2
$\begingroup$

I am trying to derive the Lagrangian multiplier statistic (GMM version) under a restriction. The question is given below

The quadratic form is given by $Q_n(\theta,\alpha)=[m(\theta)', (m^a(\theta)-\alpha)']\begin{pmatrix} W_{11} & W_{12} \\ W_{21} & W_{22} \\ \end{pmatrix} \begin{pmatrix} m(\theta)\\ m^a(\theta)-\alpha\\ \end{pmatrix},$ where $\theta, \alpha$ are vectors.

Now they ask to verify that the Lagrangian Multiplier (Gradient) Statistic for testing the null hypothesis $H_0: \alpha=0,$ based on the moment condition \begin{pmatrix} m(\theta)\\ m^a(\theta)-\alpha\\ \end{pmatrix}can be expressed as $$n(m(\tilde{\theta})', m^a(\tilde{\theta})'){W_n}^{-1}S{\tilde{V}}^{-1}S'{W_n}^{-1} \begin{pmatrix} m(\tilde{\theta} )\\ m^a(\tilde{\theta})-\alpha\\ \end{pmatrix},$$where \tilde{\theta} is the restricted estimate of $\theta$, ${W_n}$ is a consistent estimate of $\bar{\Sigma}$, $S'=[0,I]$, and $$\tilde{V}=S'W^{-1}_nS-S'W^{-1}_n\tilde{D}(\tilde{D}'W^{-1}_n\tilde{D})^{-1}\tilde{D}'W^{-1}_nS$$with $\tilde{D}'=[\partial m(\tilde{\theta})'/\partial \theta, \partial m^a(\tilde{\theta})'/\partial \theta ].$

My attempt is:

Now we can rewrite the $H_0:\alpha=0$ as $S[\theta,\alpha]=0.$ To minimize $Q_n(\theta,\alpha)$ subject to $H_0$ we set up Lagrangian $L=G_n(\theta,\alpha)-2\lambda'S[\theta,\alpha].$ The first order conditions are $\begin{pmatrix} \partial m(\theta)'/\partial \theta & \partial m^a(\theta)'/\partial \theta \\ 0 & -I \\ \end{pmatrix} \begin{pmatrix} W_{11} & W_{12} \\ W_{21} & W_{22} \\ \end{pmatrix} \begin{pmatrix} m(\theta)\\ m^a(\theta)-\alpha\\ \end{pmatrix} = S\lambda$ and $\alpha =0$. Solving this simultaneously we can get the foc $\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}W^{-1}_n\begin{pmatrix} m(\tilde{\theta})\\ m^a(\tilde{\theta})\\ \end{pmatrix} =S\tilde{\lambda}. $

Now impose this into the form of Lagrangian statistic, we get $TLM =n{\tilde{\lambda}}'S'[\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}W^{-1}_n\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}]^{-1}S\tilde{\lambda}. $ If we replace $S\tilde{\lambda}$ by $\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}W^{-1}_n \begin{pmatrix} m(\tilde{\theta})\\ m^a(\tilde{\theta})\\ \end{pmatrix}$, then it suffices to show that $\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}[\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}W^{-1}_n\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}]^{-1}\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}=S{\tilde{V}}^{-1}S'$. I calculated that $\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}W^{-1}_n\begin{pmatrix} \tilde{D}' \\ -S' \\ \end{pmatrix}=\begin{pmatrix} S'W^{-1}_nS & \tilde{D}'W^{-1}_nS \\ S'W^{-1}_n\tilde{D} & \tilde{D}'W^{-1}_n\tilde{D} \\ \end{pmatrix}$.

I tried to get the inverse of the above square matrix and multiplied by both sides. However, couldn't get the required form. I am wondering if my constrained GMM estimate is wrong, and the foc is incorrect. Or maybe there is something that I missed out. Anyone can help me? Thanks in advance.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.