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I am looking for an example (if it exists) of a bi-dimensional process say $(W^1_t,W^2_t)$ where the margins are plain uni-dimensional Brownian motions but for which we can assert that it is not a 2 dimensional Brownian motion.

I mean this in the extended sense by allowing to enter the definition of two dimensional BM any processes the margin of which are correlated Brownian motions or otherwise said that there exists a $\rho \in (-1,1)$ such that $<W^1_t,W^2_t>\not= \rho.t$.

So for this it would suffice to prove that $<W^1_t,W^2_t>\not= \rho.t$ for all $\rho \in (-1,1)$ or alternatively that the bivariate process is not a local martingale (by Lévy's characterization). The thing is that I am looking for a constructive example.

Best regards

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    $\begingroup$ What exactly do you mean by "constructive" example? $\endgroup$
    – saz
    Commented Oct 5, 2016 at 14:31
  • $\begingroup$ Well I mean an explicit construction that would allow for example to have access to the finite dimensional distribution of the process, or a path by path construction, or recursive definition, or a discretization scheme known to converge to a process with the desired property or any way that would exclude getting the result from an abstract existence theorem such as a representation theorem that tells you that such process exists but that do not allow to exhibit a concrete example. I am not sure if the concept is fully rigorously definable but I hope you get the gist of it. Regards $\endgroup$
    – TheBridge
    Commented Oct 5, 2016 at 19:13

1 Answer 1

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For a one-dimensional Brownian motion $(W_t)_{t \geq 0}$, the reflected process

$$B_t := \begin{cases} W_t, & t \leq 1, \\ W_1-(W_t-W_1), & t>1 \end{cases} \tag{1}$$

defines a one-dimensional Brownian motion.

$\hspace{80pt}$enter image description here

Since $\mathbb{E}(W_s W_t) = \min\{s,t\}$, we have

$$\mathbb{E}(W_t B_t) = \mathbb{E}(W_t^2)=t \quad \text{for all $t \leq 1$}$$

and

$$\mathbb{E}(W_t B_t) =2 \mathbb{E}(W_1 W_t) - \mathbb{E}(W_t^2) = 2-t \quad \text{for all $t >1$}.$$

This means, in particular, that there does not exist $\varrho \in [-1,1]$ such that $\langle W,B \rangle_t = \varrho t$, and so the two-dimensional process $(W_t,B_t)_{t \geq 0}$ is not a two-dimensional Brownian motion (in the sense of the definition given in the OP).

More generally, consider

$$B_t := \int_0^t f(s) \, dW_s$$

for some Borel-measurable function $f:[0,\infty) \to \mathbb{R}$ such that $|f(s)|=1$ for all $s \geq 0$. Then

$$\langle W,B \rangle_t = \int_0^t f(s) \, ds.$$

If we choose $f(s) := 1_{[0,1]}(s)-1_{(1,\infty)}(s)$, then we get the reflected Brownian motion $(1)$.

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  • $\begingroup$ Thank you very nice example and generalization, I wish I could up vote twice. Regards $\endgroup$
    – TheBridge
    Commented Oct 5, 2016 at 21:00
  • $\begingroup$ @TheBridge You are welcome. :) $\endgroup$
    – saz
    Commented Oct 6, 2016 at 5:29

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