Assume the Law of Iterated Logarithm upper bound
\begin{equation}
\limsup_{t\to \infty} \frac{B_t} {\psi(t)} \leq 1
\end{equation}
holds (almost surely). Now I have seen in a proof of the lower bound that for $t$ large enough
\begin{equation}
B_t \geq -2 \ \psi(t)
\end{equation}
is true because of the above mentioned upper bound. Can somebody explain this to me? Why wouldn't $B_t \geq – \psi(t)$ hold?
Recall that $\psi(t) = \sqrt{2t\log \log t}$.
Best Answer
Remember that by the definition of the limit supremum, we have $$\limsup_{t\to\infty} f(t) \leq M \iff \left(\forall \epsilon > 0, \exists T : t > T \implies f(t) \leq M + \epsilon\right)$$ In particular, note that we cannot conclude $f(t) \leq M$ for any $t.$ (e.g. note that $\limsup_{n\to\infty} 1 + \frac{1}{n} \leq 1$)
In your context, this means we can set $\epsilon = 1$ to see that for $t$ large enough, $$\frac{B_t}{\psi(t)} \leq 1 + 1 = 2 \implies B_t \leq 2\psi(t)$$ almost surely.
Then since $-B_t$ is a brownian motion, we can replace $B_t$ with $-B_t$ to conclude $$-B_t \leq 2\psi(t) \iff B_t \geq -2\psi(t)$$ almost surely, for $t$ large enough.