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.


1 Answer 1


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.

  • $\begingroup$ Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way. $\endgroup$
    – Monty
    Nov 13, 2018 at 8:47
  • $\begingroup$ @Monty Thank you; I fixed it. $\endgroup$
    – saz
    Nov 13, 2018 at 8:54
  • $\begingroup$ 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). $\endgroup$
    – 0xbadf00d
    Feb 21, 2019 at 21:38
  • $\begingroup$ 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.) $\endgroup$
    – 0xbadf00d
    Feb 21, 2019 at 21:45
  • 1
    $\begingroup$ @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.. $\endgroup$
    – saz
    Feb 22, 2019 at 12:33

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