Conditional probability over two different $\sigma$-algebra

Question

Let $\Omega = \{a,b,c\}$ be the set of outcomes of an experiment. Let $\mathbb P$ be a probability measure defined by setting $$ \mathbb P (\{a\}) = 1/2, \mathbb P (\{b\}) = 1/4,\mathbb P (\{c\}) = 1/4. $$ We also define the real-valued random variable X by $$X(a)=0, X(b)=X(c)=2$$ Consider the two sub sigma-algebras of $$\mathcal F_1 = \{\emptyset,\Omega,\{a\},\{b,c\}\},\mathcal F_2 = \{\emptyset,\Omega,\{a,b\},\{c\}\}$$ We need to compute $$\mathbb E[\mathbb E[X|\mathcal F_1]|\mathcal F_2] \ and \ \mathbb E[\mathbb E[X|\mathcal F_2]|\mathcal F_1]$$ Can someone explain what is partial average in this case? I'm really a little bit confused of this concept.

Answer 1

First note that $X$ is $\mathcal F_1$-measurable, so $\mathbb E[X|\mathcal F_1]=X$. So, we need to find $$ \mathbb E[X|\mathcal F_2] \text{ and } \mathbb E[\mathbb E[X|\mathcal F_2] |\mathcal F_1]$$

The conditional expectation $\mathbb E[X|\mathcal F_2]$ is $\mathcal F_2$-measurable random variable: $$ \mathbb E[X|\mathcal F_2]=\begin{cases}c_1, & \omega=a \text{ or } \omega=b, \cr c_2, & \omega=c. \end{cases} $$ Here $c_1=\mathbb E[X|\{a,b\}]$ and $c_2=\mathbb E[X|\{c\}]$ - ordinary conditional expectations with respect to events. $$ c_1 = \mathbb E[X|\{a,b\}] = \frac{\mathbb E[X;\{a,b\}]}{\mathbb P(\{a,b\})}=\frac{0\cdot 1/2+2\cdot 1/4}{1/2+1/4} = \frac23, $$ $$ c_2 = \mathbb E[X|\{c\}] = \frac{\mathbb E[X;\{c\}]}{\mathbb P(\{c\})}=\frac{2\cdot 1/4}{1/4} = 2. $$ So $$ Y=\mathbb E[X|\mathcal F_2]=\begin{cases}\dfrac23, & \omega=a \text{ or } \omega=b, \cr 2, & \omega=c. \end{cases} $$ Next you need to repeat the similar calculations to obtain $$\mathbb E[\mathbb E[X|\mathcal F_2] |\mathcal F_1] = \mathbb E[Y|\mathcal F_1]=\begin{cases}\dfrac23, & \omega=a , \cr \dfrac43, & \omega=b \text{ or }\omega=c. \end{cases}$$