Properties

Conditional expectation inherits the structural properties of the integral (linearity, monotonicity, monotone convergence) and adds a few of its own (tower property, take-out-what-is-known, independence). All the proofs reduce to checking the two conditions in the definition: $\cG$-measurability and the integration identity on $\cG$-events.

Throughout this page, $(\Omega, \cF, \Pr)$ is a probability space, $\cG \subseteq \cF$ is a sub-$\sigma$-field, and "$\E(\cdot \mid \cG)$" refers to any version of the conditional expectation.

Linearity

Theorem: Linearity

Statement

For integrable $X, Y$ and any $a \in \R$,

$$\E(a X + Y \mid \cG) \;\stackrel{\text{a.s.}}{=}\; a \, \E(X \mid \cG) + \E(Y \mid \cG).$$

Proof

The right-hand side is $\cG$-measurable (a linear combination of $\cG$-measurable functions). For any $A \in \cG$,

$$\int_A \big(a \E(X \mid \cG) + \E(Y \mid \cG)\big) \, d\Pr \;=\; a \int_A X \, d\Pr + \int_A Y \, d\Pr \;=\; \int_A (a X + Y) \, d\Pr,$$

using linearity of the integral on both sides. So the right-hand side satisfies both defining conditions for $\E(a X + Y \mid \cG)$, and by uniqueness the two agree a.s.

Monotonicity

Theorem: Monotonicity

Statement

If $X \le Y$ a.s. with $X, Y$ integrable, then

$$\E(X \mid \cG) \;\le\; \E(Y \mid \cG) \quad \text{a.s.}$$

Proof

Mirror the uniqueness argument. Suppose

$$A \;:=\; \{\E(X \mid \cG) > \E(Y \mid \cG)\}$$

has $\Pr(A) > 0$. Since both conditional expectations are $\cG$-measurable, $A \in \cG$. Applying the integration identity on $A$,

$$\int_A \E(X \mid \cG) \, d\Pr \;=\; \int_A X \, d\Pr \;\le\; \int_A Y \, d\Pr \;=\; \int_A \E(Y \mid \cG) \, d\Pr,$$

so $\int_A \big(\E(X \mid \cG) - \E(Y \mid \cG)\big) \, d\Pr \le 0$. But the integrand is strictly positive on $A$, and $\Pr(A) > 0$, so the integral is strictly positive. Contradiction. Hence $\Pr(A) = 0$.

Conditional monotone convergence

Theorem: Conditional MCT

Statement

If $0 \le X_n \uparrow X$ a.s. with $\E[X] < \infty$, then

$$\E(X_n \mid \cG) \;\uparrow\; \E(X \mid \cG) \quad \text{a.s.}$$

Proof

Let $Y_n := X - X_n$ and $Z_n := \E(Y_n \mid \cG)$. By assumption $Y_n \ge 0$ and $Y_n \downarrow 0$ a.s. By linearity,

$$Z_n \;=\; \E(X \mid \cG) - \E(X_n \mid \cG),$$

so $\E(X_n \mid \cG) \uparrow \E(X \mid \cG)$ is equivalent to $Z_n \downarrow 0$ a.s.

Step 1: $\int_A Z_n \, d\Pr \to 0$ for every $A \in \cG$. Apply the integration identity:

$$\int_A Z_n \, d\Pr \;=\; \int_A Y_n \, d\Pr.$$

Since $0 \le Y_n \le X$ and $\E[X] < \infty$, the dominated convergence theorem gives $\int_A Y_n \, d\Pr \to 0$.

Step 2: $Z_n$ has an a.s. limit. Monotonicity applied to $Y_{n+1} \le Y_n$ gives $Z_{n+1} \le Z_n$ a.s. So $Z_n$ is non-increasing in $n$ and bounded below by $0$ (apply monotonicity again to $Y_n \ge 0$). Hence $Z_n \downarrow Z_\infty$ a.s. for some $Z_\infty \ge 0$.

Step 3: $Z_\infty = 0$ a.s. Since $Z_n \le Z_1 \le \E(X \mid \cG)$, and $\E[\E(X \mid \cG)] = \E[X] < \infty$, the $Z_n$ are uniformly dominated by an integrable function. Dominated convergence applied again gives

$$\int_A Z_n \, d\Pr \;\longrightarrow\; \int_A Z_\infty \, d\Pr.$$

Combining with Step 1, $\int_A Z_\infty \, d\Pr = 0$ for every $A \in \cG$. Since $Z_\infty$ is $\cG$-measurable and non-negative, this forces $Z_\infty = 0$ a.s.

The analogous conditional Fatou and conditional dominated convergence theorems follow by the same machine (apply the unconditional versions inside $\int_A \cdot \, d\Pr$ and translate via the integration identity).

Independence

Theorem: Independence Drops Out

Statement

If $X$ is integrable and $X$ is independent of the $\sigma$-field $\sigma(Y)$, then

$$\E(X \mid Y) \;\stackrel{\text{a.s.}}{=}\; \E[X].$$

Proof

The constant function $\E[X]$ is measurable in every $\sigma$-field, including $\sigma(Y)$, so condition (1) holds. For condition (2), recall that every set in $\sigma(Y)$ has the form $A = \{Y \in C\}$ for some Borel $C \subseteq \R$. Then

$$\int_A X \, d\Pr \;=\; \E[X \, \mathbb{1}_{\{Y \in C\}}] \;=\; \E[X] \cdot \E[\mathbb{1}_{\{Y \in C\}}] \;=\; \E[X] \cdot \Pr(A) \;=\; \int_A \E[X] \, d\Pr,$$

where the second equality is independence of $X$ and $\mathbb{1}_{\{Y \in C\}}$ (the indicator is a Borel function of $Y$). So $\E[X]$ satisfies both defining conditions for $\E(X \mid Y)$.

