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 :)

  • 1
    $\begingroup$ Why can you "remove one $\mathbb{E}$ and $\mathcal{G}$"? That's exactly what you want to prove, isn't it? $\endgroup$
    – saz
    Mar 1 '15 at 14:08
  • $\begingroup$ Because if I write down $\mathbb E(X|\mathfrak{G})$, then $\mathbb E(X|\mathfrak{G})$ is $\mathfrak{G}$-measurable and the integral of it along a $\mathfrak{G}$-measurable set does agree with the integral of $X$ along the same $\mathfrak{G}$-measurable set. This is part of our definition of conditional expectation. $\endgroup$ Mar 1 '15 at 14:21

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 $$\mathbb{E}(Z \mid \mathcal{H}) = \mathbb{E}(\mathbb{E}(Z \mid \mathcal{G}) \mid \mathcal{H}).$$

  • $\begingroup$ Can you look at my prove? I posted it as an answer. $\endgroup$ Mar 1 '15 at 15:23
  • $\begingroup$ Why identity (3) holds? $\endgroup$
    – Hermi
    Sep 15 '20 at 1:29
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    $\begingroup$ @Hermi By the definition of the conditional expectation, it holds that $$\int_G \mathbb{E}(Z \mid \mathcal{G}) \, d\mathbb{P} = \int_G Z \, d\mathbb{P}$$ for all $G \in \mathcal{G}$. Since $\mathcal{H} \subseteq \mathcal{G}$ this implies in particular$$\int_H \mathbb{E}(Z \mid \mathcal{G}) \, d\mathbb{P} = \int_H Z \, d\mathbb{P}$$ for all $H \in \mathcal{H}$. $\endgroup$
    – saz
    Sep 15 '20 at 5:10

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.

  • $\begingroup$ How do you arrive at the first equation? $\mathbb{E}(\mathbb{E}(Z \mid \mathcal{G}) \mid \mathcal{H})$ being the conditional expectation of $\mathbb{E}(Z \mid \mathcal{G})$ implies $$\int_H \mathbb{E}(\mathbb{E}(Z \mid \mathcal{G}) \mid \mathcal{H}) \, d\mathbb{P} = \int_H \mathbb{E}(Z \mid \color{red}{\mathcal{G}}) \, d\mathbb{P}.$$ $\endgroup$
    – saz
    Mar 1 '15 at 15:29
  • $\begingroup$ Yes, you are right. It should be $\mathcal{G}$. I got more problems with this than I thought :/ $\endgroup$ Mar 1 '15 at 15:31
  • $\begingroup$ But $H\in \mathcal{H}\subset \mathcal{G}$. So this is not a big problem? $\endgroup$ Mar 1 '15 at 15:33
  • 1
    $\begingroup$ No, it is not a big problem. Your proof is now exactly the same as my proof. $\endgroup$
    – saz
    Mar 1 '15 at 15:37
  • 1
    $\begingroup$ Alright, thanks a lot! :) $\endgroup$ Mar 1 '15 at 15:39

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