
\documentclass[11pt]{amsart}
\usepackage{latexsym,amssymb}
\usepackage{graphics}
\usepackage{url}
\usepackage{harvard}
\usepackage{eulervm}
\usepackage{epsfig,subfigure}
\usepackage{setspace}
%\usepackage{natbib}
\usepackage{tikz}
%\usepackage[affil-it]{authblk}
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
            \node[shape=circle,draw,inner sep=1pt] (char) {#1};}}
\usepackage{graphicx}

\setlength{\textheight}{205mm}
\setlength{\oddsidemargin}{10mm}
\setlength{\evensidemargin}{10mm}
\setlength{\topmargin}{0mm}
\setlength{\textwidth}{6in}

\newtheoremstyle{normal}% name
{2ex}               % Space above, empty = `usual value'
{3ex}               % Space below
{}                  % Body font
{}                  % Indent amount (empty = no indent, \parindent = para indent)
%{\scshape} % Thm head font
{\bfseries} % Thm head font
{}                  % Punctuation after thm head
{2pt}   % Space after thm head.
{\thmname{#1}\thmnumber{ #2.} \thmnote{(#3)}}% Thm head spec


\theoremstyle{normal}
\newtheorem{definition}{Definition}
\newtheorem{remark}[definition]{Remark}
\newtheorem{example}{Example}[section]
\newtheorem{cond}[definition]{Condition}


%\newtheorem{example}{Example}
%\newtheorem{remark}{Remark}
\renewcommand{\abstractname}{Abstract}
%\def\argmax{\text{argmax}}
\newtheorem{lemma}{Lemma}
\newtheorem{assumption}{Assumption}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newtheorem{proposition}{Proposition}
%\newtheorem{definition}{Definition}


\renewenvironment{proof}{\noindent {\bf Proof.\ }}{\hfill{\rule{2mm}{2mm}}}
\renewenvironment{remark}{\noindent {\bf Remark.\ }} {\hfill{ \rule{0mm}{0mm}}}
%\renewenvironment{example}{\noindent {\bf Example.\ }}



\DeclareMathOperator{\Var}{Var}
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\Cov}{Cov}
\DeclareMathOperator{\ran}{ran}
\DeclareMathOperator{\BL}{BL}
\def\FF{\mathcal F}
\def\E{\mathbb E}
\def\PP{\mathbb P}
\def \eps {\epsilon}
\title{Supplement to ``Neyman's $C(\alpha)$ test for unobserved heterogeneity" }
\author{Jiaying Gu \\ University of Illinois at Urbana-Champaign}
%\date{\today}
\begin{document}
\bibliographystyle{et2}  % replace econometric by et2 to conform econometric theory house style citation
%\citationstyle{econometrica}
\renewcommand*{\harvardand}{and}

\maketitle

\onehalfspacing
\begin{appendix}


\section{Power Comparison for Example 4.2}
\noindent Here we conduct a power comparison between the $C(\alpha)$ test and the \citeasnoun{Breusch79} LM test described in Example 4.2. Two different alternative $\beta_{i}$ distributions are considered. The first one assumes $\beta_{i}$ takes value $0$ for $i = 1, \dots, N/2$ and $c N^{-1/4}$ for $i = N/2+1, \dots, N$.  We let  $c$ take 51 distinct values equally spaced from $0$ to $\sqrt{50}$. The second case assumes $\beta_i \sim \mathcal{N}(0,\sigma^2)$ with $\sigma$ taking 21 distinct values from $0$ to $1$. For simplicity, we consider the case with dimension two, where both $x$ covariates are standard normal variables. The sample size is fixed at 400 with 10000 replications. Figure \ref{LMtest} presents the power curve for the 10 \% nominal level. In both cases, the $C(\alpha)$ test dominate the power curve of the LM test based on the usual $\chi^2$ asymptotics. \\


\begin{figure}[h!]
\centering 
\includegraphics[scale = 0.5]{LMtest_power_uncolor.pdf}
 \caption{Power Comparison of Slope Heterogeneity Test for Linear Regression Model: The left figure corresponds to the first experiment and the right to the second. The dotted curve corresponds to the power curve of the $C(\alpha)$ test based on the mixture of $\chi^2$ asymptotics and the solid curve for the LM test based on the $\chi^{2}_{3}$ asymptotics. }
\label{LMtest}
\end{figure}




\section{Computational Details in Examples 4.3}
\noindent \textbf{Joint test for Gaussian panel data model.}
The information matrix for $(\xi, \theta)=(\xi_{1}, \xi_{2}, \mu_{0}, \sigma_{0}^{2})$ is 
\[
I = \begin{pmatrix}
I_{\xi \xi} & I_{\xi \theta}\\
I_{\theta \xi} & I_{\theta \theta}
\end{pmatrix} = 
\frac{NT}{\sigma_{0}^{4}} \begin{pmatrix}
2T & \sigma_{0}^{2} & 0 & 1 \\
\sigma_{0}^{2} & (T+3)\sigma_{0}^{4}/2 & 0 & \sigma_{0}^{2}/2 \\
0 & 0 & \sigma_{0}^{2} & 0\\
1 &\sigma_{0}^{2}/2 & 0 & 1/2
\end{pmatrix}
\]
We further find 
\[
I_{\xi.\theta} = I_{\xi \xi}-I_{\xi \theta}I_{\theta \theta}^{-1}I_{\theta \xi} =
\begin{pmatrix}
2NT(T-1) / \sigma_{0}^{4}& 0 \\
0 & NT(T/2+1)
\end{pmatrix}
\]
and 
\[
I_{\xi \theta} I_{\theta \theta}^{-1} = \begin{pmatrix} 0 & 2 \\ 0 & \sigma_{0}^{2} \end{pmatrix}. 
\]\\

