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\begin_body

\begin_layout Title
Problem Set 9
\end_layout

\begin_layout Author
Daryl Larsen
\end_layout

\begin_layout Section
Overidentifying restrictions test for nonlinear GMM
\end_layout

\begin_layout Standard

\series bold
(a) Find efficient weight matrix and asymptotic variance
\end_layout

\begin_layout Standard
The efficient weight matrix for nonlinear GMM is 
\begin_inset Formula $A^{*\prime}A^{*}=Var\big(g(W_{i},\theta_{0})\big)^{-1}=\Sigma_{0}^{-1}$
\end_inset

.
 The asymptotic variance of the GMM estimator is 
\begin_inset Formula 
\[
(\Gamma_{0}^{\prime}A^{\prime}A\Gamma_{0})^{-1}\Gamma_{0}^{\prime}A^{\prime}A\Sigma_{0}A^{\prime}A\Gamma_{0}(\Gamma_{0}^{\prime}A^{\prime}A\Gamma_{0})^{-1}
\]

\end_inset


\end_layout

\begin_layout Standard
with the efficient weight matrix, this becomes
\begin_inset Formula 
\begin{align*}
 & =(\Gamma_{0}^{\prime}\Sigma_{0}^{-1}\Gamma_{0})^{-1}\Gamma_{0}^{\prime}\Sigma_{0}^{-1}\Sigma_{0}\Sigma_{0}^{-1}\Gamma_{0}(\Gamma_{0}^{\prime}\Sigma_{0}^{-1}\Gamma_{0})^{-1}\\
 & =(\Gamma_{0}^{\prime}\Sigma_{0}^{-1}\Gamma_{0})^{-1}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard

\series bold
(b) Show that when the model holds, 
\begin_inset Formula $J_{n}\rightarrow_{d}\chi_{l-d}^{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $J_{n}=nQ_{n}^{*}(\hat{\theta}_{n}^{*}),\text{ where }Q_{n}^{*}(\theta)=\big(n^{-1}\sum_{i=1}^{n}g(W_{i},\theta)\big)^{\prime}A_{n}^{*\prime}A_{n}^{*}\big(n^{-1}\sum_{i=1}^{n}g(W_{i},\theta)\big)$
\end_inset

.
\end_layout

\begin_layout Section
Efficient IVs for nonlinear GMM
\end_layout

\begin_layout Standard

\series bold
(a) Find asymptotic variance of 
\begin_inset Formula $\hat{\theta}_{n}^{h}$
\end_inset


\end_layout

\begin_layout Standard
The criterion function for this GMM is
\begin_inset Formula 
\[
Q_{n}(\theta)=\frac{1}{2}||A_{n}n^{-1}\sum_{i=1}^{n}g(W_{i},\theta)h(Z_{i})||^{2}
\]

\end_inset

 and 
\begin_inset Formula 
\[
\sqrt{n}(\hat{\theta}_{n}-\theta_{0})=-(\Gamma_{n}^{\prime}A^{\prime}A\Gamma_{n})^{-1}\Gamma_{n}^{\prime}A^{\prime}A\big(n^{-1/2}\sum_{i=1}^{n}g(W_{i},\theta_{0})h(Z_{i})\big)+o_{p}(1)
\]

\end_inset

 Since by assumption, 
\begin_inset Formula 
\[
n^{-1/2}\sum_{i=1}^{n}g(W_{i},\theta_{0})h(Z_{i})\rightarrow_{d}N(0,\Omega_{n})
\]

\end_inset

it follows that 
\begin_inset Formula 
\[
\sqrt{n}(\hat{\theta}_{n}-\theta_{0})\rightarrow_{d}N(0,(\Gamma_{n}^{\prime}A^{\prime}A\Gamma_{n})^{-1}\Gamma_{n}^{\prime}A^{\prime}A\Omega_{n}A^{\prime}A\Gamma_{n}(\Gamma_{n}^{\prime}A^{\prime}A\Gamma_{n})^{-1})
\]

\end_inset

 then if we choose 
\begin_inset Formula $A^{\prime}A=\Omega_{n}^{-1},$
\end_inset


\begin_inset Formula 
\[
\sqrt{n}(\hat{\theta}_{n}-\theta_{0})\rightarrow_{d}N(0,(\Gamma_{n}^{\prime}\Omega_{n}^{-1}\Gamma_{n})^{-1})
\]

