advanced30 min studyLesson 674 in Mathematics

Strong and weak solutions

Separate a solution built from given noise from a solution specified jointly with its probability space.

On this page

What you’ll understand

  • State what is fixed in strong and weak existence questions.
  • Distinguish pathwise uniqueness from uniqueness in law.
  • Verify a solution formula and its measurability with respect to the driving noise.

An SDE is an integral equation together with information and probability requirements. Its coefficient symbols alone do not specify whether the noise and probability space are given in advance. Strong and weak solutions answer different existence questions, while pathwise uniqueness and uniqueness in law answer different comparison questions. The adjectives describe these probabilistic structures; a weak solution of an SDE is not the same concept as a distributional weak solution of a partial differential equation.

A strong solution uses prescribed randomness

Given an initial variable and a Brownian motion on a filtered probability space satisfying the usual conditions, a strong solution is constructed as a nonanticipating measurable function of that initial data and the driving Brownian path. In particular, it is adapted to their completed generated filtration. We assume the initial variable is independent of future Brownian increments. The integral equation holds almost surely for every time, using continuous versions so that equality at a countable dense set extends to all times.

A process merely adapted to some larger filtration containing extra independent randomness is not automatically strong in this sense. That extra randomness may be permitted in a weak solution, but the strong existence question asks whether it is necessary. For deterministic coefficients and deterministic initial state, an explicit expression involving only past Brownian values or stochastic integrals gives a natural route to checking strong existence. It still needs to satisfy the equation and its integrability requirements, rather than merely look nonanticipating.

Xt=X0+∫0tb(s,Xs) ds+∫0tσ(s,Xs) dWs,X=Φ(X0,W) nonanticipativelyX_t=X_0+\int_0^t b(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dW_s,\qquad X=\Phi(X_0,W)\text{ nonanticipatively}
Strong existence asks for a solution from the prescribed initial data and Brownian driver.↔ Scroll wide equations horizontally.

A weak solution includes its probability space

A weak solution consists of a probability space, a suitable filtration, a Brownian motion for that filtration, and an adapted continuous process satisfying the same integral equation. These objects are chosen together. Weak existence can therefore hold even when no solution can be constructed as a function of a specified Brownian driver alone. Changes of measure and martingale problems are useful weak-existence tools because they construct a joint probabilistic system without first supplying a pathwise solution map.

Uniqueness in law means that any two weak solutions with the same prescribed initial distribution have the same distribution as processes on continuous-path space. This is stronger than having equal one-time marginals but weaker than saying two coupled sample paths coincide. Pathwise uniqueness compares two solutions on the same space with the same Brownian driver and the same initial variable, and requires them to be indistinguishable. The shared driver is essential; two independent solutions of a well-posed equation need not be equal.

pathwise uniqueness: P(Xt=Yt for every t≥0)=1for the same X0,W\text{pathwise uniqueness: }\mathbb P(X_t=Y_t\text{ for every }t\geq0)=1\quad\text{for the same }X_0,W
The comparison is made under a common coupling, rather than only between marginal distributions.↔ Scroll wide equations horizontally.

Existence and uniqueness interact

For standard finite-dimensional Itô SDEs with Borel coefficients, the Yamada–Watanabe principle says that weak existence together with pathwise uniqueness yields strong existence and uniqueness in law, under the usual solution framework. This is a theorem connecting two separately established properties, not a definition that makes them interchangeable. Discontinuous or degenerate coefficients can exhibit distinctions invisible in the globally Lipschitz setting, so conclusions should be matched to the precise hypotheses actually proved.

When comparing candidate formulas, first identify whether they use the same driver and initial value. Next check the integral identity. Finally decide whether the desired conclusion concerns a measurable construction, a coupling identity, or a law on path space. This order avoids a frequent mistake: observing that two Gaussian variables share a distribution and treating that observation as a proof that they agree almost surely. Equal laws allow many couplings, including independent and negatively correlated ones.

An explicit mean-reverting strong solution

For deterministic xx, verify that Xt=e−3tx+2∫0te−3(t−s) dWsX_t=e^{-3t}x+2\int_0^t e^{-3(t-s)}\,dW_s is a strong solution of dXt=−3Xt dt+2 dWtdX_t=-3X_t\,dt+2\,dW_t. Find its mean and variance.

  1. Rewrite the expression as e−3t(x+2∫0te3s dWs)e^{-3t}(x+2\int_0^t e^{3s}\,dW_s). The integrand is deterministic and square integrable on every finite horizon, so the process uses only the given Brownian past.

  2. The ordinary product rule with the finite-variation factor gives drift −3Xt-3X_t and Brownian coefficient two. The initial value is xx, confirming the integral equation and strong construction.

  3. The stochastic term has mean zero. Its variance is 4∫0te−6(t−s)ds=23(1−e−6t)4\int_0^t e^{-6(t-s)}ds=\tfrac23(1-e^{-6t}). Thus the mean is e−3txe^{-3t}x, with a Gaussian law at each time because the integrand is deterministic.

The result

The formula establishes strong existence and gives the complete one-time Gaussian law. A separate common-driver comparison proves pathwise uniqueness in the practice problem.

Explore the model

Interactive lab

Interactive lab · probability

Equal laws can hide different couplings

Pause/Reset, then set both probability to 0.5 and trials to 20. Predict that K and 20−K have the same binomial bars. On the same outcome define X=2K−20 and Y=−X: they have equal distributions but usually differ. Reset and set both probability to 0.25 and trials to 20; the reflection symmetry no longer gives the same law.

Equal laws can hide different couplingsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe exact finite distribution illustrates the difference between a law and a coupling. It does not construct an SDE solution or test strong existence.

Check your understanding

What is the correct comparison in pathwise uniqueness?

Try it yourself

Let X and Y solve the worked mean-reverting equation with the same Brownian motion but initial values x and y. Determine their difference, then state what happens when the initial values are equal. Does the argument compare independent drivers?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Subtract the two integral equations before introducing any distributional calculation.

Compare with the solution

The Brownian integrals cancel under the common driver. The difference solves Dt=x−y−3∫0tDsdsD_t=x-y-3\int_0^tD_sds, hence Dt=e−3t(x−y)D_t=e^{-3t}(x-y). When x=yx=y it is identically zero almost surely, establishing pathwise uniqueness for this equation. With independent drivers an additional noise difference remains, so the cancellation and the resulting identity do not apply. Equal one-time laws of independently driven copies would not change that fact.

Take it with you

  • Strong existence fixes the driver; weak existence chooses a full probabilistic setup.
  • Pathwise uniqueness uses a common coupling, while uniqueness in law compares path distributions.
  • An explicit law does not by itself establish a strong solution map.

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