The Tower Property

Why Successive Conditioning Works

Conditional expectation represents the information about an integrable random variable that is available through a specified sigma-algebra. When information is organized in levels, it is natural to condition first on a finer level and then summarize what remains at a coarser level. The tower property says that this successive calculation gives the same result as conditioning directly on the coarser information.

The order matters. If $\mathcal H\subseteq\mathcal G$, then $\mathcal G$ contains at least as much information as $\mathcal H$. The tower property starts with conditioning on $\mathcal G$, the finer sigma-algebra, and then conditions that result on $\mathcal H$.

Central idea: Averaging a conditional expectation over a coarser level of information recovers the conditional expectation at that coarser level.

Definition and Characterization

Let $X$ be an integrable random variable on a probability space $(\Omega,\mathcal F,\mathbb P)$, and let $\mathcal G\subseteq\mathcal F$ be a sigma-algebra. A conditional expectation of $X$ given $\mathcal G$, denoted $\mathbb E[X\mid\mathcal G]$, is an integrable, $\mathcal G$-measurable random variable $Y$ such that, for every $A\in\mathcal G$,

\int_A Y\,d\mathbb P=\int_A X\,d\mathbb P.

Such a random variable exists and is unique up to almost-sure equality. In particular, the defining property identifies its integrals over every event that can be distinguished using $\mathcal G$. We use this characterization to prove the tower property.

Two conditional expectations that are equal almost surely are regarded as versions of the same conditional expectation. Thus identities between conditional expectations below are understood to hold almost surely.

Theorem (Tower Property): Let $X$ be integrable, and let $\mathcal H$ and $\mathcal G$ be sigma-algebras satisfying $\mathcal H\subseteq\mathcal G\subseteq\mathcal F$. Then $$ \mathbb E\!\left[\mathbb E[X\mid\mathcal G]\mid\mathcal H\right] = \mathbb E[X\mid\mathcal H] \quad\text{almost surely}. $$

Proof. Set $Y=\mathbb E[X\mid\mathcal G]$ and $Z=\mathbb E[Y\mid\mathcal H]$. Since $Y$ is integrable, $Z$ is a well-defined conditional expectation. It is $\mathcal H$-measurable and integrable. For any $A\in\mathcal H$, the inclusion $\mathcal H\subseteq\mathcal G$ ensures $A\in\mathcal G$. Apply the defining integral property first to $Z=\mathbb E[Y\mid\mathcal H]$, and then to $Y=\mathbb E[X\mid\mathcal G]$:

\int_A Z\,d\mathbb P = \int_A Y\,d\mathbb P = \int_A X\,d\mathbb P.

Thus $Z$ is $\mathcal H$-measurable and has the required integrals over every $A\in\mathcal H$. It satisfies the defining characterization of $\mathbb E[X\mid\mathcal H]$. Uniqueness up to almost-sure equality gives the stated identity. $\square$

The proof uses only the defining property and uniqueness of conditional expectation. The key step is that every event in the coarser sigma-algebra is also an event in the finer one, so the integral identity for conditioning on $\mathcal G$ applies to all the tests required for conditioning on $\mathcal H$.

Successive levels: If $\mathcal H_1\subseteq\mathcal H_2\subseteq\cdots\subseteq\mathcal H_n$, then for integrable $X$, $$ \mathbb E\!\left[\mathbb E[X\mid\mathcal H_n]\mid\mathcal H_{n-1}\right] = \mathbb E[X\mid\mathcal H_{n-1}] \quad\text{almost surely}. $$ Applying this identity repeatedly gives $\mathbb E[X\mid\mathcal H_1]$.

Worked Examples

Worked Example: Moving from a Fine Partition to a Coarse Partition

Let $\Omega=\{(a,b,c):a,b,c\in\{0,1\}\}$, with all eight outcomes equally likely. Let $\mathcal G$ record the first two coordinates $(a,b)$, and let $\mathcal H$ record only the first coordinate $a$. Then $\mathcal H\subseteq\mathcal G$.

Define $X$ by the following values, grouped according to $(a,b)$: when $(a,b)=(0,0)$, the values for $c=0,1$ are $0,2$; when $(0,1)$, they are $4,6$; when $(1,0)$, they are $2,4$; and when $(1,1)$, they are $8,10$. Given $(a,b)$, the two possible outcomes have equal probability. Hence the conditional averages are

\mathbb E[X\mid\mathcal G]= \begin{cases} 1,&(a,b)=(0,0),\\ 5,&(a,b)=(0,1),\\ 3,&(a,b)=(1,0),\\ 9,&(a,b)=(1,1). \end{cases}

Given $a=0$, the two values of $b$ are equally likely, so the average of the $\mathcal G$-conditional values is $(1+5)/2=3$. Given $a=1$, that average is $(3+9)/2=6$. Therefore,

\mathbb E\!\left[\mathbb E[X\mid\mathcal G]\mid\mathcal H\right] = \begin{cases} 3,&a=0,\\ 6,&a=1. \end{cases}

Directly averaging $X$ over the four outcomes with $a=0$ gives $(0+2+4+6)/4=3$. Averaging over the four outcomes with $a=1$ gives $(2+4+8+10)/4=6$. Thus direct conditioning on $\mathcal H$ gives the same result as successive conditioning.

