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Denis Serre
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Is this isa Brownian motion?

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Is this is Brownian motion?

I am building a 2D stochastic process as follows. I start with a point $P_0=(0,0)$. Then $P_k=(X_k,Y_k)$ is defined as follows, for $k>0$: \begin{align} X_k & =X_{k-1}+R_k \cos(2\pi\theta_k) \\ Y_k & =Y_{k-1}+R_k \sin(2\pi\theta_k) \end{align} where the $\theta_k$'s are uniformly and independently distributed (iid) on $[0,1]$, the $R_k$'s are also iid, and independent from the $\theta_k$'s. The $R_k$'s are generated as follows:

$$R_k=\frac{1}{\lambda}\Big[-\log(1-U_k)\Big]^c,$$

where $\lambda>0$ and $U_k$ is uniform on $[0,1]$. If $c=1$, $R_k$ has an exponential distribution of parameter $\lambda$. If $c>0$, $R_k$ has a Weibull distribution; if $c<0$, $R_k$ has a Fréchet distribution. In all cases, $E[R_k]=\Gamma(1+c)/\lambda$ if $c>-1$, otherwise the expectation is infinite. It is possible to rescale the process, in the same way a random walk is rescaled to become at the limit, a Brownian motion.

My question is this: will my simulation always result in a Brownian motion regardless of the parameter values? My hope is that the answer is sometimes yes, sometimes no. For standardization purposes, assume that $\lambda=\Gamma(1+c)$ if $c>-1$. I am interested in finding parameter values such that the resulting process is not Brownian. What if $c=4$? Below is a realization with $10^4$ points and $c=4$. It does not look Brownian, it consists of well separated clusters, typical for a large value of $c$. And my goal is to illustrate clustering techniques (in particular, identifying the number of clusters) for processes that are Brownian-related, but exhibiting a much stronger cluster structure with well separated clusters.

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