Theorem: Suppose X1,,Xn are iid with density p(x|θ) and that derivatives up to second order can be passed under the integral sign in dP(x|θ). Then

1nu(θ)dN(0,I(θ))

Proof:

Vu(θ)=I(θ)DefIn{u¯(θ)Eu(θ)}dN(0,I(θ))CLT (iid)1nu(θ)dN(0,I(θ))Eu(θ)=0;u¯=u/n

Corollaries

The above result can be stated in a variety of ways.

Corollary (#1): Under the same conditions as the original theorem, if I(θ) is positive definite, then

1nI1/2(θ)u(θ)dN(0,I).

Corollary (#2): Under the same conditions as the first corollary,

In(θ)1/2u(θ)dN(0,I).

Proof:

11nIn(θ)pI(θ)LLN (iid)In(θ)1/2u(θ)=[n1nIn(θ)]1/2u(θ)=[1nIn(θ)]1/2I(θ)1/2A1nI(θ)1/2u(θ)b

At this point, note that ApI by 1 and CMT, while bdN(0,I) by our first corollary. Thus, the result

In(θ)1/2u(θ)dN(0,I).

holds by Slutsky’s theorem.

Finally, note that Corollary #2 holds for any consistent estimator of I(θ). For example, if 1nIn(θ^)pI(θ), we would also have

In(θ^)1/2u(θ)dN(0,I)

by the same proof.