I'm reading a book, and they say Brownian Motion is martingale then show it with the following calculation:
Suppose $(B_t)$ is brownian motion which generates the filtration $\mathcal F_t$ (for all $B_s$ such that $s \leq t$). Then we have: $$E[|B_t|]^2 \leq E[|B_t|^2] = |B_0|^2 + nt$$ and if $s \geq t$ then they do a calculation to show $E[B_s|\mathcal F_t] = B_t$
- Why are they showing $s \leq t$ case? Isn't the defining property of martingale that in the $s \geq t$ case, $E[B_s|\mathcal F_t] = B_t$?
- The steps to obtain the inequality are a bit unclear to me.