The smoothing property of conditional expectation (the version I learned) states that for a random variable $X$, if $\mathcal{G}$ and $\mathcal{H}$ are two linear spaces with $\mathcal{H}\subseteq\mathcal{G}$, then $E[E[X|\mathcal{G}]|\mathcal{H}]=E[X|\mathcal{H}]$.
Then what if $\mathcal{H}$ is not a subset of $\mathcal{G}$? Which estimator will result in a smaller MSE, $E[E[X|\mathcal{G}]|\mathcal{H}]$ or $E[X|\mathcal{H}]$? My current guess is $E[X|\mathcal{H}]$, as I think projecting directly to $\mathcal{H}$ should have smaller error than projecting first to $\mathcal{G}$ and then to $\mathcal{H}$.
Any answers or comments would be appreciated.
When $X$ is $\mathcal H$ measurable and independent of $\mathcal G$, $E[E[X|\mathcal{G}]|\mathcal{H}]$ becomes $EX$ and $E[X|\mathcal H]$ becomes $X$. So no inequality between these holds in general.
In detail, $E[X|\mathcal G]=EX$ by independence so $E[E[X|\mathcal G]|\mathcal H]$ is also $EX$. Next, $E[X|\mathcal H]=X$ because $X$ so already measuable w.r.t. $\mathcal H$.