# 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 $$E(E(X|F_1)|F_2) \neq E(E(X|F_2)|F_1)$$

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, 2022 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, 2022 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, 2022 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, 2022 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, 2022 at 16:55

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, 2022 at 18:56
• You re the only one that answered thk you May 30, 2022 at 10:05

Here an other answer. I hope it is true.

So let $$\Omega=\left \{ a,b,c \right \}$$ be our probabilisable space and a r.v. $$X(\omega)=1_{\omega=c}$$. So obviously $$\sigma(X)=\left \{ (a,b) ,(c), \Omega, \varnothing \right \}$$.

1)Now let define:
$$F_1=\left \{\varnothing, \Omega, (a), (b,c) \right \}$$
$$F_2=\left \{\varnothing, \Omega, (b), (a,c) \right \}$$
Obiously none $$F_1$$ or $$F_2$$ is include into the other.

2)We beguin by computing:
2.1)
$$Z_1=E(X|F_1)$$, so $$Z_1$$ can be written as follow: $$Z_1=\alpha1_{\omega=a}+\beta1_{\omega=b \cup c}$$ and by definition $$Z_1$$ must verify: $$\forall A \in F_1,E(Z_1.1_A)=E(X.1_A)$$ If it is true $$\forall A$$ so it is true too in particular for $$A=a$$ pe $$A=b \cup c$$
-For $$A=a$$ we get:
$$E(Z_1.1_A)=\int_{\omega \in \Omega}^{}Z_1.1_ad\mathbb{P}=\alpha.\mathbb{P}(\left \{ \omega \in \Omega:Z_1(\omega)=\alpha \right \})=0=\int_{\omega \in \Omega}^{}1_c.1_ad\mathbb{P}=E(X.1_A)$$
-For $$A=(b,c)$$ $$E(Z_1.1_A)=\int_{\omega \in \Omega}^{}Z_1.1_{b,c}d\mathbb{P}=\beta.\mathbb{P}(\left \{ \omega \in \Omega:Z_1(\omega)=\beta \right \})=\mathbb{P}(\left \{ \omega \in \Omega:X(\omega)=1 \right \})=\int_{\omega \in \Omega}^{}1_c.1_{b,c}d\mathbb{P}=E(X.1_A)$$
So $$\alpha=0, \beta=P(c)$$ and $$Z_1(\omega)=1_{(b,c)}P(c)/P(b,c)$$
2.2)
By doing exactly the same thing for $$Z_2=E(X|F_2)$$ we get: $$Z_2=1_{(a,c)}P(c)/P(a,c)$$

3)Similar as before but now we are looking for:
3.1)

$$Z_{12}=E(E(X|F_1)|F_2)=E(p(c)/p(b,c)1_{(b,c)}|F_1)=p(c)/p(b,c)E(1_{(b,c)}|F_2)$$. So we want to know: $$E(1_{(b,c)}|F_2)$$ so by proceeding as before and according to the definition we are looking for $$Z_{12}$$ that looks like this:
$$Z_{12}=\alpha.1_{(b)}+\beta.1_{(a,c)}$$
$$E(Z_{12}.1_{(b)})=\alpha P(b)=P(b)=E(1_{b})=E(1_{(b,c)}.1_{(b)})$$ So $$\alpha=1$$
$$E(Z_{12}.1_(a,c))=\beta P(a,c)=P(c)=E(1_{c})=E(1_{(b,c)}.1_{(a,c)})$$ So $$\beta=P(c)/P(a,c)$$
3.2)

$$Z_{21}=E(E(X|F_2)|F_1)=E(p(c)/p(a,c)1_{(a,c)}|F_1)=p(c)/p(a,c)E(1_{(a,c)}|F_1)$$. So we want to know: $$E(1_{(a,c)}|F_1)$$. By proceeding as before and according to the definition we are looking for $$Z_{21}$$ that looks like this:
$$Z_{21}=\alpha.1_{(a)}+\beta.1_{(b,c)}$$
$$E(Z_{21}.1_{(a)})=\alpha P(a)=P(a)=E(1_{a})=E(1_{(a,c)}.1_{(a)})$$ So $$\alpha=1$$
$$E(Z_{21}.1_(b,c))=\beta P(b,c)=P(c)=E(1_{c})=E(1_{(b,c)}.1_{(a,c)})$$ So $$\beta=P(c)/P(b,c)$$

4)In conclusion we get:
$$Z_{12}=\frac{P(c)}{P(b,c)}(1_{(b)}+1_{(a,c)}\frac{P(c)}{P(a,c)})$$ wich is different from $$Z_{21}=\frac{P(c)}{P(a,c)}(1_{(a)}+1_{(b,c)}\frac{P(c)}{P(b,c)})$$

Rem: For exemple: $$P(a,c)=P(a \cup c)$$ and $$1_{(a)}$$ is the indicator function.