How to Prove the Main Properties of Conditional Expectation

Conditional expectation is defined relative to the information represented by a sigma-algebra. Its familiar rules—such as linearity and the tower property—can all be proved from the same two conditions. The key technique is to show that a proposed random variable satisfies the definition, then use uniqueness.

Definition and uniqueness

Let $X$ be integrable and let $\mathcal{G}$ be a sigma-algebra. A conditional expectation of $X$ given $\mathcal{G}$, written $\mathbb{E}[X\mid\mathcal{G}]$, is an integrable, $\mathcal{G}$-measurable random variable $Z$ such that

$$\mathbb{E}[\mathbf{1}_A Z]=\mathbb{E}[\mathbf{1}_A X] \quad\text{for every }A\in\mathcal{G}.$$

Here $\mathbf{1}_A$ is the indicator of the event $A$. The condition says that $Z$ and $X$ have the same average over every event whose occurrence can be determined from the information in $\mathcal{G}$. Conditional expectations are unique almost surely: two random variables satisfying the definition can differ only on an event of probability zero.

That uniqueness is what makes the proofs short. To establish an identity, show its right-hand side is measurable with respect to the required sigma-algebra and has the required integrals.

Linearity

For integrable $X,Y$ and constants $a,b$, let $U=\mathbb{E}[X\mid\mathcal{G}]$ and $V=\mathbb{E}[Y\mid\mathcal{G}]$. The random variable $aU+bV$ is $\mathcal{G}$-measurable and integrable. For every $A\in\mathcal{G}$,

$$\begin{aligned} \mathbb{E}[\mathbf{1}_A(aU+bV)] &=a\mathbb{E}[\mathbf{1}_A U]+b\mathbb{E}[\mathbf{1}_A V]\\ &=a\mathbb{E}[\mathbf{1}_A X]+b\mathbb{E}[\mathbf{1}_A Y]\\ &=\mathbb{E}[\mathbf{1}_A(aX+bY)]. \end{aligned}$$

Thus $aU+bV$ satisfies the definition of the conditional expectation of $aX+bY$. By uniqueness,

$$\mathbb{E}[aX+bY\mid\mathcal{G}] =a\mathbb{E}[X\mid\mathcal{G}]+b\mathbb{E}[Y\mid\mathcal{G}] \quad\text{almost surely}.$$

Positivity and monotonicity

Suppose $X\geq 0$ almost surely, and set $Z=\mathbb{E}[X\mid\mathcal{G}]$. If $Z$ were negative on a set of positive probability, then for some positive integer $n$ the event $A=\{Z\leq -1/n\}$ would have positive probability. This event belongs to $\mathcal{G}$, so the defining property gives

$$\mathbb{E}[\mathbf{1}_A Z]=\mathbb{E}[\mathbf{1}_A X]\geq 0.$$

But $Z\leq -1/n$ on $A$, making the left side at most $-\mathbb{P}(A)/n<0$. This contradiction shows $\mathbb{E}[X\mid\mathcal{G}]\geq0$ almost surely.

If $X\leq Y$ almost surely, then $Y-X\geq0$. Apply positivity and linearity to get

$$\mathbb{E}[X\mid\mathcal{G}]\leq\mathbb{E}[Y\mid\mathcal{G}] \quad\text{almost surely}.$$

Taking out known factors

If $H$ is bounded and $\mathcal{G}$-measurable, then

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

To prove this, first take $H=\mathbf{1}_B$ for $B\in\mathcal{G}$. For every $A\in\mathcal{G}$, the intersection $A\cap B$ is also in $\mathcal{G}$, so

$$\mathbb{E}[\mathbf{1}_A\mathbf{1}_B X] =\mathbb{E}[\mathbf{1}_A\mathbf{1}_B\mathbb{E}[X\mid\mathcal{G}]].$$

This proves the identity for indicators, and linearity proves it for simple $H$. A bounded measurable $H$ can be approximated uniformly by simple $\mathcal{G}$-measurable functions; passing to the limit gives the stated result. Boundedness ensures the products in this argument are integrable.

The tower property

Let $\mathcal{H}\subseteq\mathcal{G}$ be sigma-algebras. Then

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

Indeed, the left side is $\mathcal{H}$-measurable. For every $A\in\mathcal{H}$, the defining property applied twice gives

$$\mathbb{E}\big[\mathbf{1}_A\mathbb{E}[\mathbb{E}[X\mid\mathcal{G}]\mid\mathcal{H}]\big] =\mathbb{E}[\mathbf{1}_A\mathbb{E}[X\mid\mathcal{G}]] =\mathbb{E}[\mathbf{1}_A X],$$

because $A\in\mathcal{G}$ as well. The left side therefore satisfies the definition of $\mathbb{E}[X\mid\mathcal{H}]$, and uniqueness proves the claim. Taking $\mathcal{H}$ to be the trivial sigma-algebra yields the law of total expectation, $\mathbb{E}[\mathbb{E}[X\mid\mathcal{G}]]=\mathbb{E}[X]$.

Two other useful consequences follow directly from the definition. If $X$ is already $\mathcal{G}$-measurable, then $\mathbb{E}[X\mid\mathcal{G}]=X$. If $X$ is independent of $\mathcal{G}$, then $\mathbb{E}[X\mid\mathcal{G}]=\mathbb{E}[X]$.

Worked example: conditioning on a partition

Suppose the outcomes $1,2,3,4,5,6$ are equally likely, and $X$ takes values $1,2,6,0,3,9$ on those outcomes, respectively. Let $\mathcal{G}$ reveal only whether the outcome is in $A=\{1,2,3\}$ or $B=\{4,5,6\}$. A $\mathcal{G}$-measurable random variable is constant on each block of this partition. The defining averages give

$$\mathbb{E}[X\mid\mathcal{G}] =\begin{cases} (1+2+6)/3=3,&\text{on }A,\\ (0+3+9)/3=4,&\text{on }B. \end{cases}$$

The expectation of this conditional expectation is $(3+3+3+4+4+4)/6=3.5$. Directly, $\mathbb{E}[X]=(1+2+6+0+3+9)/6=3.5$ as well, illustrating the tower property.

Common mistakes

• Treating equality as pointwise. Conditional-expectation identities generally hold almost surely, not necessarily at every outcome.
• Using the factor rule without measurability. The factor must be measurable with respect to the conditioning sigma-algebra; boundedness is a convenient sufficient condition for the proof above.
• Confusing conditioning with a number. Given a nontrivial sigma-algebra, the conditional expectation is usually a random variable, not one fixed average.
• Skipping uniqueness. Matching the integrals alone is not enough: the proposed variable must also be measurable with respect to the conditioning sigma-algebra.