# Corrigendum to Pustejovsky and Tipton (2018), redux

A revised version of Theorem 2

$\def\Pr{{\text{Pr}}} \def\E{{\text{E}}} \def\Var{{\text{Var}}} \def\Cov{{\text{Cov}}} \def\bm{\mathbf} \def\bs{\boldsymbol}$ In my 2018 paper with Beth Tipton, published in the Journal of Business and Economic Statistics, we considered how to do cluster-robust variance estimation in fixed effects models estimated by weighted (or unweighted) least squares. As explained in my previous post, we were recently alerted that Theorem 2 in the paper is incorrect as stated. It turns out, the conditions in the original version of the theorem are too general. A more limited version of the Theorem does actually hold, but only for models estimated using ordinary (unweighted) least squares, under a working model that assumes independent, homoskedastic errors. In this post, I’ll give the revised theorem, following the notation and setup of the previous post (so better read that first, or what follows won’t make much sense!).

### 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 $$\bm{A}_i \bm{\ddot{U}}_i = \bm{\tilde{A}}_i \bm{\ddot{U}}_i - \bm{T}_i \bm{M_T} \bm{T}_i' \bm{\ddot{U}}_i = \bm{\tilde{A}}_i \bm{\ddot{U}}_i$$.