E[E[A|B]|C], law of total expectation with two conditional expectations 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.
 A: 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 $G \in \mathcal{G}_2$.
