Right-Continuous Versions of Martingales

Why a Version Matters in Continuous Time

A martingale is defined through integrability, adaptation, and conditional expectations at each pair of fixed times. Those conditions are statements up to almost-sure equality at each time; they do not, by themselves, prescribe good paths simultaneously at every time. Yet many continuous-time arguments use path properties such as right-continuity, left limits, or a first time a path crosses a level. The regularization theorem bridges this gap: under the usual conditions on a filtration, a martingale has a version with right-continuous paths and left limits.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) satisfying the usual conditions: the filtration is right-continuous, so \(\mathcal{F}_t=\bigcap_{s>t}\mathcal{F}_s\), and \(\mathcal{F}_0\) contains every subset in \(\mathcal{F}\) of every \(\mathbb{P}\)-null set. The space may carry a Brownian motion \(B_t\), but the regularization result below concerns martingales in general and does not require Brownian motion or a stochastic integral.

Definition: A process \(M=(M_t)_{t\ge0}\) is a martingale if each \(M_t\) is integrable and \(\mathcal{F}_t\)-measurable, and \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely whenever \(0\le s\le t\). A process is cadlag if its paths are right-continuous and have finite left limits at every positive time. A process \(\widetilde M\) is a modification of \(M\) if \(\mathbb{P}(\widetilde M_t=M_t)=1\) for each fixed \(t\). They are indistinguishable if there is one event of probability one on which \(\widetilde M_t=M_t\) for every \(t\ge0\).

The quantifiers in these definitions matter. For a modification, the probability-one event may depend on \(t\). Indistinguishability requires a single event to work for all times. A cadlag modification is not a claim that the original process already has cadlag paths: it is a process with the same random variable at each fixed time, almost surely, whose paths have the desired regularity.

The Right-Continuous Regularization Theorem

Theorem (Doob regularization theorem for martingales): Suppose the filtration satisfies the usual conditions and \(M\) is a martingale indexed by \(t\ge0\), with \(\mathbb{E}[|M_t|]<\infty\) for every \(t\). Then \(M\) has an adapted cadlag modification \(\widetilde M\). In particular, for each fixed \(t\), \(\mathbb{P}(\widetilde M_t=M_t)=1\).

This is a standard regularization result. Its full proof uses a nontrivial upcrossing estimate on countable dense sets of times and a path-regularity construction; that technical estimate is not proved here. The statement is given with the hypotheses needed for its martingale form. In this case no additional continuity assumption on \(t\mapsto\mathbb{E}[M_t]\) is needed: the martingale property gives \(\mathbb{E}[M_t]=\mathbb{E}[M_0]\) for every \(t\). The stronger theorem for submartingales has additional hypotheses, so it should not be inferred from this martingale statement.

The theorem gives a cadlag modification, not necessarily an indistinguishable version of the original process. Once a cadlag version has been chosen, however, it is unique up to indistinguishability among cadlag modifications. This uniqueness is particularly useful: it lets us choose the regular version without making later pathwise conclusions depend on which cadlag version was selected.

Worked Example: A Martingale with a Deterministic Information Jump

Let \(Y\) take the values \(2\) and \(-2\), each with probability \(1/2\). Set \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(t<1\), and \(\mathcal{F}_t=\sigma(Y)\) for \(t\ge1\). This filtration is complete if the probability space is completed, and it is right-continuous. Define \(M_t=\mathbb{E}[Y\mid\mathcal{F}_t]\).

For \(t<1\), the information is trivial and \(\mathbb{E}[Y]=\tfrac12(2)+\tfrac12(-2)=0\), so \(M_t=0\). For \(t\ge1\), \(Y\) is known, so \(M_t=Y\). Thus the process is \(M_t=0\) for \(t<1\) and \(M_t=Y\) for \(t\ge1\). Its paths are right-continuous, with left limit \(0\) at time \(1\).

To check the martingale property, take \(s\le t\). If \(s\ge1\), both values equal \(Y\). If \(t<1\), both values are zero. If \(s<1\le t\), then \(\mathbb{E}[M_t\mid\mathcal{F}_s]=\mathbb{E}[Y\mid\mathcal{F}_s]=0=M_s\). These cases cover every pair \(s\le t\), and each \(M_t\) is integrable because \(\mathbb{E}[|M_t|]\) is either \(0\) or \(\tfrac12(2)+\tfrac12(2)=2\).

Check: at \(t=1\), the value is \(Y\), not its pre-jump conditional mean. The left limit is \(0\), while right-continuity requires the value at time \(1\) to equal the limit from times above \(1\), which it does.

Why Cadlag Modifications Are Unique

Proposition: If two cadlag processes \(X\) and \(Y\) are modifications of the same process, then they are indistinguishable.

Proof. Let \(\mathbb{Q}_+\) denote the nonnegative rational numbers. For each \(q\in\mathbb{Q}_+\), the modification property gives \(\mathbb{P}(X_q=Y_q)=1\). Since \(\mathbb{Q}_+\) is countable, the event \(A=\bigcap_{q\in\mathbb{Q}_+}\{X_q=Y_q\}\) has probability one. Fix an outcome \(\omega\in A\) for which both paths are cadlag. For any \(t\ge0\), choose rational numbers \(q_n>t\) with \(q_n\downarrow t\). Then \(X_{q_n}(\omega)=Y_{q_n}(\omega)\) for every \(n\). Right-continuity gives \(X_t(\omega)=\lim_n X_{q_n}(\omega)=\lim_n Y_{q_n}(\omega)=Y_t(\omega)\). This also works at \(t=0\). Thus equality holds for every \(t\) on one event of probability one, which is indistinguishability. \(\square\)

The argument explains why right-continuity is enough for this uniqueness result: equality on a countable dense set can be extended to every time by taking rational times down to the time in question. Left limits are part of the cadlag path property, but are not needed for this particular proof.

