ forces its square above $(\lambda +c{)}^2$. The bound $(v+c^2)/(\lambda +c{)}^2$ is minimized by $c=v/\lambda$. This extends the Cantelli inequality from one random variable to a martingale maximum, without requiring independent increments.

For a nonnegative martingale, $T_R=\inf \{n:M_n\ge R\}$ satisfies $\mathbb{P}(T_R<\infty )\le \mathbb{E}{M}_0/R$. Stop at $T_R\wedge n$ and use nonnegativity before passing to the increasing event. A limit of thresholds from below also gives the bound for $\{{\sup }_n{M}_n\ge R\}$ even when the supremum is not attained.

For independent centered random variables and $S_m={\sum }_{j\le m}{X}_j$,Apply the Doob maximal inequality for a nonnegative submartingale to $(S_m+c{)}^2$ and optimize at $c=\mathrm{Var}(S_n)/x$.

For independent centered square-integrable variables, if $S_n={\sum }_{m\le n}{X}_m$ and $V_n=\mathrm{Var}(S_n)$, thenis a martingale.

If additionally $|X_m|\le K$, thenStop the square martingale for independent centered sums at the first exit from $[-x,x]$ or at time $n$.

For an integrable random variable $Z$ and an increasing filtration $({\mathcal{F}}_n)$, the process $M_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$ is a martingale. The tower property of conditional expectation gives the martingale identity.

Expose the independent edge indicators of a binomial random graph in a fixed order, and set $M_i=\mathbb{E}[Z\mid \text{the first }i\text{ indicators}]$ for an integrable statistic $Z$. This is a martingale from $\mathbb{E}Z$ to $Z$. If changing one edge changes $Z$ by at most $c$, coupling the unexposed indicators gives $|M_i-M_{i-1}|\le c$, permitting the Azuma-Hoeffding inequality.

If $Y_1,Y_2,\dots$ are independent integrable random variables with $\mathbb{E}[Y_i]=1$, thenis a martingale for ${\mathcal{F}}_n=\sigma (Y_1,\dots ,Y_n)$, because independence gives $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n\mathbb{E}[Y_{n+1}]=M_n$.

Suppose $(X_n)$ are identically distributed independent random variables with $X_n\ge 0$, $\mathbb{E}{X}_n=1$, and $X_1$ not constant almost surely. For each $0<r<1$, strict concavity of $t^r$ givesThe moment is finite because $t^r\le 1+t$ on $[0,\infty )$. Let $Y_n={\prod }_{k=1}^n{X}_k$. Independence and Markov's inequality giveThe series of the latter probabilities is finite. The First Borel-Cantelli lemma, applied for the countable choices $\epsilon =1/m$, proves $Y_n\to 0$ almost surely. This proof requires neither existence nor integrability of $\log X_1$.

Although this nonnegative martingale has $\mathbb{E}{Y}_n=1$ at every time, its terminal expected value is zero. It is therefore not uniformly integrable. The contrast is useful: bounded expected values ensure almost sure convergence here, but do not allow preservation of the expected value through the limit.

An adapted integrable process $(M_t{)}_{t\ge 0}$ is a continuous-time martingale when $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$ for every $s\le t$.

For finitely many independent uniform arrival times on $[0,1]$, and fixed positive weights, normalize the unarrived mass by the remaining time. Conditional on not having arrived by $s$, each arrival remains uniform on $(s,1)$, giving the martingale identity. The process is càdlàg, nonnegative, and has constant mean equal to the total weight. It vanishes eventually before time one on each typical path, so it is not uniformly integrable when the total weight is positive.

For the uniform-arrival residual mass martingale with total weight $0<S<1$, the first crossing of level one happens continuously: its only jumps are downward, and an arrival almost surely misses the finitely many deterministic possible crossing times. Stop at that crossing or at $r<1$. The stopped values lie between zero and one, and their limit as $r\uparrow 1$ is the crossing indicator. Optional stopping and dominated convergence identify its probability as $S$. At $S=1$ the strict boundary inequality already fails at time zero.

A martingale whose values over its whole time interval form a uniformly integrable family has an integrable terminal limit and is closed by that limit. Conversely, conditional expectations of an integrable terminal variable form a uniformly integrable martingale, by uniform integrability of conditional expectations. For continuous martingales, one may equivalently use uniform integrability of the family stopped at finite stopping times.

