One way to generate Brownian motion is as follows: Define a waiting time probability distribution $\psi(t)$ and step length probability distribution $\lambda(x)$. Require also that $\langle \psi \rangle = \tau$, $\langle \lambda \rangle = 0$, and $\langle \lambda^2 \rangle = \sigma^2$. That is, the waiting time distribution has finite mean and the step length is symmetric about zero with finite variance. Then, we can generate a sequence of time steps $\delta t_i \sim \psi$, a sequence of step lengths $\delta x_i \sim \lambda$. We can then define the trajectory $x(\sum^N \delta t_i) = \sum^N \delta x_i$.
If one is simulating the process, convenient choices include $\psi(t) = \delta(t - \tau)$, and $\lambda(x) = \frac{1}{2}( \delta(x - \sigma ) + \delta(x + \sigma ))$. That is, at regular time intervals you pick whether the particle goes left or right, with equal probability. So far so good.
Now for simplicity, let anomalous diffusion be a process such that $\langle \Delta x^2 (t) \rangle \propto t^\alpha$, for $0 < \alpha < 1$.
To generate anomalous diffusion, it must be that the mean waiting time diverges $\langle \psi \rangle \to \infty$ or that the step length variance diverges $\langle \lambda^2 \rangle \to \infty$ or, I guess, possibly both. I know one way to do this: Let $\langle \psi \rangle$ have a Pareto distribution $\psi(t) = \frac{\alpha \tau^\alpha}{t^{1+\alpha}}$ for $t > \tau$, so that it has a diverging mean. We can then pick any $\lambda$ with finite variance, like the one in the previous paragraph.
My question is whether it is possible to require that $\psi(t) = \delta(t - \tau)$ and still get anomalous diffusion. Obviously $\lambda$ would need to have diverging variance, but I have no idea beyond that. I suppose $\lambda(x_i)$ could even depend on past history. That is, I'm looking to pick steps at regular time intervals $\Delta t$ from a certain distribution $\lambda$ such that the resulting trajectories display anomalous diffusion.
