I am trying to understand a paper related to brownian motions, and right in the begginining it is claimed that $$X_t=\sqrt{t}\bigg( G_1 \cos ( W_{a + \ln t}) + G_2 \sin( W_{a + \ln t})\bigg)$$ has the same distribution as a standard Brownian motion, for $t \geq \exp(-a), a \geq 0$, $G_i$'s standard normaly distributed and where $W_t$ is a standard Brownian motion. It is assumed that the $G_i$'s and $W_t$ are all mutually independent.
To see that the expected value is $0$ is straightforward, but of course this does not suffice. My attempt then was to try to compute the distribution of $X_t$ using conditional probabilities and taking advantage of the independence, but I have read however that the sine and cosine of normal distributions are not easily derived. So probably this was not the author's approach.
Is there maybe some "general fact" I can use to prove this claim?
Thanks!
It suffices to compute the characteristic function of $Y_t:=X_t/\sqrt{t}$ and check that it is equal to $e^{-t^2/2}$. This is since $X_t \sim B_t \iff X_t/\sqrt{t} \sim B_1$ which is of course a standard normal variable.
Write $Y_t$ as the sum $X_1 + X_2$ and note that conditional on $W_t$, $X_1 \sim N(0, \cos^2(W_t)$) and $X_2 \sim N(0, \sin^2(W_t)$) and that both are independent. Using iterated expectations and the fact that the expected value of a product of independent random variables is the product of the corresponding expectations we have:
$$ \begin{equation} \mathbb{E}[e^{it Y_t}] = \mathbb{E}[\mathbb{E}[e^{it Y_t}|W_t]] = \mathbb{E}[\mathbb{E}[e^{it X_1}e^{it X_2}|W_t]] = \mathbb{E}[\mathbb{E}[e^{it X_1}|W_t]\mathbb{E}[e^{it X_2}|W_t]] \\ = \mathbb{E}[e^{-t^2\cos^2(W_t)/2} e^{-t^2\sin^2(W_t)/2}] = \mathbb{E}[e^{-t^2/2}] = e^{-t^2/2} \end{equation} $$