Let $p > 0$ and $W$ be a standard Brownian motion. I want to investigate the following limits $$L(p) \equiv \lim_{t \rightarrow \infty} \frac{W_t^p}{t}$$

Obviously if $p = 1$, then the SLLN for BM implies that $L(1) = 0$. $L(0)$ is also trivially $0$. For $p \in (0,1)$ we see $$\frac{W_t^p}{t} = \left(\frac{W_t}{t^{1/p}}\right)^p = \left(\frac{W_t}{t}\right)^p \cdot \frac{1}{t^{1-p}} \rightarrow 0$$ so that $L(p) = 0 \quad \forall p \in [0,1]$.

Now if $p \in (1,2)$, we see that $$\begin{aligned} E \left(\sup_{2^n \leq t \leq 2^{n+1}} \frac{|W_t|^p}{t} \right) \leq E \left(\sup_{2^n \leq t \leq 2^{n+1}} \frac{|W_t|^2}{t^{2/p}} \right)^{p/2} \\ \leq 2^{-n} E \left(\sup_{2^n \leq t \leq 2^{n+1}} |W_t|^2\right)^{p/2} \\ \leq C 2^{-n} E \left( W_{2^{n+1}}^2\right)^{p/2} \propto 2^{-n(1-p/2)} \end{aligned}$$ where the first inequality is Holder's, the second is trivial, the third is Doob's martingale inequality. I have abused notation with the constant. Now $2^{1-p/2} > 1$ so that $$\begin{aligned} \sum_{n =1}^\infty P\left(\sup_{2^n \leq t \leq 2^{n+1}} \frac{|W_t|^p}{t} > \epsilon \right) \leq C \sum_n \frac{1}{(2^{1-p/2})^n} < \infty\end{aligned}$$ and from Borel-Cantelli we see immediately that $L(p) = 0 \quad \forall p \in (1,2)$.

I don't know how to address the case where $p \ge 2$. Does anyone have any ideas?

  • 2
    $\begingroup$ Law of Iterated Logarithm fo BM would be useful. $\endgroup$ May 27, 2021 at 8:25
  • $\begingroup$ Just to check then: $L(p)$ does not exist for $p \ge 2$ with $\limsup |W_t^p|/t = \infty$ and $\liminf |W_t^p|/t = 0$? Since by definition of $\limsup$ $$\forall T \ge 0, \exists t \ge T \text{ s.t. } 1/2 \leq \frac{|W_t|^p}{(2t \log \log (t))^{p/2}} \implies \frac{|W_t|^p}{t} \ge \frac{(2t \log \log(t))^{p/2}}{2t}$$ at arbitrarily large times. The $\liminf$ follows since $\liminf |W_t| = 0$. Would this be the right logic? $\endgroup$
    – qp212223
    May 27, 2021 at 16:57


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