# Law of Iterated Logarithm Upper Bound Implication

Assume the Law of Iterated Logarithm upper bound $$$$\limsup_{t\to \infty} \frac{B_t} {\psi(t)} \leq 1$$$$ holds (almost surely). Now I have seen in a proof of the lower bound that for $$t$$ large enough $$$$B_t \geq -2 \ \psi(t)$$$$ is true because of the above mentioned upper bound. Can somebody explain this to me? Why wouldn't $$B_t \geq - \psi(t)$$ hold?

Recall that $$\psi(t) = \sqrt{2t\log \log t}$$.

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.