# Exemple where tower property of conditional expectation is NOT verify

Question:
Let $$\Omega=\{a,b,c\}$$. Give an example for $$X, F_1, F_2$$ in which $$E(E(X|F_1)|F_2) \neq E(E(X|F_2)|F_1)$$

My answer:
I am not at all sure of my answer. If you have any shorter and nicer answer i will be happy to read it.

-Let define:
(a) $$F=B(\Omega ), \; F_1=\left \{ \Omega;\left \{ \emptyset \right \} ;\left \{ a \right \};\left \{ b;c \right \}\right \}, \; F_2=\left \{ \Omega;\left \{ \emptyset \right \} ;\left \{ b \right \};\left \{ a;c \right \}\right \}$$. By def: $$Z_{12}=E(X|F_1)$$ is a rv $$F_1$$ measurable, $$Z_{21}=E(X|F_2)$$ is a rv $$F_2$$ measurable.
(b) X a bijective measurable function from $$(\Omega ; B(\Omega )) \rightarrow (\left \{ 1;2;3 \right \}; B(\left \{ 1;2;3 \right \}))$$

-Proof: $$Z_{12} \neq Z_{21} \; a.s$$
By absurd, we assume that: $$Z_{12} = Z_{21} \; a.s \; \Rightarrow E(Z_{12}) = E(Z_{21})$$.
(i) But on the other side we have: $$E(Z_{12}|F_1) = Z_{12}$$ because is $$F_1$$ measurable.
(ii) And by the absurd assumption: $$E(Z_{21}|F_1) = E(Z_{21}) = E(Z_{12})$$
So we get from (i)+(ii): $$Z_{12}= E(Z_{21})$$ Wich is not necessarly always true.
And of course $$Z_{12} \neq Z_{21} \; a.s \; \Rightarrow E(Z_{12}) \neq E(Z_{21})$$

-Now from what we just wrotte above:
$$E(E(X|F_1)|F_2)=E(Z_{12}|F_2)=E(Z_{12}) \neq E(Z_{21})=E(Z_{21}|F_1)=E(E(X|F_2)|F_1)$$

-Q.E.D

• Here are some concerns. (1) The Borel sigma-algebra is the one generated by open intervals. I don't know how you define intervals in a finite set $\{a,b,c\}$. You probably meant the powerset, not the Borel sigma-algebra. (2) The empty set is $\emptyset$, not $\{\emptyset\}$. (continued) May 23 at 16:27
• (3) Where do you get this definition of the tower property from? I think the tower rule is $E[E[\dots|F_1]] = E[E[\dots|F_2]]$, and not $E[E[\dots|F_1]|F_2] = E[E[\dots|F_2]|F_1]$. You can find something reminiscent of what you wrote here, but those expressions assume $F_1 \subseteq F_2 \subseteq F$, which clearly breaks in your example. To sum up, I am not sure what exactly you are trying to explore, disprove or find a counter-example to. Maybe you could give more context and/or references to your source? May 23 at 16:28
• @paperskilltrees thk for your comment. 1-I mean the sigma algebra generated by all the singleton 2- I am exactly trying to find, for the given univers, an exemple where tower property is not verify and to know if the exemple i bring is correct. 3- the def is from here. en.wikipedia.org/wiki/Conditional_expectation 4-i cannot be more precise that in the question May 23 at 16:44
• Re (3): as far as I can see, the property is $F_1 \subseteq F_2 \subseteq F \implies E[E[X|F_2]|F_1]=E[X|F_1]$. Are you trying to find an example for when $E[E[X|F_2]|F_1] \neq E[X|F_1]$? May 23 at 16:49
• @paperskilltrees No i am trying to find $F_1$ and $F_2$ s.t.$E(E(X|F1)|F2)≠E(E(X|F2)|F1)$ with $F_1$ and $F_2$ as i wish but must be sigma algebra on the given univers. May 23 at 16:55

## 1 Answer

Take this with a grain of salt, as I am also a learner of Probability.

Your example is correct in the design (take $$F_{1,2}$$ such that $$F_1 \not\subseteq F_2, F_2 \not\subseteq F_1$$), but is wrong in the implementation. In particular, you write:

$$Z_{12}=E(E(X|F_1)|F_2)$$ is a rv $$F_{1}$$ measurable

which I believe is false. While $$E[E[X|F_1]|F_2]$$ is $$F_2$$-measurable by definition, it does not have to be $$F_1$$-measurable.

Let's unwrap your example using the notation $$Z = (Z(a), Z(b), Z(c))$$ for the values of r.v. $$Z$$ on $$\Omega = \{a, b, c\}$$. I use the uniform probability measure $$P(a)=P(b)=P(c)=1/3$$ to compute the expectations.

• $$X=(1,2,3)=E[X|X]$$, $$\quad E[X]=(2,2,2)$$,
• $$E[X|F_1]=(1,\frac{5}{2},\frac{5}{2})$$, $$\quad E[X|F_2]=(2,2,2)$$,
• $$E[E[X|F_1]|F_2]=(\frac{7}{4}, \frac{5}{2}, \frac{7}{4})$$, $$\quad E[E[X|F_2]|F_1]=(2, 2, 2)$$.

Clearly, $$E[E[X|F_1]|F_2] \neq E[E[X|F_2]|F_1]$$. So you have your concrete counter-example. Done. Now let's check some other statements.

$$Z_{12}\neq Z_{21}$$ a.s.

While true in our particular example, the truthfulness of this statement depends on the choice of the measure (and other things). Had we chosen $$P$$, s.t. $$P(c)=1, P(a)=P(b)=0$$, we would have $$X \equiv (1,2,3) = (0,0,3)$$ (almost everywhere), and all the above expectations would also be a.e. equal to it.

$$E(Z_{12}|F_1)=Z_{12}$$

This does not hold, $$(\frac{7}{4}, \frac{17}{8}, \frac{17}{8}) \neq (\frac{7}{4}, \frac{5}{2}, \frac{7}{4})$$, and illustrates that $$Z_{12}=E(E(X|F_1)|F_2)$$ does not have to be $$F_1$$-measurable. Note that $$E(Z_{21}|F_2)=Z_{21}$$ holds by coincidence.

I could not come up with anything as abstract as the proof you attempted, nor did I find a way to remedy it. But if you need just one simple example, this should suffice.

• Thk for your answer. It's an error of typing, thk to have noted it, i of course meaned $Z_{12}=E(X|F_1)$ is a rv $F_1$ measurable, $Z_{21}=E(X|F_2)$ is $F_2$ measurable. I ll read your answer when i ll have time May 23 at 18:56
• You re the only one that answered thk you May 30 at 10:05