Brownian motion increments - are they random variables or random processes

Question

If $W_t$ is a Brownian motion process and $0 \le t_1 \le t_2$ then is the increment $W_{t2} - W_{t1}$ a random variable or a random process? My lectures say "random variable" but I believe it makes more sense to call this a random process because any segment of a Brownian motion process is still a Brownian motion process.

Answer 1

If $t_1$ and $t_2$ are fixed it is a random variable. If you consider it as a mapping from $\mathbb{R}^2$ $(t_1,t_2)\mapsto W_{t_2}-W_{t_1}$, so that $t_1$ and $t_2$ are variables, then you can see it as a process on $\mathbb{R}^2$.

In practice you can see a random process as a set of random variables which are labelled on some space (in your case by $(t_1,t_2)\in\mathbb{R}^2$). So even when you look at $W_t$ itself, if you consider it as a mapping $t\mapsto W_t$ it is a random process, if you consider it at a fixed time $t$ it is a random variable.

Answer 2

The increment is not only a random variable but it is a normally distributed random variable.

I think your confusion is on the difference between an increment between two points of the path and a subset of the path itself. The increments are normally distributed random variables, and independent ones if the increments are disjoint, but if you take a subset of any part of the path that path itself is another Brownian motion.