Difference between weak ( or martingale ) and strong solutions to SDEs

Hi Im fairly new to SDE theory and am struggling with the difference between a weak ( or martingale ) solution and a strong solution to an SDE :

$$d(X_{t})=b(t,X_{t})dt + \sigma(t,X_{t})dW_{t}$$

Are these two differences and what do they really mean in detail?

1. For a strong solution we are given an initial value, whereas for weak solutions only a probability law?

2. For strong solutions we know what probability space we are working in and have a Brownian Motion $$W$$ in that space. For a weak solution we can only say that there exists some probability space where the SDE holds (with a new brownian motion in the space).

As you can tell I am confused with this topic some clarifications would be amazing.

The main difference between weak and strong solutions is indeed that for strong solutions we are given a Brownian motion on a given probability space whereas for weak solutions we are free to choose the Brownian motion and the probability space.

Definition: Let $$(B_t)_{t \geq 0}$$ be a Brownian motion with admissible filtration $$(\mathcal{F}_t)_{t \geq 0}$$. A progressively measurable process $$(X_t,\mathcal{F}_t)$$ is a strong solution with initial condition $$\xi$$ if $$X_t-X_0 = \int_0^t \sigma(s,X_s) \, dB_s + \int_0^t b(s,X_s) \, ds, \qquad X_0 =\xi \tag{1}$$ holds almost surely for all $$t \geq 0$$.

Definition: A stochastic process $$(X_t,\mathcal{F}_t)$$ on some probability space $$(\Omega,\mathcal{F},\mathbb{P})$$ is called a weak solution with initial distribution $$\mu$$ if there exists a Brownian motion $$(B_t)_{t \geq 0}$$ on $$(\Omega,\mathcal{F},\mathbb{P})$$ such that $$(\mathcal{F}_t)_{t \geq 0}$$ is an admissible filtration, $$\mathbb{P}(X_0 \in \cdot) = \mu(\cdot)$$ and $$X_t-X_0 = \int_0^t \sigma(s,X_s) \, dB_s + \int_0^t b(s,X_s) \, ds$$ holds almost surely for all $$t \geq 0$$.

As a consequence of these definitions, we have to consider different notions of uniqueness. For strong solutions we are typically looking for pathwise unique solutions, i.e. if $$(X_t^{(1)})_{t \geq 0}$$ and $$(X_t^{(2)})_{t \geq 0}$$ are strong solutions to $$(1)$$ with the same initial condition, then pathwise uniqueness means $$\mathbb{P} \left( \sup_{t \geq 0} |X_t^{(1)}-X_t^{(2)}|=0 \right)=1.$$ As the following simple example shows it doesn't make sense to talk about pathwise uniqueness of weak solutions.

Example 1: Let $$(W_t^{(1)})_{t \geq 0}$$ and $$(W_t^{(2)})_{t \geq 0}$$ be two Brownian motions (possibly defined on different probability spaces), then both $$X_t^{(1)} := W_t^{(1)}$$ and $$X_t^{(2)} := W_t^{(2)}$$ are weak solutions to the SDE $$dX_t = dB_t, \qquad X_0 = 0$$ Why? According to the definition we are free choose the driving Brownian motion, so we can set $$B_t^{(1)} := W_t^{(1)}$$ and $$B_t^{(2)} := W_t^{(2)}$$, respectively, and then $$dX_t^{(i)} = dB_t^{(i)} \quad \text{for i=1,2}.$$

What do we learn from this? Since weak solutions might be defined on different probability spaces, there is no (immediate) way to compute probabilities of the form $$\mathbb{P}(X_t^{(1)}=X_t^{(2)})$$ for two weak solutions $$(X_t^{(1)})_{t \geq 0}$$ and $$(X_t^{(2)})_{t \geq 0}$$, and therefore we cannot even attempt to talk about pathwise uniqueness. For the same reason, it doesn't make sense to talk about pointwise initial conditions $$\xi$$ for weak solutions (... for this we would need to fix some probability space on which $$\xi$$ lives...); instead we only prescribe the initial distribution of $$X_0$$.

The next example shows that we cannot expect to have pathwise uniqueness even if the weak solutions are defined on the same probability space.

Example 2: Let $$(W_t)_{t \geq 0}$$ be a Brownian motion. It follows from Example 1 that $$X_t^{(1)} := W_t$$ and $$X_t^{(2)} := -W_t$$ are weak solutions to the SDE $$dX_t = dB_t, \qquad X_0 =0.$$ Clearly, $$\mathbb{P}(X_t^{(1)} = X_t^{(2)}) = \mathbb{P}(W_t=0)=0$$.

The "good" notion of uniqueness for weak solutions is weak uniqueness, i.e. uniqueness in distribution (= the solutions have the same finite-dimensional distributions).

Typically it is much easier to prove the existence (and/or uniqueness of) a weak solution the the existence (and/or uniqueness) of a strong solution.

Example 3: The SDE $$dX_t = - \text{sgn}\,(X_t) \, dB_t, \qquad X_0 = 0 \tag{2}$$ has a weak solution but no strong solution.

Let's prove that the SDE has a weak solution. Let $$(X_t,\mathcal{F}_t)_{t \geq 0}$$ be some Brownian motion and define $$W_t := -\int_0^t \text{sgn} \, (X_s) \, dX_s.$$ It follows from Lévy's characterization that $$(W_t,\mathcal{F}_t)$$ is also a Brownian motion. Since $$dW_t = - \text{sgn} \, (X_t) \, dX_t$$ implies $$dX_t = - \text{sgn} \, (X_t) \, dW_t$$ this means that $$(X_t)_{t \geq 0}$$ is a weak solution to $$(2)$$. For a proof that a strong solution does not exist see e.g. Example 19.16 in the book by Schilling & Partzsch on Brownian motion.

Let me finally mention that weak solutions are closely related to martingale problems; in this answer I tried to give some insights on the connection between the two notions.

• Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way. Nov 13, 2018 at 8:47
• @Monty Thank you; I fixed it.
– saz
Nov 13, 2018 at 8:54
• I've seen that some authors use the notions of weak and strong solutions in a slightly different way. Sometimes $X$ is only called a strong solution if it is adapted with respect to the filtration generated by $B$. And usually the probability space and filtration are part of a weak solution (If I understand your definiton correctly, you assume that at least the probability space is given beforehand). Feb 21, 2019 at 21:38
• Oh, and are you implicitly assuming the filtrations to be complete? Otherwise, there might be a problem in defininig the stochastic integral as a local martingale. (Just asking cause I'm currently trying to figure out how the notions of weak/storng solutions are commonly used in the literature.) Feb 21, 2019 at 21:45
• @0xbadf00d If you have (deterministic) Lipschitz coefficients and a deterministic initial condition, then the solution is measurable wrt to the canonical filtration... that's what they show there. It doesn't contradict what I was saying..
– saz
Feb 22, 2019 at 12:33