A discrete approach to the chaotic representation property

— In continuous time, let (Xt)t>0 be a normal martingale (i.e. a process such that both Xt and X 2 t − t are martingales). One says that X has the chaotic representation property if L2 ( σ(X) ) is the (direct) Hilbert sum ⊕ p∈N χp(X), where χp(X) is the space of all p-fold iterated stochastic integrals ∫ 0<t1<...<tp f(t1, . . . , tp) dXt1 . . . dXtp with f square-integrable (χp(X) is called the p th chaotic space; by convention, χ0(X) is the one-dimensional space of deterministic random variables). An open problem is to characterize those processes X. Instead of working in continuous time, we shall address an analogue of this problem where the time-axis is the set Z of signed integers; in this setting, we shall give a sufficient (but probably far from necessary) condition for the chaotic representation property to hold. Notation and preliminaries We shall use the set Z of all signed integers as our time-axis; the set of all finite subsets of Z will be denoted by P. For m and n in Z, we shall have to do with the following “intervals”: cem,nce = {k ∈ Z : m < k 6 n} ; cen,∞bd = {k ∈ Z : n < k} ; ce−∞, nce = {k ∈ Z : k 6 n} . ce−∞, nbd = {k ∈ Z : k < n} . Given a filtration F = (Fn)n∈Z, a process X = (Xn)n∈Z is adapted (respectively predictable ) if for each n the random variable Xn is Fn-measurable (respectively Fn−1-measurable); a stopping time is an F∞-measurable random variable T with values in Z∪{+∞}, such that for each n ∈ Z the event {T =n} (or, for that matter, {T 6n}) belongs to Fn; notice that the value −∞ is not allowed to stopping times. An empty sum ∑ i∈∅ xi is always null, an empty product ∏ i∈∅ xi is always 1. With Z as the time-axis, the analogue of a normal martingale is no longer a martingale, but a sort of normalized martingale increment: Definition. — On a probability space (Ω,A,P), let F = (Fn)n∈Z be a filtration. A process X = (Xn)n∈Z is a novation (more precisely: an F-novation) if, for each time n ∈ Z, Xn belongs to L(Fn) and verifies EbdXn|Fn−1ce = 0 ; (N1) EbdX n|Fn−1ce = 1 . (N2)