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For continuous random variables $X, Y$ the conditional expectation $\mathbb{E}[X | Y]$ is itself a random variable. I understood this in the sense that for a realisation of $Y$ we can say

$$ \mathbb{E}[X | Y=y] = \int_{-\infty}^{\infty}xf_{X|Y}(x|y)dx = \frac{1}{f_Y(y)}\int_{-\infty}^{\infty}xf_{X,Y}(x,y)dx. $$

And so I imagine a random elementary event happening on $\Omega$ that gives a corresponding result to $Y$ and allows conditioning $X$ as above.

The signature would then be something like this

$$\mathbb{E}[X | Y]: \Omega \to \mathbb{R}, \quad \omega \mapsto \mathbb{E}[X | Y(\omega)]$$

The conditional expectation here is random because $Y$ is a random variable that gives an output to the random occurrences on the magical space $\Omega$.

However in more advanced courses and textbooks I studied on the matter the conditional expectation is often introduced via sub-$\sigma$-algebras and then defined by some characterisation like this:

Let $X \in L_1(\Omega, \mathcal{A}, \mathbb{P})$ and $Y \in L_1(\Omega, \mathcal{F}, \mathbb{P})$ where $\mathcal{F}$ is a sub-$\sigma$-algebra of $\mathcal{A}$. Then

$$ Y = \mathbb{E}[X | \mathcal{F}] \quad \iff \quad \forall F \in \mathcal{F}: \mathbb{E}[\mathbb{1}_F X] = \mathbb{E}[\mathbb{1}_F Y]. $$

These concepts supposedly coincide as $\mathbb{E}[X | Y] = \mathbb{E}[X | \sigma(Y)]$ and $\mathbb{E}[X | \mathcal{F}]$ is understood to be a random variable as well. But the interpretation gets lost for me.

The above definition of $\mathbb{E}[X | Y]$ is coherent with the common idea that $\Omega$ is the set of outcomes of some (real) random experiment. Selecting an outcome $Y(\omega)$ allows for the calculation of $\mathbb{E}[X | Y=y]$.

But for $\mathbb{E}[X | \mathcal{F}]$ it is not at all clear to me how this concept is "calculated". What is $\mathbb{E}[X | \mathcal{F}](\omega)$? Can you give a non-trivial (non discrete) example of a random experiment and a continuous random variable such that this expression is understandable and has meaning in the model of that random experiment?

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    $\begingroup$ Just as $X$ is actually a function, $X:\,\omega\mapsto X(\omega)$, $\mathbb{E}[X | \mathcal{F}]$ is also a function (of $\omega$) that is furthermore $\mathcal{F}$-measurable and the closest to $X$. The solution does not have to be a constant function. See stats.stackexchange.com/a/230559/7224 $\endgroup$
    – Xi'an
    Commented Aug 11, 2022 at 11:54
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    $\begingroup$ $\Omega$ is far from being some "magical space:" it has an extremely concrete meaning. It can represent all individuals in a population, for instance. Each individual $\omega\in\Omega$ has a definite numerical property $X(\omega).$ That's obviously not random. What is random is the process of selecting one individual from the population. The probability distribution $\mathbb P$ describes the relative chances of selection, thereby endowing $X$ with a distribution, etc. $\endgroup$
    – whuber ♦
    Commented Aug 11, 2022 at 15:45
  • $\begingroup$ @Xi'an I understand that it is also a function of $\omega$, but it is not at all clear to me what that law is. I.e. given an $\omega$, what do you do with it? $\endgroup$
    – lpnorm
    Commented Aug 12, 2022 at 8:03
  • $\begingroup$ @whuber This is clear from basic probability theory. But don't we all have the implicit understanding that elements of $\Omega$ are picked at random? If you eliminate that intuition then all that is left is deterministic relations between deterministic sets and their measures. Sorry if "magical space" offended you, but I don't see how that hyperbole is relevant to the question. $\endgroup$
    – lpnorm
    Commented Aug 12, 2022 at 8:11
  • $\begingroup$ I cannot discern a definite answerable question. What kind of responses are you hoping for? $\endgroup$
    – whuber ♦
    Commented Aug 12, 2022 at 12:44

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