A càdlàg martingale is a continuous-time martingale represented by a càdlàg process: its paths are right-continuous and have finite left limits at positive times, on one common probability-one event. This path regularity permits approximation of stopping times from above.

A continuous martingale is a continuous-time martingale whose paths are continuous outside one null event. Continuity does not remove the integrability or conditional-expectation requirements in the martingale definition.

A continuous martingale $M$ is L2-bounded when ${\sup }_{t\ge 0}\mathbb{E}|M_t{|}^2<\infty$. The L2 martingale convergence theorem gives a terminal value $M_{\infty }$ with convergence in $L^2$ and almost sure convergence. The resulting conditional expectation identity is $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$. Boundedness on each separate finite horizon is a weaker condition. The subspace with $M_0=0$ is commonly denoted ${\mathcal{M}}_0^2$.

Continuous square-integrable martingales on a fixed finite horizon form a complete space under the displayed terminal norm, modulo indistinguishability of stochastic processes. A terminal-norm Cauchy sequence admits a subsequence whose successive differences have norms below $2^{-n}$. The Doob L2 maximal inequality and Markov's inequality make the probabilities that these differences have suprema above $2^{-n/2}$ summable. The Borel-Cantelli first lemma then gives an almost surely uniformly convergent telescoping series, so the limit has continuous paths.

Conditional-expectation contractivity gives convergence in $L^2$ at every fixed time, and passage to the limit in the martingale identities proves that the limit is a martingale. Adaptedness can be retained by taking the pointwise finite subsequence limit when it exists, and zero otherwise. The original Cauchy sequence converges in terminal norm to this limit. This completeness allows the Itô isometry to extend stochastic integration from simple predictable integrands to the full square-integrable integrand space while retaining a continuous martingale version.

For a continuous local martingale starting at zero, finite expected terminal quadratic variation is equivalent to being an L2-bounded continuous martingale. Stop at bounded path and bracket levels and use $\mathbb{E}(M_t^{\tau }{)}^2=\mathbb{E}[M{]}_{t\wedge \tau }$. Uniform integrability removes the localization; Doob L2 maximal inequality justifies convergence of squares. The terminal identity is $\mathbb{E}{M}_{\infty }^2=\mathbb{E}[M{]}_{\infty }$. For nonzero initial values one must additionally assume $M_0\in L^2$.

For a L2-bounded continuous martingale, the terminal and maximal norms satisfyThe first inequality follows from almost sure convergence. Apply the Doob L2 maximal inequality on $[0,T]$ and then the monotone convergence theorem as $T\to \infty$ for the second. These are equivalent norms on the space modulo indistinguishability of stochastic processes; neither inequality needs a zero initial value.

An adapted process $M$ is a local martingale when there are stopping times $T_n\uparrow \infty$ almost surely such that each stopped process $M_{t\wedge T_n}$ is a martingale. Every martingale is local, while a positive local martingale is a supermartingale.

A uniformly bounded local martingale is a true uniformly integrable martingale: a localizing sequence gives uniformly dominated stopped values, so their conditional identities pass to the limit. A deterministic bound on each compact time interval gives a true martingale on each such interval, even when the bound grows with time.

An integrable discrete-time local martingale is a true martingale. Its stopped increments have the form $1_{\{\tau \ge t\}}(X_t-X_{t-1})$, and the dominated convergence theorem removes a localizing sequence from the conditional increment identity.

The Fatou lemma makes every time value of a nonnegative discrete-time local martingale integrable, assuming the usual integrable initial value. The integrable discrete-time result then makes it a true martingale.

A strict local martingale is a local martingale that is not a true martingale. The reciprocal of a three-dimensional Bessel process is an example: it is a nonnegative continuous local martingale whose expectation decreases strictly.

For planar Brownian motion started at zero, $\log |B_t|$, considered from time one onward, is a local martingale because $\log |x|$ is harmonic off the polar point zero. Every fixed-time absolute moment is finite, but $\mathbb{E}{X}_t=\frac{1}{2}\log t+\mathbb{E}\log |B_1|$ increases. It is therefore not a true martingale. Finite moments at individual times do not establish class DL.

// Target: probability-and-statistics.bigb

A localizing sequence for a local martingale $M$ is an increasing sequence of stopping times $T_n\uparrow \infty$ almost surely such that every stopped process $M^{T_n}$ is a true martingale. One can often choose the times so that the stopped processes are bounded.

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.