Explicitly compute conditional expectation of a random variable on the discrete probability space

Question

I am working Exercise 4.1.5 on Durrett $3^{rd}$, which states as follows:

Give an example on $\Omega=\{a,b,c\}$ in which $$\mathbb{E}(\mathbb{E}(X|\mathcal{F}_{1})|\mathcal{F}_{2})\neq\mathbb{E}(\mathbb{E}(X|\mathcal{F}_{2})|\mathcal{F}_{1}).$$

I know that to achieve this, we need $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ not including each other, otherwise if $\mathcal{F}_{1}\subset\mathcal{F}_{2}$, then both side will be $\mathbb{E}(X|\mathcal{F}_{1})$, since the smaller $\sigma-$algebra always wins.

So I did this:

$\Omega:=\{a,b,c\}$, and consider $$\mathcal{F}_{1}:=\sigma(\{a\})=\Big\{\varnothing, \{a,b,c\}, \{a\}, \{b,c\}\Big\},$$ $$\mathcal{F}_{2}:=\sigma(\{b\})=\Big\{\varnothing, \{a,b,c\}, \{b\}, \{a,c\}\Big\}.$$

Then I need to compute $\mathbb{E}(X|\mathcal{F}_{1})$, but how could I compute this? I tried to wikipedia this but it seems that it does not give a general way to compute conditional expectation with condition upon a $\sigma-$algebra.

Do I directly plug in $\mathbb{E}(X(a)|\mathcal{F}_{1})$? Then do I need to assign values to $X(a), X(b)$ and $X(c)$?

Thank you!

Answer 1

Recall that the conditional expectation $\mathbb{E}(X \mid \mathcal{G})$ is measurable with respect to the sub $\sigma$-algebra $\mathcal{G}$. Consequently, it suffices to find a random variable $X$ such that

$$\mathbb{E} \big( \mathbb{E}(X \mid \mathcal{F}_1) \mid \mathcal{F}_2 \big)$$

is not $\mathcal{F}_1$-measurable. This implies, then, automatically that

$$\mathbb{E} \big( \mathbb{E}(X \mid \mathcal{F}_1) \mid \mathcal{F}_2 \big) \neq \mathbb{E} \big( \mathbb{E}(X \mid \mathcal{F}_2) \mid \mathcal{F}_1 \big). \tag{1}$$

(Just suppose that both conditional expectations were equal, then the left-hand side would be $\mathcal{F}_1$-measurable in contradiction to our choice of $X$.)

Take

$$X=m 1_{\{a\}} + M 1_{\{b,c\}}.$$

Clearly, $X$ is $\mathcal{F}_1$-measurable and so $\mathbb{E}(X \mid \mathcal{F}_1) = X$. To compute the conditional expectation wrt $\mathcal{F}_2$ we can use the following general statement.

Let $X$ be an integrable random variable and $\mathcal{F}$ a $\sigma$-algebra generated by a partition $(F_j)_{j \in \mathbb{N}}$, i.e. $\Omega = \bigcup_j F_j$, the sets $F_j$ are disjoint and $\mathcal{F}=\sigma(F_j, j \geq 1)$. Then $$\mathbb{E}(X \mid \mathcal{F}) = \sum_{j: \mathbb{P}(F_j)>0} \frac{\mathbb{E}(X 1_{F_j})}{\mathbb{P}(F_j)} 1_{F_j}$$

Since $\mathcal{F}_2=\sigma(F_1,F_2)$ for $F_1=\{b\}$, $F_2=\{a,c\}$, we get

\begin{align*} \mathbb{E}(X \mid \mathcal{F}_2) &=\frac{ \mathbb{E}(X 1_{\{b\}})}{\mathbb{P}(\{b\})} 1_{\{b\}} + \frac{\mathbb{E}(X 1_{\{a,c\}})}{\mathbb{P}(\{a,c\})} 1_{\{a,c\}} \\ &= M 1_{\{b\}} + \frac{m \mathbb{P}(\{a\})+M \mathbb{P}(\{c\})}{\mathbb{P}(\{a,c\})} 1_{\{a,c\}}.\end{align*}

Since $\{b\} \notin \mathcal{F}_1$, this fails to be $\mathcal{F}_1$-measurable if, say, $M>0$. Consequently, we can simply take

$$X = M 1_{\{b,c\}}$$

and, by construction, this random variable satisfies $(1)$.