Intuition for $E[E[X\vert \mathcal{H}]\vert \mathcal{G}]\not= E[X\vert \mathcal{G}]$ where $\mathcal{G}\supset\mathcal{H}$?

Question

This is just the general Law of iterated Expectations but with the larger sigma-algebra in the outer (instead of the inner) expectation.

I am wondering if anyone can provide some intuition for why this is not true?

Mathematically I am thinking about it as the outer expectation must be equal to the inner expectation (which equals $E[X]$) for all subsets of $\mathcal{G}$. But this leaves some subsets of $\mathcal{H}$ where the inner and outer expectation can disagree (and thus the outer one is not necessarily $E[X]$ on those sets).

Thanks.

Answer 1

Since $E(X\mid \mathcal{H})$ is $\mathcal{H}$ measurable and thus $\mathcal{G}$ measurable, \begin{align*} E\left(E(X\mid\mathcal{H})\mid\mathcal{G}\right) = E(X\mid\mathcal{H}), \end{align*} which is generally not equal to $E(X\mid\mathcal{G})$, if $\mathcal{H}$ and $\mathcal{G}$ are not equal.