I have some understanding issues following the proof of the tower property of conditional expectation.
The Theorem is the following:
Let $F_0, F_1$ be $\sigma$-fields with $F_0 \subseteq F_1 \subseteq F$ and $X$ a random variable with $X \geq 0$. Then,
$$ \mathbb{E}[\mathbb{E}[X|F_1]|F_0] = \mathbb{E}[X|F_0] \qquad(1)$$
and
$$\mathbb{E}[\mathbb{E}[X|F_0]|F_1] = \mathbb{E}[X|F_0] \qquad(2)$$
$P$ almost surely.
The proof given states:
For $Y_0\geq0$, $Y_0$ measurable with respect fo $F_0$, $\mathbb{E}[Y_0\mathbb{E}[X|F_1]] = \mathbb{E}[Y_0X]$ and this proves (1).
Question (only regarding (1))
First point I do not get. $Y_0$ is by definition $F_0$ measurable.
So, I would assume that we have
$$\mathbb{E}[Y_0\mathbb{E}[X|F_1]|F_0].$$ Then, I assume that we plug in the defining property of the conditional expectation which yields
$$\mathbb{E}[Y_0X|F_0]$$
–> Why is this possible? For plugging $X$ in, $Y_0$ has to be $F_1$ measurable, but it is only $F_0$ measurable?
–> And then, where goes "|$F_0$"? I don't understand why this part vanishes…
So, I am kind of lost. Thanks a million for any help and further explenations! 🙂
Best Answer
$F_0 \subset F_1$. So the fact that $Y_0$ is $F_0$ measurable implies that it is also $F_1$ measurable: $Y_o^{-1}(E) \in F_0 \subset F_1$ for any Borel set $E$ in $\mathbb R$.
Second question: $E(E(Z_0|F_0))=EZ$ for any $Z$. This is immediate from the definition of conditional expectation: $\int_{A} E(Z|F_0) dP =\int_{A} ZdP$ for any $A \in F_0$. Take $A=\Omega$ to get $\int_{\Omega} E(Z|F_0) dP =\int_{\Omega} ZdP$. This means $E(E(Z|F_0))=EZ$.