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.


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.