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I'm quite new in probability, and I would like to understand better the concept of filtration. So let $(\Omega ,\mathcal F,\mathbb P)$ a probability space, $(\mathcal F_n)_{n=1}^\infty $ a filtration and $(X_n)_{n\in\mathbb N}$ a stochastic process (adapted to the filtration).

Question : Let $m>n$ and to simplify, $(X_n)$ describe the a tossing dice and $X_n$ is the result at the $n-$th toss. My teacher always say : "$\mathbb E[X_m\mid \mathcal F_n]$ is the expectation of $X_m$ when $\mathcal F_n$ is reveal."

I'm not sure exactly how to interpret this. I know that $\mathbb E[X_m\mid \mathcal F_n]=\mathbb E[X_m\mid X_n]$ (it's a bit more clear for the notation).

  • Does it mean that : $\mathbb E[X_m\mid X_n]$ is the expectation of $X_m$ just before we played $X_n$ (i.e. just before tossing the dice a $n-$th time, and thus we don't know the result of $X_n$, but we know the results of all the $n-1$ first tosses) OR is it the expectation of $X_m$ when $X_n$ has been already played (i.e. we know the result of the $n-$th toss, but not the one of the $n+1$-th toss) ?
Pierre
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    You might want to take a look at this related question. – saz Mar 09 '19 at 14:11
  • @saz: Unfortunately I can't comment under your answer. But I have one question : When you say : "Then {=1} is not contained in (). Why? Once we have observed (), we cannot tell whether ∈{=1}". Do you fix an $\omega $ ? If yes, this depend on $\omega $. Since if $R(\omega )=0$ or $R(\omega )=2$ we know if $\omega \in {U=1}$ or not... May be you wanted to say : when we see $R(\omega )$ for all $\omega$... – Pierre Mar 09 '19 at 16:13
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    $\omega$ is fixed but arbitrary, i.e. we observe $R$ many times (leading to observations $R(\omega_1),R(\omega_2),\ldots$). If ${U=1}$ was $\sigma(R)$-measurable this would mean that for any of these observations we can decide whether $\omega_i \in {U=1}$ ... but we can't because sooner or later we will observe $R(\omega)=1$ and then we can't tell whether $\omega \in {U=1}$ or not. – saz Mar 09 '19 at 17:22
  • @saz : I see now, thank you. If I understand well, there are $\omega $ s.t. we know if $\omega \in A$ or not for all $A\in \sigma (R)$, but there are also $\omega \in \Omega $ and $A\in \sigma (R)$ s.t. we don't know if there are in $A$ or not. Correct ? – Pierre Mar 09 '19 at 20:00
  • I'm a bit confused because you are writing "for all $A \in \sigma(R)$". In the given example, i.e. $A={U=1}$, it is true that there are (as you say) some $\omega$ s.t. we can decide if $\omega \in A$ or not and there are some $\omega$ s.t. we can't decide whether $\omega \in A$. Hence, $A \notin \sigma(R)$. (Because if $A \in \sigma(R)$, then you should be able to tell (with probability $1$) whether $\omega \in A$ or $\omega \notin A$). – saz Mar 09 '19 at 20:09

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