I am playing a bit with properties of conditional expectation and I am wondering about the following problem.
Let $(\Omega,\Sigma, \mathbb P)$ be a probability space, let $X$ be a real random variable taking support in $[0, 1]$. Let also $\mathcal F$ and $\mathcal G$ be two sub $\sigma$-algebras of $\Sigma$. Let also $\mathcal H$ be the smallest sub $\sigma$-algebra of $\Sigma$ that contains both $\mathcal F$ and $\mathcal G$, I want to know if the following is true.
$\mathbb E[\mathbb E[X|\mathcal F]|\mathcal G]=\mathbb E[X|\mathcal G]$ a.s. $\Leftrightarrow$ $\mathbb E[X|\mathcal H]=\mathbb E[X|\mathcal F]$ a.s.
Now it is easy to prove $\Leftarrow$, since $\mathbb E[X|\mathcal G]=\mathbb E[\mathbb E[X|H]|G]=\mathbb E[\mathbb E[X|\mathcal F]|\mathcal G]$. I cannot prove the other direction, maybe someone has an idea.
Best Answer
Here is a counter-example to the direction $(\Rightarrow)$:
Let $\Omega = \{1, 2, 3, 4\}$ with $\Sigma = 2^{\Omega}$, and let
$$\mathcal{F} = \sigma(\{1,2\}, \{3,4\}) \qquad\text{and}\qquad \mathcal{G}=\sigma(\{1,4\}, \{2,3\}). $$
Also, let $X : \Omega \to [0, 1]$ be defined by
$$ X(\omega) = c\omega $$
where $c$ is a small positive number so that $0 < c < 4c \leq 1$. Then, with the uniform measure $\mathbf{P}$ on $(\Omega, \Sigma)$ that assigns the probability $\frac{1}{4}$ to each singleton event of $\Sigma$, we get
\begin{align*} \mathbf{E}[X\mid\mathcal{F}] &= \mathbf{E}[X \mid \{1,2\}]\mathbf{1}_{\{1,2\}} + \mathbf{E}[X \mid \{3,4\}]\mathbf{1}_{\{3,4\}} \\ &= \frac{3c}{2}\mathbf{1}_{\{1,2\}} + \frac{7c}{2}\mathbf{1}_{\{3,4\}}. \end{align*}
By a similar argument, we can also check that
\begin{align*} \mathbf{E}[X\mid\mathcal{G}] = \frac{5c}{2} = \mathbf{E}[\mathbf{E}[X\mid\mathcal{F}]\mid\mathcal{G}]. \end{align*}
On the other hand, by noting that $\mathcal{H} = \sigma(\mathcal{F}, \mathcal{G}) = \Sigma$, we get
\begin{align*} \mathbf{E}[X\mid\mathcal{H}] = X \neq \mathbf{E}[X\mid\mathcal{F}]. \end{align*}