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Answer by bakula for Conditional expectation with respect to a $\sigma$-algebra

  1. One can think of the "information carried by a $\sigma$-algebra" as a measure of "roughness" of the events we are considering.

For example, think of the probability space modelling the outcome of rolling a dice: $$X = \{1,2,3,4,5,6\}.$$ The standard $\sigma$-algebra we could use here is just the power set of $X$; that is, we can consider any subset of $X$ as an event. However, if we are interested only in the parity of the outcome, then a smaller $\sigma$-algebra will suffice:$$\mathcal{F} = \{\emptyset, \{1,3,5\}, \{2,4,6\}, X\}.$$ If we restrict ourselves to this $\sigma$-algebra, we can still measure the probability of events which we can express in terms of parity, e.g. "the outcome was an even number". On the other hand, if we wanted to consider a question such as "was the outcome a prime number?", we would have to refine our $\sigma$-algebra in such a way that would make the subset $\{2,3,5\}$ measurable.

  1. As you said, the intuition behind the concept of conditional expectation can be drawn from (1). If $X$ is a random variable, we may (informally) think of $\mathbb{E}[X|\mathcal{F}]$ as the random variable which describes (imitates) $X$ as close as possible, while having only events from $\mathcal{F}$ in its "output".

Note: The "closeness" mentioned in (2) can actually be formalized in the following way: if we restrict our attention to the (Hilbert) space $L^2$ of square-integrable variables, then $\mathbb{E}[X|\mathcal{F}]$ is the orthogonal projection of $X$ onto the subspace consisting of $\mathcal{F}$-measurable functions.


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