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We have the following property of conditional expectations from the point of veiw of measure theory ($X$ and $Y$ are two random variables):

$$\mathbb{E}(XY\mid\mathcal{G})=X\operatorname{\mathbb{E}}(Y\mid\mathcal{G}) $$

when $X$ is $\mathcal{G}$-measurable.

Shreve calls this property: Taking out what is known.

Now, I know that how this works in statistics. Also, I know the definition of measurable function in Measure Theory. But I don't get the intuition here. Since when "measurable" is the same as "known"? I always thought that "measurable" random variable was one that can be assigned probabilities and events to. This wouldn't make it necessarily known. Can you explain?

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    $\begingroup$ If you condition on $\mathcal{G}$ you are fixing the randomness that could happen from events there. If you fix this knowledge, since $X$ is $\mathcal{G}$ measurable, you can fully determine how $X$ would behave and what values it would take on. Effectively, it is a constant so you can take it out. $\endgroup$
    – James Yang
    Commented Oct 2, 2018 at 16:39

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You can think of a $\sigma$-field as providing information about the outcome of your experiment. A more refined $\sigma$-field gives you more granular information. (Another interpretation is that the $\sigma$-field captures all the questions you can answer about the experiment, i.e. the events that you are allowed to say for sure whether they have occurred or not.)

For example, consider the sample space $\Omega = \{1,2,3,4\}$. One $\sigma$-field would be the collection of all $2^3=8$ subsets of $\Omega$. Let us call this $\mathcal F$. Another $\sigma$-field could be $$ \mathcal G = \{ \emptyset, \{1,2\},\{3,4\}, \{1,2,3,4\}\}. $$ This second $\sigma$-field has limited information about your experiment. If you only know $\mathcal G$, you cannot differentiate between $1$ and $2$ or similarly between $3$ and $4$. Roughly speaking, you can say for example whether $\{1,2\}$ has occurred but cannot say which element (take this last sentence with a grain of salt.)

A function which is measurable with respect to $\mathcal G$ has to be constant over $\{1,2\}$ and also constant over $\{3,4\}$. For example, $f(1)=f(2) = a$ and $f(2)=f(3)=b$. Then, knowning $\mathcal G$ determines the function in that sense (there would be no ambituity for the value of the function over the most granular subsets in $\mathcal G$.)

On the other hand $h(i) = i$ for $i=1,2,3,4$ is not measurable w.r.t. to $\mathcal G$ since it takes different values over say $\{1,2\}$, hence differentiates between the elements of $\Omega$ more than what $\mathcal G$ allows. Knowing $\mathcal G$ leaves some ambiguity in the value of the function, i.e. $f$ is not deterministic given $\mathcal G$ (but it is given $\mathcal F$).

PS. Also in the above example, if $a \neq b$, then knowing the value of the function $f$ gives the same amount of information about the experiment as the $\sigma$-filed $\mathcal G$. In the language of measure theory, this is captured by saying that $\mathcal G$ is the $\sigma$-field generated by $f$. In the language of probability, you can think of "knowing $\mathcal G$" as "observing the (value of) random variable $f$".

PS2. When you take the conditional expectation $\hat h = E[h \mid \mathcal G]$, you would be taking a (weighted) average of the values of $h$ over those granular subsets $\{1,2\}$ and $\{3,4\}$ and set $\hat h$ to be constant over those sets with those average values, i.e., $$\hat h (1) = \hat h(2) = \frac{p_1}{p_1+p_2} h(1) + \frac{p_2}{p_1+p_2} h(2)$$ and $\hat h (3) = \hat h(4) = \frac{p_3}{p_3+p_4} h(3) + \frac{p_4}{p_3+p_4} h(4)$ (Check! I might be wrong about the weights. If the weights are correct you should also see how $E[E[h \mid \mathcal G]] = E[h]$).

Now you can see that $E [f \mid \mathcal G] = f$ because $f$ is already constant over those sets, hence forming that average doesn't change anything.

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