Showing $E(E(\xi|\mathcal G)|\mathcal H)=E(\xi|\mathcal H)$ if $\mathcal H\subset \mathcal G$

Question

Showing $E(E(\xi|\mathcal G)|\mathcal H)=E(\xi|\mathcal H)$ if $\mathcal H\subset \mathcal G$

by Definition I have;

$\displaystyle\int_B E(\xi|\mathcal G)\ dP=\int_B\xi\ dP$ for any $B\in\mathcal G$, also

$\displaystyle\int_B E(\xi|\mathcal H)\ dP=\int_B\xi\ dP$ for any $B\in\mathcal H$

both combined gives;

$\displaystyle\int_B E(\xi|\mathcal G)\ dP=\displaystyle\int_B E(\xi|\mathcal H)\ dP$ for any $B\in\mathcal H\subset\mathcal G$ and using the definition again gives;

$\displaystyle\int_B E(E(\xi|\mathcal G)|\mathcal H)\ dP=\displaystyle\int_B E(\xi|\mathcal H)\ dP$

now how does the equality follow ? Is there a uniqueness argument, if so then they are almost surely equal, am I wrong ?

Answer 1

Yes, you do need uniqueness.

$E(\xi|\mathcal H)$ is the a.s. unique random variable, such that

$$\int_B E(\xi|\mathcal H) dP = \int_B\xi dP,$$ for all $B\in \mathcal{H}$.

But $E(E(\xi|\mathcal G)|\mathcal H)$ also has this property. Indeed, for all $B\in\mathcal{H}$, we have $$\int_B E(E(\xi|\mathcal G)|\mathcal H) dP = \int_B E(\xi|\mathcal{G})dP=\int_B\xi dP,$$ where the last equality follows from $\mathcal{H}\subseteq\mathcal{G}$.

Thus $E(E(\xi|\mathcal G)|\mathcal H) = E(\xi|\mathcal{H})$ almost surely.