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Second moment is the expected predictable quadratic variation

Statement

Assume AC. For a real square-integrable martingale M and every n0, E[Mn2]=E[M02]+E[Mn], with all three terms finite. In particular, M0=0 gives E[Mn2]=E[Mn].

Facts & Assumptions

Given: The hypotheses and conventions in the statement.

[F1]

The square-minus-bracket process is a martingale with initial M0 squared. Square minus predictable quadratic variation is a martingale.

[F2]

Conditional expectation is linear, order preserving and expectation preserving. Basic algebra and order properties of conditional expectation.

[F3]

Finite linear combinations remain integrable and their integrals are linear. The Lebesgue integral is linear on L1(μ).

[F4]
[F5]

For a square-integrable martingale, predictable quadratic variation is an integrable finite sum of conditional square increments. Predictable quadratic variation in discrete time.

Proof

technique · direct
1.1

By [F1], Zn=Mn2Mn is integrable and E[ZkFk1]=Zk1 for each k1. Expectation preservation gives EZk=EZk1. Induction over the finitely many times up to n yields EZn=EZ0=EM02, also at n=0. The invocations of F1 and F2 are made under the AC assumption F4.

givenF1F2F4
2.1

For the finite integral linearity used here, augment each finite disjoint display of a nonnegative simple function by its zero-coefficient complement. Pairwise intersections of two augmented displays partition the whole space and have equal coefficients on nonempty cells, so finite additivity and 0(+)=0 prove representation independence. Common refinements give simple addition and monotonicity, while scalar zero is direct and positive scalars are termwise. Supremum over simple minorants, increasing simple approximation and the sets {fjcs} for 0<c<1 give monotone convergence and nonnegative additivity. Positive/negative and real/imaginary decompositions then give finite L1 linearity. The square is integrable by the square-integrability hypothesis, and the bracket is integrable by [F5], so this local linearity gives EZn=EMn2EMn. Rearranging the finite equality from step 1.1 proves the formula. If M0=0 its second moment is zero. At n=0 the bracket is zero and the equation reads EM02=EM02.

F1F3F5step 1.1construct

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