solution2 最近の更新履歴 Econometrics Ⅰ 2016 TA session

Solution#2

Shouto Yonekura

August 1, 2016

(1)

First we can decompose ˆβ as follows:

β = (Xˆ ^′X)⁻¹X^′y

= (X^′X)⁻¹X^′(Xβ + u)

= β + (X^′X)⁻¹X^′u

= β + (_n¹X^′X)^{−1 1}_nX^′u

= β + Q⁻¹_xxQxu. Using the LLN, we can easily that Qxu_→p0 since

1 n^X

′u →^pX′E[u]

= 0.

Suppose that Qxxconverges to some finite matrix Mxx. Then QxxQxu→p0. Hence we get ˆ_{β →}pβ.

(2) (3)

Suppose that assumptions A1 ∼ A5 hold. In addtion to these, E[u⁴i^{] and (xix}

′

i⁾²are finite for any i. Where xi_{∈ R}ⁿ. By the triangle inequality and Jensen’s inequality, we can show that

k E[x^ix′i^u2i] k²≤ E[k x^ix′i^u2i k^2]

= E[k x²i^u2i k] =k xik^2E[u2i^].

Using Cauchy–Schwarz inequality, we will obtain

k xik^2E[u2i] ≤ (k xik^4)1/2(E[u4i^])1/2

< ∞.

Therefore, we finally get _√n^{1 P}_ixiui_→dN (0, Ω) by CLT. Next, we can rewriten^√n( ˆβ − β) as follows:

√n( ˆ_{β − β) =}^√n(X^′X)⁻¹X^′u,

= (_n¹X^′X)⁻¹_√n¹ X^′u,

= Q⁻¹_xx√n¹ X^′u.

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From Prop11.11,_√n¹ X^′uui→dN (0, Ω).Suppose that Qxxconverges to some finite matrix Mxx. Then using Slutsky’s theorem, we obtain that Q⁻¹_xx√n¹ X^′_{u →}dM_xx⁻¹N (0, Ω). Hence we can finally show that:

√n( ˆ_{β − β) →}dNk(0, M_xx⁻¹ΩM_xx⁻¹) , (M_xx^′ = Mxx)

= Nk(0, V ) Q.E.D.

(4)

Suppose that Xi_∼iid f (xi; θ). Then the likelihood function is given by L(θ) :=^Qn_i=1f (θ; xi)

(5)

First we can show that

l^′′(θ; X) = ∂θ(^∂θ_{f (θ;X)}^{f (θ;X)})

= ^∂

2 θ2^{f (θ;X)}

f (θ;X) −^∂θf (θ;X)^{f (θ;X)}

²

holds. Multiplying both side by f (θ; X) and integrating with respect to x, we get E[l^′′(θ; X)] =^{´ ∂θ22}_{f (θ;X)}^{f (θ;X)}f (θ; X)dx −^∂θ_{f (θ;X)}^{f (θ;X)2f (θ; X)dx}

=´ ∂_θ²2f (θ; X)dx − I(θ)

∂_θ²2´ f (θ; X)dx − I(θ)

= −I(θ).

Therefore, I(θ) = −E[∂θ²²lnf (θ; X)] holds. Let T (X) be an unbiased estimator of θ. Next we get

∂θE[T (X)] = ∂θ´ T (X)f (x; θ)dx

⇐⇒ ^{1 = ∂}θ´ T (X)f (x; θ)dx

=´ ∂θT (X)f (x; θ)dx

=´ T (X)∂θlnfX(x; θ)fX(x; θ)dx E[T (X)l^′(θ)].

Since E[l^′(θ)] = 0 (Prop9.2,) this can be rewriten as follows:

E[T (X)l^′(θ)] = E[(T (X) − θ)l^′(θ)]

= Cov(T (X), l^′(θ)).

2

This leads to

1 = Cov((T (X), l^′(θ)))²≤ V [T (X)]V [l^′(θ)] , (−1 ≤ ^{Cov(X, Y )} pV [X]pV [Y ] ^{≤ 1)}

= V [T (X)]I(θ). Therefore, V [T (X)] ≥ 1/I(θ) holds. Q.E.D.

(6)

Assume that parameter space Θ is a compact set. Let θ0 be a true value of parameter θ. Then for any θ0_{6= θ, we}

have by Jensen’s inequality

Eθ0

hln_{f (x;θ}^{f (x;θ)}

0)

i

≤ lnEθ0

hln_{f (x;θ}^{f (x;θ)}

0)

i= 0,

since

Eθ0

h_{f (x;θ)}

f (x;θ0)

i=^´ _{f (x;θ}^{f (x;θ)}₀₎f (x; θ0)dx

=´ f (x; θ)dx

= 1,

in the case of continious X. (with an analogous story in the case of discrete X). Fix δ > 0. Let a1:= Eθ0

hln^{f (x;θ}_{f (x;θ}^0−δ)₀₎ ⁱ< 0, a2:= Eθ0

hln^{f (x;θ}_{f (x;θ}^0+δ)₀₎ ⁱ< 0. Usinng LLN and continuity of l(θ) := lnL(θ), we get

l(θ0−δ)−l(θ0) n →^pa1,

thus l(θ0− δ) < l(θ0) for any sufficiently large n. Similaly, we have l(θ0+ δ) < l(θ0) for any sufficiently large n. Since l(θ) is a continious function with respect to θ and Θ is a compact set, there exists ˆθ such that miximizes l(θ) on the interval (θ0− δ, θ0+ δ). This menas that ˆθ is a consistent estimator.

Next we assume that l(θ) is twice continiously defferentiable on a negihrhood of θ0. We have already shown that there exixts ˆθ such that l^′(ˆθ) = 0 and ˆ_{θ →}pθ0. A Taylor expansion gives that

−l^{′(θ0) = l′(ˆ}θ) − l^{′(θ0) = (ˆ}θ − θ^{0)l””(θ#),} where θ^#lies on between θ0 and ˆθ. Therefore

(ˆ_{θ − θ}0) = −_l””^l′(θ_(θ⁰#⁾₎,

3

and we can rewrite this as follows:

pnI(θ0)(ˆ_{θ − θ}0) = ^l

′(θ₀)

√nI(θ0) l^′′(θ₀) l^′′(θ^#)



−^l

′′(θ₀) nI(θ0)

−1

.

First, usging CLT, we get

l^′(θ0)

√nI(θ0) ^{→d N (0, 1).}

Next, since l(θ) is twice continiously defferentiable on a negihrhood of θ0 , LLN and continious mapping theorem yield

n⁻¹l^′′(θ0) n⁻¹l^′′(θ^#)^→p

E[l^′′(θ0)] E[l^′′(θ0)]

= ^{−I (θ0)}

−I (θ0)^{= 1.}

Finally, LLN also gives

n⁻¹l^′′(θ0) →p−I(θ0). Thus



−^l

′′(θ0) nI(θ0)



→p1. From these results, we can conclude that

√n(ˆ_{θ − θ}0) → N(0, I(θ^0)−1).

Q.E.D.

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