By Maxime Bocher

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Xn , xn + 1 ) = 1 f (y1 , . . , yn , max {xi }), 1≤i≤n+1 n+1 where (y1 , . . , yn ) is an arbitrary permutation of the remaining xj when the maximum value xˆj := max1≤i≤n+1 {xi } is removed. This maximum is obtained for a unique ˆj, for almost all x ∈ Rn+1 with respect to Lebesgue measure. 54 2 Framework Hence the Itˆo integral of Y (t) is Y (t)dB(t) R ∞ fn (x1 , . . , xn , t)dB ⊗n (x) dB(t) = n=0 R Rn = ··· n! n=0 · fˆn (x1 , . . (n + 1) n=0 −∞ −∞ −∞−∞ ∞ fˆn (x1 , . . , xn , xn+1 )dB ⊗(n+1) (x1 , .

13) R Proof Suppose Y (t) has the expansion ∞ Y (t) = ˆ 2 (Rn ) for all n. fn (x, t)dB ⊗n (x); fn (·, t) ∈ L n = 0 Rn Since Y (t) is adapted we know that fn (x1 , x2 , . . e. Therefore, the symmetrization fˆn (x1 , . . , xn , t) of fn (x1 , . . , xn , t) satisﬁes (with xn + 1 = t) fˆn (x1 , . . , xn , xn + 1 ) = 1 f (y1 , . . , yn , max {xi }), 1≤i≤n+1 n+1 where (y1 , . . , yn ) is an arbitrary permutation of the remaining xj when the maximum value xˆj := max1≤i≤n+1 {xi } is removed. This maximum is obtained for a unique ˆj, for almost all x ∈ Rn+1 with respect to Lebesgue measure.

X∈ / F }. ˆ 2 (Rn ), we have Proof We ﬁrst observe that for all n and all f ∈ L f (x)dB ⊗n (x) Ft E Rn ∞ tn t2 f (t1 , . . , tn )dB(t1 )dB(tn ) Ft = E n! −∞ −∞ −∞ t tn t2 f (x)χ[0,t]n (x)dB ⊗n (x). f (t1 , . . , tn )dB(t1 )dB(tn ) = = n! 0 0 Rn 0 Therefore we get Y (t) is Ft -adapted ⇔ E [Y (t)|Ft ] = Yt ⎡ ∞ ⇔ Rn ∞ ∞ Rn ⇔ fn (x, t)χ[0,t]n (x) = fn (x, t) by the uniqueness of the expansion. 3. Suppose Y (t) is a stochastic process with E[Y 2 (t)] < ∞ for all t and with the Hermite chaos expansion cα (t)Hα (ω).