Worked Example: Unequal Probabilities Within a Coarse Event

Let $\Omega=\{1,2,3,4\}$ with probabilities $\mathbb P(\{1\})=1/2$, $\mathbb P(\{2\})=1/4$, and $\mathbb P(\{3\})=\mathbb P(\{4\})=1/8$. Let $\mathcal H$ have atoms $\{1,2\}$ and $\{3,4\}$, and let $\mathcal G$ have atoms $\{1\}$, $\{2\}$, and $\{3,4\}$. Then $\mathcal H\subseteq\mathcal G$. Set $X(1)=0$, $X(2)=8$, $X(3)=4$, and $X(4)=12$.

Because $X$ is constant on each atom of $\mathcal G$, $\mathbb E[X\mid\mathcal G]$ has values $0,8,8$ on $\{1\},\{2\},\{3,4\}$, respectively. On the coarse event $\{1,2\}$, the conditional probabilities of outcomes 1 and 2 are $2/3$ and $1/3$. Therefore,

\mathbb E[\mathbb E[X\mid\mathcal G]\mid\mathcal H] = \frac{0\cdot(1/2)+8\cdot(1/4)}{1/2+1/4} = \frac{8}{3} \quad\text{on }\{1,2\}.

On $\{3,4\}$, the two outcomes have equal conditional probability, so the conditional average is $(4+12)/2=8$. Directly conditioning $X$ on $\mathcal H$ gives these same values: $8/3$ on $\{1,2\}$ and $8$ on $\{3,4\}$. The weighting is important: averaging the two values on $\{1,2\}$ equally would be incorrect because outcomes 1 and 2 do not have equal probability.

Worked Example: Successive Conditional Probabilities

Consider eight equally likely cards, divided into four groups according to color (red or black) and label (A or B). Each color-label group contains two cards. Suppose the number of aces in the groups red-A, red-B, black-A, and black-B is, respectively, $1,0,0,2$. Let $X$ be the indicator that the selected card is an ace. Let $\mathcal G$ record both color and label, and let $\mathcal H$ record only color.

Within the four groups, the probabilities of an ace are $1/2,0,0,1$. Thus $\mathbb E[X\mid\mathcal G]$ takes those four values. Given red, labels A and B are equally likely, so the average of the two conditional probabilities is $(1/2+0)/2=1/4$. Given black, it is $(0+1)/2=1/2$. Therefore,

\mathbb E[\mathbb E[X\mid\mathcal G]\mid\mathcal H] = \begin{cases} 1/4,&\text{the card is red},\\ 1/2,&\text{the card is black}. \end{cases}

These are also the direct probabilities of drawing an ace given the color: among the four red cards there is one ace, and among the four black cards there are two. Since $X$ is an indicator, the calculation is a conditional-probability version of the tower property.

Ordinary Expectation as the Coarsest Level

The trivial sigma-algebra $\{\varnothing,\Omega\}$ records no event information beyond certainty and impossibility. Conditioning on it produces the constant ordinary expectation. In particular, the tower property implies that taking an ordinary expectation after conditioning on $\mathcal G$ recovers the original expectation.

Corollary: For every integrable $X$ and every sigma-algebra $\mathcal G\subseteq\mathcal F$, $$ \mathbb E\!\left[\mathbb E[X\mid\mathcal G]\right]=\mathbb E[X]. $$

Proof. Apply the tower property with $\mathcal H=\{\varnothing,\Omega\}$. A conditional expectation given this sigma-algebra is constant, and its integral over $\Omega$ must equal $\int_\Omega X\,d\mathbb P$. That constant is therefore $\mathbb E[X]$. The tower identity gives the result. $\square$

Why the Direction Cannot Be Reversed

The inclusion $\mathcal H\subseteq\mathcal G$ is not a cosmetic assumption. It makes every test event available at the finer level also available at the coarser level. If one conditions first on $\mathcal H$, then conditions that result on the finer $\mathcal G$, no additional detail about $X$ is restored: $\mathbb E[X\mid\mathcal H]$ is already $\mathcal H$-measurable, and hence $\mathcal G$-measurable. Its conditional expectation given $\mathcal G$ is itself.

For example, in the first worked example, $\mathbb E[X\mid\mathcal H]$ is 3 when $a=0$ and 6 when $a=1$. Conditioning this on $\mathcal G$ leaves those values unchanged, while $\mathbb E[X\mid\mathcal G]$ takes the more detailed values 1, 5, 3, and 9. Coarsening information can average away detail; conditioning the coarsened result cannot reconstruct it.

Common pitfall: With $\mathcal H\subseteq\mathcal G$, the tower property is $$ \mathbb E[\mathbb E[X\mid\mathcal G]\mid\mathcal H] = \mathbb E[X\mid\mathcal H], $$ not the same expression with the sigma-algebras reversed.

Check Your Understanding

Use the defining integral property and the inclusion of sigma-algebras to answer the following.

In the tower property, why must every $A\in\mathcal H$ also belong to $\mathcal G$?
Which integrability assumption on $X$ is required for the stated theorem?
For nested sigma-algebras $\mathcal H_1\subseteq\mathcal H_2\subseteq\mathcal H_3$, what does repeated conditioning on these levels reduce to?
Why is an ordinary average of values on a coarse event incorrect in the second worked example?
What does the tower property give when the coarser sigma-algebra is $\{\varnothing,\Omega\}$?

The Tower Property

What You'll Learn

Why Successive Conditioning Works

Definition and Characterization