I have a question on how to interpret conditional expectation its properties geometrically. There are two properties of conditional expectation in particular that I’m trying to interpret:

Given a probability space $(\Omega, \mathcal{G}, \mathbb{P})$

  • If $\mathcal{H} \subset \mathcal{G}$ then $\mathbb{E}[\mathbb{E}[X|\mathcal{H}] | \mathcal{G}] = \mathbb{E}[ X|\mathcal{H}]$. (tower rule)

  • If $ X \mathrel{\unicode{x2AEB}} G$ then $\mathbb{E}[X|\mathcal{G}] = \mathbb{E}[ X]$.

I interpret the tower rule geometrically using linear algebra as follows:

  • $H \subset G$ is interpreted as $\mathcal{H}$ is a subspace contained in the subspace $\mathcal{G}$.

  • $\mathbb{E}[X|\mathcal{H}]$ is the projection of $X$ onto $\mathcal{H}$.

  • So, $\mathbb{E}[\mathbb{E}[X|\mathcal{H}] | \mathcal{G}] = \mathbb{E}[ X|\mathcal{H}]$ can be interpreted as the projection of X onto $\mathcal{H}$, projected onto $\mathcal{G}$, is the same as the projection of X onto $\mathcal{H}$.

This makes sense. However, the reason I don't like this interpretation is that it is also true if I swap $\mathcal{H}$ and $\mathcal{G}$. So, this doesn't use the property that $\mathcal{H} \subset \mathcal{G}$. Can someone help me come up with a better interpretation of this?

related: Question about conditional expectation as projection


1 Answer 1


Let's start with a little Linear Algebra warm-up. Let $V$ be a vector space (which might be infinite dimensional, in that case we will require it to be a Hilbert space), and let $W \subset U$ be two (closed) sub-spaces of $V$. Denote by $P_W$ and $P_U$ the orthogonal projections on $W$ and $U$, respectively. It is straightforward to check $$P_W\cdot P_U = P_U\cdot P_W = P_W.$$ That is, the operators commute and their product is precisely $P_W$.

Back to the probabilistic setting. We consider the Hilbert space $L^2(\mathcal{G})$ of square integrable random variables measureable with respect to $\mathcal{G}$. Now suppose that there exists $\mathcal{H}_1 \subset \mathcal{H}_2 \subset \mathcal{G}$, and consider the closed subspaces $L^2(\mathcal{H}_1)$, $L^2(\mathcal{H}_2)$ of square integrable random variables with respect to $\mathcal{H}_1$ and $\mathcal{H}_2$. It is clear that $L^2(\mathcal{H}_1) \subset L^2(\mathcal{H}_2)$, and by the previous comment $$P_{L^2(\mathcal{H}_1)}\cdot P_{L^2(\mathcal{H}_2)} = P_{L^2(\mathcal{H}_2)}\cdot P_{L^2(\mathcal{H}_1)}= P_{L^2(\mathcal{H}_1)}.$$

Now, we can interpert $P_{L^2(\mathcal{H}_1)}(X)$ as $\mathbb{E}\left[X|\mathcal{H}_1\right]$ and $P_{L^2(\mathcal{H}_2)}(X)$ as $\mathbb{E}\left[X|\mathcal{H}_2\right]$ to get the desired result.

  • $\begingroup$ but this also seems to imply we can swap $\mathcal{H}$ and $\mathcal{G}$ on the LHS of the tower rule and get the same result? $\endgroup$
    – makansij
    Oct 24, 2018 at 5:37
  • 2
    $\begingroup$ Yes, this is exactly what it means. If $\mathcal{H} \subset \mathcal{G}$ then $\mathbb{E}[\mathbb{E}[X|\mathcal{H}]|\mathcal{G} ]= \mathbb{E}[\mathbb{E}[X|\mathcal{G}]|\mathcal{H} ] =\mathbb{E}[X|\mathcal{H} ]$. In other words, the smaller sigma algebra 'wins'. $\endgroup$
    – Cain
    Oct 24, 2018 at 18:02
  • $\begingroup$ Interesting. Now I understand why I've heard the Tower Rule interpreted as you can always condition on more information. It is because $\mathcal{H}$ is less information than $\mathcal{G}$. So, if you know $E[X|\mathcal{H}]$, then if you condition that on more information $E[E[X\\mathcal{H}|\mathcal{G}]$, then it doesn't change the result! $\endgroup$
    – makansij
    Oct 26, 2018 at 3:57
  • 1
    $\begingroup$ Yup. Actually, in this case, it follows from the fact that if $X$ is measurable with respect to $\mathcal{G}$, then $\mathbb{E}[X|\mathcal{G}] = X$. (Here $\mathbb{E}[X|\mathcal{H}]$ is measurable with respect to $\mathcal{H}$, and thus, by inclusion, also with respect to $\mathcal{G}$). $\endgroup$
    – Cain
    Oct 28, 2018 at 5:54

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