# Confused about the Tower Property of expectations

Let $$(\Omega, \mathcal{A}, \mathbb{P})$$ be a probability space, $$X$$ be integrable w.r.t. this space, and $$\mathcal{G} \subset \mathcal{F} \subset \mathcal{A}$$ be sigma algebras.

Then by the tower property, we have that $$\mathbb{E}[\mathbb{E}[X|\mathcal{F}]| \mathcal{G}] = \mathbb{E}[\mathbb{E}[X|\mathcal{G}]| \mathcal{F}] = \mathbb{E}[X|\mathcal{G}]$$.

However, if I take $$\mathcal{F} = \mathcal{A}$$, then the above means $$\mathbb{E}[\mathbb{E}[X]| \mathcal{G}] = \mathbb{E}[\mathbb{E}[X|\mathcal{G}]] = \mathbb{E}[X|\mathcal{G}]$$. Considering the first term equals $$\mathbb{E}[X]$$ since it's an expectation of constant, we have that:

$$\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X|\mathcal{G}]] = \mathbb{E}[X|\mathcal{G}]$$. But how can $$\mathbb{E}[X] = \mathbb{E}[X|\mathcal{G}]?$$

• $E[X|\mathcal{A}]=X$. To see this directly, note that trivially $E[\mathbf{1}_AX]=E[\mathbf{1}_AX]$ for any $A\in \mathcal{A}$. Sep 1 at 18:36
• @Snoop Thank you! This explains
– Tom
Sep 1 at 18:55
• The simple rule is that the smaller the $\sigma$-algebra ${\cal G}$ is the closer $\mathbb E[X|{\cal G}]$ is to the constant $\mathbb E[X]\,.$ Sep 1 at 19:04