\end_inset


\end_layout

\begin_layout Standard

\series bold
(b) Find asymptotic variance of 
\begin_inset Formula $\hat{\theta}_{n}^{*}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\hat{\theta}_{n}^{*}$
\end_inset

 is the GMM estimator corresponding to the unconditional moment restriction
 using 
\begin_inset Formula $h_{*}(Z_{i})$
\end_inset

 as the instrument: 
\begin_inset Formula $E(g(W_{i},\theta_{0})h_{*}(Z_{i}))=0$
\end_inset

.
 Let 
\begin_inset Formula $\Gamma_{n}^{*}$
\end_inset

 be the 
\begin_inset Formula $\Gamma_{n}$
\end_inset

 moment associated with this instrument.
 Then
\begin_inset Formula 
\begin{align*}
\Gamma_{n}^{*} & =\frac{\partial}{\partial\theta^{\prime}}E(g(W_{i},\theta_{0})h_{*}(Z_{i}))\\
 & =\frac{\partial}{\partial\theta^{\prime}}E\big[g(W_{i},\theta_{0})\frac{1}{E(g^{2}(W_{i},\theta_{0})|Z_{i})}E\big(\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i}\big)\big]\\
 & =E\big[\frac{\partial}{\partial\theta^{\prime}}g(W_{i},\theta_{0})\frac{1}{E(g^{2}(W_{i},\theta_{0})|Z_{i})}E\big(\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i}\big)\big]\\
 & =E\big[\frac{1}{E(g^{2}(W_{i},\theta_{0})|Z_{i})}E(\frac{\partial g(W_{i},\theta_{0})}{\partial\theta^{\prime}}|Z_{i})E(\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i})\big]
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Similarly, let 
\begin_inset Formula $\Omega_{n}^{*}$
\end_inset

 be the 
\begin_inset Formula $\Omega_{n}$
\end_inset

 moment associated with the instrument:
\begin_inset Formula 
\begin{align*}
\Omega_{n}^{*} & =E\big[g^{2}(W_{i},\theta_{0})h_{*}(Z_{i})h_{*}(Z_{i})^{\prime}\big]\\
 & =E\big[E[g^{2}(W_{i},\theta_{0})|Z_{i}]h_{*}(Z_{i})h_{*}(Z_{i})^{\prime}\big]\\
 & =E\big[E[g^{2}(W_{i},\theta_{0})|Z_{i}]\frac{1}{E(g^{2}(W_{i},\theta_{0})|Z_{i})}E(\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i})\big(\frac{1}{E(g^{2}(W_{i},\theta_{0})|Z_{i})}E[\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i}]\big)^{\prime}\big]\\
 & =E\big[\frac{1}{E(g^{2}(W_{i},\theta_{0})|Z_{i})}E[\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i}]E[\frac{\partial g(W_{i},\theta_{0})}{\partial\theta^{\prime}}|Z_{i}]^{\prime}\big]\\
 & =E\big[\frac{1}{E(g^{2}(W_{i},\theta_{0})|Z_{i})}E[\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i}]E[\frac{\partial g(W_{i},\theta_{0})}{\partial\theta^{\prime}}|Z_{i}]\big]\\
 & =\Gamma_{n}^{*}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
then the asymptotic variance of 
\begin_inset Formula $\hat{\theta}_{n}^{*}$
\end_inset

 becomes 
\begin_inset Formula 
\begin{align*}
(\Gamma_{n}^{*\prime}\Omega_{n}^{*-1}\Gamma_{n}^{*})^{-1} & =(\Gamma_{n}^{*\prime}\Gamma_{n}^{*-1}\Gamma_{n}^{*})^{-1}\\
 & =(\Gamma_{n}^{*\prime})^{-1}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard

\series bold
(c) Show 
\begin_inset Formula $h_{*}(Z_{i})$
\end_inset

 is the efficient instrument
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V^{h}-V^{*}=(\Gamma_{n}^{\prime}\Omega_{n}^{-1}\Gamma_{n})^{-1}-(\Gamma_{n}^{*\prime})^{-1}
\]

\end_inset


\end_layout

\begin_layout Standard
Now let there exist 
\begin_inset Formula $\epsilon$
\end_inset

 and 
\begin_inset Formula $e$
\end_inset

 such that 
\begin_inset Formula $E(\epsilon e^{\prime})=E(ee^{\prime})$
\end_inset

 which implies 
\begin_inset Formula $E(\epsilon\epsilon^{\prime})-E(ee^{\prime})\geq0$
\end_inset

