# On Tanaka's SDE

Throughout this question, for an arbitrary real valued continuous time stochastic process $$Y$$, I define $$\{\mathcal{F}_t^Y\}_{t \ge 0}$$ to be the filtration generated by $$Y$$. I also define $$\text{sgn}(x) = \mathbf{1}(x > 0) - \mathbf{1}(x \leq 0)$$

The standard argument that Tanaka's SDE has no strong solutions for a particular probability space is as follows, but I do not understand the last part of it: it suffices to consider a probability space $$(\Omega, \mathcal{F}, P)$$ on which there exists a Brownian motion $$B$$, and we define our filtration $$\{\mathcal{F}_t\}_{t \ge 0}$$ to be that which is generated by $$B$$. If $$dX_t = \text{sgn}(X_t) dB_t$$ has a strong solution, then $$B_t = \int_0^t \text{sgn}(X_s)dX_s = |X_t| - L_t^0$$ where $$L_t^0$$ is the local time of $$X$$ about $$0$$ at time $$t$$. It is well known that $$L_t^0 = |X_t| - \lim_{h \rightarrow 0} \frac{1}{h}\text{Leb}\left( \{s \in [0,t] : \left|X_s \right| \leq h \right)$$ and thus $$\mathcal{F}_t^{X} \subseteq \mathcal{F}_t^B \subseteq \mathcal{F}_t^{|X|}$$ which is a contradiction, since $$\text{sgn}(X_t) \text{ cannot possibly be } \mathcal{F}_t^{|X|} \text{ measurable.} \qquad \qquad \textbf{(1)}$$ Intuitively, $$\textbf{(1)}$$ is true, but how does one rigorously go about showing this?

I try to do so by contradiction, but it leads me nowhere after a bit. If $$\text{sgn}(X_t)$$ were $$\mathcal{F}_t^{|X|}$$ measurable, then there would exist some functional $$f$$ defined on the space of continuous functions on $$[0,t]$$ such that $$\text{sgn}(X_t) = f(|X_s|, s \leq t)$$ I know this is intuitively obvious to be a contradiction, but I cannot formally show it. Any help would be appreciated!

So my original answer did not take into account the fact that there could exist a transformation of the entire path $$(|X_{s}|)_{0\leq s\leq t}$$. Instead, I think I have found a way to prove that for a Brownian motion $$X$$, we cannot have the inclusion $$\mathcal{F}_{t}^{X}\subseteq\mathcal{F}_{t}^{|X|}$$. Since the solution to Tanaka's SDE is a Brownian motion, this will suffice.

Define the $$\sigma$$-algebra $$\mathcal{D}_{t} = \lbrace F\in \mathcal{F}_{t}^{|X|}\mid \mathbb{E}\left[ X_{t}\mathbb{1}_{F} \right] = 0 \rbrace$$ That this is in fact a $$\sigma$$-algebra follows by linearity of the integral and that $$X_{t}\in L^{1}$$ with expectation 0.

Let $$B\in\mathcal{B}(\mathbb{R})$$ and $$s\leq t$$. Then $$\lbrace |X_{s}|\in B\rbrace \in \mathcal{D}_{t}$$ since by the reflection principle $$\mathbb{E}\left[ X_{t}\mathbb{1}_{\lbrace |X_{s}|\in B\rbrace } \right] = 0$$ This means that $$\mathcal{G}_{t}=\mathcal{F}_{t}^{|X|}$$, at least if we assume the filtration $$\mathcal{F}_{t}^{|X|}$$ to be complete.

But since $$\mathbb{E}\left[ X_{t}\mathbb{1}_{\lbrace X_{t}>0\rbrace} \right] > 0$$ we have $$\lbrace X_{t}>0\rbrace\notin\mathcal{F}_{t}^{|X|}$$, but naturally we have $$\lbrace X_{t}>0 \rbrace\in\mathcal{F}_{t}^{X}$$ and thus we must have $$\mathcal{F}_{t}^{X}\not\subseteq\mathcal{F}_{t}^{|X|}$$ as desired.

This lines up with the intuition that knowing something about $$|X_{t}|$$ does not tell us about the sign of $$X_{t}$$.

• It's obvious that there is no $f$ such that $f(|X_t|) = X_t$, but slightly less obvious there is no function of the entire path $(|X_s|)_{s \le t}$ that gives $X_t$. I agree that it's intuitively obvious, but I think the question asker was asking for a formal proof of that fact. Jul 22, 2022 at 5:34
• You are right, I did not think of the fact that the entire path should be considered. I have tried my hand at a new approach for the proof. Jul 22, 2022 at 12:34
• This is great, thanks. I think another way to show this is as follows: If $X_t = f(|X_s|, s \leq t)$ almost surely (on our original fixed probability space $(\Omega, \mathcal{F}, P)$), then we may define the pushforward measures for $X$ and $-X$ on the space of continuous functions, which must be equal since $X$ and $-X$ are equal in law. Hence $-X_t = f(|X_s|, s \leq t)$ almost surely as well, but this is an obvious contradiction for any $t > 0$. Jul 22, 2022 at 20:42
• Another question I have is, in the non-Brownian case with no additional moment assumptions/symmetry assumptions, how would we show this formally? For an arbitrary stochastic process $X$ with $P(X_t < 0) > 0$ for some $t \ge 0$, how does one show that $$\mathcal{F}_t^X \not \subseteq \mathcal{F}_t^{|X|} \quad ?$$ Jul 22, 2022 at 20:44
• @qp212223 I don't think that's true in general. For example, a deterministic process would have $\mathcal F_t^X = \mathcal F_t^{|X|}$. As another (slightly less trivial) example, you could have a process that only takes on values in $[1,2] \cup [-5,-3]$. Jul 22, 2022 at 21:58