Mathematics.

probability and integration

Conditional Expectation

Measure Theory90 minDifficulty8 out of 10

You should know: lebesgue integral, probability measure

Overview

Conditional expectation is one of the central constructions in modern probability theory. Given a probability space (Ω, F, P) and a sub-σ-algebra G ⊆ F, the conditional expectation E[X | G] of an integrable random variable X is the essentially unique G-measurable random variable that agrees with X on every G-measurable set in the sense of integration. This generalises the elementary notion of conditioning on an event to a fully measure-theoretic framework, and is the foundation of martingale theory, Bayesian statistics, and stochastic filtering.

Intuition

Think of G as encoding the information you are given. E[X | G] is the best L²-prediction of X given only that information — it is the orthogonal projection of X onto the closed subspace L²(Ω, G, P) of G-measurable square-integrable functions. When G = {∅, Ω} you know nothing and E[X | G] = E[X]; when G = F you know everything and E[X | G] = X.

Formal Definition

Definition

Let (Ω, F, P) be a probability space, G ⊆ F a sub-σ-algebra, and X ∈ L¹(Ω, F, P). The conditional expectation E[X | G] is any random variable Z satisfying:

Z is G-measurableZ \text{ is } \mathcal{G}\text{-measurable}

Z is measurable with respect to the sub-σ-algebra G

CE-1
AZdP=AXdPAG\int_A Z \, dP = \int_A X \, dP \quad \forall A \in \mathcal{G}

Partial averaging property

CE-2

Notation

NotationMeaning
E[XG]E[X \mid \mathcal{G}]Conditional expectation of X given sub-σ-algebra G
E[XY]E[X \mid Y]Conditional expectation given the σ-algebra generated by Y
E[XA]E[X \mid A]Elementary conditional expectation given event A (when P(A) > 0)

Proofs

Existence and uniqueness of conditional expectation
  1. Define ν(A)=AXdP for AG\text{Define } \nu(A) = \int_A X \, dP \text{ for } A \in \mathcal{G}(ν is a signed measure on (Ω, G))
  2. νPG\nu \ll P|_{\mathcal{G}}(ν is absolutely continuous with respect to P restricted to G, since P(A)=0 implies ∫_A X dP = 0)
  3. ZL1(Ω,G,P):ν(A)=AZdP  AG\exists Z \in L^1(\Omega, \mathcal{G}, P) : \nu(A) = \int_A Z \, dP \; \forall A \in \mathcal{G}(Radon–Nikodym theorem applied to ν and P|_G)
  4. If Z also satisfies the conditions, then A(ZZ)dP=0  AG\text{If } Z' \text{ also satisfies the conditions, then } \int_A (Z - Z') dP = 0 \; \forall A \in \mathcal{G}(By subtraction of the averaging equations)
  5. Z=Z a.s.Z = Z' \text{ a.s.}(Since Z - Z' is G-measurable and integrates to 0 over every G-set, taking A = {Z > Z'} gives P(Z > Z') = 0 and symmetrically P(Z < Z') = 0)

Properties

Tower property

E[E[XG]H]=E[XH] when HGE[E[X \mid \mathcal{G}] \mid \mathcal{H}] = E[X \mid \mathcal{H}] \text{ when } \mathcal{H} \subseteq \mathcal{G}

Condition: H ⊆ G ⊆ F

Linearity

E[aX+bYG]=aE[XG]+bE[YG]E[aX + bY \mid \mathcal{G}] = a\, E[X \mid \mathcal{G}] + b\, E[Y \mid \mathcal{G}]

Condition: a, b ∈ R, X, Y ∈ L¹

Independence

E[XG]=E[X] when XGE[X \mid \mathcal{G}] = E[X] \text{ when } X \perp \mathcal{G}

Condition: X independent of G

Taking out what is known

E[ZXG]=ZE[XG] when Z is G-measurableE[ZX \mid \mathcal{G}] = Z \cdot E[X \mid \mathcal{G}] \text{ when } Z \text{ is } \mathcal{G}\text{-measurable}

Condition: Z G-measurable and bounded (or appropriate integrability)

Jensen's inequality

φ(E[XG])E[φ(X)G]\varphi(E[X \mid \mathcal{G}]) \leq E[\varphi(X) \mid \mathcal{G}]

Condition: φ convex, X, φ(X) ∈ L¹

Monotone convergence

XnX    E[XnG]E[XG] a.s.X_n \nearrow X \implies E[X_n \mid \mathcal{G}] \nearrow E[X \mid \mathcal{G}] \text{ a.s.}

Condition: Xₙ ≥ 0 non-decreasing

Theorems

Theorem 1: L² projection characterisation
E[XG]=argminZL2(Ω,G,P)E[(XZ)2]E[X \mid \mathcal{G}] = \underset{Z \in L^2(\Omega,\mathcal{G},P)}{\arg\min}\, E[(X - Z)^2]
Theorem 2: Conditional expectation as Radon–Nikodym derivative
E[XG]=d(XP)dPGE[X \mid \mathcal{G}] = \frac{d(X \cdot P)}{dP}\bigg|_{\mathcal{G}}

Applications

Risk-neutral pricing: asset price processes are martingales under the risk-neutral measure, exploiting the tower property of conditional expectation.

