Conditional Expectation: What happens if you take conditional expectation on trivial sigma field? Consider for the trvial $\sigma$ - field $\mathcal{F}_0 = \{\emptyset , \Omega\}$,
What is Conditional expectation of the following in the following cases when $A = \emptyset$ and $A = \Omega$ ???
? Can someone please help me fill in the ?? below, as this would help improve my understanding a lot ?
$$\int_? E[X | \mathcal{F}_0]1_A dP =  ?    \;\; \forall A \in \mathcal{F}_0$$
Question 2: And What if I just condition on the $\sigma$-field, 
$$ \int_? E[X | \mathcal{F}] dP = ? $$
For the second question, I guess it is = X right? Since X is already $\mathcal{F}$- measurable by definition of random variable, if given the $\sigma$ - feld $\mathcal{F}$, everything is known, there is no randomness in X.
 A: The conditional expectation over the trivial sigma field would be the Expectation of the random variable. To see this, let $\mathcal{F}$ be the trivial sigma field. Then by def of conditional expectation,
$$E[E[X|\mathcal{F}]1_A] = E[X1_A] \quad \forall A \in \mathcal{F}$$
But A can take on only two values i.e. $\{\emptyset,\Omega\}$.
Hence substitute $A=\Omega$ to get $E[X|\mathcal{F}] = E[X]$ as a candidate solution which works.
A: As Stefan Hansen commented, I think you mean $E[X|\mathcal{F}]$ and $E[X|\{\emptyset,\Omega\}]$. Note that by definition, we have: For a r.v. $X$ (integrable) on a probability space $(\Omega,\mathcal{F},P)$ and a sub $\sigma-$algebra $\mathcal{G}$ of $\mathcal{F}$, the conditional expectation is the unique $\mathcal{G}$ measurable and integrabel r.v. $Z$ such that
$$E[X\mathbf1_G]=E[Z\mathbf1_G]$$
for all $G\in \mathcal{G}$. Usually one writes $Z:=E[X|\mathcal{G}]$. It follows immediately, in your case, that $E[X|\mathcal{F}]=X$ and $E[X|\{\emptyset,\Omega\}]=E[X]$. For the first, just check the definition.
For the second you may use that if you have $G_i\in \mathcal{F}$ for $1\le i\le N\le \infty$ pairwise disjoint with $P(G_i)>0$ and $\bigcup G_i=\Omega$ and let $\mathcal{G}=\sigma(G_i,1\le i\le N)$ then $E[X|\mathcal{G}]=\sum_{i=1}^NE[X|G_i]\mathbf1_{G_i}$. Apply it for $N=1$, i.e. $G_1=\Omega$.
EDIT: By the special structure of $\mathcal{G}$ you know $E[X|F_i]=E[X\mathbf1_{G_i}]/P(G_i)$
A: Let $(\Omega,\mathcal{F},P)$ be a probability space. Suppose $X$ is an integrable random variable and let $\mathcal{G}$ be a sub-sigma-field of $\mathcal{F}$. The conditional expection $E[X\mid\mathcal{G}]$ is the unique random variable that satisfies:
1) $E[X\mid\mathcal{G}]$ is $\mathcal{G}$-measurable.
2) $\int_A E[X\mid\mathcal{G}]\,\mathrm dP = \int_A X\,\mathrm dP$ for all $A\in\mathcal{G}$.
I'm assuming you want to find expressions for $E[X\mid\mathcal{F}]$ and $E[X\mid\{\emptyset,\Omega\}]$. For the first conditional expectation, try showing that $X$ satisfies 1) and 2), and for the last conditional expectation, try with $E[X]$.

One some times sees $E[X\mid\mathcal{G}]$ as our best guess of $X$ given the information contained in $\mathcal{G}$. Try holding this up with the expectations when $\mathcal{G}=\mathcal{F}$ and $\mathcal{G}=\{\emptyset,\Omega\}$.
