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STOCH3 · 1

Conditional expectation

9 concepts · 17 questions

Information as a σ-algebra, conditional expectation given a partition and in general, its defining property, uniqueness, the main rules, conditional Jensen, and conditional expectation as an orthogonal projection.

Overview 1The plan of this section

E[X∣Y]\mathbb E[X\mid Y] was defined as a function of YY. Martingale theory needs conditioning on "everything known at time nn", which is best encoded as a σ-algebra.

  1. Information as a σ-algebra (Def. 1.1).
  2. Conditional expectation (Def. 1.2), uniqueness (Prop. 1.3) and its rules (Thm 1.4).
  3. Conditional Jensen (Prop. 1.5) and the projection picture (Thm 1.6), with examples (Example 1.7).

E[X∣G]\mathbb E[X\mid\mathcal G]: the best guess of XX using only the information in G\mathcal G.

Definition 1.1Information as a σ-algebra

A σ-algebra G\mathcal G on Ω\Omega is a collection of events containing Ω\Omega and closed under complements and countable unions. For a countable partition {Bi}\{B_i\} of Ω\Omega, σ({Bi})\sigma(\{B_i\}) consists of all unions of blocks. For a random variable XX, σ(X)={{X∈A}:A Borel}\sigma(X) = \{\{X\in A\}: A\text{ Borel}\}; if XX is discrete it is generated by the partition {X=x}\{X = x\}. A random variable ZZ is G\mathcal G-measurable if {Z≤z}∈G\{Z\le z\}\in\mathcal G for all zz; for a partition this means ZZ is constant on each block. Think of G\mathcal G as "what you know".

G\mathcal G = information; G\mathcal G-measurable = determined by that information (constant on blocks of a partition).

Definition 1.2Conditional expectation

Let E∣X∣<∞\mathbb E|X| < \infty. A conditional expectation of XX given G\mathcal G is a random variable ZZ with (i) ZZ G\mathcal G-measurable and integrable, and (ii) E[X1A]=E[Z1A]\mathbb E[X\mathbf 1_A] = \mathbb E[Z\mathbf 1_A] for every A∈GA\in\mathcal G. We write Z=E[X∣G]Z = \mathbb E[X\mid\mathcal G]. For a partition {Bi}\{B_i\} with P(Bi)>0\mathbb P(B_i) > 0,

E[X∣G]=∑iE[X1Bi]P(Bi)1Bi,\mathbb E[X\mid\mathcal G] = \sum_i\frac{\mathbb E[X\mathbf 1_{B_i}]}{\mathbb P(B_i)}\mathbf 1_{B_i},

the average of XX over whichever block occurred. (For general G\mathcal G, existence follows from the Radon–Nikodym theorem, or from the projection picture (see Theorem 1.6) and approximation.)

E[X∣G]\mathbb E[X\mid\mathcal G]: G\mathcal G-measurable, and has the same integral as XX over every A∈GA\in\mathcal G. On a partition: block averages.

Proposition 1.3Uniqueness of conditional expectation

If ZZ and Z′Z' both satisfy Definition 1.2, then Z=Z′Z = Z' almost surely.

The defining property pins E[X∣G]\mathbb E[X\mid\mathcal G] down almost surely.

Proof

Theorem 1.4Rules for conditional expectation

For integrable X,YX, Y and σ-algebras H⊆G\mathcal H\subseteq\mathcal G: (a) linearity: E[aX+bY∣G]=aE[X∣G]+bE[Y∣G]\mathbb E[aX + bY\mid\mathcal G] = a\mathbb E[X\mid\mathcal G] + b\mathbb E[Y\mid\mathcal G]; (b) positivity: X≥0X\ge0 ⇒ E[X∣G]≥0\mathbb E[X\mid\mathcal G]\ge0; (c) E[E[X∣G]]=EX\mathbb E[\mathbb E[X\mid\mathcal G]] = \mathbb E X; (d) taking out what is known: if YY is G\mathcal G-measurable and XYXY integrable, E[XY∣G]=YE[X∣G]\mathbb E[XY\mid\mathcal G] = Y\mathbb E[X\mid\mathcal G]; (e) tower: E[E[X∣G]∣H]=E[X∣H]\mathbb E[\mathbb E[X\mid\mathcal G]\mid\mathcal H] = \mathbb E[X\mid\mathcal H]; (f) independence: if XX is independent of G\mathcal G, E[X∣G]=EX\mathbb E[X\mid\mathcal G] = \mathbb E X.

Linear, positive, tower, take out what is known, independent ⇒ constant. Prove each by checking the defining property.

Proof

Proposition 1.5Conditional Jensen inequality

If ϕ\phi is convex and X,ϕ(X)X, \phi(X) are integrable, then ϕ(E[X∣G])≤E[ϕ(X)∣G]\phi(\mathbb E[X\mid\mathcal G])\le\mathbb E[\phi(X)\mid\mathcal G] a.s. In particular ∣E[X∣G]∣≤E[∣X∣∣G]|\mathbb E[X\mid\mathcal G]|\le\mathbb E[|X|\mid\mathcal G] and E[X∣G]2≤E[X2∣G]\mathbb E[X\mid\mathcal G]^2\le\mathbb E[X^2\mid\mathcal G].

ϕ(E[X∣G])≤E[ϕ(X)∣G]\phi(\mathbb E[X\mid\mathcal G])\le\mathbb E[\phi(X)\mid\mathcal G] for convex ϕ\phi.

Proof

Theorem 1.6Conditional expectation as orthogonal projection

If EX2<∞\mathbb E X^2 < \infty, then E[X∣G]\mathbb E[X\mid\mathcal G] is the orthogonal projection of XX onto the closed subspace of G\mathcal G-measurable square-integrable variables (inner product E[UV]\mathbb E[UV]): it minimises E[(X−Y)2]\mathbb E[(X - Y)^2] over all such YY.

E[X∣G]\mathbb E[X\mid\mathcal G] = least-squares best predictor using G\mathcal G.

Proof

Example 1.7Computing conditional expectations

  • Symmetry: if X1,…,XnX_1, \dots, X_n are i.i.d. and S=∑XiS = \sum X_i, then E[X1∣S]=Sn\mathbb E[X_1\mid S] = \frac Sn, since all E[Xi∣S]\mathbb E[X_i\mid S] are equal and sum to SS.
  • Random walk: for Sn=X1+⋯+XnS_n = X_1 + \cdots + X_n with i.i.d. steps of mean μ\mu and m<nm < n: E[Sn∣Fm]=Sm+(n−m)μ\mathbb E[S_n\mid\mathcal F_m] = S_m + (n - m)\mu, where Fm=σ(X1,…,Xm)\mathcal F_m = \sigma(X_1, \dots, X_m).
  • Bivariate normal: E[Y∣X]=μY+ρσYσX(X−μX)\mathbb E[Y\mid X] = \mu_Y + \rho\frac{\sigma_Y}{\sigma_X}(X - \mu_X), the regression line.

Use symmetry, independence and "take out what is known".

Check 1.8Check: computing E[X | G]

  • Split XX into a part known given G\mathcal G (take it out) and a part independent of G\mathcal G (replace by its mean).
  • Use symmetry for exchangeable variables.
  • Verify a guess ZZ: is it G\mathcal G-measurable, and does E[Z1A]=E[X1A]\mathbb E[Z\mathbf 1_A] = \mathbb E[X\mathbf 1_A] for generating sets AA?
  • Sanity: E[E[X∣G]]=EX\mathbb E[\mathbb E[X\mid\mathcal G]] = \mathbb E X.

Known parts out, independent parts averaged, guess and verify; expectations must match.