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 $$Z_t = \exp \left(\frac{\alpha}{\sigma} \int_0^t \operatorname{sgn} (\widetilde W_s – c_0^\sigma) d \widetilde W_s + \frac{\alpha^2 t}{2\sigma^2}\right), \quad c_0^\sigma \equiv -\frac{X_0}{\sigma}$$
Define
$\tau \equiv \inf \{t \ge n : |X_t|^p/t > \epsilon\}, \quad \quad B_t \equiv -\int_0^t \operatorname{sgn}(\widetilde W_s – c_0^\sigma) d\widetilde W_s, \quad \quad c \equiv \alpha/\sigma$
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.
Best Answer
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.