# 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:

$$Y_e - Y_1 = \sqrt{e}(B_1\cos(W_1)+G\sin(W_1))-B_1 = B_1(\sqrt{e} \cos(W_1) -1) + G \sin(W_1).$$ 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?

• Not completely sure, but you might be able to use Eulers identity and the fact that $E[e^{itW_1}]$ is the characteristic function of $W_1$ (again, just brainstorming). Commented Sep 6, 2020 at 4:47

Suppose $$X$$ is Gaussian with mean 0. Then $$E(X^4) = 3E(X^2)^2$$.
Lemma. Suppose that $$Y$$ is a random variable independent of $$X$$, and $$X$$ is $$\mathcal N(0,1)$$. Then $$XY$$ is a Gaussian if and only if $$Y^2$$ is constant almost surely.
Proof: Note that $$E(XY) = 0$$, $$E((XY)^2) = E(Y^2)$$, and $$E((XY)^4) = 3 E(Y^4)$$. So if $$XY$$ is Gaussian, then $$E(Y^4) = E(Y^2)^2$$. This implies $$\text{Var}(Y^2) = 0$$, and hence $$Y^2$$ is constant almost surely. The converse is obvious. $$\square$$
Now consider $$Y_t - Y_1$$ for $$t > 1$$. It can be written as a $$\mathcal N(0,1)$$ Gaussian multiplied by $$V = \sqrt{(\sqrt t\cos W_{\log t}-1)^2 + (\sqrt t\sin W_{\log(t)})^2} = \sqrt{1+t-2\sqrt t\cos W_{\log t})} .$$ Thus $$Y_t - Y_1$$ is Gaussian if and only if $$1+t-2\sqrt t\cos W_{\log t}$$ is constant almost surely. And this is not the case.