Let $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0},P)$ a filtered probability space satisfying the usual conditions. Is there a real-valued stochastic process $X=(X_t)_{t\geq0}$ satisfying the conditions
- $\forall 0\leq s<t<\infty: X_t-X_s$ is independent of $\mathcal F_s$ and
- $\forall 0\leq s<t<\infty: X_t-X_s$ is uniformly distributed on the intervall $[-(t-s),t-s]$ ?
My question is closely related to Lévy processes and obviously motivated by the Browian motion, where in the second condition a normal distribution is required. But I'm not happy about the fact, that the paths are not of finite variation. So, if there is this process $X$, what can we say about the paths (differentiability, ...)? $X$ should has definitely paths of finite variation and be continuous - provided it exists. If there is no such process $X$, might there be another distribution for condition (2) such that the process exists (aside from the normal distribution, where the result is well known)?
Let $s < u < t$. We have $X_t - X_s = (X_t - X_u) + (X_u - X_s)$, but the uniform distribution on $[-(t-s), t-s]$ is not a sum of two independent uniform distributions.
Indeed, in order for (1) to be true, the distribution of each $X_t - X_s$ has to be infinitely divisible. This is what leads to the classification of Lévy processes and the distributions of their increments. In particular, the only such processes which is continuous are Brownian motions, and none of them are differentiable (except, trivially, the constant process).