Worked Examples

  1. G has exactly two atoms: A₁ = [0, 1/2) and A₂ = [1/2, 1]. E[X|G] must be G-measurable, so it is constant on each atom.

    E[XG](ω)=c11[0,1/2)(ω)+c21[1/2,1](ω)E[X \mid \mathcal{G}](\omega) = c_1 \mathbf{1}_{[0,1/2)}(\omega) + c_2 \mathbf{1}_{[1/2,1]}(\omega)
  2. Apply the partial averaging condition on A₁:

    01/2c1dω=01/2ωdω    c112=18    c1=14\int_0^{1/2} c_1 \, d\omega = \int_0^{1/2} \omega \, d\omega \implies c_1 \cdot \tfrac{1}{2} = \tfrac{1}{8} \implies c_1 = \tfrac{1}{4}
  3. Apply on A₂:

    1/21c2dω=1/21ωdω    c212=38    c2=34\int_{1/2}^1 c_2 \, d\omega = \int_{1/2}^1 \omega \, d\omega \implies c_2 \cdot \tfrac{1}{2} = \tfrac{3}{8} \implies c_2 = \tfrac{3}{4}
  4. Therefore:

    E[XG](ω)=141[0,1/2)(ω)+341[1/2,1](ω)E[X \mid \mathcal{G}](\omega) = \tfrac{1}{4}\mathbf{1}_{[0,1/2)}(\omega) + \tfrac{3}{4}\mathbf{1}_{[1/2,1]}(\omega)

Answer: E[X|G](ω) = 1/4 on [0,1/2) and 3/4 on [1/2,1], i.e. the conditional mean of X on each atom.

Practice Problems

Difficulty 7/10

Prove the tower property: if H ⊆ G ⊆ F, then E[E[X|G]|H] = E[X|H] a.s.

Difficulty 8/10

Let X ~ Exp(1) and let G = σ({X > 1}). Compute E[X | G] explicitly.

Difficulty 9/10

Prove Jensen's inequality for conditional expectation: if φ is convex and X, φ(X) ∈ L¹, then φ(E[X|G]) ≤ E[φ(X)|G] a.s.

Common Mistakes

Common Mistake

E[X|G] is a number

E[X|G] is a random variable (a G-measurable function on Ω), not a single number. E[X|Y=y] for a fixed y is a number, but E[X|Y] is a function of Y.

Common Mistake

Conditioning on an event with probability zero is straightforward

E[X|Y=y] for a continuous Y is defined via the regular conditional distribution, not by dividing by P(Y=y) = 0.

Historical Background

The elementary notion of conditional probability P(A|B) = P(A∩B)/P(B) dates to the 17th century. The rigorous measure-theoretic definition of conditional expectation as a Radon–Nikodym derivative was given by Andrei Kolmogorov in his 1933 monograph Grundbegriffe der Wahrscheinlichkeitsrechnung, which placed all of probability theory on a measure-theoretic foundation.

  1. 1933

    Kolmogorov defines conditional expectation via the Radon–Nikodym theorem

    Andrei Kolmogorov

  2. 1953

    Doob's Stochastic Processes systematises martingales built from conditional expectations

    Joseph Doob

  3. 1960s

    Stochastic filtering (Kalman–Bucy) exploits conditional expectation in continuous time

    Rudolf Kalman, Richard Bucy

Summary

  • E[X|G] is the unique G-measurable integrable random variable satisfying ∫_A E[X|G] dP = ∫_A X dP for all A ∈ G.
  • Existence follows from the Radon–Nikodym theorem; uniqueness holds almost surely.
  • Key properties: tower property, linearity, Jensen's inequality, taking out what is known.
  • In L², E[X|G] is the orthogonal projection of X onto L²(Ω, G, P).
  • Conditional expectation is the backbone of martingale theory and Bayesian inference.

References

  1. BookDurrett, R. (2019). Probability: Theory and Examples (5th ed.). Cambridge University Press.
  2. BookWilliams, D. (1991). Probability with Martingales. Cambridge University Press.