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Let $(M_t)_t$ a stochastic process. Let $(\mathcal F_t)_t$ an adapted filtration. The process $(M_t)_t$ is called a martingales if

1) $\mathbb E|M_t|<\infty $

2) $\mathbb E[M_t\mid \mathcal F_s]=M_s$ where $s\leq t$.

My problem is the 2) in the definition. Indeed, since the filtration is adapted, then $M_t$ is $\mathcal F_t-$measurable for all $t$. By definition of a filtration, $\mathcal F_s\subset \mathcal F_t$ provided $s\leq t$. Now, I know that if $X$ is $\mathcal G-$measurable, then,$$\mathbb E[X\mid \mathcal G]=X\mathbb E[\boldsymbol 1\mid \mathcal G]=X.$$

Now, I'm trying to do a link of this and the fact that $\mathbb E[M_t\mid \mathcal F_s]=M_s$, but I don't see. So, obviously $M_t$ is not $\mathcal F_s-$measurable, otherwise, we would have that $\mathbb E[M_t\mid \mathcal F_s]=M_t$, but is there an idea near that behind the martingale ?

Bernard
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idm
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  • Yes some processes are such that (2) holds and some are such that it does not. The former, if integrable, are called martingales and the latter are not. Is this your question? – Did Aug 01 '16 at 11:11
  • First, why the downvote ? I the question idiot ? @Did: It's in fact not really my question. I'm trying to understand what is the intuition behind this definition (and in particular what does $\mathbb E[M_t\mid \mathcal F_s]=M_s$ mean exactly). – idm Aug 01 '16 at 13:35
  • Unfortunately, "what does X mean exactly?" is, nearly typically, NARQ. – Did Aug 01 '16 at 13:38
  • @Did: What do you mean by NARQ ? – idm Aug 01 '16 at 13:41
  • http://meta.math.stackexchange.com/questions/3012/common-acronyms-used-on-main-and-meta – Did Aug 01 '16 at 13:48

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