# How to geometrically interpret conditional expectation property tower rule??

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?

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.

• 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? Oct 24, 2018 at 5:37
• 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'.
– Cain
Oct 24, 2018 at 18:02
• 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! Oct 26, 2018 at 3:57
• 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 $\mathcal{G}$).
– Cain
Oct 28, 2018 at 5:54