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I was reading this answer that gives some intuition about what conditional expectation is. I can more or less understand it now but at some point in the response it is written :

A single numerical response is not enough because the particular piece of information that I will give you is itself random. In fact, your response is necessarily a function of this particular piece of information. Mathematically, this is reflected in the requirement that ${\mathbb E}(X\ |\ {\cal F})$ must be $\cal F$ measurable.

How can I see that this mathematically translate to ${\mathbb E}(X\ |\ {\cal F})$ is $\cal F$ measurable?

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roi_saumon
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2 Answers2

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Let's assume that $\mathcal{F}$ is the sigma algebra generated by event $A$. Then $$\mathbb{E}(X|\mathcal{F})(\omega) = C_1\cdot \mathbb{1}_A(\omega) + C_2\cdot \mathbb{1}_{A^C}(\omega)$$ Hence, if we know whether or not event $A$ happened, we know this value. That is essentially what the measurability means intuitively.

Does that make sense?

Stan Tendijck
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  • Does measurability intuitively means this? I just know the definition that tells that the inverse image of a set should be in $\cal{F}$, or something like this. But how to link this definition with what you say that measurability intuitively means? – roi_saumon Feb 26 '19 at 14:45
  • So are we not talking about the same notion of measurable function? I am a bit confused... – roi_saumon Feb 26 '19 at 14:59
  • This is more like the intuitive sense, not the formal sense. – Stan Tendijck Feb 26 '19 at 16:14
  • @StanTendijck But how do you link both, intuitive and formal sense? I don't get it... – roi_saumon Feb 27 '19 at 13:29
  • I don't think I can help you any further. I believe your problem is: why is the conditional expectation measurable? Not per se what is the conditional expectation. In the more formal sense, the expression I gave you is measurable since it consists of a linear combination of indicator functions with sets, elements of the sigma algebra. If you have a different set $B$ and you say that this event happened then we don't know per se whether $A$ happened or not and you can't conclude stuff about the above expression. – Stan Tendijck Feb 27 '19 at 14:02
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In the context of that answer, $\mathbf E(X|\mathcal F)$ is the best guess of a random variable $X$ in view of the information contained in $\mathcal F$.

In some cases (as in the parity example in the answer) the best estimate of $X$ given $\mathcal F$ is not a fixed number but it depends on the specific value in $\mathcal F$. In the die example

$$\mathbf E(Die|Parity) = \begin{cases}4 & \text{if $Parity=Even$},\\ 3& \text{if $Parity = Odd$},\end{cases}$$

This means that $\mathbf{E}(X|\mathcal F)$ is a random variable, however its value can be uniquely determined using information in $\mathcal F$ (that is, wether the die rolled is odd or even); hence, $\mathbf{E}(X|\mathcal F)$ has to be $\mathcal F$-measurable.

Riccardo Sven Risuleo
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