Consider the SDE: $$dX_t = -\alpha \text{sgn}(X_t) dt + \sigma dW_t \quad \quad \alpha, \sigma > 0$$ where $\text{sgn}(x) = \mathbf{1}(x > 0) - \mathbf{1}(x \leq 0)$ and $W$ is standard Brownian motion. It is stated without proof in a textbook I am reading (Stochastic Portfolio Theory by Fernholz, p. 96) that for any weak solution, $p > 0$, $$ \textbf{(1)} \quad \lim_{t \rightarrow \infty} \frac{X_t^p}{t} = 0 \quad \text{a.s.}$$ I'd like to prove this, but I am hopelessly lost and would appreciate any help.

I tried yet another approach, and this one might be better:

I have managed to (partially) solve this - I just need some help bounding the expectation given below. I assume that $\exp (\alpha|X_0|/\sigma^2)$ is integrable for the sake of my sanity.

Fix $n \in \mathbb{N}, \epsilon > 0$ and let's try to compute $$p_n \equiv P \left( \sup_{n \leq t \leq n+1} \frac{|X_t|^p}{t} > \epsilon \right)$$

By Kolmogorov's consistency theorem, there exists a probability measure $Q$ such that $\forall A \in \mathcal{F}_t$, $$Q(A) = E_P \left(\mathbf{1}_A Z_t \right), \quad Z_t \equiv \exp \left( \frac{\alpha}{\sigma}\int_0^t \text{sgn}(X_s) dW_s - \frac{\alpha^2 t}{2\sigma^2}\right)$$ so that by Girsanov's theorem, $$\widetilde W_t \equiv \frac{X_t - X_0}{\sigma} = W_t - \frac{\alpha}{\sigma} \int_0^t \text{sgn}(X_s)ds \quad \text{is } Q \text{ Brownian motion}$$

Notice that $Z$ may be written as $$Z_t = \exp \left(\frac{\alpha}{\sigma} \int_0^t \operatorname{sgn} (\widetilde W_s - c_0^\sigma) d \widetilde W_s + \frac{\alpha^2 t}{2\sigma^2}\right), \quad c_0^\sigma \equiv -\frac{X_0}{\sigma}$$


$\tau \equiv \inf \{t \ge n : |X_t|^p/t > \epsilon\}, \quad \quad B_t \equiv -\int_0^t \operatorname{sgn}(\widetilde W_s - c_0^\sigma) d\widetilde W_s, \quad \quad c \equiv \alpha/\sigma$

and denote by $L_t^x$ the local time of $\widetilde W_t$ about the point $x$. Note that $|X_t|^p/t = \sigma^p|\widetilde W_t - c_0^\sigma|^p/t$ and $B$ is $Q$-Brownian motion. We have

$$\begin{aligned} p_n &\leq P(\tau \leq n+1) = E_P( \mathbf{1}(\tau \leq n+1) Z_{n+1}/Z_{n+1}) = E_Q(\mathbf{1}(\tau \leq n+1) Z_{n+1}^{-1}) \\ &= E_Q \left ( \mathbf{1}(\tau \leq n+1) \exp \left(c B_{n+1} - \frac{c^2(n+1)}{2} \right) \right) \\ &\stackrel{\text{(i)}}{=} E_Q \left ( \mathbf{1}(\tau \leq n+1) \exp \left(c B_\tau - \frac{c^2\tau}{2} \right) \right) \\ &\stackrel{\text{(ii)}}{\leq} E_Q \left(\mathbf{1}(\tau \leq n+1) \exp \left( c\sup_{0 \leq s \leq \tau} \left(B_s - \frac{cs}{2} \right) + \frac{c^2 \tau}{2} + c|c_0^\sigma| - \frac{c(\epsilon n)^{1/p}}{\sigma} - \frac{c^2 \tau}{2} \right) \right) \\ &\leq \exp \left( - \frac{\alpha(\epsilon n)^{1/p}}{\sigma^2} \right) E_Q \left( E_Q \left( \exp \left( c\sup_{0 \leq s \leq \tau} \left(B_s - \frac{cs}{2} \right) \right) \bigg| \mathcal{F_0} \right) e^ { \frac{\alpha|X_0|}{\sigma^2}}\right)\\ &\stackrel{\text{(iii)}}{=} ...\end{aligned}$$ where

$\text{(i)}$ follows from optional sampling for martingales.

