why is the expected value of a Wiener Process = 0?

Question

This section of wikipedia says that the expected value of a Wiener Process is equal to 0.

Why is that?

Answer 1

In the characterizations at Wikipedia,

W_t has independent increments with W_t−W_s ~ N(0, t−s) (for 0 ≤ s < t), .

(mean of the normal distributed increments is 0)

Lévy characterization that says that the Wiener process is an almost surely continuous *martingale with W_0 = 0*

(martingale has expected increment zero)

spectral representation as a sine series whose coefficients are independent N(0, 1) random variables.

(coefficients have a mean of zero)

scaling limit of a [symmetric] random walk,

(random walk goes up and down with equal probability)

Answer 2

Because by the third part of the definition there (under "Characterizations of the Wiener process"), $W_t = W_t - W_0$ is normally distributed with mean $0$ and variance $t$.