What Is the Tower Property of Conditional Expectation?

The tower property describes how conditional expectation behaves when one information set is contained inside another. It lets you simplify repeated conditioning: if you first condition on finer information and then on coarser information, the final result is the same as conditioning on the coarser information directly.

This identity is also called the law of iterated expectations. It is useful in probability, statistics, and stochastic processes, where calculations often involve several levels of information.

Statement and notation

Let $X$ be an integrable random variable, meaning $\mathbb{E}[|X|]<\infty$. Let $\mathcal{H}$ and $\mathcal{G}$ be sigma-algebras such that $\mathcal{H}\subseteq\mathcal{G}$. The inclusion means that $\mathcal{G}$ contains at least as much information as $\mathcal{H}$.

The tower property states that

$$\mathbb{E}\!\left[\mathbb{E}[X\mid\mathcal{G}]\mid\mathcal{H}\right] =\mathbb{E}[X\mid\mathcal{H}] \quad\text{almost surely.}$$

Conditional expectations are defined only up to changes on events of probability zero, so the equality is almost sure rather than necessarily pointwise everywhere.

The same statement is often written using random variables. If $Y$ and $Z$ represent information and $Y$ is determined by $Z$, then

$$\mathbb{E}\!\left[\mathbb{E}[X\mid Y,Z]\mid Y\right] =\mathbb{E}[X\mid Y].$$

Here, knowing both $Y$ and $Z$ is at least as informative as knowing $Y$ alone.

Why the property holds

Write $U=\mathbb{E}[X\mid\mathcal{G}]$ and $V=\mathbb{E}[U\mid\mathcal{H}]$. By definition, $V$ is measurable with respect to $\mathcal{H}$. To show that it equals $\mathbb{E}[X\mid\mathcal{H}]$, check the defining integral property of conditional expectation.

For every event $A\in\mathcal{H}$, the inclusion $\mathcal{H}\subseteq\mathcal{G}$ ensures that $A$ also belongs to $\mathcal{G}$. Therefore,

$$\begin{aligned} \int_A V\,d\mathbb{P} &=\int_A U\,d\mathbb{P}\\ &=\int_A X\,d\mathbb{P}. \end{aligned}$$

The first equality follows because $V$ is the conditional expectation of $U$ given $\mathcal{H}$. The second follows because $U$ is the conditional expectation of $X$ given $\mathcal{G}$. Thus $V$ is $\mathcal{H}$-measurable and has the required integrals over every event in $\mathcal{H}$. By uniqueness of conditional expectation up to almost-sure equality, $V=\mathbb{E}[X\mid\mathcal{H}]$.

Worked finite example

Suppose the sample space has eight equally likely outcomes, $\omega_1,\ldots,\omega_8$, and $X$ takes the following values. Let $\mathcal{H}$ distinguish only whether the outcome is among the first four or the last four. Let $\mathcal{G}$ distinguish the four pairs shown below. Thus $\mathcal{H}\subseteq\mathcal{G}$.

$\mathcal{H}$ block	$\mathcal{G}$ block	Values of $X$	$\mathbb{E}[X\mid\mathcal{G}]$
$\{\omega_1,\ldots,\omega_4\}$	$\{\omega_1,\omega_2\}$	$0,2$	$1$
$\{\omega_1,\ldots,\omega_4\}$	$\{\omega_3,\omega_4\}$	$4,6$	$5$
$\{\omega_5,\ldots,\omega_8\}$	$\{\omega_5,\omega_6\}$	$10,12$	$11$
$\{\omega_5,\ldots,\omega_8\}$	$\{\omega_7,\omega_8\}$	$14,16$	$15$

Given the first $\mathcal{H}$ block, the two possible $\mathcal{G}$-conditional values, 1 and 5, are equally likely. Their average is 3. Given the second block, the corresponding values 11 and 15 average to 13. Therefore,

$$\mathbb{E}\!\left[\mathbb{E}[X\mid\mathcal{G}]\mid\mathcal{H}\right] =\begin{cases} 3, & \omega\in\{\omega_1,\ldots,\omega_4\},\\ 13, & \omega\in\{\omega_5,\ldots,\omega_8\}. \end{cases}$$

Conditioning $X$ directly on $\mathcal{H}$ gives the same block averages: the first four values average to 3, and the last four average to 13. This verifies the tower property in a concrete case.

Useful consequences

If $\mathcal{H}$ is the trivial sigma-algebra, conditioning on it gives an ordinary expectation. The tower property then yields the law of total expectation:

$$\mathbb{E}\!\left[\mathbb{E}[X\mid\mathcal{G}]\right]=\mathbb{E}[X].$$

Also, if $X$ is already measurable with respect to $\mathcal{H}$, then $\mathbb{E}[X\mid\mathcal{H}]=X$ almost surely. In that case, conditioning on any finer sigma-algebra does not change $X$ either.

Common mistakes

• Reversing the order: The tower property requires the outer conditioning to use the coarser sigma-algebra. If $\mathcal{H}\subseteq\mathcal{G}$, then $\mathbb{E}[\mathbb{E}[X\mid\mathcal{H}]\mid\mathcal{G}]=\mathbb{E}[X\mid\mathcal{H}]$, because the inner result is already $\mathcal{G}$-measurable. It does not generally become $\mathbb{E}[X\mid\mathcal{G}]$.
• Ignoring nesting: There is no general tower simplification for two unrelated sigma-algebras. Check that one information set is contained in the other.
• Expecting equality at every outcome: Conditional expectations can differ on probability-zero events while still representing the same conditional expectation. The standard identity is almost sure.
• Forgetting integrability: The usual statement assumes $X$ is integrable. Without this condition, the conditional expectations in the formula may not be defined as finite integrable random variables.