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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.
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    $\begingroup$ For point 2, it follows from the variance of $|Bt|$. The variance is always positive. For point 1, I don't see where they are showing a case $s\leq t$. Maybe you're confused with their remark about the filtration. $\endgroup$
    – Raskolnikov
    Jun 15, 2018 at 17:16
  • $\begingroup$ Could you be more explicit about point 2? $\endgroup$
    – yoshi
    Jun 15, 2018 at 18:25

2 Answers 2

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

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1) They never showed $s \leq t$ case. If you’re refering to the definition of $\mathcal{F}_t$, that has nothing to do with the inequality. A stochastic process is defined with respect to a filtration.

2) $f(x) = x^2$ is convex so by Jensen’s inequality, $f(E(|B_t|)) \leq E(f(|B_t|))$

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