18. Comments on Fisher's Information (Feb7)

# 18. Comments on Fisher's Information (Feb7) - Statistics...

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Statistics 358b : Comments on Fisher’s Information February 9, 2011 We showed under the regularity conditions, f being pdf (or pmf and sum- mation accordingly) E θ 0 [ log ( f ( X ; θ 0 )) ∂θ ] = R [ log ( f ( x ; θ 0 )) ∂θ ] f ( x ; θ 0 ) dx = R [ 1 f ( x, θ 0 ) ∂f ( x ; θ 0 ) ∂θ ] f ( x ; θ 0 ) dx = R ∂f ( x ; θ 0 ) ∂θ dx = ± R f ( x ; θ ) dx ∂θ ² ² ² ² θ = θ 0 = (1) ∂θ = 0 Notice the integration is done with respect to θ 0 , and this ﬁnal statement is therefore true for all θ 0 . Thus in our statement we can also write R [ log ( f ( x ; θ )) ∂θ ] f ( x ; θ ) dx = 0 (1) for all θ . Deﬁne I( θ ) in both the 1-D and higher dimensional parameter case, that is as a matrix. Re : Lemma 8.5.2 A proof. Remark : Equation (1) is true for all θ but it is not true that Λ( θ ) ∂θ = E θ 0 [ log ( f ( X ; θ )) ∂θ ] = R [ log ( f ( x ; θ )) ∂θ ] f ( x ; θ 0 ) dx is equal to 0 for all θ . In each integral the integrand is a product of two func- tions, one with parameters θ and θ 0 (not necessarily matching) and in the other integral with parameters θ and θ (necessarily matching). If we calculate the 1

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Stat 358 : Fisher’s Information 2 derivative is this second integral, which is valid, it is not the case that the derivative is equal to 0. However if we diﬀerentiate (1) with respect to θ , it is the case that this derivative is equal to 0. The correct method is therefore to diﬀerentiate (1). For real θ ∂θ E θ [ log ( f ( X ; θ )) ∂θ ] = ∂θ R [ log ( f ( x ; θ )) ∂θ ] f ( x ; θ ) dx = ∂θ 0 = 0 This last line is not true if we calculate E θ 0 [ log ( f ( X ; θ )) ∂θ ] = R [ log ( f ( x ; θ )) ∂θ ] f ( x ; θ 0 ) dx even though the diﬀerentiation is valid. Remark : A variance matrix (also called a variance-covariance or covariance
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18. Comments on Fisher's Information (Feb7) - Statistics...

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