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Consider the stochastic differential equation $$ dX_t = b(t,X_t)\,dt + \sigma(t,X_t) \, dB_t, \quad t \geq 0. $$ where $b:[0, \infty) \times \mathbb{R}^n \to \mathbb{R}^n$, $\sigma: [0,\infty) \times \mathbb{R}^n \to \mathbb{R}^{n \times d}$ are measurable and $B$ is a standard Brownian Motion in $\mathbb{R}^d$.

In dimension $n=1$, it is a well-known result by Yamada and Watanabe that for pathwise uniqueness, we do not need both $b$ and $\sigma$ to be uniformly Lipschitz continuous in the second variable (the usual condition in arbitrary dimensions), but it suffices for $b$ to be Lipschitz and $\sigma$ to only be Hölder-$\frac{1}{2}$-continuous. Actually, this is a special case (insert $\kappa(u) = L|u|$ and $\rho(u) = C|u|^{1/2}$ to recover it) of the slightly more general

Theorem. Assume $\exists \gamma >0$ a constant and $\kappa, \rho:[0, \gamma] \to [0, \infty)$ with \begin{align}\kappa(0) = 0, \quad |b(t,x)-b(t,y)| &\leq \kappa(|x-y|), \\ | \sigma(t,x) - \sigma(t,y) | &\leq \rho(|x-y|) \end{align} for all $t \in [0, \infty), x,y \in \mathbb{R}^n $ with $ |x-y| \leq \gamma$.

Assume $\rho$ and $\kappa$ are non-decreasing and strictly positive on $(0, \gamma]$, $\kappa$ is concave, and $$\int_0^{\gamma} \frac{du}{\kappa(u)} = \infty = \int_0^{\gamma} \frac{du}{\rho^2(u)}. $$

Then for every initial distribution $\mu$, pathwise uniqueness of the SDE holds.

The theorem in this (and even more general) form can for example be found at the beginning of Altay and Schmock.


In general dimensions $n \geq 3$ ($n=2$ too by Swart), it was proved by Yamada and Watanabe in their second(!) paper from 1971 that this is not true. In general, $\rho$ has to satisfy (some technical conditions and) $\int_0^{\gamma} \frac{u}{\rho(u)^2} du = \infty$, which again is nearly Lipschitz continuity. If not, so if $$\int_0^{\gamma} \frac{u}{\rho(u)^2} du < \infty$$ and $\rho$ is subadditive, then pathwise uniqueness (and also uniqueness in law) does not hold. This is fulfilled by $$\rho(u) = |u|^{\alpha}, \quad \alpha <1.$$

Now this raises the

Question. Why does (essentially) Hölder-$\frac{1}{2}$ continuity not suffice in the multidimensional case?

Sure, Yamada and Watanabe provide a counterexample, but for me it is not intuitively clear why it only works in $n \geq 2$ dimensions.

Also, I thought that the Hölder-$\frac{1}{2}$ continuity in one dimension is essentially due to that of the driving Brownian Motion (locally for every $\beta < \frac{1}{2}$). Is that true, and why does this fail for $n \geq 2$?

Alex
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1 Answers1

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Sure, Yamada and Watanabe provide a counterexample, but for me it is not intuitively clear why it only works in $n \geq 2$ dimensions.

I will go over their example for $n\geq 3$. (For $n=2$, they suggest using a similar example by further inserting $\log(1/\xi)$).

We go over the counterexample by Yamada-Watanabe in On the uniqueness of solutions of stochastic differential equations II. Consider function $\rho:\mathbb{R}^{+}\to \mathbb{R}^{+}$ satisfying $\rho(0)=0$

$$\int_{0}^{\epsilon}\rho^{-2}(\xi)\xi d \xi<\infty\text{ and }\rho(\xi_1+\xi_2)\leq \rho(\xi_1)+\rho(\xi_2).$$

So their counterexample is quite generic because there are many such functions $\rho$.

Fix $n\geq 3$. Then we let for $x\in \mathbb{R}^{n}$

$$\sigma_{i,j}(t,x):=\delta_{ij}\rho(|x|),i,j=1,...,n.$$

By the above subadditivity this implies $$|\sigma_{i,j}(t,x)-\sigma_{i,j}(t,y)|\leq \rho(|x-y|).$$

The claim is that $$\left\{\begin{matrix}dx_{t}=\sigma(x_{t})dB_{t}\\x_{0}=0 \end{matrix}\right.$$

has multiple solutions. First it has the zero solution. But by the Dubins-Schwartz, we will obtain yet another nonzero solution.

