Two Different Meanings of Equality
For stochastic processes, the statement “the two processes are equal almost surely” needs a time qualification. At a fixed time \(t\), the equality \(X_t=Y_t\) is an equality of random variables and may hold outside a null set that depends on \(t\). Indistinguishability is stronger: outside one null set, the paths agree at every time. The distinction matters in stochastic differential equations, where solutions are continuous processes and uniqueness is ultimately a statement about whole paths.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration. The discussion below concerns processes on this space; it does not require a stochastic integral. For a fixed finite horizon \(T>0\), assume \(X=(X_t)_{0\le t\le T}\) and \(Y=(Y_t)_{0\le t\le T}\) are real-valued processes. When we assume continuous paths, this means that there is an event of probability one on which the entire path \(t\mapsto X_t\), respectively \(t\mapsto Y_t\), is continuous.
In the first definition, the exceptional null set may vary with \(t\). In the second, one event must work for all times simultaneously. Indistinguishability implies equality almost surely at each fixed time: for any fixed \(t\), the event \(A\) is contained in \(\{\omega:X_t(\omega)=Y_t(\omega)\}\). The converse is not true for arbitrary processes.
Why Continuity and a Countable Dense Set Are Enough
Proof. For each \(q\in D\), let \(A_q=\{\omega:X_q(\omega)=Y_q(\omega)\}\). By assumption, \(\mathbb{P}(A_q)=1\). Let \(C_X\) and \(C_Y\) be probability-one events on which the respective paths are continuous. Because \(D\) is countable, the event
has probability one: its complement is a countable union of probability-zero events. Fix \(\omega\in A\). The two paths are continuous and agree at every point of \(D\). For any \(t\in[0,T]\), density gives a sequence \(q_n\in D\) with \(q_n\to t\). Continuity then gives \(X_t(\omega)=\lim_{n\to\infty}X_{q_n}(\omega)=\lim_{n\to\infty}Y_{q_n}(\omega)=Y_t(\omega)\). Thus the same probability-one event \(A\) works for every \(t\), which is indistinguishability. If equality is known at every fixed time, it is in particular known at every \(q\in D\), so the result applies. \(\square\)
The proof uses two separate facts. Countability lets us intersect the events of equality without losing probability one. Continuity then extends equality from the dense set to all times. Neither step can simply be omitted: an uncountable intersection of probability-one events need not have probability one, and agreement on a dense set need not control a discontinuous path.
Worked Example: Upgrading Rational-Time Equality
Suppose \(X\) and \(Y\) are continuous adapted processes on \([0,1]\), and a prior estimate gives \(\mathbb{E}[|X_q-Y_q|^2]=0\) for every rational \(q\in[0,1]\). Show that the processes are indistinguishable.
Fix one such \(q\), and put \(Z=|X_q-Y_q|^2\). This is nonnegative and has expectation zero. For each positive integer \(n\), Markov's inequality gives \(\mathbb{P}(Z\ge 1/n)\le n\mathbb{E}[Z]=0\). Since \(\{Z>0\}=\bigcup_{n=1}^{\infty}\{Z\ge1/n\}\), countable subadditivity yields \(\mathbb{P}(Z>0)=0\). Therefore \(\mathbb{P}(X_q=Y_q)=1\) for each rational \(q\in[0,1]\).
The rationals in \([0,1]\) are countable and dense, and both processes have continuous paths almost surely. The theorem therefore gives one probability-one event on which \(X_t=Y_t\) for every \(t\in[0,1]\). Check: the conclusion is stronger than the input at any single time: the input gives a possibly different exceptional null set for each rational, while the proof takes only a countable intersection and then uses continuity to cover all real times.
Continuous Versions and SDE Uniqueness
Here \(X\) is a modification of a process \(Z\) if \(\mathbb{P}(X_t=Z_t)=1\) for each fixed \(t\). The corollary says that continuous modifications cannot differ in their whole-path behavior.
Proof. Fix \(t\). Since \(X_t=Z_t\) almost surely and \(Y_t=Z_t\) almost surely, the intersection of those two probability-one events also has probability one. On that intersection \(X_t=Y_t\), so \(\mathbb{P}(X_t=Y_t)=1\). This holds for every fixed \(t\). The theorem applied to \(X\) and \(Y\) now gives indistinguishability. \(\square\)
This distinction clarifies the usual pathwise formulation of uniqueness for SDEs. On a common filtered probability space, two candidate solutions are driven by the same Brownian motion and have the same initial condition. If a uniqueness argument proves \(X_t=Y_t\) almost surely for each fixed \(t\), and both solutions have almost surely continuous paths, the theorem upgrades that conclusion to indistinguishability. Thus one must check the continuity hypothesis when passing from fixed-time equalities to a whole-path uniqueness statement. This is a statement about solutions on the same space and noise; it does not by itself identify it with uniqueness in law, which compares probability distributions.
