Symmetric matrix notes

# Symmetric matrix notes - Lemma 1 Consider a positive denite...

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Lemma 1. Consider a positive defnite symmetric matrix A . ( Det A ) 1 2 = c n i S n - 1 ds [ < s,As > ] n 2 Proof: ( Det A ) 1 2 = (2 π ) n 2 i R n e <x,Ax> 2 dx = (2 π ) n 2 i S n - 1 e r 2 <s,As> 2 r n 1 dsdr = c n i S n - 1 ds [ ] n 2 It is thereFore su±cient to estimate For each k , sup x : b x |b =1 E [ < x,Ax > k ] C k to yield an estimate oF the Form E [( Det A ) k ] B k Weak formulation. With out estimates. Let us note that the Malliavin covariance A ( t ) has the Form A ( t ) = i t 0 B ( s,t, ω ) a ( x ( s )) B ( ) ds with B ( ) being the Jacobian oF the map R n R n that maps the initial point x = x ( s,ω ) in R n to x ( t ) R n . The problem with A ( t ) is that B ( ) is NOT progressively measurable. Moreover intrinsically A ( t ) is a quadratic Form on the cotangent space at x ( t, ω ). We can instead look at B (0 , t, ω ) 1 A ( t ) B (0 , t, ω ) 1 . Since Det B (0 , t,ω ) and its inverse will have moments oF all orders, we can try to show that C ( t ) = B (0 , t, ω ) 1 A ( t ) B (0 , t, ω ) 1 has a nice inverse. C ( t ) has a better representation, B (0 , s,ω ) being progressively measur- able. C ( t ) = i t 0 B (0 ) 1 a ( x ( s )) B (0 ) 1 ds Suppose C (1) x = 0 For some x = { x i } . Then denoting by h i,j ( ) = ( B (0 ) 1 ) i,j we have s i,j h i,j ( ) x i σ j,k ( x ( s )) 0 1

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for k = 1 , 2 , . . .n . If we think of σ j,k ( x ) as vector Felds X k ( x ), then s i x i q i,k ( s,ω ) 0 for k = 1 , 2 , . . .n, where s i x i q i,k ( ) = < x,h ( ) X k ( x ( s )) > Now ξ ( s ) = ( ) X k ( x ( s )) > is a semi-martingale and we can compute the Stratonovich di±erential d ( h ( ) X k ( x ( s )) = [ d h ( )] X k ( x ( s )) + h ( )[ d X k ( x ( s ))] = s r e k,r ( ) r ( s ) + e k, 0 ( ) ds One can compute easily e k,r ( s ) = h ( s )[ X r X k + X k X r ]( x ( s )) = h ( s )[ X k , X r ]( x ( s )) and e k, 0 ( ) = h ( s )[ X 0 X k + X k X 0 ]( x ( s )) = h ( s )[ X k , X 0 ]( x ( s ))
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