.
 Now I want to find 
\begin_inset Formula $\epsilon$
\end_inset

 and 
\begin_inset Formula $e$
\end_inset

 such that 
\begin_inset Formula $E(ee^{\prime})=V^{*}$
\end_inset

and 
\begin_inset Formula $E(\epsilon\epsilon^{\prime})=V^{h}$
\end_inset

.
\end_layout

\begin_layout Standard
Let 
\begin_inset Formula $e\equiv V^{*}g(W_{i},\theta_{0})h_{*}(Z_{i})$
\end_inset

.
 Then 
\begin_inset Formula 
\begin{align*}
E[ee^{\prime}] & =E[V^{*}g(W_{i},\theta_{0})h_{*}(Z_{i})h_{*}(Z_{i})^{\prime}g(W_{i},\theta_{0})^{\prime}V^{*\prime}]\\
 & =V^{*}E[g^{2}(W_{i},\theta_{0})h_{*}(Z_{i})h_{*}(Z_{i})^{\prime}]V^{*}\\
 & =V^{*}V^{*-1}V^{*}\\
 & =V^{*}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Now let 
\begin_inset Formula $\epsilon\equiv V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}g(W_{i},\theta_{0})h(Z_{i})$
\end_inset

.
 Then
\begin_inset Formula 
\begin{align*}
E[\epsilon\epsilon^{\prime}] & =E[V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}g(W_{i},\theta_{0})h(Z_{i})h(Z_{i})^{\prime}g(W_{i},\theta_{0})\Omega_{n}^{-1}\Gamma_{n}V^{h}]\\
 & =V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}E[g^{2}(W_{i},\theta_{0})h(Z_{i})h(Z_{i})^{\prime}]\Omega_{n}^{-1}\Gamma_{n}V^{h}\\
 & =V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}\Gamma_{n}V^{h}\\
 & =V^{h}V^{h-1}V^{h}\\
 & =V^{h}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align*}
E[\epsilon e^{\prime}] & =E[V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}g(W_{i},\theta_{0})h(Z_{i})h_{*}(Z_{i})^{\prime}g(W_{i},\theta_{0})V^{*}]\\
 & =V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}E[g^{2}(W_{i},\theta_{0})h(Z_{i})h_{*}(Z_{i})^{\prime}]V^{*}\\
 & =V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}E\big[g^{2}(W_{i},\theta_{0})h(Z_{i})\frac{1}{E[g^{2}(W_{i},\theta_{0})|Z_{i}}E[\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i}]^{\prime}\big]V^{*}\\
 & =V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}E\big[E[g^{2}(W_{i},\theta_{0})|Z_{i}]\frac{h(Z_{i})}{E[g^{2}(W_{i},\theta_{0})|Z_{i}]}E[\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i}]^{\prime}\big]V^{*}\\
 & =V^{h}\Gamma_{n}^{\prime}\Omega_{n}^{-1}E\big[E[h(Z_{i})\frac{\partial g(W_{i},\theta_{0})}{\partial\theta}|Z_{i}]^{\prime}\big]V^{*}\\
 & =V^{h}V^{h-1}V^{*}\\
 & =V^{*}\\
 & =E[ee^{\prime}]
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Then 
\begin_inset Formula $V^{h}-V^{*}\geq0$
\end_inset


\end_layout

\begin_layout Standard

\series bold
(d) Steps for constructing a feasible version of 
\begin_inset Formula $\hat{\theta}_{n}^{*}$
\end_inset

 when 
\begin_inset Formula $h_{*}(Z_{i})$
\end_inset

 is unknown
\end_layout

\begin_layout Standard
First, create a consistent but inefficient estimator 
\begin_inset Formula $\hat{\theta}_{n}$
\end_inset

 using 
\begin_inset Formula $h(Z_{i})$
\end_inset

 as the instrument.
 Since 
\begin_inset Formula $h(Z_{i})$
\end_inset

 is known, it can be used.
 We can then estimate 
\begin_inset Formula $E[g^{2}(W_{i},\hat{\theta}_{n})|Z_{i}]$
\end_inset

 and 
\begin_inset Formula $E[\frac{\partial g^{2}(W_{i},\hat{\theta}_{n})}{\partial\theta}|Z_{i}]$
\end_inset

, which are all we need to construct 
\begin_inset Formula $\hat{h}_{*}(Z_{i})$
\end_inset

.
 
\end_layout

\end_body
\end_document