probability sigma algebra

Question

I am trying to get $E(X|F)$. But I am not sure if I am right.

The only single variables available in $F$ are $\{1\}$ and $\{ 2\}$. So, the probability of 2 possible events out of 6 events in F is $\frac{2}{6}=1/3$.

Am I right? If not right, how do we get $ E(X|F)$ ? I would appreciate your help.

Answer 1

Let $\mathcal F$ be the $\sigma$-algebra generated by the set of events $F$. Then $\mathcal F$ is the $\sigma$-algebra with the atoms $\{1\}$, $\{2\}$, $\{3,4\}$, since these are in $F$, and since all other events in $F$ are unions of some of the atoms. Then the expression $Y=E[X|\mathcal F]$ is an $\mathcal F$-measurable random variable, it is enough to describe it on the atoms (it is constant on each atom), and we have (assuming uniform probability distribution on $\{1,2,3,4\}$):

• $Y(1)$ is the mean of $X$ on $\{1\}$, so it is $Y(1)=X(1)=1$,
• $Y(2)$ is the mean of $X$ on $\{2\}$, so it is $Y(2)=X(2)=2$,
• $Y(3)=Y(4)$ is the mean of $X$ on $\{3,4\}$, so it is $Y(3)=Y(4)=\frac 12(X(3)+X(4))=\frac 12(3+4)=\frac 72$.

(Just a) ... check: We have $E[X]=\frac 14(1+2+3+4)=\frac {10}4=\frac 52$, and $E[Y]=\frac 14\cdot 1+\frac 14\cdot 2+\frac 12\cdot\frac 72=\frac 14(1+2+7)=\frac {10}4=\frac 52$, so $E[X]=E[Y]$.