Proof of the tower property for conditional expectations

Question

Let $Z$ be a $\mathfrak{F}$-measurable random variable with $\mathbb E(|Z|)<\infty$ and let $\mathfrak{H}\subset \mathfrak{G}\subset \mathfrak{F}$.

Show that then $\mathbb E(\mathbb E(Z|\mathfrak{G})|\mathfrak{H})=\mathbb E(Z|\mathfrak{H})=\mathbb E(\mathbb E(Z|\mathfrak{H})|\mathfrak{G})$.

The second equality is clear to me since the random variable $E(Z|\mathfrak{H})$ is $\mathfrak{H}$-measurable, hence its also $\mathfrak{G}$-measurable, so we can take it out.

If I want to prove the first equality I have to show that $E(Z|\mathfrak{G})$ and $Z$ do agree on $\mathfrak{H}$-measurable sets, right? Because I have to show that $Z$ and $\mathbb E(Z|\mathfrak{G})$ are equal if I condition both with the sigma algebra $\mathfrak{H}$.

Is this argumentation correct?

For the proof: $\int_\Omega\mathbb E(Z|\mathfrak{G})\mathbb 1_HdP=\int_\Omega Z\mathbb 1_H$ just because $E(Z|\mathfrak{G})$ is $\mathfrak{H}$ measurable, hence "I can remove one $\mathbb E$ and $\mathfrak{G}$ and still have equality".

I know I wrought much for a simple task, but I just want to understand it correctly, thanks :)

Answer 1

Basically, your idea is correct, but you really should try to write this up more formally; otherwise it is hard to tell what you did.

By definition, $Y = \mathbb{E}(Z \mid \mathcal{A})$ if and only if $Y$ is $\mathcal{A}$-measurable and

$$\int_A Y \, d\mathbb{P} = \int_A Z \, d\mathbb{P} \qquad \text{for all} \, \, A \in \mathcal{A}. \tag{1}$$

Now, in the given framework, we have

$$\int_H \mathbb{E}(Z \mid \mathcal{H}) \, d\mathbb{P} = \int_H Z \, d\mathbb{P} \qquad \text{for all} \, \, H \in \mathcal{H}. \tag{2}$$

Moreover, since $\mathcal{H} \subseteq \mathcal{G}$,

$$\int_H \mathbb{E}(Z \mid \mathcal{G}) \, d\mathbb{P} = \int_H Z \, d\mathbb{P} \qquad \text{for all} \, \, H \in \mathcal{H}. \tag{3}$$

Combining $(2)$ and $(3)$ yields

$$\int_H \mathbb{E}(Z \mid \mathcal{H}) \, d\mathbb{P} = \int_H \mathbb{E}(Z \mid \mathcal{G})\, d\mathbb{P}.$$

Now it follows from the definition $(1)$ that $$\mathbb{E}(Z \mid \mathcal{H}) = \mathbb{E}(\mathbb{E}(Z \mid \mathcal{G}) \mid \mathcal{H}).$$

Answer 2

Prove of $\mathbb E(\mathbb E(Z|\mathcal{G})|\mathcal{H})=\mathbb E(Z|\mathcal{H})$:

$\mathbb E(\mathbb E(Z|\mathcal{G})|\mathcal{H})$ is a conditional expectation of $E(Z|\mathcal{G})$, hence $\int_{H}\mathbb E(\mathbb E(Z|\mathcal{G})|\mathcal{H})d P=\int_H\mathbb E(Z|\mathcal{G})dP$ for all $H\in \mathcal{H}$.

$\mathbb E(Z|\mathcal{G})$ is a conditional expectation of $Z$, hence $\int_G\mathbb E(Z|\mathcal{G})dP=\int_GZdP$ for all $G\in \mathcal{G}$.

Combining both equalities and using that $H\in \mathcal{H}$ yields: $\int_{H}\mathbb E(\mathbb E(Z|\mathcal{G})|\mathcal{H})d P=\int_HZdP$ as desired.