Edit: Question Resolved. See below.

As a part of my bachelor thesis, I have to work through a paper about fake Brownian motion by Oleszkiewicz. In this paper he defines a stochastic process.

Let $G_1, G_2, W_t$ be independent. Here we have $G_1$, $G_2$ $\sim \mathcal{N}(0,1)$ and $W_t$ Brownian motion. For $a\geq 0$ and $t\geq e^{-a}$ define the filtration $\mathcal{F}_t^{(a)}=\sigma(G_1,G_2,(W_s)_{0\leq s\leq a+\ln t}),$ and the process $$X_t^{(a)} = \sqrt{t}\left(G_1\cos{W_{a+\ln t}}+ G_2\sin{W_{a+\ln t}}\right).$$

Now Oleskiewicz claims,

(1) $X_t^{(a)}$ is a continuous martingale,

(2) $X_t^{(a)}\sim \mathcal{N}(0,t)$,

(3) $X_e^{(a)}-X_1^{(a)}$ is not gaussian,

without actually proofing any of these statements.

While I managed to proof the first two statements, the proofs are rather long and involved, so I wanted to ask if I am missing something obvious. Furthermore, I am absolutely clueless regarding the third statement.

I hope you can help me.

  • $\begingroup$ The question has been resolved. Professor Oleszkiewicz took the time to answer it himself, after I sent him an E-mail. $\endgroup$
    – Alex
    Commented Jun 10, 2014 at 21:55
  • $\begingroup$ Would you mind posting the answer? I found the question interesting. $\endgroup$
    – Ian
    Commented Jun 11, 2014 at 0:10

1 Answer 1


This is the answer he send me regarding these increments.

"As for the increments, $X^{(a)}_e-X^{(a)}_1= (\sqrt{e}\cos W_{a+1}-\cos W_a)G_1+(\sqrt{e}\sin W_{a+1}-\sin W_a)G_2.$ Since the pair $(W_a, W_{a+1})$ is independent of the pair $(G_1, G_2)$, this increment has the same distribution as $[(\sqrt{e}\cos W_{a+1}-\cos W_a)^2+ (\sqrt{e}\sin W_{a+1}-\sin W_a)^{2}]^{1/2} G$, where $G$ is $\mathcal{N}(0,1)$ independent of the pair $(W_a, W_{a+1})$.

Now, if $Z$ and $G$ are independent, $G$ is $\mathcal{N}(0,1)$-distributed and $ZG$ is Gaussian then $|Z|$ must be constant a.s. -- indeed, $\mathbb{E}ZG=\mathbb{E}Z \mathbb{E}G=\mathbb{E}Z \cdot 0=0$, so that $ZG$ is a centered Gaussian random variable; now it suffices to notice that $\mathbb{E}G^4=(\mathbb{E}G^2)^2, \mathbb{E}(ZG)^4=3(E(ZG)^2)^2$, so that $EZ^4=(EZ^2)^2$ and hence $Var(Z^2)=0$, i.e., $Z^2$ is constant a.s.

However, the expression standing inside $[...]^{1/2}$ obviously is not constant a.s., as it is equal to $e+1-2\sqrt{e}\cos(W_{a+1}-W_a)$, and $W_{a+1}-W_{a}$ is a $\mathcal{N}(0,1)$ Gaussian random variable."

It's a quite impressive and interesting approach.


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