As we have remarked in Section 3, the diagonality of $I_{\xi.\theta}$ provides much convenience for finding the optimal test statistics. Denote 
\[
T_{n} := \begin{pmatrix} t_{1n}\\t_{2n} \end{pmatrix}
= I_{\xi.\theta}^{-1/2} \begin{pmatrix}
\sum_{i}v_{i1} - 2 \sum_{i} v_{4i} \\
\sum_{i}v_{2i} - \sigma_{0}^{2} \sum_{i} v_{4i}
\end{pmatrix} 
 = \begin{pmatrix}
(2NT(T-1)/\sigma_{0}^{4})^{-1/2} \left (\sum_{i} (\frac{\bar y_{i.}-\mu_{0}}{\sigma_{0}^{2}/T})^{2} - NT/\sigma_{0}^{2} \right) \\
(NT(T/2+1))^{-1/2} \left ( \sum_{i} (Z_{i} - T/2)^{2}-NT/2 \right )
\end{pmatrix}
\]
Replacing $(\mu_{0}, \sigma_{0}^{2})$ by their MLEs yields the joint $C(\alpha)$ test. 

\section{Claim in Section 5}
\noindent Here we provide the detail derivation for the claim in Section 4 that the reparameterization adopted in \citeasnoun{Chesher84} and \citeasnoun{Cox83} for heterogeneity test requires extra moment conditions on $U$ for second derivative of log density with respect to the test parameter to be bounded. 

\begin{proposition}
For iid random variable $Y_{1}, \dots, Y_{n}$ each with density function $\int p(y; \lambda_{0} + \tau \sqrt{\eta} u_{i})dF(u_{i})$, where $U_{i}$ is a random variable with zero mean and unit variance. The second-order derivative of the log density with respect to $\eta$ evaluated under $\eta=0$ is unbounded unless $\mathbb{E}(U^{3})=0$ and $\mathbb{E}(U^4)<\infty$. 
\end{proposition}
\noindent \textbf{Proof.}
Denote the log density as $l=\log \int p(y; \lambda_{0}+\tau \sqrt{\eta} u_{i})dF(u_{i})$. The first order derivative with respect to $\eta$ is
\[
\left. \nabla_{\eta}l \right |_{\eta=0} = \frac{\tau \int \nabla_{\lambda}p(y; \lambda_{0}) u dF(u)}{2 \sqrt{\eta} \int p(y; \lambda_{0})dF(u)}
=\frac{\tau^{2}}{2} \mathbb{E}(U^{2}) \frac{\nabla_{\lambda}^{2} p(y; \lambda_{0})}{p(y; \lambda_{0})}
\]
The last step is obtained by applying the l'H\^opital's rule. \\

The second order derivative is 
\[
\begin{array}{ll}
\left. \nabla_{\eta} ^{2} l \right |_{\eta=0} &= \left. \frac{\tau^{2} \sqrt{\eta} \int \nabla_{\lambda}^{2}p(y; \lambda_{0})u^{2}dF(u) - \tau \int \nabla_{\lambda}p(y; \lambda_{0})udF(u)}{4\eta \sqrt{\eta} \int p(y; \lambda_{0})dF(u)} \right |_{\eta=0}- \left (\left. \nabla_{\eta}  l \right |_{\eta=0} \right)^{2}\\
&= \left. \frac{\tau^{3} \int \nabla_{\lambda}^{3} p(y; \lambda_{0})u^{3}dF(u)}{12 \sqrt{\eta} \int p(y; \lambda_{0})dF(u)} \right|_{\eta=0} - \left (\left. \nabla_{\eta}  l \right |_{\eta=0} \right)^{2}
\end{array}
\]
Provided that $\nabla_{\lambda}^{3} p(y; \lambda_{0})$ is not degenerately zero,  $\nabla_{\eta}^{2}l$ is unbounded unless $\mathbb{E}(U^3)=0$ and $\mathbb{E}(U^{4})<\infty$ so that we can apply l'H\^opital's rule again and get 
\[
\left. \nabla_{\eta}^{2}l\right |_{\eta=0} = \frac{\tau^{4}}{12} \left [ \mathbb{E}(U^4) \frac{\nabla_{\lambda}^{4} p(y; \lambda_{0})}{p(y; \lambda_{0})} - 3 \mathbb{E}(U^2)^2\Big(\frac{\nabla_{\lambda}^{2}p(y; \lambda_{0})}{p(y; \lambda_{0})}\Big)^2 \right] < \infty
\] \hfill{\rule{2mm}{2mm}} 
\bibliography{Calpha}
\end{appendix}
\end{document}