# Fake Brownian motion – not Gaussian

Let $$G$$ be a standard normal random variable and define two standard Brownian motions $$(W_t)_{t ge 0}$$, $$&$$ $$(B_t)_{t ge 0}$$. Assume $$G, (B_t)$$ and $$(W_t)$$ are independent.

Moreover, define that process $$Y_t$$ by
$$Y_t = begin{cases} B_t, & 0 le t le 1 \ sqrt{t}big(B_1 cos(W_{log t})+ G sin(W_{log t})big) & t ge 1 end{cases}$$

Show that $${Y_t : t ge 0 }$$ is not Brownian motion by proving that it is not Gaussian (this is called fake Brownian motion).

My attempt:

$$Y_e – Y_1 = sqrt{e}(B_1cos(W_1)+Bsin(W_1)-1) = B_1(sqrt{e} cos(W_1) -1) + G sin(W_1)$$. I know that any linear combination of independent normal random variables is also normal. However, $$cos(a)$$ and $$sin(a)$$ are not linear transformations. I’m not quite sure how to prove that this isn’t Gaussian because I don’t know the distribution of $$cos(W_1)$$ and $$sin(W_1)$$. Is there another way to show this?