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Let $(W_t)_{0\leq t\leq 1}$ be a Wiener process defined up to time $1$ on some probability space. Consider the random vector

$$\left(W_{1},\int_0^1 \operatorname{sgn}(W_s) \, dW_s\right)=:(W_1,X_1)$$ where the integral expression is an Ito integral.

Using properties of the properties of the Ito integral and the P. Levy characterization of Brownian motion, it is not hard to show that both marginals are $\mathcal{N}(0,1)$.

Furthermore, it is claimed that this random vector is not bivariate Gaussian.

I'm having trouble seeing this last bit. My attempt so far is as follows:

Assume the contrary. Then since

\begin{align*} \operatorname{E}\left[W_1\int_0^1\operatorname{sgn}(W_s)\,dW_s\right] & = \operatorname{E}\left[\left(\int_0^1 \, dW_s\right)\cdot\left(\int_0^1 \operatorname{sgn}(W_s) \, dW_s \right)\right]\\[6pt] & = \operatorname{E}\left[\int_0^1 \operatorname{sgn}(W_s) \, ds\right] \\[6pt] &=\int_0^1 \operatorname{E} \left[\operatorname{sgn}(W_s)\right] \, ds\\[6pt] &=0, \end{align*} where we use the Ito isometry, Fubini-Tonelli, and the symmetry of the normal distribution, we would then have that $W_1$ and $X_1$ are uncorrelated hence independent $\mathcal{N}(0,1)$ random variables. I am not sure where to rigorously go from here.

One thing that's clear to me from John Dawkins comment is that the process $(W_{t},X_{t})_{0\leq t\leq 1}$ is not Gaussian, since $$aW_{t}+X_{t}=\int_{0}^{t}(a+\text{sgn}(W_{s}))dW_{s}$$ and therefore $$\langle{aW+X}\rangle_{t}=\int_{0}^{t}(a+\text{sgn}(W_{s}))^{2}ds,$$ which is not deterministic. Since Gaussian martingales have deterministic quadratic variation, the process is not Gaussian. So there is some collection of times $t_{1}<\cdots<t_{n}$ such that the finite-dimensional distribution is not multivariate Gaussian. But why does it follow that $(W_{1},X_{1})$ is not Gaussian?

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    $\begingroup$ Hint: To check the bivariate Gaussian property, you need to show that $\alpha W_1+\beta X_1$ has the normal distribution, for all real $\alpha$ and $\beta$. $\endgroup$ – John Dawkins May 12 '16 at 20:44
  • $\begingroup$ @JohnDawkins: Thank you for your comment. I'm aware that that it is equivalent to having bivariate distribution. But it isn't evident to me why the r.v.s fail this property. $\endgroup$ – Nik Quine May 12 '16 at 21:16
  • $\begingroup$ @JohnDawkins: If you have a moment, I would appreciate it if you would take a look my claimed solution. $\endgroup$ – Nik Quine May 13 '16 at 0:59
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Let $X_{t}$ and $W_{t}$ be defined as above. Let's apply Ito with $f(x,y)=xy^{2}$ to get the semimartingale decomposition of $f(X_{t},W_{t})=X_{t}W_{t}^{2}$.

\begin{align*} X_{t}W_{t}^{2}&=\int_{0}^{t}W_{s}^{2}\text{sgn}(W_{s})dW_{s}+2\int_{0}^{t}X_{s}W_{s}dW_{s}\\ &+\frac{1}{2}\left[4\int_{0}^{t}W_{s}\underbrace{d\langle{X,W}\rangle_{s}}_{=\text{sgn}(W_{s})ds}+2\int_{0}^{t}X_{s}ds\right]\\ \end{align*} Taking the expectation of both sides and using the martingale property, the first two terms have zero expectation, and we obtain $$\mathbb{E}[X_{t}W_{t}^{2}]=\mathbb{E}\left[\int_{0}^{t}(2|W_{s}|+X_{s})ds\right]=2\int_{0}^{t}\mathbb{E}[|W_{s}|]ds \tag{1}$$

If $X_{1}$ and $W_{1}$ were independent, then $\mathbb{E}[X_{1}W_{1}^{2}]=\mathbb{E}[X_{1}]\mathbb{E}[W_{1}^{2}]=0$. But clearly the RHS of (1) is nonzero at $t=1$.

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  • $\begingroup$ This looks fine; $X_1$ and $W_1$ are not independent. In particular (being uncorrelated) they are not jointly normal. $\endgroup$ – John Dawkins May 13 '16 at 20:11

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