Corrigendum to Pustejovsky and Tipton (2018), redux

Theorem 2, revised

Consider the model \[ \bm{y}_i = \bm{R}_i \bs\beta + \bm{S}_i \bs\gamma + \bm{T}_i \bs\mu + \bs\epsilon_i, \tag{1} \] where \(\bm{y}_i\) is an \(n_i \times 1\) vector of responses for cluster \(i\), \(\bm{R}_i\) is an \(n_i \times r\) matrix of focal predictors, \(\bm{S}_i\) is an \(n_i \times s\) matrix of additional covariates that vary across multiple clusters, and \(\bm{T}_i\) is an \(n_i \times t\) matrix encoding cluster-specific fixed effects, all for \(i = 1,...,m\). Let \(\bm{U}_i = \left[ \bm{R}_i \ \bm{S}_i \right]\) be the set of predictors that vary across clusters and \(\bm{X}_i = \left[ \bm{R}_i \ \bm{S}_i \ \bm{T}_i \right]\) be the full set of predictors. Let \(\bm{\ddot{U}}_i = \left(\bm{I} - \bm{T}_i \bm{M}_{\bm{T}}\bm{T}_i'\right) \bm{U}_i\) be an absorbed version of the focal predictors and the covariates. The cluster-robust variance estimator for the coefficients of \(\bm{U}_i\) is \[ \bm{V}^{CR2} = \bm{M}_{\bm{\ddot{U}}} \left(\sum_{i=1}^m \bm{\ddot{U}}_i' \bm{W}_i \bm{A}_i \bm{e}_i \bm{e}_i' \bm{A}_i \bm{W}_i \bm{\ddot{U}}_i \right) \bm{M}_{\bm{\ddot{U}}}, \tag{2} \] where \(\bm{A}_1,...,\bm{A}_m\) are the CR2 adjustment matrices.

If we assume a working model in which \(\bs\Psi_i = \sigma^2 \bm{I}_i\) for \(i = 1,...,m\) and estimate the model by ordinary least squares, then the CR2 adjustment matrices have a fairly simple form: \[ \bm{A}_i = \left(\bm{I}_i - \bm{X}_i \bm{M_X} \bm{X}_i'\right)^{+1/2}, \tag{3} \] where \(B^{+1/2}\) is the symmetric square root of the Moore-Penrose inverse of \(\bm{B}\). However, this form is computationally expensive because it involves the full set of predictors, \(\bm{X}_i\), including the cluster-specific fixed effects \(\bm{T}_i\). If the model is estimated after absorbing the cluster-specific fixed effects, then it would be convenient to use the adjustment matrices based on the absorbed predictors only, \[ \bm{\tilde{A}}_i = \left(\bm{I}_i - \bm{\ddot{U}}_i \bm{M_\ddot{U}} \bm{\ddot{U}}_i'\right)^{+1/2}. \tag{4} \] The original version of Theorem 2 asserted that \(\bm{A}_i = \bm{\tilde{A}}_i\), which is not actually the case. However, for ordinary least squares with the independent, homoskedastic working model, we can show that \(\bm{A}_i \bm{\ddot{U}}_i = \bm{\tilde{A}}_i \bm{\ddot{U}}_i\). Thus, it doesn’t matter whether we use \(\bm{A}_i\) or \(\bm{\tilde{A}}_i\) to calculate the cluster-robust variance estimator. We’ll get the same result either way, but \(\bm{\tilde{A}}_i\) is bit easier to compute.

Here’s a formal statement of Theorem 2:

Let \(\bm{L}_i = \left(\bm{\ddot{U}}'\bm{\ddot{U}} - \bm{\ddot{U}}_i'\bm{\ddot{U}}_i\right)\) and assume that \(\bm{L}_1,...,\bm{L}_m\) have full rank \(r + s\). If \(\bm{W}_i = \bm{I}_i\) and \(\bs\Phi_i = \bm{I}_i\) for \(i = 1,...,m\), then \(\bm{A}_i \bm{\ddot{U}}_i = \bm{\tilde{A}}_i \bm{\ddot{U}}_i\), where \(\bm{A}_i\) and \(\tilde{\bm{A}}_i\) are as defined in (3) and (4), respectively.

