6
$\begingroup$

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 $\mathscr F_{.}$.


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


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

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

$\endgroup$
  • 1
    $\begingroup$ I am joining you in this question and I do not pretend to have the answer. May be that we can see $\color{blue}{\text{Y}}$ as a universal event ? E.g. if I expect to expect $\color{green}{\text{something}}$ in this place, then I also expect that $\color{green}{\text{something}}$ in the univers ? $\endgroup$ – keepAlive Dec 7 '16 at 16:01
  • 1
    $\begingroup$ I don't see the connection. This talk about what you expect to expect tomorrow can be interpreted psychologically (e.g., if you're manic-depressive, you may expect that tomorrow you will expect a better result than you're expecting right now), then it's not (at least not purely) mathematics; or if not, then it's trivial: If you expect your future self to think like you're thinking right now, then you trivially expect all your future expectations to be identical to your current expectations. I don't see the analogy with non-trivial mathematics that you seem to see. $\endgroup$ – joriki May 29 '18 at 15:16
  • 1
    $\begingroup$ A heuristic that does not involve time could be pixelization en.wikipedia.org/wiki/Pixelization. To pixelize an image, you divide it into subregions and for each subregion you make it's color to be the average color in the region. If you do pixelization in two steps, it's the same as just doing one large pixelization. And when you expect something, maybe what you are doing is pixelizing your future knowledge? $\endgroup$ – Mark Sep 8 '18 at 9:59
  • 1
    $\begingroup$ I don't think so. In venn diagram, you are restricting to a particular sub-event. In conditioning, you are restricting to a coarser sigma algebra. But the set of possible outcomes $\omega \in \Omega$ remains unchanged $\endgroup$ – Mark Sep 8 '18 at 10:25
  • 1
    $\begingroup$ Okay I'll try to clarify in an answer. I'm not sure about part 2 yet though. $\endgroup$ – Mark Sep 8 '18 at 10:29
4
+100
$\begingroup$

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

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.

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 $t$.

$\endgroup$
  • $\begingroup$ Thanks Mark! I'll analyse later. $\endgroup$ – BCLC Sep 8 '18 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.