[Math] Brownian Motion is Martingale

brownian motionprobability theorystochastic-calculusstochastic-integralsstochastic-processes

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 $E[B_s|\mathcal F_t] = B_t$

  1. Why are they showing $s \leq t$ case? Isn't the defining property of martingale that in the $s \geq t$ case, $E[B_s|\mathcal F_t] = B_t$?
  2. The steps to obtain the inequality are a bit unclear to me.

Best Answer

$\text{Var}(X) := E[(X-E(X))^2] = E(X^2)-E(X)^2$. From the definition, you easily see that the variance is positive, hence $E(X^2)\geq E(X)^2$.

Related Question