I am currently on a quest to understand the conditional expectation $\mathbb{E}[X \mid \mathcal{H}]$ defined on the probability space $(\Omega, \mathcal{F}, P)$ with $\mathcal{H} \subset \mathcal{F}$ and $X$ a random variable.
I know that if $\mathcal{H} = \{\emptyset, \Omega\}$ (i.e. no information) then $\mathbb{E}[X \mid \mathcal{H}] = \mathbb{E}[X]$, and if $\mathcal{H} = \mathcal{F}$ (i.e. perfect information) we have $\mathbb{E}[X \mid \mathcal{H}] = X$. It seems to me then that the conditional expectation is a device to "recover" $X$ given imperfect information. Is this intuition reasonable?
The diagram below seems to justify my intuition:
(Here we have that $\mathcal{B,C}$ are algebras generated by certain intervals, and $\mathcal{A}$ the entire algebra.)
The above diagram used the Lebesgue measure on $[0,1]$. I am now interested in generating similar diagrams for arbitrary probability measures. Is this possible? That is to say, given a random variable given "explicitly", say $X(\omega) = \omega^2$ and a PDF $f$, can one construct a diagram as above?
Will the diagram depend on what probability measure I choose?
I tried my hand at constructing one but ended up in an odd situation in which I had to calculate not $\int t f(t) dt$ but $\int t^2 f(t)^2 dt$ over each interval to get the correct result. I assume this is because I'm somehow treating $\omega$ as my r.v.
Let $\Omega = [0,1]$ with the $\sigma$-field of Borel sets and $\mathbb{P}$ the Lebesgue measure on $[0,1]$. And let $$ \xi(x) = 2x^2,\qquad \eta(x)=\begin{cases}1 & \text{if } x\in[0,\frac 1 3],\\2 & \text{if } x\in(\frac 1 3,\frac 2 3),\\0 & \text{if } x\in[\frac 2 3,1].\end{cases} $$
Then, for $x\in[0,\frac 1 3]$ $$ \mathbb{E}[\xi|\eta](x) = \mathbb{E}\Big[\xi|[0,\frac 1 3]\Big](x) = \frac{1}{\frac{1}{3}}\int_0^\frac{1}{3}2x^2{\rm d}\!x = \frac{2}{27} ; $$ for $x\in(\frac 1 3,\frac 2 3)$ $$ \mathbb{E}[\xi|\eta](x) = \mathbb{E}\Big[\xi|(\frac 1 3,\frac 2 3)\Big](x) = \frac{1}{\frac{1}{3}}\int_\frac{1}{3}^\frac{2}{3}2x^2{\rm d}\!x = \frac{14}{27} ; $$ for $x\in[\frac 2 3,1]$ $$ \mathbb{E}[\xi|\eta](x) = \mathbb{E}\Big[\xi|[\frac 2 3,1]\Big](x) = \frac{1}{\frac{1}{3}}\int_\frac{2}{3}^1 2x^2{\rm d}\!x = \frac{38}{27}. $$
(extract from Zdzislaw Brzezniak, Tomasz Zastawniak: "Basic Stochastic Processes: A Course Through Exercises", Example 2.3 - p. 20)