$\text{(ii)}$ is because, on the event $\tau \leq n+1$, $$L_\tau^{c_0^\sigma} = \max \{- |c_0^\sigma| + \sup_{0 \leq s \leq \tau} B_s , 0 \} = |\widetilde W_\tau - c_0^\sigma| - |c_0^\sigma| + B_\tau \\ \text{(see, for example, Kallenberg (2021) p. 662)} \\ \begin{aligned} \implies B_\tau \leq \max \{ \sup_{0 \leq s \leq \tau }B_s, |c_0^\sigma| \} - \frac{(\epsilon \tau)^{1/p}}{\sigma} \leq \sup_{0 \leq s \leq \tau }B_s + |c_0^\sigma| - \frac{(\epsilon n)^{1/p}}{\sigma} \end{aligned}$$

$\text{(iii)}$ Not sure where to go from here, but I know if the conditional expectation can be bounded then we are done (in that case the $p_n$ are summable and we can use Borel Cantelli to get almost sure convergence). I know that the $\sup$ process (replacing $\tau$ with $\infty$) has an exponential distribution precisely with parameter $c$, so its MGF doesn't exist evaluating at $c$. Obviously I have been loose with bounds in some places, but I don't know where I can change this effectively in order to avoid this problem.

  • $\begingroup$ Hm. I'm pretty sure this kind of (mean-reverting, Markov type) SDE should have been exhaustively resolved somewhere. $\endgroup$ May 29, 2021 at 2:13
  • $\begingroup$ @ParesseuxNguyen I think I managed to fix it, but I've been trying to figure this out for so long that I have no brain cells left. Is what I have below valid? $\endgroup$
    – qp212223
    May 29, 2021 at 2:44
  • $\begingroup$ It's gonna take a lot of time. If you want, I can try to check it with you as a friend. Firstly, would you mind telling what is $\tilde{W}, c^{\sigma}_0$ and why do you have $|X_t|= |\tilde{W}-c^{\sigma}_0|$ $\endgroup$ May 29, 2021 at 13:56
  • $\begingroup$ Thanks!! Sorry about the confusing definitions; I edited the post so many times and deleted a bunch of my original work because it didn't work. Unfortunately part of that was in the definition of $\widetilde W_t$. I have re-included it in the post - hopefully it is clear now $\endgroup$
    – qp212223
    May 29, 2021 at 21:35
  • 1
    $\begingroup$ @Tobsn no, $B_t^p /t$ does not tend to zero for $p > 2$ for example (use the LIL if you want to prove this) $\endgroup$
    – qp212223
    Jun 4, 2021 at 6:32

2 Answers 2


I think I have FINALLY fixed it. From inequality $\text{(ii)}$, instead write $$\begin{aligned} p_n &\leq \exp \left(- \frac{\alpha (\epsilon n)^{1/p}}{\sigma^2} \right)E_Q \left(\exp \left(c \sup_{0 \leq s \leq n+1}B_s + c |c_0^\sigma| - \frac{c^2 \tau}{2} \right) \right) \\ &\leq \exp \left(- \frac{\alpha (\epsilon n)^{1/p}}{\sigma^2} \right) E_Q \left( E_Q \left( \exp \left(c \sup_{0 \leq s \leq n+1}B_s \right) \bigg | \mathcal{F}_0 \right) \exp \left( c |c_0^\sigma| - \frac{c^2 n}{2} \right) \right) \\ &\stackrel{\text{(iii)}}{=} \exp \left(- \frac{\alpha (\epsilon n)^{1/p}}{\sigma^2} \right) E_Q \left( E_Q \left( \exp \left(c |B_{n+1}| \right) \bigg | \mathcal{F}_0 \right) \exp \left( c|c_0^\sigma| - \frac{c^2 n}{2} \right) \right) \\ &\leq 2 \exp \left(- \frac{\alpha (\epsilon n)^{1/p}}{\sigma^2} \right) E_Q \left( \exp \left(\frac{c^2(n+1)}{2}+ \frac{\alpha|X_0|}{\sigma^2} - \frac{c^2 n}{2} \right) \right) \\ &= 2e^{c^2/2} \cdot \exp \left( - \frac{\alpha (\epsilon n)^{1/p}}{\sigma^2} \right) \end{aligned}$$

