Diagonal-term contribution to Wald's second equation: applying Wald's equation
to the squared sequence Y_i = X_i ^ 2 yields
E[∑_{i < T} X_i^2] = E[X_0^2] · E[T].
A single cross term E[X_j · X_i · 1_{T > i}] (with j < i) vanishes when the
X_i are i.i.d. mean-zero and X_i is independent of (X_j, 1_{T > i}).
The full cross-term contribution to E[(S_T)^2] vanishes:
E[∑_{i < T} ∑_{j < i} X_j X_i] = 0 under the i.i.d./independence/mean-zero
assumptions, given suitable measurability and summability bookkeeping.
Algebraic identity expanding the square of a finite sum into diagonal and
cross-term contributions: (∑_{i<n} f i)^2 = ∑_{i<n} (f i)^2 + 2 ∑_{i<n} ∑_{j<i} f j · f i.
The stopped sum of squares ω ↦ ∑_{i < T(ω)} X_i(ω)^2 is integrable, under
i.i.d./independence assumptions with E[X_0^2] < ∞ and E T < ∞.
The "predictable" filtration associated to a sequence X : ℕ → Ω → ℝ:
ℱ_k = σ(X_0, …, X_{k-1}), i.e. the supremum of the σ-algebras generated by
X_j for j < k. This makes each X_j measurable strictly before time j+1.
Instances For
Wald's second equation. Let X_i be i.i.d. with E[X_i] = 0 and
E[X_i^2] = σ^2 ∈ (0, ∞). If T is a stopping time (for the predictable
filtration generated by X) with E[T] < ∞, then
E[S_T^2] = σ^2 · E[T], where S_T = ∑_{i < T} X_i.