# Example for $E[E(X\mid\mathcal{G_1})\mid\mathcal{G_2}] \ne E[E(X\mid\mathcal{G_2})\mid\mathcal{G_1}]$

If $$X$$ is an integrable random variable on probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$ and $$\mathcal{G_1}, \mathcal{G_2}$$ sub sigma fields of $$\mathcal{F}$$ then how can we find an example where $$\Omega = \{a, b, c\}$$ in which $$E[E(X\mid\mathcal{G_1})\mid\mathcal{G_2}] \ne E[E(X\mid\mathcal{G_2})\mid\mathcal{G_1}].$$

I would really appreciate if you could analytically show the steps of conclusion. I am self learning.

$$\def\Ω{{\mit Ω}}\def\F{\mathscr{F}}\def\G{\mathscr{G}}\def\emptyset{\varnothing}$$Take $$\F = 2^\Ω$$ and denote $$p(ω) = P(\{ω\})$$ for $$ω \in \Ω = \{a, b, c\}$$. If$$\G_1 = \{\Ω, \emptyset, \{a\}, \{b, c\}\},\ \G_2 = \{\Ω, \emptyset, \{b\}, \{a, c\}\},$$ then for any random variable $$X$$ on $$(Ω, \F)$$,$$\begin{gather*} E(X \mid \G_1)(ω) = \begin{cases} X(a); & ω \in \{a\}\\ \dfrac{p(b) X(b) + p(c) X(c)}{p(b) + p(c)}; & ω \in \{b, c\} \end{cases},\\ E(X \mid \G_2)(ω) = \begin{cases} X(b); & ω \in \{b\}\\ \dfrac{p(a) X(a) + p(c) X(c)}{p(a) + p(c)}; & ω \in \{a, c\} \end{cases}, \end{gather*}$$ which implies$$\begin{gather*} E(E(X \mid \G_1) \mid \G_2)(ω) = \begin{cases} \dfrac{p(b) X(b) + p(c) X(c)}{p(b) + p(c)}; & ω \in \{b\}\\ \small\dfrac{p(a)(p(b) + p(c)) X(a) + p(b)p(c) X(b) + (p(c))^2 X(c)}{(p(a) + p(c))(p(b) + p(c))}; & ω \in \{a, c\} \end{cases},\\ E(E(X \mid \G_2) \mid \G_1)(ω) = \begin{cases} \dfrac{p(a) X(a) + p(c) X(c)}{p(a) + p(c)}; & ω \in \{a\}\\ \small\dfrac{p(a)p(c) X(a) + p(b)(p(a) + p(c)) X(b) + (p(c))^2 X(c)}{(p(a) + p(c))(p(b) + p(c))}; & ω \in \{a, c\} \end{cases}. \end{gather*}$$ In order to have $$E(E(X \mid \G_1) \mid \G_2) ≠ E(E(X \mid \G_2) \mid \G_1)$$, it suffices to make$$E(E(X \mid \G_1) \mid \G_2)(a) ≠ E(E(X \mid \G_2) \mid \G_1)(a),$$ i.e.$$\frac{p(a)(p(b) + p(c)) X(a) + p(b)p(c) X(b) + (p(c))^2 X(c)}{(p(a) + p(c))(p(b) + p(c))} ≠ \frac{p(a) X(a) + p(c) X(c)}{p(a) + p(c)},$$ which can be simplified as $$X(b) ≠ X(c)$$ assuming $$p(ω) > 0$$ for $$ω \in \Ω$$.
To summarize, it suffices to take $$p(a) = p(b) = p(c) = \dfrac{1}{3}$$ and $$X = I_{\{c\}}$$.