Using the general measure-theoretic definition of conditional expectation, let $X$ be a random variable measurable with respect to the $\sigma$-algebra $\mathscr{F}$, and let $\mathcal{G} \subset \mathscr{F}$ and $\mathcal{H} \subset \mathscr{F}$ be two $\sigma$-algebras contained in the larger $\sigma$-algebra $\mathscr{F}$. Note that the intersection of $\sigma$-algebras is again a $\sigma$-algebra.

Question: Is it true in general that $$\mathbb{E}[\mathbb{E}[X|\mathcal{G}]|\mathcal{H}] = \mathbb{E}[X | \mathcal{G} \cap \mathcal{H}] = \mathbb{E}[\mathbb{E}[X|\mathcal{H}]|\mathcal{G}]\,? $$

Note: A pointer to a reference would be more than sufficient for an accepted answer. If you know of a brief and simple counterexample, of course, feel free to post it. /Note

I know that it is true in special cases. For example, the tower property, the case when $\mathcal{G} \subset \mathcal{H}$, is an example. Also the property that $\mathbb{E}[X|\mathcal{A}]=\mathbb{E}[X]$ when $\sigma(X)$ and $\mathcal{A}$ are independent, because their independence implies that their intersection is the trivial $\sigma$-algebra, with respect to which the conditional expectation of any random variable is just its expectation in the conventional sense.

This question and this one seem to discuss what appears to also be a special case, which is according to Billingsley is apparently true. See also this related question.

It almost seems to follow from the definition of conditional expectation, but the definition one usually sees, e.g. on Wikipedia, does not seem to be applicable to the case when $\mathcal{G} \not\subset \mathcal{H}$ and $\mathcal{H} \not\subset\mathcal{G}$.

Useless comments: It would be helpful to know whether or not this is the case, so as to better be able to avoid false identities which contradict this property (in case the property is actually, in fact, true). E.g. if one denotes the join of two $\sigma$-algebras, $\sigma(\mathcal{G} \cup \mathcal{H})$ (the smallest $\sigma$-algebra containing both $\mathcal{G}$ and $\mathcal{H}$) by $\mathcal{G} \lor \mathcal{H}$, it is extremely tempting for me to write stupid things like $$"\mathbb{E}[\mathbb{E}[X|G]|\mathcal{H}] = \mathbb{E}[X| \mathcal{G} \lor \mathcal{H}] = \mathbb{E}[\mathbb{E}[X|\mathcal{H}]|\mathcal{G}] " \,, $$ since in the case that $\mathcal{G} = \sigma(Y)$ and $\mathcal{H} = \sigma(Z)$, so that $\mathcal{G}\lor \mathcal{H} = \sigma(Y,Z)$, it corresponds to the very tempting (in general) false identity $$"\mathbb{E}[\mathbb{E}[X|Y]|Z] = \mathbb{E}[X|Y,Z] = \mathbb{E}[\mathbb{E}[X|Z]|Y] ]" \,. $$

  • 1
    $\begingroup$ I have a doubt, if projection on $A$, then on $B$ is the same point, as first on $B$, then on $A$? Cause Conditional expectation is a projection of $X$ onto $L^2(\mathcal{G})$ if in $L^2$, and the outer equalities hold $\endgroup$ – dEmigOd Feb 5 '18 at 21:49
  • 1
    $\begingroup$ Ah no, what I said before is wrong, I forgot that even in finite dimensions orthogonal projections need not commute. $\endgroup$ – Ian Feb 5 '18 at 22:14

Not equal in general.

For a simple counterexample, assume that $\mathcal G=\sigma(X)$ and $\mathcal H=\sigma(X+Y)$, where $(X,Y)$ is i.i.d. and integrable.

Then $E(X\mid \mathcal G)=X$, $E(Y\mid \mathcal G)=E(Y)$ and $E(X\mid \mathcal H)=\frac12(X+Y)$ hence $E(E(X\mid \mathcal G)\mid \mathcal H)=\frac12(X+Y)$ while $E(E(X\mid \mathcal H)\mid \mathcal G)=\frac12(X+E(Y))$.

Edit: The properties shown above are elementary. Slightly less direct is the fact that, with this choice of $\mathcal G$ and $\mathcal H$, the sigma-algebra $\mathcal G\cap \mathcal H$ is trivial, hence $E(X\mid \mathcal G\cap\mathcal H)=E(X)$. Thus, in general, the conditional expectation $E(X\mid \mathcal G\cap\mathcal H)$ "in the middle", is neither $E(E(X\mid \mathcal G)\mid \mathcal H)$, nor $E(E(X\mid \mathcal H)\mid \mathcal G)$.

  • 1
    $\begingroup$ Finite dimensional version: orthogonally project $(1,0)^T$ onto the span of itself and the span of $(1,1)^T$, in each order. In one case you get a nonzero vector on the x axis, in one case you get a nonzero vector on the line $y=x$, and in the last case you just get 0. This has an analogue with conditional expectations of discrete variables. $\endgroup$ – Ian Feb 5 '18 at 22:22

Consider the following elementary example: $\Omega=\{1,2,3,4\}$, $X(\omega)=\omega$. Assume that $X$'s distribution is uniform. Now, $$\mathscr F=2^{\Omega}.$$Let $$\mathscr H=\{\Omega,\emptyset,\{1,2\},\{3,4\}\}$$ snd $$\mathscr G=\{\Omega,\emptyset,\{1,2,3\},\{4\}\}.$$

We have $$E[X\mid \mathscr H\ ](\omega)=\begin{cases}\frac32&\text{ for }&\omega=1,2\\\frac72&\text{ for }& \omega=3,4\end{cases}$$and$$E[X\mid \mathscr G\ ](\omega)=\begin{cases}2&\text{ for }&\omega=1,2,3\\4&\text{ for }& \omega=4\end{cases}.$$


$$E[E[X\mid \mathscr H\ ]\mathscr G\ ](\omega)=\begin{cases}\frac{13}6&\text{ for }&\omega=1,2,3\\\frac72&\text{ for }& \omega=4\end{cases}.$$ and

$$E[E[X\mid \mathscr G\ ]\mathscr H\ ](\omega)=\begin{cases}2&\text{ for }&\omega=1,2\\3&\text{ for }& \omega=3,4\end{cases}.$$


This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .