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For a standard Brownian motion $(B_t)_{t \geq 0}$, I want to show that $$\limsup _{t \to \infty} \frac{B_t}{\sqrt{t \ln t}} \leq 1 \ \ \text{a.s.}$$ using the fact that $\frac{\exp(B_t ^2 / (1+2t))} { \sqrt{1+2t}}$ is a martingale.

It suffices to show $\limsup _{t \to \infty} \frac{B_t ^2}{t \ln t} \leq 1 \ \ \text{a.s.}$ and I know that $\frac{B_t ^2}{t \ln t} \leq \frac{B_t ^2} {1+2t}$ for large $t$, but I don't see any further.

Any help is appreciated.

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  • $\begingroup$ How do you propose to use that (positive) martingale? $\endgroup$ – John Dawkins Nov 8 '19 at 19:13
  • $\begingroup$ @JohnDawkins Maybe Stopping theorem ? $\endgroup$ – izimath Nov 8 '19 at 19:15
  • $\begingroup$ Suggestion: the martingale converges a.s. to some random variable $Z$ with values in $[0,\infty)$. $\endgroup$ – John Dawkins Nov 9 '19 at 19:12
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Thanks to @John Dawkins, I have come up with an answer.

By the martingale convergence theorem, our martingle converges pointwise to a non-negative real valued random variable $Z$ a.s. Note that

$$\frac {e^{B_t^2/(1+2t)}}{\sqrt{1+2t}} = \exp \bigg({\frac{t \ln t}{1+2t} \left( \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t}\right) \bigg)}$$

On the event $Z > 0$ (and where the martingale converges to $Z$, of course)

$$\lim _{t \to \infty}\frac{t \ln t}{1+2t} \left( \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t}\right) \bigg) \in \mathbb R$$

which implies $\lim_{t \to \infty} \frac{B_t ^2}{t \ln t} = 1$. On the event $Z=0$

$$\lim _{t \to \infty}\frac{t \ln t}{1+2t} \Big( \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t}\Big) \bigg) = -\infty$$

which implies $\limsup_{t \to \infty} \frac{B_t ^2}{t \ln t} \leq 1$ (otherwise, there exists $\delta > 0$, such that $\frac{B_t ^2}{t \ln t} >1+\delta$ infinitely often as $t \to \infty$, so $ \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t} >0$ infinitely often for large $t$ with positive probability). Note that $$\frac{\ln(1+2t)}{\ln t} \to 1$$ using L'Hospital's rule.

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  • $\begingroup$ Hi unless mistaken on the event $Z=0$, you need to have $\limsup_{t \to \infty} \frac{B_t ^2}{t \ln t} < 1$, otherwise if $\limsup_{t \to \infty} \frac{B_t ^2}{t \ln t} = 1$ you are reduced to the previous situation where $\lim _{t \to \infty}\frac{t \ln t}{1+2t} \Big( \frac{B_t^2}{t \ln t}- \frac{(1+2t)\ln(1+2t)}{2t \ln t}\Big) \bigg) $ is finite. $\endgroup$ – TheBridge Nov 12 '19 at 13:15
  • $\begingroup$ @TheBridge Maybe your implication is correct, but my answer is still okay in that case too. $\endgroup$ – izimath Nov 20 '19 at 13:40
  • $\begingroup$ absolutely right $\endgroup$ – TheBridge Nov 20 '19 at 20:27
  • $\begingroup$ Hi I am able to prove that $M_t= \frac{e^{B_t ^2 / (1+2t)}}{\sqrt{1+2t}}$ is a local martingale (via Itô's lemma I find $\frac{dM_t}{M_t}=\frac{2B_t dB_t}{(1+2t)}$) but $E[M_t^2]=+\infty$ as soon as $t>\frac{1}{2}$ this is only a sufficient condition so this is not no go for your proof, but I was wondering how do you get that $M_t$ is a "true" martingale ? regards $\endgroup$ – TheBridge Jan 2 '20 at 14:59
  • $\begingroup$ @TheBridge You can directly check the definition of a martigale by considering independent increment property. $\endgroup$ – izimath Jan 2 '20 at 15:08

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