2
$\begingroup$

I am reading Kline, Saggio, Solvsten 2020 and am confused about some basic econometric stuffs in this paper.

They begin their introduction as below:

"""

Consider the linear model $$y_{i}=x_{i}^{\prime} \beta+\varepsilon_{i} \quad(i=1, \ldots, n)$$, where regressors $x_{i} \in \mathbb{R}^{k}$ are non-random, design matrix $S_{x x}=\sum_{i=1}^{n} x_{i} x_{i}^{\prime}$ has full rank, unobserved errors $\left\{\varepsilon_{i}\right\}_{i=1}^{n}$ are mutually independent and obey $\mathbb{E}\left[\varepsilon_{i}\right]=0$, but may possess observation-specific variances $\mathbb{E}\left[\varepsilon_{i}^{2}\right]=\sigma_{i}^{2}$.

Our object of interest is a quadratic form $\theta=\beta^{\prime} A \beta$ for some known non-random symmetric matrix $A \in \mathbb{R}^{k \times k}$ of rank $r$.

Following Searle, Casella, McCulloch (2009), when $A$ is positive semi-definite, $\theta$ is a variance component, while when $A$ is non-definite, $\theta$ may be referred to as a covariance component.

Note that linear restrictions on the parameter vector $\beta$ can be formulated in terms of variance components: for a non-random vector $v$, the null hypothesis $v^{\prime} \beta=0$ is equivalent to the restriction $\theta=0$ when $A=v v^{\prime}$.

"""

What do the "variance component", "covariance component", and "linear restrictions" here mean?

In section 2, the authors show a simple example of the model.

"""

Consider a data set composed of observations on $N$ groups with $T_{g}$ observations in the $g$ th group. The "analysis of covariance" model posits that outcomes can be written $$y_{g t}=\alpha_{g}+x_{g t}^{\prime} \delta+\varepsilon_{g t} \quad\left(g=1, \ldots, N, t=1, \ldots, T_{g} \geq 2\right)$$, where $\alpha_{g}$ is a group effect and $x_{g t}$ is a vector of strictly exogenous covariates.

The variability in outcomes attributable to $\alpha_g$ can be written $\sigma_{\alpha}^{2}=\frac{1}{n} \sum_{g=1}^{N} T_{g}\left(\alpha_{g}-\bar{\alpha}\right)^{2}$ where $\bar{\alpha}=\frac{1}{n} \sum_{g=1}^{N} T_{g} \alpha_{g}$.

This model can be aligned with the notation of the preceding section by letting $i=$ $i(g, t)$, where $i(\cdot, \cdot)$ is bijective, with inverse denoted $(g(\cdot), t(\cdot))$, and defining $y_{i}=y_{g t}$, $\varepsilon_{i}=\varepsilon_{g t}$ $x_{i}=\left(d_{i}^{\prime}, x_{g t}^{\prime}\right)^{\prime}, \quad \beta=\left(\alpha^{\prime}, \delta^{\prime}\right)^{\prime}, \quad \alpha=\left(\alpha_{1}, \ldots, \alpha_{N}\right)^{\prime} \quad \text { and } \quad d_{i}=\left(\mathbf{1}_{\{g=1\}}, \ldots, \mathbf{1}_{\{g=N\}}\right)^{\prime}$

To represent the target parameter in this notation, we write $\sigma_{\alpha}^{2}=\beta^{\prime} A \beta$, where $A=\left[\begin{array}{cc} A_{d}^{\prime} A_{d} & 0 \\ 0 & 0 \end{array}\right] \quad \text { for } A_{d}=\frac{1}{\sqrt{n}}\left(d_{1}-\bar{d}, \ldots, d_{n}-\bar{d}\right) \text { and } \bar{d}=\frac{1}{n} \sum_{i=1}^{n} d_{i}$.

"""

I am confused about why and how the $A$ should be written as this. The $A_d$ seems to be a $N \times n$ matrix, and thus $A_{d}^{\prime} A_{d}$ would be a $n \times n$ matrix, and I cannot see why $\sigma_{\alpha}^{2}=\beta^{\prime} A \beta$. It would make more sense if it is $A_{d} A_{d}^{\prime}$ but still I struggle to show $\beta^{\prime} A \beta$ equivalent to $\frac{1}{n} \sum_{g=1}^{N} T_{g}\left(\alpha_{g}-\bar{\alpha}\right)^{2}$.

Also I want to estimate the variability in outcomes attributable to the covariance $\sigma_{\alpha, X}$ between $\alpha_g$ and $x_{g t}^{\prime} \delta$. I think this can be done by constructing a new $A$, and I guess it would be something like $A= \frac{1}{2} \left[\begin{array}{cc} 0 & A_{x}^{\prime} A_{d} \\ A_{d}^{\prime} A_{x} & 0\end{array}\right]$ where $A_{x}$ is constructed similarly to $A_{d}$. But I am not sure if this is correct and what is the general rule to construct $A$ of the quadratic forms in various potential cases.

Update: I use software to do some trials and find $A=\left[\begin{array}{cc} A_{d} A_{d}^{\prime} & 0 \\ 0 & 0 \end{array}\right]$ and $A= \frac{1}{2} \left[\begin{array}{cc} 0 & A_{d} A_{x}^{\prime} \\ A_{x} A_{d}^{\prime} & 0\end{array}\right]$ do give the correct variance and covariance.

$\endgroup$

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.