If $E[B_{t}]=0$ then why is $E[B_{t}^{2}]=t$

Question

Let B be a brownian motion. I know that a brownian motion includes the fact that for a family $(B_{t})_{ t \in [0,\infty[}$, the increments have the normal distribution: $B_{t}-B_{s}$ ~ $\mathcal{N}(0, t-s)$. So then why would $E[B_{t}^{2}]=t$ if $E[B_{t}]=0$? since $B_{t}-B_{0}=B_{t}$ as $B_{0}=0$ a.s.

Answer 1

The facts you've agreed to show that $B_t$ is normally distributed with mean $\mu = 0$ and variance $\sigma^2 = t-0 = t.$

In other words, $E[B_t] = \mu = 0$ and $$E[B_t^2] = \sigma^2 + \mu^2 = t + 0^2 = t.$$

Answer 2

One formulation of the variance is $\mathbb V[B_{t}] = \mathbb E[B_{t} ^2] - (\mathbb E[B_{t}]^2)$. The second term is 0 as the mean is 0, as you stated. The left term however remains. Does that answer your question?

Answer 3

$\mathsf E[B_t]=0$ suggests $B_t$ means be non-negative or non-positive.

However $B_t^2$ may only be non-negative, so its expectation shall be positive (since $B_t$ is often non-zero).