Can we state the distribution of this Brownian motion as follows?

Question

Consider the following Brownian motion:

$W(e^{2t})$

Where $t\in[0,\infty)$. What can we then say about how $W(t)$ is distributed? Could we say:

$W(e^{2t})\sim \sqrt{e^{2t}}N(0,1)$?

Answer 1

I suppose that what you meant to say is that $(W_t)_{t\ge0}$ is a Brownian Motion and you're asking for the distribution of $\bar W_t := W_{e^{2t}}$ for all $t\ge 0$.

First note that $\bar W$ is not a Brownian Motion, as for all $0\le s\le t, \ \ \mathbb E [\bar W_t \bar W_s]=e^{2t} \wedge e^{2s} \ne t \wedge s$

However, $W$ is a Brownian Motion so it holds that for all $\bar t\ge 0$, $W_{\bar t} \sim \mathcal N(0,\bar t)$. In particular, for $\bar t := e^{2t}$, $W_{e^{2t}} \sim \mathcal N(0,e^{2t})$.

So you were indeed correct.