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Let's say we are tossing two fair coins. Let $X$ be the random variable representing the number of heads.

$X$ is defined on $(\Omega, F, P)$ where $\Omega = \{HH,HT,TH,TT\}$.

Say if we want to find $E[X|G]$ where $G$ is a $\sigma$-algebra which contains the information about the first coin toss. It is clear $G=\{\emptyset, \Omega,\{HH,HT\},\{TT,TH\} \}$, where $G \subset F$. By definition, we know that $E[X|G]$ is a random variable mapping $\Omega \rightarrow \mathbb{R} $

Intuitively we know that the conditional expectation of X given the first toss is a $H$ is $1.5$, and the conditional expection of X given the first toss is a $T$ is $0.5$

However, when we are evaluating this quantity formally, using $E[X|G](\omega)$, wouldn't this mean we already know $\omega$? Why isn't $E[X|G](HH)$ simply $2$?

I do not understand why $E[X|G]$ takes the input $\omega \in \Omega$, instead of a set $s \in G$?

Any help is greatly appreaciated!

Just a thought (can someone confirm if this is correct?): By evaluating $E[X|G](HH)$ we do not actually know $\omega = HH$, but simply that $\omega \in \{HH,HT\}$ since all we are given is $G$

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Let $Y$ denote the number of heads thrown at the first toss.

$\mathbb E[X\mid G]:\Omega\to\mathbb R$ has the following properties:

  • it is measurable wrt $G$ or equivalently some $f:\mathbb R\to\mathbb R$ exists with $\mathbb E[X\mid G]=f(Y)$
  • for every $B\in G$ we have: $\mathbb E[\mathbb E[X\mid G]\mathbf1_B]=\mathbb EX\mathbf1_B$

Combining the bullets we find that $$\mathbb Ef(Y)\mathbf1_B=\mathbb EX\mathbf1_B\text{ for every }B\in G$$

Substituting $B=\{HH,HT\}$ in the second bullet we find:$$\frac14f(1)+\frac14f(1)+\frac140+\frac140=\frac142+\frac141+\frac140+\frac140$$ telling us that $f(1)=\frac32$.

Substituting $B=\{TT,TH\}$ in the second bullet we find:$$\frac140+\frac140+\frac14f(0)+\frac14f(0)=\frac140+\frac140+\frac140+\frac141$$ telling us that $f(0)=\frac12$.


$\mathbb E[X\mid G]$ is defined for every $\omega\in\Omega$ and is prescribed as follows:

  • $HH\stackrel{Y}{\to}1\stackrel{f}{\to}\frac32$
  • $HT\stackrel{Y}{\to}1\stackrel{f}{\to}\frac32$
  • $TT\stackrel{Y}{\to}0\stackrel{f}{\to}\frac12$
  • $TH\stackrel{Y}{\to}0\stackrel{f}{\to}\frac12$
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  • $\begingroup$ Thanks a lot for the answer, this was very helpful! I have a quick question regarding $f(Y)$. Is there a way to evaluate this quality without setting $E[X|G] = f(Y)$? $\endgroup$ – Putter Lertplakorn Oct 17 '18 at 15:59
  • $\begingroup$ The essence is that $\mathbb E[X|G]$ is constant on the sets $\{HH,HT\}$ and $\{TT,TH\}$. That comes to same as that it factors through $Y$. I used that in my answer but you can also focus on "being constant on..." itself. $\endgroup$ – drhab Oct 18 '18 at 8:51

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