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 $\mathbb{E}[X] = \mathbb{E}[X|\mathcal{G}]?$

  • 3
    $\begingroup$ $E[X|\mathcal{A}]=X$. To see this directly, note that trivially $E[\mathbf{1}_AX]=E[\mathbf{1}_AX]$ for any $A\in \mathcal{A}$. $\endgroup$
    – Snoop
    Sep 1 at 18:36
  • $\begingroup$ @Snoop Thank you! This explains $\endgroup$
    – Tom
    Sep 1 at 18:55
  • 1
    $\begingroup$ The simple rule is that the smaller the $\sigma$-algebra ${\cal G}$ is the closer $\mathbb E[X|{\cal G}]$ is to the constant $\mathbb E[X]\,.$ $\endgroup$
    – Kurt G.
    Sep 1 at 19:04


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