Is this a Brownian motion?

Question

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.

Answer 1

I vote for Mateusz Kwaśnicki. The condition for whether the random walk you generate this way scales towards Brownian motion under taking long times and rescaling is whether or not the variance is finite. You already know how to take the expectation, your formula is correct. $$ \int_0^1 \left(-\ln(1-x)\right)^c\, dx\, =\, \Gamma(1+c)\, . $$ So if $\lambda=1/\Gamma(1+c)$, then $\mathbf{E}[R_k]=1$. Note that you still have $$ \mathbf{E}[X_k]\, =\, \mathbf{E}[X_{k-1}]\, ,\ \mathbf{E}[Y_k]\, =\, \mathbf{E}[Y_{k-1}]\, , $$ because of the $\theta_k$ randomization. Then by the same integral formula $$ \mathbf{E}[R_k^2]\, =\, \frac{1}{\lambda^2}\, \int_0^1 \left(-\ln(1-x)\right)^{2c}\, dx\, =\, \frac{\Gamma(1+2c)}{\Gamma(1+c)^2}\, . $$ So, as long as $-1/2<c$ then this is finite. (It is infinite if $c\leq -1/2$ and the Gamma formula above should not be used.) So if you take $$ \widetilde{B}^{(n)}_t\, =\, \frac{\Gamma(1+c) \sqrt{2}}{\sqrt{n\Gamma(1+2c)}}(X_{\lfloor nt\rfloor},Y_{\lfloor nt\rfloor}) $$ then this should converge in distribution to a process $B_t = (B_t^{(1)},B_t^{(2)})$ where $B_t^{(i)}$ for $i=1,2$ are two independent Brownian motions normalized so that $\mathbf{E}[(B_t^{(i)})^2]=t$.

Even if you started with a Levy process, if you truncated the increments at say integer times, cutting-it-off if its magnitude is greater than a given finite value, and you then took a long time limit with the Brownian scaling, even that would rescale to Brownian motion. That is what happens when you truncate, for example.