Worked Example: Two Cadlag Versions That Differ on a Null Set

Let \(N\in\mathcal{F}_0\) satisfy \(\mathbb{P}(N)=0\), and set \(X_t=0\) and \(Y_t=\mathbf{1}_N\) for every \(t\ge0\). Both processes are adapted and have constant, hence cadlag, paths. For every fixed \(t\), \(\mathbb{P}(X_t=Y_t)=\mathbb{P}(N^c)=1\), so \(Y\) is a modification of \(X\).

They are also indistinguishable: on the single event \(N^c\), which has probability one, \(X_t=Y_t=0\) for every \(t\). On \(N\) they differ at every time, but that event has probability zero. This illustrates that indistinguishability permits differences on one exceptional null set; it does not require pointwise equality on all of \(\Omega\).

Check: the event used here is \(N^c\), independent of \(t\). That is stronger than separately observing \(\mathbb{P}(X_t=Y_t)=1\) at each fixed time.

Modification Does Not Mean Indistinguishability

The distinction cannot be dropped merely because the process is a martingale. The following example has a modification with excellent paths, but the original process has a spike at a random time. At each fixed time the spike is invisible to probability, while pathwise it remains visible.

Worked Example: A Random-Time Spike

Take \(U\) uniformly distributed on \([1,2]\), and let \(\mathcal{F}_t=\mathcal{F}\) for every \(t\ge0\), where \(\mathcal{F}\) is the completed sigma-field generated by \(U\). This is a complete, right-continuous filtration. Set \(X_t=\mathbf{1}_{\{t=U\}}\). For each fixed \(t\), \(\mathbb{P}(U=t)=0\), so \(\mathbb{E}[|X_t|]=0\) and \(X_t=0\) almost surely.

For \(s\le t\), the conditional expectation \(\mathbb{E}[X_t\mid\mathcal{F}_s]\) is zero almost surely, since \(X_t=0\) almost surely. Also \(X_s=0\) almost surely. Therefore \(\mathbb{E}[X_t\mid\mathcal{F}_s]=X_s\) almost surely, and \(X\) is a martingale under the definition at each pair of fixed times. The identically zero process \(\widetilde X_t=0\) is a cadlag modification of \(X\).

For every outcome with \(U\in[1,2]\), the path \(t\mapsto\mathbf{1}_{\{t=U\}}\) takes value \(1\) at \(t=U\) and value \(0\) at times immediately to its right. It is not right-continuous at \(U\). Consequently, \(X\) and \(\widetilde X\) are not indistinguishable: their paths differ at the random time \(U\) for every outcome. There is no contradiction with the theorem, which promises a cadlag modification, not cadlag paths for the original process.

Check: for any particular \(t\), the event \(\{U=t\}\) has probability zero. But the event that the two processes agree at every time is empty, since every path has a spike at its own time \(U\). Fixed-time almost-sure equality does not imply equality simultaneously over an uncountable time set.

Using the Regular Version Correctly

After choosing a cadlag modification, continuous-time arguments can be formulated using its paths. For example, the value just after a deterministic time is determined by right-continuity, and values on a countable dense time set determine the full path. The proposition above makes these statements independent of the choice of cadlag modification, up to one null event.

There is also a useful preservation fact. If \(M\) is a martingale and \(\widetilde M\) is an indistinguishable adapted process, then \(\widetilde M\) is a martingale. Indeed, for each fixed \(t\), indistinguishability implies \(\widetilde M_t=M_t\) almost surely, so \(\widetilde M_t\) is integrable. For fixed \(s\le t\), conditional expectations preserve almost-sure equality, giving \(\mathbb{E}[\widetilde M_t\mid\mathcal{F}_s]=\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s=\widetilde M_s\) almost surely. Adaptation is assumed in this statement; it should not be omitted when replacing a process.

Common pitfall. A tempting but invalid argument is: “For every \(t\), \(M_t=\widetilde M_t\) almost surely; therefore, outside one null set, the paths agree for every \(t\).” The first assertion gives events \(A_t=\{M_t=\widetilde M_t\}\), each of probability one. Their intersection over uncountably many \(t\) need not have probability one. The random-time spike example gives the correction: equality at every fixed time holds almost surely, but simultaneous path equality fails. The cadlag uniqueness proposition succeeds because it first intersects over only countably many rational times and then uses right-continuity to extend equality to all times.

In applications, keep three claims separate: a process is a martingale at each fixed pair of times; it has a cadlag modification under the theorem's hypotheses; and that modification is indistinguishable from another cadlag modification. Only the last claim provides simultaneous equality of paths, and it requires the path regularity used in the uniqueness proof.

Check Your Understanding

For each question, distinguish carefully between fixed-time statements and statements about all times at once.

State the two usual conditions on the filtration and the conclusion of the martingale regularization theorem.
Can a martingale with a cadlag modification have non-cadlag original paths? Explain using the random-time spike construction.
Why does equality at every nonnegative rational time imply equality at every time for two cadlag processes?
In the deterministic information-jump example, verify the conditional-expectation identity when \(s<1\le t\).
What is the difference between two processes being modifications of one another and being indistinguishable?