Question

Let $(\Omega, \mathcal{F}, P)$ be a probability space, on which $\mathcal{F}_t$ is filtration satisfying general conditions. $W_t$ is a standard Brownian motion. By law of iterated logarithm, one has $$ P\Big(\lim_{t\to \infty} \frac{W_t}{t} = 0\Big) = 1 $$ This implies $$ P\Big(\lim_{t\to \infty} \frac{t + W_t}{t} = 1\Big) = 1. \quad (1)$$ We have contradiction in this below. By Girsanov theorem, there exists a probability measure $Q$ equivalent to $P$, such that $t+ W_t$ is a Brownian motion w.r.t. $Q$. By law of iterated logarithm, $$ Q\Big(\lim_{t\to \infty} \frac{t + W_t}{t} = 0\Big) = 1, $$ which implies $$ P\Big(\lim_{t\to \infty} \frac{t + W_t}{t} = 0\Big) = 1, \quad (2)$$ since $P$ is equivalent to $Q$. Why does this argument leads to a contradiction between (1) and (2)?

Answer 1

Girsanov's theorem tells you that on any finite interval $[0,T]$ you can find an equivalent probability that makes $t+B_t$ a Brownian motion. You just gave the proof that it cannot be done on the whole real line.

Answer 2

I am afraid that (1) is not correct, since the law of the iterated logarithm states that $$|B_t|\approx \sqrt{2t\log\log(1/t)}$$ as $t\to0$. This yields $$\frac{|B_t|}{t}\to\infty$$ as $t\to0$