Proof

We can prove this revised Theorem 2 by showing how \(\bm{A}_i\) can be constructed in terms of \(\bm{\tilde{A}}_i\) and \(\bm{T}_i\). First, because \(\bm{T}_i'\bm{T}_k = \bm{0}\) for any \(i \neq k\), it follows that \(\bm{T}_i \bm{M_T} \bm{T}_i'\) is idempotent, i.e., \[ \bm{T}_i \bm{M_T} \bm{T}_i' \bm{T}_i \bm{M_T} \bm{T}_i' = \bm{T}_i \bm{M_T} \bm{T}_i'. \]

Next, denote the thin QR decomposition of \(\bm{\ddot{U}}_i\) as \(\bm{Q}_i \bm{R}_i\), where \(\bm{Q}_i\) is semi-orthogonal \((\bm{Q}_i'\bm{Q}_i = \bm{I})\) and \(\bm{R}_i\) has the same rank as \(\bm{\ddot{U}}_i\). Next, let \(\bm{\tilde{B}}_i = \bm{I}_i - \bm{\ddot{U}}_i \bm{M_\ddot{U}} \bm{\ddot{U}}_i'\) and observe that this can be written as \[ \tilde{\bm{B}}_i = \bm{I}_i - \bm{Q}_i \bm{Q}_i' + \bm{Q}_i \left(\bm{I} - \bm{R}_i \bm{M}_{\bm{\ddot{U}}} \bm{R}_i'\right)\bm{Q}_i'. \] It can then be seen that \[ \bm{\tilde{A}}_i = \tilde{\bm{B}}_i^{+1/2} = \bm{I}_i - \bm{Q}_i \bm{Q}_i' + \bm{Q}_i \left(\bm{I} - \bm{R}_i \bm{M}_{\bm{\ddot{U}}} \bm{R}_i'\right)^{+1/2} \bm{Q}_i'. \] It follows that \(\bm{\tilde{A}}_i \bm{T}_i = \bm{T}_i\) because \(\bm{Q}_i'\bm{T}_i = \bm{0}\). Further, \(\bm{\tilde{B}}_i \bm{T}_i = \bm{T}_i\) as well.

Now, let \(\bm{B}_i = \left(\bm{I}_i - \bm{X}_i \bm{M_X} \bm{X}_i'\right)\) and observe that this can be written as \[ \bm{B}_i = \bm{I}_i - \bm{\ddot{U}}_i \bm{M_{\ddot{U}}}\bm{\ddot{U}}_i' - \bm{T}_i \bm{M_T}\bm{T}_i' = \bm{\tilde{B}}_i - \bm{T}_i \bm{M_T}\bm{T}_i' \] because \(\bm{\ddot{U}}_i'\bm{T}_i = \bm{0}\).

We then construct the full adjustment matrix \(\bm{A}_i\) as \[ \bm{A}_i = \tilde{\bm{A}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'. \tag{5} \] Showing that \(\bm{B}_i \bm{A}_i \bm{B}_i \bm{A}_i = \bm{B}_i\) will suffice to verify that \(\bm{A}_i\) is the symmetric square root of the Moore-Penrose inverse of \(\bm{B}_i\). Because \(\bm{T}_i \bm{M_T} \bm{T}_i'\) is idempotent, \(\bm{\tilde{B}}_i \bm{T}_i = \bm{T}_i\), and \(\bm{\tilde{A}}_i \bm{T}_i = \bm{T}_i\), we have \[ \begin{aligned} \bm{B}_i \bm{A}_i \bm{B}_i \bm{A}_i &= \left(\tilde{\bm{B}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'\right) \left(\tilde{\bm{A}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'\right)\left(\tilde{\bm{B}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'\right) \left(\tilde{\bm{A}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'\right) \\ &= \left(\tilde{\bm{B}}_i\tilde{\bm{A}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'\right)\left(\tilde{\bm{B}}_i\tilde{\bm{A}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'\right) \\ &= \left(\tilde{\bm{B}}_i\tilde{\bm{A}}_i\tilde{\bm{B}}_i\tilde{\bm{A}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'\right) \\ &= \left(\tilde{\bm{B}}_i - \bm{T}_i \bm{M_T}\bm{T}_i'\right) \\ &= \bm{B}_i. \end{aligned} \]

From the representation of \(\bm{A}_i\) in (5), it is clear that \(\bm{A}_i \bm{\ddot{U}}_i = \bm{\tilde{A}}_i \bm{\ddot{U}}_i - \bm{T}_i \bm{M_T} \bm{T}_i' \bm{\ddot{U}}_i = \bm{\tilde{A}}_i \bm{\ddot{U}}_i\).