where $\text{(iii)}$ follows since $\sup_{s \leq n+1} B_s \stackrel{\text{d}}{=}|B_{n+1}|$ and the subsequent inequality is trivially bounding the MGF for the folded normal.

The final equality term is summable in $n$, and so we may finally apply Borel-Cantelli. If someone would be willing to comment and see if this is valid I would be tremendously grateful.


Your answer, though potentially correct, does not highlight the particular structure of your SDE. So here's another answer, for the pleasure of showing how pretty the theory of these equations can get.

I will work under the following simplification: replace $sgn(x)$ with a smooth strictly increasing function $f(x)$ such that $f(x) = sgn(x)$ for $x\not\in(-1,1)$. In this way we can consider strong solutions instead of weak ones and the argument gets a bit cleaner. I trust you will be able to translate this to your setting without further problems.

Now, take $F(x)$ such that $f(x) = \partial_x F(x)$ and, for example, $F(0)=0$. Then the measure $$\mu( d x) = \frac{1}{Z} e^{-\frac{ 2 \alpha}{\sigma^2}F(x)} d x, \qquad Z = \int_{\mathbb{R}} e^{- \frac{ 2 \alpha}{\sigma^2}F(x)} d x$$ is invariant under the flow of the SDE $$d X_t = - \alpha f(X_t) dt + \sigma dW_t.$$ To see this check that $e^{-\frac{ 2 \alpha}{\sigma^2}F(x)}$ is a time-independent solution to the Kolmogorov forward equation.

Hence, let $Y_t$ be a solution to the same SDE of $X_t$ with the same driving noise but an independent initial condition $Y_0\sim \mu$. In particular the law of this solution at any given time does not change and is equal to $\mu$. Then compute: $$d|X_t- Y_t|^2 = -2\alpha(X_t - Y_t)(f(X_t) - f(Y_t)) \leq 0.$$ An excursion: Here your SDE is quite special since for "typical" potentials $F$ (say $F(x) \geq x^2$) one would actually expect the last inequality to be strict and bounded by $-(X_t-Y_t)^2$, so that actually the two paths converge exponentially fast to each other. But in your case you expect that most of the time $=0$ holds (so the two paths still converge to each other, but the distance is strictly decreasing only when $X$ and $Y$ are on the opposite sides of zero).

In any case, in view of the previous inequality it suffices to show that $$\lim_{t \to \infty} \frac{Y_t^p}{t} = 0 \ \ a.s..$$ So define $p_n$ just as in your question, only with $X_t$ replaced by $Y_t$. We have, by invariance: $$p_n \leq \Phi(\varepsilon n), \qquad \Phi(x) = \mathbb{P}\left(\sup_{0 \leq t \leq 1} |Y_t|^p \geq x\right).$$ Clearly we are interested in the decay of $\Phi$ in $x$, as we want to apply the Borel-Cantelli lemma. Observe that $$\sup_{0 \leq t \leq 1} |Y_t|^p \leq (|Y_0| + \alpha + \sup_{0 \leq t \leq 1}|W_t|)^p.$$ From the integrability properties of $\mu$ and $\sup_{0\leq t\leq 1} |W_t|$ we find that for any $q\in \mathbb{N}$ $$\mathbb{E}\left[ \sup_{0 \leq t \leq 1} |Y_t|^{q p} \right]< \infty.$$ By the Markov inequality $\Phi$ decays polynomially of any order as $x \to \infty,$ so the the result follows by Borel-Cantelli.


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