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Kernel conditional expectations for square-integrable Markov observables

Statement

Assume AC. Let (Xk)k≥0 be a stationary time-homogeneous Markov chain with probability kernel P, invariant law π, and natural filtration Fk=σ(X0,…,Xk). Let g∈L2(π) (The space Lp(μ) as the quotient by null functions) and fix a finite real measurable representative, also denoted by g. Define Pg(x)=∫g(y)P(x,dy) and P(g2)(x)=∫g(y)2P(x,dy) on the π-full measurable set where the respective integrals are finite, and set these functions to zero on the exceptional set. Then, for each k≥0, E[g(Xk+1)∣Fk]=Pg(Xk). E[g(Xk+1)2∣Fk]=P(g2)(Xk)a.s. Consequently, if Dk+1=g(Xk+1)−Pg(Xk), then (Dk+1) is a square-integrable martingale-difference sequence and E[Dk+12∣Fk]=P(g2)(Xk)−(Pg(Xk))2a.s.

Facts & Assumptions

Given: AC, a stationary chain X with kernel P and invariant law π, its natural filtration, and a real measurable representative g with π(g2)<∞.

[A1]

AC is the choice principle used by the library's conditional-expectation class convention. (The Axiom of Choice)

[F1]

For a bounded measurable h and a P-chain, E[h(Xm+n)∣Fm]=Pnh(Xm); in particular, at n=1 the conditional mean is the kernel integral. (Chapman-Kolmogorov equations)

[F2]

Increasing nonnegative inputs have increasing conditional-expectation versions converging almost surely to the conditional expectation of their limit. (Conditional monotone convergence)

[F3]

Invariance is the event identity ∫P(x,A) π(dx)=π(A) for every measurable A. (Invariant and stationary distribution for a Markov kernel)

[F4]

Nonnegative increasing functions pass through an integral by monotone convergence. (Monotone convergence for the integral)

[F5]

Kernel integration of a nonnegative product-measurable function is measurable, with infinity permitted. (Measurability of integration against a kernel)

[F6]

Conditional expectation is linear and preserves order. (Basic algebra and order properties of conditional expectation)

[F7]

Conditional Jensen gives ∣E[Z∣G]∣2≤E[∣Z∣2∣G] for square-integrable Z. (Conditional jensen inequality)

[F8]

On a probability space, Hölder gives π(∣g∣)≤π(g2)1/2<∞. (Holder's inequality for integrals, including the endpoint cases)

[F9]

An integrable Fk-measurable factor can be taken out of a conditional expectation when the products are integrable. (Taking out what is known)

[F10]

Products of square-integrable random variables are integrable by Hölder's inequality. (Holder's inequality for integrals, including the endpoint cases)

Proof

1.1A1F1F5given

For bounded measurable h, [F1] with n=1 gives E[h(Xk+1)∣Fk]=Ph(Xk). Kernel measurability [F5] makes the right side measurable.

2.1F2F3F4F6F8step 1.1

Write g=g+−g−. By [F8], both parts are integrable. For N≥1 let gN±=g±∧N. These functions are bounded and increase to g±. Step 1.1 gives the conditional kernel identity for each gN±; [F2] makes the conditional expectations increase to E[g±(Xk+1)∣Fk], while [F4] makes PgN±(Xk)↑Pg±(Xk). Invariance [F3], first for nonnegative simple functions and then by [F4], gives E[Pg±(Xk)]=π(g±)<∞; hence both limits are finite almost surely. Subtracting the two finite identities using [F6] proves E[g(Xk+1)∣Fk]=Pg(Xk).

2.2F2F3F4step 1.1

Apply the same argument to the nonnegative integrable function g2: the truncations g2∧N increase to g2, and invariance gives E[P(g2)(Xk)]=π(g2)<∞. Thus E[g(Xk+1)2∣Fk]=P(g2)(Xk). These identities do not depend on the chosen representative: if two representatives differ on a measurable π-null set A, invariance gives π(P1A)=0, hence P(x,A)=0 for π-almost every x; their kernel integrals agree there, and their evaluations at each Xk agree almost surely.

3.1F3F6F7F9F10step 2.1step 2.2∎

By [F7] and step 2.2, (Pg(Xk))2≤P(g2)(Xk) almost surely, so stationarity and [F3] imply E[(Pg(Xk))2]≤π(g2); therefore Dk+1∈L2. Step 2.1, the preceding square-integrability bound, conditional linearity and [F9] for the Fk-measurable factor Pg(Xk) give E[Dk+1∣Fk]=0. The product g(Xk+1)Pg(Xk) is integrable by the two L2 bounds and [F10], so [F9] also gives its conditional expectation as (Pg(Xk))2. Expanding the square and using step 2.2 now yields E[Dk+12∣Fk]=P(g2)(Xk)−(Pg(Xk))2. Thus the sequence is a square-integrable martingale-difference sequence.

Source notes

Jones §2 defines the kernel operator Pg and states the conditional one-step kernel convention. The bounded-to-square-integrable extension and the exceptional-set version are proved here; no pointwise identity is asserted outside a π-full set.

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