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?
Best Answer
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.