Which Brownian motion property is the most important?

Question

Which Brownian motion property is the most important?

A standard Brownian motion is a stochastic process $(W_t, t\geqslant 0)$ indexed by nonnegative real numbers t with the following properties:

1. $W_0=0$;
2. With probability 1, the function $t \to W_t$ is continuous in t;
3. The process $(W_t, t\geqslant 0)$ has stationary, independent increments;
4. The increment $W_{t+s}-W_s$ has the $\mathrm{NORMAL}(0, t)$ distribution.

Answer 1

I agree with @FooBar that the first property, i.e. that the initial point equals $0$ a.s., is not an important one - in the sense that this property is not needed in order to characterize the Brownian motion. In fact, the shifted process $(x+W_t)_{t \geq 0}$ is called Brownian motion started at $x \in \mathbb{R}$, and there are further generalizations which basically allow to consider arbitrary initial distributions.

Concerning the continuity of the sample paths: Any process $(W_t)_{t \geq 0}$ which satisfies the properties 3+4 has a modification which has (a.s) continuous sample paths. This follows from the Kolmogorov-Chentsov theorem. (Obviously, this does not mean that the continuity is not important.)

Concerning the stationary independent increments: If a process $(L_t)_{t \geq 0}$ has stationary independent increments (and is stochastically continuous), then it is called a Lévy process. This is a rather important class of stochastic processes and, in particular the sample path properties differ from these of the Brownian motion. Most importantly, the sample paths can have jumps. In fact, the Brownian motion is the only Lévy process with continuous sample paths. This means that the properties 2+3 imply property 4.

This does not answer the question which property is the most important one; but if I would have to strip down the definition to two properties, I guess I would choose property 3 and 4.

Answer 2

I would just exclude $W_0$. The initial value is quite irrelevant and is typically chosen to be zero only for normalization purposes. If you compare it to discrete-time, the Brownian motion is the equivalent of an iid $N(0,\sigma)$ process.

Then, it depends on the context, whether you care most about it being continuous, independent or normal.