The reading: if $Y$ carries no information about $X$, then conditioning on $Y$ is the same as not conditioning at all.

Self-measurability

Theorem: Self-Measurability

Statement

If $X$ is integrable and $\cG$-measurable, then

$$\E(X \mid \cG) \;\stackrel{\text{a.s.}}{=}\; X.$$

Proof

$X$ itself is $\cG$-measurable, and trivially $\int_A X \, d\Pr = \int_A X \, d\Pr$ for every $A \in \cG$. Both conditions are satisfied with $Y = X$.

This is the opposite extreme of the independence case: if $X$ is already measurable in $\cG$, then $\cG$ carries all the information about $X$ and conditioning is the identity.

Tower property

Theorem: Tower Property

Statement

Let $\cF_1 \subseteq \cF_2 \subseteq \cF$ be sub-$\sigma$-fields. For any integrable $X$,

$$\begin{aligned} \E\!\big( \E(X \mid \cF_1) \,\big|\, \cF_2 \big) &\;\stackrel{\text{a.s.}}{=}\; \E(X \mid \cF_1), \\ \E\!\big( \E(X \mid \cF_2) \,\big|\, \cF_1 \big) &\;\stackrel{\text{a.s.}}{=}\; \E(X \mid \cF_1). \end{aligned}$$

Iterated conditional expectations collapse to the smaller $\sigma$-field, regardless of the order.

Proof

(a) Inner $\cF_1$, outer $\cF_2$. $\E(X \mid \cF_1)$ is $\cF_1$-measurable, hence $\cF_2$-measurable (since $\cF_1 \subseteq \cF_2$). By self-measurability, conditioning a $\cF_2$-measurable function on $\cF_2$ returns the function itself.

(b) Inner $\cF_2$, outer $\cF_1$. Take any $A \in \cF_1$. Since $\cF_1 \subseteq \cF_2$, also $A \in \cF_2$. Applying the integration identity at each level,

$$\int_A \E(X \mid \cF_1) \, d\Pr \;=\; \int_A X \, d\Pr \;=\; \int_A \E(X \mid \cF_2) \, d\Pr.$$

The right-hand side is $\int_A Y \, d\Pr$ with $Y = \E(X \mid \cF_2)$, which is the integration identity defining $\E(Y \mid \cF_1)$. Combined with the $\cF_1$-measurability of $\E(X \mid \cF_1)$, uniqueness gives $\E(\E(X \mid \cF_2) \mid \cF_1) = \E(X \mid \cF_1)$ a.s.

Specializing to $\cF_1 = \{\emptyset, \Omega\}$ (the trivial $\sigma$-field) gives the law of iterated expectations:

$$\E\!\big[ \E(X \mid \cG) \big] \;=\; \E[X].$$

Averaging the conditional expectation against the full distribution recovers the unconditional expectation.

Take out what is known

Theorem: Take-Out-What-Is-Known

Statement

If $X$ is $\cG$-measurable with $\E|X| < \infty$ and $\E|XY| < \infty$, then

$$\E(XY \mid \cG) \;\stackrel{\text{a.s.}}{=}\; X \, \E(Y \mid \cG).$$

A $\cG$-measurable factor can be pulled out of the conditional expectation as if it were a constant.

Proof

Check the two defining conditions for $X \, \E(Y \mid \cG)$ to be $\E(XY \mid \cG)$.

(1) Measurability. $X$ and $\E(Y \mid \cG)$ are both $\cG$-measurable, hence so is their product.

(2) Integration identity. We show

$$\int_A X \, \E(Y \mid \cG) \, d\Pr \;=\; \int_A XY \, d\Pr \qquad \text{for every } A \in \cG, \tag{$*$}$$

using the standard machine.

Indicator case: $X = \mathbb{1}_B$ for $B \in \cG$. For any $A \in \cG$, $A \cap B \in \cG$, so

$$\int_A \mathbb{1}_B \, \E(Y \mid \cG) \, d\Pr \;=\; \int_{A \cap B} \E(Y \mid \cG) \, d\Pr \;=\; \int_{A \cap B} Y \, d\Pr \;=\; \int_A \mathbb{1}_B Y \, d\Pr,$$

which is $(*)$ for $X = \mathbb{1}_B$.

Simple case. By linearity, $(*)$ extends to any $\cG$-measurable simple function $X = \sum_i a_i \mathbb{1}_{B_i}$.

Non-negative case. For $\cG$-measurable $X \ge 0$, pick a sequence of $\cG$-measurable simple functions $X_n \uparrow X$ and apply monotone convergence on both sides of $(*)$ (use $X_n Y^+ \uparrow X Y^+$ and $X_n Y^- \uparrow X Y^-$ separately, and combine).

General case. Decompose $X = X^+ - X^-$ and combine by linearity. The integrability hypotheses $\E|X| < \infty$ and $\E|XY| < \infty$ ensure each step produces a finite quantity.

Note that this is the same indicator-simple-non-negative-general progression used to define the Lebesgue integral itself. Every “extend to integrable $f$” argument in measure theory looks the same.