Intuition for Conditional Expectation

Question

It seems like NNT aka Nero in The Black Swan (2007) is giving the law of iterated expectations that involve filtrations in a heuristic way by matching the everyday usage of the word 'expect' with the mathematical definition of expectation (a Riemann integral or sum in elementary probability theory; a Lebesgue or Riemann-Stieltjes integral in advanced probability theory).

I'm guessing the correspondence between the precise and the heuristic is as follows:

Heuristic:

$\text{If I expect to expect} \ \color{green}{\text{something}} \ \text{at} \ \color{red}{\text{some date in the future}},$

$\text{then I already expect that} \ \color{green}{\text{something}} \ \text{at} \ \color{purple}{\text{present}}.$

Precise in the case of one non-trivial $\sigma-$algebra,

$$E[E[\color{green}{X}|\color{red}{\mathscr F_t}]] = E[\color{green}{X}|\color{purple}{\mathscr F_0}] (= E[\color{green}{X}])$$

Or

Precise in the case of two non-trivial $\sigma-$algebras,

$$E[E[\color{green}{X}|\color{red}{\mathscr F_{t+1}}]|\color{purple}{\mathscr F_t}] = E[\color{green}{X}|\color{purple}{\mathscr F_t}]$$

where $\color{green}{X}$ is a random variable in $(\Omega, \mathscr F, \mathbb P)$ with filtration $\{\mathscr F_t\}_{t\in I}$ where $I \subseteq \mathbb R$

An example I thought of for second case

I currently expect to expect tomorrow at 1pm that someone will try to prank me tomorrow at 3pm if and only if I currently expect someone to prank me tomorrow at 3pm.

Where 3pm refers to the larger $\mathscr F_{.}$ and 1pm refers to the smaller $\mathscr F_{.}$.

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1. Anything wrong? If so, please explain why, and suggest how it may be improved.

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2. How to similarly heuristically explain law of iterated expectation when we don't have filtrations?

For example

$$E[E[\color{green}{X}|\color{blue}{Y}]] = E[\color{green}{X}]$$

$\text{If I expect to expect} \ \color{green}{\text{something}}$ _____ $\color{blue}{(?)}$_____,

$\text{then I (?)expect that} \ \color{green}{\text{something}} $ _____ $(?)$ _____

What I tried:

I guess we can consider X as payoff of playing one game out of Y possible games.

So the amount we expect to win is equal to the (probabilistically) weighted average of the amounts we expect to win in each of the Y games.

But I wanted to use similar language to the one with filtrations so I'm looking for something like

If I expect to expect to win 5 dollars (something something) then I expect to win 5 dollars.

Of course without the something something we have simply

$E[E[X]] = E[X]$

Answer 1

Here I will clarify on my pixelization intuition that I started in the comment.

Part 1. Filtrations

Suppose a random variable $\color{green}{X}$ measureable with respect to $\sigma$-algebra $F$. You could make the analogy that $X$ is a high resolution color image and it is displayed on the high resolution monitor $F$, and the pixels are the elements of $F$.

Now maybe you have a friend with a lower resolution monitor, $\color{red}{G} \subset F$, and so it will be impossible for your friend to view $X$. As a compromise he can view $E[X|G]$ so that for each pixel $g \in G$ we have $\int_g E[X|G] \,dp = \int_g X\, dp$.

So at least the color on each of his pixels $g$ is the average of the colors inside the corresponding pixels $\cup_{x \subset g} x$ on your monitor.

In the worst case, his monitor only has one pixel (the $\sigma$ algebra $G=\{ \emptyset, \Omega\})$ and in that case the best we can do is show him the single color $E[X|G] = \int_{\Omega}X\,dp = E[X]$.

Now if your friend sends his blurry picture $E[X|G]$ to his friend with even worse monitor $\color{purple}{H}$, the result would be the same as if you sent the original image $X$ yourself. In other words $E[E[\color{green}{X}|\color{red}{G}]|\color{purple}{H}] = E[\color{green}{X}|\color{purple}{H}]$ (Averaging an average is just an average).

Part 2. Without Filtrations

Now let's say you have yet another friend who has a monitor that's worse than yours but he doesn't know how much worse. But he tells you his monitor can display the image $Y$ at full resolution, but it pixelates any image sharper than $Y$.

You can reason that his monitor's pixels are the minimal $\sigma$ algebra with which $Y$ is measureable. In which case your image $X$ will appear on his monitor as $E[X|Y]$

In your template:

If I pixelize an image $\color{green}{X}$ so it is displayable on a monitor which can display $\color{blue}{Y}$, and then I pixelize that image completely, it is the same as pixelizing $\color{green}{X}$ completely.

Part3. Time

At this point, we can build an analogy on top of an analogy. Suppose at time $\infty$ there will be some important event $X$, which we will be able to view with perfect clarity using our knowledge $F_{\infty}$. Until then, say at time $t$, our mental images of $X$ will only be low resolution $E[X|F_t]$ where $F_t$ encodes the coarseness of our mental pixels at time $t$.