Worked Example: Two Continuous Processes with a Null Exceptional Initial Value
Let \(U\) be uniform on \([0,1]\), independent of \(B\), and include \(U\) in \(\mathcal{F}_0\). Take the filtration generated by \(U\) and the Brownian motion up to the current time, with the usual completion. Define \(A=\mathbf{1}_{\{U<1/2\}}\), \(C=\mathbf{1}_{\{U\le1/2\}}\), and, for \(0\le t\le1\), set \(X_t=tA\) and \(Y_t=tC\).
Both processes are adapted and have continuous paths because each is a random constant multiplied by \(t\). At \(t=0\), \(X_0=Y_0=0\) for every outcome. For fixed \(t>0\), their difference is \(X_t-Y_t=t(A-C)\). The indicators differ exactly when \(U=1/2\), and a uniform random variable assigns probability zero to a singleton: \(\mathbb{P}(U=1/2)=0\). Hence \(\mathbb{P}(X_t=Y_t)=1\) for every fixed \(t\).
For any outcome with \(U\ne1/2\), \(A=C\), so the paths agree for every \(t\). The event \(\{U\ne1/2\}\) has probability one, and therefore the processes are indistinguishable. Check: at \(t=1\), the difference is \(A-C\), which is nonzero only at \(U=1/2\); the same exceptional event is the only possible pathwise discrepancy at any time in \([0,1]\).
Why the Claim Fails Without Continuity
Worked Example: A Random-Time Spike
Let \(T>0\), let \(U\) be uniform on \([0,T]\), and define \(X_t=\mathbf{1}_{\{U=t\}}\) and \(Y_t=0\) for \(0\le t\le T\). One may place \(U\) in \(\mathcal{F}_0\) and take it independent of the Brownian motion, so that \(B\) remains Brownian relative to the filtration enlarged by \(U\). The process is measurable: the set of pairs \((t,u)\) with \(u=t\) is a Borel subset of \([0,T]^2\).
For any fixed \(t\in[0,T]\), \(X_t=1\) exactly on the event \(\{U=t\}\). Since \(U\) is uniform, \(\mathbb{P}(U=t)=0\), including at the endpoints. Thus \(\mathbb{P}(X_t=Y_t)=1\) for every fixed \(t\). But for every outcome \(\omega\), taking \(t=U(\omega)\) gives \(X_t(\omega)=1\ne0=Y_t(\omega)\). Consequently there is no outcome on which the paths agree at every time, and the processes are not indistinguishable. Check: the spike has height \(1\) at its random time, so \(\sup_{0\le t\le T}|X_t-Y_t|=1\) for every outcome, even though each fixed-time difference is zero almost surely. The paths are not continuous.
Common Mistake and Practical Test
A common but invalid calculation is to write, “For every \(t\), \(\mathbb{P}(X_t=Y_t)=1\), so \(\mathbb{P}(X_t=Y_t\text{ for all }t)=1\).” This treats an uncountable intersection of probability-one events as though it were a countable one. The random-time-spike example disproves the conclusion: every fixed-time equality holds almost surely, but every path has a time at which equality fails.
The correction is to identify what extra structure is available. If the processes have almost surely continuous paths, it is enough to verify equality almost surely at a countable dense set, such as the rational times in a finite interval. Intersect the corresponding countably many probability-one events, and then use continuity. If continuity is absent, fixed-time equality alone does not justify a whole-path conclusion.
| Statement | Exceptional set | What it implies |
|---|---|---|
| \(\mathbb{P}(X_t=Y_t)=1\) for each fixed \(t\) | May depend on \(t\) | Equality of the random variables at each chosen time |
| Equality almost surely on a countable dense set, with continuous paths | One countable union of null sets | Indistinguishability |
| Indistinguishability | One null set for all times | Equality of paths outside that set |
Check Your Understanding
For each question, specify whether continuity is needed and whether the exceptional event may depend on time.
- Why does equality almost surely at every rational time not, by itself, imply indistinguishability for arbitrary processes?
- State the countable-dense-set argument that upgrades equality to indistinguishability when both processes have almost surely continuous paths.
- If \(X\) and \(Y\) are continuous modifications of the same process, why does \(\mathbb{P}(X_t=Y_t)=1\) hold for each fixed \(t\)?
- For the random-time-spike example, compute \(\mathbb{P}(X_t\ne0)\) at a fixed \(t\), and compare it with \(\sup_{0\le s\le T}|X_s|\).
- In a pathwise uniqueness argument for continuous SDE solutions, what additional step is needed if the estimate first proves equality only at each fixed time?