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N as avar n n dg d n 2 c n and then we define an

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̂ n as Avar n ̂ n  dg d ̂ n 2 c ̂ n , and then we define an estimator of the asymptotic variance of ̂ n as Avar ̂ n dg d ̂ n 2 c ̂ n / n 50
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EXAMPLE : Returning to the Poisson example with exp we take ̂ n X ̄ n and ̂ n exp X ̄ n .Now g exp and g 1 exp . It follows that Avar n ̂ n  exp  2 Avar n X ̄ n  exp 2 and so the asymptotic standard deviation of ̂ n is exp / n . It is not true that Var n ̂ n  exp 2 . We do not know Var n ̂ n  . 51
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The delta method can be applied to vectors of parameters. Let g : Θ m be continuously differentiable. We can write g g 1 g 2 g m where g j g j 1 , 2 ,..., p . 52
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Let g denote its m p Jacobian, which we can write g g 1 g 2 g m where g j is the 1 p gradient of g j . 53
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A similar mean value expansion argument [for each row of g ] delivers n ̂ n g n ̂ n o p 1 G n ̂ n o p 1 where G ≡∇ g is m p . It follows that n ̂ n a ~ Normal 0 , G C G where C Avar n ̂ n  54
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The asymptotic variance of n ̂ n is Avar n ̂ n  G C G G  Avar n ̂ n  G and so Avar ̂ n G C G / n . This is what we would get if we (naively) treat ̂ n as a linear function of ̂ n : ̂ n G ̂ n Var ̂ n C / n 55
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EXAMPLE :Let , where , 0. Define as the proportionate difference between and : 1 So g g , / 1 and g 1 , g  2 and we write g 1 ,  2 56
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If C is the 2 2 asymptotic variance of n ̂ n , C c 11 c 12 c 12 c 22 then Avar n ̂ n  1 ,  2 C  1 ,  2 2 c 11 2 4 c 22 2  3 c 12
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n as Avar n n dg d n 2 c n and then we define an estimator...

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