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!


1 Answer 1


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 $X_{t}$.

  • 3
    $\begingroup$ 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. $\endgroup$ Commented Jul 22, 2022 at 5:34
  • $\begingroup$ 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. $\endgroup$ Commented Jul 22, 2022 at 12:34
  • $\begingroup$ 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 $t > 0$. $\endgroup$
    – qp212223
    Commented Jul 22, 2022 at 20:42
  • $\begingroup$ 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 $$\mathcal{F}_t^X \not \subseteq \mathcal{F}_t^{|X|} \quad ?$$ $\endgroup$
    – qp212223
    Commented Jul 22, 2022 at 20:44
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    $\begingroup$ @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 $[1,2] \cup [-5,-3]$. $\endgroup$ Commented Jul 22, 2022 at 21:58

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