What can we say about this conditional expectation of a conditional expectation?

Question

I have just started on probability theory and I was thinking about this object.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $\mathcal{F_1},\mathcal{F_2}$ be sub $\sigma$-algebras. If we just have any random variable on the original probability space, can we say anything about $\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}]?$

Can we simplify this in some way? I know there is some sort of tower property given the inclusion, however I am explicitly not assuming it here. In particular I am conjecturing $\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}]=\mathbb{E}[X|\sigma(\mathcal{F_1},\mathcal{F_2})]$ but I am not sure if that is true?

Many thanks in advance!

Answer 1

No, you are wrong. If $\mathcal F_1 \subseteq \mathcal F_2$, then $$ \mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}] = \mathbb{E}[\mathbb{E}[X|\mathcal{F_2}]|\mathcal{F_1}] = \mathbb{E}[X|\mathcal{F_1}] . $$ We used to call this rule "coarse wins". But in this case we have $\sigma(\mathcal{F_1},\mathcal{F_2}) = \mathcal F_2$, so your conjecture makes the wrong choice.

If neither of $\mathcal F_1 , \mathcal F_2$ contains the other, then the composition $\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}]$ in general will not be of the form $\mathbb E[X | \mathcal G]$ for some sigma-algebra $\mathcal G$.

Answer 2

A simple counter-example is obatined by taking $\mathcal F_1=\{\emptyset, \Omega\}$ and $\mathcal F_1=\mathcal F$. In this case LHS is $EX$ and RHS is $X$.