Question about tower property of conditional expectation

Question

If $\mathcal{F}_1 \subset \mathcal{F}_2$

Then $\mathbb{E}(\mathbb{E}(X|\mathcal{F}_2)|\mathcal{F}_1) = \mathbb{E}(X|\mathcal{F}1)$

Do I understand correctly that if we change order of conditional expectations nothing changes:

$\mathbb{E}(\mathbb{E}(X|\mathcal{F}_1)|\mathcal{F}_2) = \mathbb{E}(X|\mathcal{F}_1)$

So we only get the information available from smaller sigma algebra

Answer 1

Your new formula is also correct because $E(X|\mathcal F_1)$ is already measurable w.r.t. $\mathcal F_2$ so conditioning it w.r.t. $\mathcal F_2$ has no effect.