Existence

With the Radon-Nikodym theorem in hand, existence of conditional expectation is a short argument. The proof has two cases: the non-negative case (a direct application of Radon-Nikodym) and the general case (split $X$ into positive and negative parts).

Proposition

Proposition: Existence

Statement

Let $(\Omega, \cF, \Pr)$ be a probability space, $\cG \subseteq \cF$ a sub-$\sigma$-field, and $X$ an integrable random variable, $\E|X| < \infty$. Then a conditional expectation $\E(X \mid \cG)$ exists: there is a $\cG$-measurable random variable $Y$ such that

$$\int_A X \, d\Pr \;=\; \int_A Y \, d\Pr \qquad \text{for every } A \in \cG.$$

Proof

Case 1: $X \ge 0$.

Let $\mu = \Pr|_\cG$, the original probability measure $\Pr$ restricted to the sub-$\sigma$-field $\cG$. So $\mu$ is a measure on $(\Omega, \cG)$. Define a second set function on $\cG$ by

$$\nu(A) \;:=\; \int_A X \, d\Pr, \qquad A \in \cG.$$

Since $X \ge 0$, the integral is non-negative; countable additivity of $\nu$ follows from the monotone convergence theorem. So $\nu$ is a measure on $(\Omega, \cG)$. It is finite because $\nu(\Omega) = \E[X] \le \E|X| < \infty$.

Absolute continuity. For any $A \in \cG$,

$$\mu(A) = 0 \;\Longleftrightarrow\; \Pr(A) = 0 \;\Longrightarrow\; \int_A X \, d\Pr = 0 \;\Longleftrightarrow\; \nu(A) = 0.$$

The middle implication uses the fact that integrating any function over a $\Pr$-null set gives zero. So $\nu \ll \mu$.

Apply Radon-Nikodym. Both $\mu$ and $\nu$ are finite, hence $\sigma$-finite, measures on $(\Omega, \cG)$. By the Radon-Nikodym theorem, there is a $\cG$-measurable function $Y = \frac{d\nu}{d\mu}$ such that

$$\nu(A) \;=\; \int_A Y \, d\mu \qquad \text{for every } A \in \cG.$$

Identify $d\mu$ with $d\Pr$ on $\cG$-events. Since $\mu = \Pr|_\cG$ agrees with $\Pr$ on every set in $\cG$, and $Y$ is $\cG$-measurable, integrating $Y$ over $A \in \cG$ against $\mu$ or against $\Pr$ gives the same value:

$$\int_A Y \, d\mu \;=\; \int_A Y \, d\Pr.$$

Chaining the identities,

$$\int_A X \, d\Pr \;=\; \nu(A) \;=\; \int_A Y \, d\mu \;=\; \int_A Y \, d\Pr \qquad \text{for every } A \in \cG.$$

So $Y$ satisfies both conditions of the definition: it is $\cG$-measurable, and it has the same integral as $X$ over every $\cG$-event. This finishes Case 1.

Case 2: general integrable $X$.

Split $X$ into positive and negative parts,

$$X \;=\; X^+ - X^-, \qquad X^+ = \max(X, 0), \quad X^- = \max(-X, 0).$$

Both $X^+$ and $X^-$ are non-negative and integrable, since $\E[X^\pm] \le \E|X| < \infty$. Case 1 applied separately to $X^+$ and $X^-$ produces $\cG$-measurable

$$Y_1 = \E(X^+ \mid \cG), \qquad Y_2 = \E(X^- \mid \cG).$$

Define $Y := Y_1 - Y_2$. Then $Y$ is $\cG$-measurable (a difference of $\cG$-measurable functions), and for every $A \in \cG$,

$$\begin{aligned} \int_A X \, d\Pr &= \int_A X^+ \, d\Pr - \int_A X^- \, d\Pr \\ &= \int_A Y_1 \, d\Pr - \int_A Y_2 \, d\Pr \\ &= \int_A Y \, d\Pr. \end{aligned}$$

The first equality uses linearity of the integral on $X = X^+ - X^-$; the second is Case 1 applied to $X^+$ and $X^-$; the third is linearity in reverse. So $Y$ is a conditional expectation of $X$ given $\cG$.

Remarks

• What integrability buys. The hypothesis $\E|X| < \infty$ enters at two points: it ensures the measure $\nu$ in Case 1 is finite (so Radon-Nikodym applies), and it makes both $\E(X^+ \mid \cG)$ and $\E(X^- \mid \cG)$ finite a.s. in Case 2 (so their difference is well-defined). Without integrability, $Y_1 - Y_2$ could produce $\infty - \infty$.

• The role of $\sigma$-finiteness in Radon-Nikodym. The theorem requires both measures to be $\sigma$-finite. Here both $\mu = \Pr|_\cG$ and $\nu$ are finite (totals equal $1$ and $\E[X^\pm]$ respectively), so $\sigma$-finiteness holds with room to spare. This is one reason conditional expectation is so well-behaved in probability: the dominating measure is a probability measure, which is finite by definition.

• Why uniqueness is a separate question. The Radon-Nikodym derivative is unique only up to $\mu$-null sets, which means the conditional expectation produced above is unique only up to $\Pr$-null sets in $\cG$. Different choices of “version” of $\E(X \mid \cG)$ all satisfy the defining identity, and the question of how to interpret this near-uniqueness is the subject of the uniqueness page.