Time transform of Brownian motion

Question

I have the following doubt. In an exercise I was given $(B_t)_{t \geq 0}$ as a standard Brownian motion (i. e. $B_t \sim N(0,t) \ \forall \ t \geq 0$). Now I have to investigate whether or not $B_{e^{2t}} \sim N(0, e^{2t}) \ \ \forall t \geq 0$. If this is indeed true (may be something related to the invertibility of the exponential function?... how can I show that process has that distribution?), the process defined by $(e^{- t} \cdot B_{e^{2t}})_{t \geq 0}$ is not a standard BM because it lacks $B_0 = 0$, right?

Answer 1

$B_t \sim N(0,t)$ for all $t >0$ so it is legitimate to replace $t$ by $e^{2t}$ and conclude that $B_{e^{2t}} \sim N(0,e^{2t})$. There is no probability theory involved in this implication.

You are right in saying that $(e^{-t}B_{e^{2t}})$ is not a standard BM.