Consider an independent Brownian motion $\beta_{s}$. We consider the time change

$$A_{t}:=\int_{0}^{t}\rho^{-2}(|\beta_{s}|)ds.$$

This defines a continuous functional: the density of $Y=|\beta_{s}|$ is What's the expectation of square root of Chi-square variable?

$$f_Y(\xi)=\frac{1}{s^{n/2}}\frac{\xi^{n-1} e^{-\xi^2/2s}}{2^{n/2-1} \Gamma\left(\frac{n}{2}\right)}\qquad \xi>0,$$

and so

$$E[A_{t}]=\omega_{n}\int_{0}^{\infty}\rho^{-2}(\xi)\left[\int_{0}^{t}\frac{1}{(2\pi s)^{n/2}}e^{-\frac{\xi^{2}}{2s}}ds\right] \xi^{n-1}d\xi.$$

The inner integral evaluates to

$$\int_{0}^{t}\frac{1}{(2\pi s)^{n/2}}e^{-\frac{\xi^{2}}{2s}}ds=c\xi^{-(n-2)}\Gamma(\frac{n}{2}-1,\frac{\xi^{2}}{2t})=c\xi^{-(n-2)}\int_{\frac{\xi^{2}}{2t}}^{\infty}y^{\frac{n}{2}-2}e^{-y}dy.$$

So we see in the second integral that we need $n\geq 3$ to get it to be finite (i.e. in n=3 we get

$$\int_{\frac{\xi^{2}}{2t}}^{\infty}y^{-1/2}e^{-y}dy$$

which is still finite even if $\frac{\xi^{2}}{2t}\approx 0$.) From here we proceed as in their proof to get boundedness

$$E[A_{t}]\leq c_{n}\int_{0}^{\infty}\rho^{-2}(\xi) \xi d\xi<\infty.$$

So

$$B_{t}:=\int_{0}^{t}\sigma^{-1}(\beta_{A_{t}^{-1}})d\beta_{A_{t}^{-1}}$$

is a Brownian motion and $\beta_{A_{t}^{-1}}$ will be a nonzero solution of the above SDE (since $A_{t}^{-1}\neq 0$ to $t>0$).

Why does Hölder-$(\frac{1}{2})^{-}$-continuity not suffice in the multidimensional case?

First to be clear there have been some extensions to Hölder-regular coefficients eg some highly cited works are

In Revuz-Yor in "The Case of Hölder Coefficients in Dimension One", they give some intuition on what is special about 1-d.

  • They use that local time in 1-d is a uniformly continuous process. (So here we are touching the deeper phenomenon of Brownian motion being recurrent in 1d, neighboorhood-reccurent in 2d and transient in $n\geq 3$).

  • They mention that even though ODEs with Hölder Coefficients can fail to have uniqueness eg. $y'=\sqrt{y}$ Uniqueness of ODEs, here we can have uniqueness because we define SDEs and Itô-integrals in terms of the quadratic variation which we know is finite and so it relaxes the integrability constraints.

More recently we also had a notion of SDEs using Rough paths that helped defined stochastic integrals pathwise Could someone explain rough path theory? More specifically, what is the higher ordered "area process" and what information is it giving us?.

Thomas Kojar
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  • Thank you for your explanation! This really sheds some light onto the counterexample. Do you also know something about the intuition of why Hölder-$\frac{1}{2}$ suffices in 1 dimension but not more? (So the third paragraph of the question) – Alex Dec 14 '23 at 09:46
  • @Alex well the above $\rho$ in the counterexample can be taken to be Hölder too. So I am not sure what you are asking. Yes in d=1 one can work with Hölder-coefficients and in higher dimensions the above provides a counterexample for n\geq 3 (for n=2 you can adjust their example as they mention). – Thomas Kojar Dec 14 '23 at 17:40
  • Then please excuse my unclear phrasing :) I meant now just the intuitive side in gerenal, not the counterexample. Like, I thought that (what only holds in 1d) Hölder suffices due to only having to match the integrator Brownian Motion. This is not true. So there must (hopefully) be a different "intuitive" reason. – Alex Dec 14 '23 at 19:18
  • @Alex In Revuz-Yor in "The Case of Hölder Coefficients in Dimension One", they give some intuition on what is special about 1-d. – Thomas Kojar Dec 14 '23 at 20:08
  • Thank you very much for the reference and added details in your answer! I will read it after the Christmas break and come back to you. Merry Christmas! – Alex Dec 20 '23 at 15:13