Symmetric matrix notes

T b 0 s 1axsb 0 s 1 ds c t 0 suppose

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Unformatted text preview: has a nice inverse. C (t) has a better representation, B (0, s, ω ) being progressively measurable. t B (0, s, ω )−1a(x(s))B (0, s, ω )−1 ∗ ds C ( t) = 0 Suppose C (1)x = 0 for some x = {xi }. Then denoting by hi,j (s, ω ) = (B (0, s, ω )−1)i,j we have hi,j (s, ω )xiσj,k (x(s)) ≡ 0 i,j 1 for k = 1, 2, . . . n. If we think of σj,k (x) as vector fields Xk (x), then i xi qi,k (s, ω ) ≡ 0 for k = 1, 2, . . . n, where xi qi,k (s, ω ) =< x, h(s, ω )Xk (x(s)) > i Now ξ (s) =< x, h(s, ω )Xk (x(s)) > is a semi-martingale and we can compute the Stratonovich differential d ◦ (h(s, ω )Xk (x(s)) = [d ◦ h(s, ω )]Xk (x(s)) + h(s, ω )[d ◦ Xk (x(s))] = r ek,r (s, ω ) ◦ dβr (s) + ek,0 (s, ω )ds One can compute easily ek,r (s) = h(s)[−Xr Xk + Xk Xr ](x(s)) = h(s)[Xk , Xr ](x(s)) and ek,0 (s, ω ) = h(s)[−X0 Xk + Xk X0 ](x(s)) = h(s)[Xk , X0 ](x(s)) One knows from the uniqueness in Doob-Meyer decomposition that if dξ = dM (t) + b(t)dt ≡ 0 then M (t) ≡ 0 and b(t) ≡ 0. Moreover if dM (t) = ej (s)dβj (s) Therefore ek,r are equal to 0. The...
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This note was uploaded on 03/19/2013 for the course MATH GA 2931 taught by Professor Varadhan during the Fall '06 term at NYU.

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