E[E[A|B]|C], law of total expectation with two conditional expectations

Question

I have a conditional expectation inside of a conditional expectation.

Let A, B, C be real-valued random variables.

I am trying to simplify E[E[A|B]|C]. I am not sure much can be done in general; however, in this case, I know that C$\in$B.

I also know from TLOTE: E[E[A|B]]= E[A].

Intuition tells me that since C is a subspace of B, then I should wind up with E[A|C], but I am short on ideas to show it mathematically.

Thank you.

Answer 1

One may show that if $\mathbb{E}[\lvert X \rvert] < \infty$, then for $\mathcal{G}_1\subset\mathcal{G}_2\subset\mathcal{F}$ it holds that $$ \mathbb{E}\!\left[\left. \mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right] \, \right| \mathcal{G}_1\right] = \mathbb{E}\!\left[X \, | \, \mathcal{G}_1\right]. $$ To proof this, note e.g. that for $G \in \mathcal{G}_1$, $$ \int_G \mathbb{E}\!\left[\left. \mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right] \, \right| \mathcal{G}_1\right] \mathrm{d}\mathbb{P} = \int_G \mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right]\mathrm{d}\mathbb{P} = \int_G X \, \mathrm{d}\mathbb{P}, $$ since $\mathbb{E}\!\left[\left. \mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right] \, \right| \mathcal{G}_1\right]$ is the conditional expectation of $\mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right]$ given $\mathcal{G}_1$, since $\mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right]$ is the conditional expectation of $X$ given $\mathcal{G}_2$, and since $\mathcal{G}_1 \subset \mathcal{G}_2$ implies $G \in \mathcal{G}_2$.