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:
$E(E(X|F_1)|F_2)=E(Z_{12}|F_2)=E(Z_{12}) \neq E(Z_{21})=E(Z_{21}|F_1)=E(E(X|F_2)|F_1)$


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    $\begingroup$ 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) $\endgroup$ May 23, 2022 at 16:27
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    $\begingroup$ (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? $\endgroup$ May 23, 2022 at 16:28
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    $\begingroup$ @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 $\endgroup$
    – X0-user-0X
    May 23, 2022 at 16:44
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    $\begingroup$ 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 $E[E[X|F_2]|F_1] \neq E[X|F_1]$? $\endgroup$ May 23, 2022 at 16:49
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    $\begingroup$ @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. $\endgroup$
    – X0-user-0X
    May 23, 2022 at 16:55

2 Answers 2


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.


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.

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    $\begingroup$ 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 $\endgroup$
    – X0-user-0X
    May 23, 2022 at 18:56
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    $\begingroup$ You re the only one that answered thk you $\endgroup$
    – X0-user-0X
    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:
$Z_1=E(X|F_1)$, 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 $Z_1(\omega)=1_{(b,c)}P(c)/P(b,c)$
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:

$Z_{12}=E(E(X|F_1)|F_2)=E(p(c)/p(b,c)1_{(b,c)}|F_1)=p(c)/p(b,c)E(1_{(b,c)}|F_2)$. 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:
$E(Z_{12}.1_{(b)})=\alpha P(b)=P(b)=E(1_{b})=E(1_{(b,c)}.1_{(b)})$ 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 $\beta=P(c)/P(a,c)$

$Z_{21}=E(E(X|F_2)|F_1)=E(p(c)/p(a,c)1_{(a,c)}|F_1)=p(c)/p(a,c)E(1_{(a,c)}|F_1)$. 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:
$E(Z_{21}.1_{(a)})=\alpha P(a)=P(a)=E(1_{a})=E(1_{(a,c)}.1_{(a)})$ 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.


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