Iterated conditioning

Question

I understand how this holds:

$\mathbb{E}(\mathbb{E}(X|\mathit{G})|\mathit{H}) = \mathbb{E}(X|\mathit{H})$

where $\mathit{H} \subset \mathit{G} $.

Then, how does this hold:

$\mathbb{E}(\mathbb{E}(X|\mathit{G})) = \mathbb{E}(X)$

as here, $\mathit{H} = \Omega$ and now $\mathit{G} \subset \Omega$ not $\Omega \subset \mathit{G}$.

Answer 1

Let $(\Omega,\mathcal{F},P)$ be a probability space.

Taking ordinary expectation does not correspond to taking the conditional expectation conditioned on $\mathcal{F}$ (which is probably what you intended to write, since conditioning on $\Omega$ does not make sense). Instead, it corresponds to conditioning on the trivial sigma-field $\{\varnothing,\Omega\}$, i.e. $$ {\rm E}[X]={\rm E}[X\mid\{\varnothing,\Omega\}] $$ for any integrable variable $X$.

In particular ${\rm E}[{\rm E}[X\mid\mathcal{G}]]={\rm E}[X]$ does not contradict the statement that $$ {\rm E}[{\rm E}[X\mid\mathcal{G}]\mid\mathcal{H}]={\rm E}[X\mid\mathcal{H}] $$ whenever $\mathcal{H}\subseteq\mathcal{G}$ since the trivial sigma-field $\{\varnothing,\Omega\}$ is a subset of every sigma-field.

Answer 2

Here $H$ and $G$ are sigma-algebras, not events.

For example $E(E(X\mid G))=E(X)$ for every sigma-algebra $G$ is the case $H=\{\varnothing,\Omega\}$ of the preceding identity [$E(E(X\mid G)\mid H)=E(X\mid H)$ if $H\subseteq G$], since for this sigma-algebra $H$, $H\subseteq G$ is always true and $E(Y\mid H)=E(Y)$ for every (integrable) random variable.