Simplified Estimation and Testing in Unbalanced Repeated Measures Designs

Abstract

In this paper we propose a simple estimator for unbalanced repeated measures design models where each unit is observed at least once in each cell of the experimental design. The estimator does not require a model of the error covariance structure. Thus, circularity of the error covariance matrix and estimation of correlation parameters and variances are not necessary. Together with a weak assumption about the reason for the varying number of observations, the proposed estimator and its variance estimator are unbiased. As an alternative to confidence intervals based on the normality assumption, a bias-corrected and accelerated bootstrap technique is considered. We also propose the naive percentile bootstrap for Wald-type tests where the standard Wald test may break down when the number of observations is small relative to the number of parameters to be estimated. In a simulation study we illustrate the properties of the estimator and the bootstrap techniques to calculate confidence intervals and conduct hypothesis tests in small and large samples under normality and non-normality of the errors. The results imply that the simple estimator is only slightly less efficient than an estimator that correctly assumes a block structure of the error correlation matrix, a special case of which is an equi-correlation matrix. Application of the estimator and the bootstrap technique is illustrated using data from a task switch experiment based on an experimental within design with 32 cells and 33 participants.

Appendices

A Simplifications

Given $\mathbf{X }_i = \mathbf{D }_i \mathbf{Z }$ and $\mathbf{D }^T_i \widetilde{\mathbf{V }}^{-1/2}_i = \mathbf{V }^{-1/2} \mathbf{D }^T_i$, (5) simplifies to

$$\begin{aligned} \hat{\varvec{\theta }}^{(0)}= \left[ \mathbf{Z }^T \mathbf{V }^{-1/2} \left( \sum _i \mathbf{D }^T_i \mathbf{R }^{-1}_i \mathbf{D }_i\right) \mathbf{V }^{-1/2} \mathbf{Z } \right] ^{-1} \left[ \mathbf{Z }^T \mathbf{V }^{-1/2} \sum _i \mathbf{D }^T_i \mathbf{R }^{-1}_i \widetilde{\mathbf{V }}^{-1/2}_i \mathbf{y }_i \right] . \end{aligned}$$

The inverse of $\mathbf{R }_i$ is given by the following

Theorem 1

Let $0 \le \rho _1 \le \rho _2 < 1$ and $T_{i,k}>0$, then the inverse of

$$\begin{aligned} \mathbf{R }_i = (1-\rho _2 ) \mathbf{I }_{T_i} + (\rho _2 - \rho _1) \mathbf{D }_i \mathbf{D }^ T_i + \rho _1 \mathbf{J }_{T_i} \end{aligned}$$

is

$$\begin{aligned} \mathbf{F }_i = \frac{1}{1-\rho _2} \mathbf{I }_{T_i} - \mathbf{D }_i \left[ \frac{\rho _2 - \rho _1}{1-\rho _2} \ \mathrm {diag}\left( \mathbf{a }_i\right) + \frac{\rho _1}{1+\rho _1 \sum _k d_{i,k}} \mathbf{a }_i\mathbf{a }^T_i\right] \mathbf{D }^T_i , \end{aligned}$$

where $\mathbf{I }_{T_i}$ is the $(T_i \times T_i)$ identity matrix, $\mathbf{D }_i = \mathrm {bdiag}(\mathbf{1 }_{T_{i,1}}, \ldots , \mathbf{1 }_{T_{i,K}})$, $\mathbf{J }_{T_{i}}$ is a $(T_i \times T_i)$ matrix of ones, $\mathbf{a }_i = (a_{i,1} , \ldots , a_{i,K})^T$ and $a_{i,k} = 1/(1-\rho _2 + (\rho _2 - \rho _1) T_{i,k})$.

Proof

Note that for $0 \le \rho _1 \le \rho _2 < 1$ and $T_{i,k}>0$ we have $1-\rho _2 + (\rho _2 - \rho _1) T_{i,k} \ne 0$: Let $\rho _2 = \rho _1 +\delta$, where $\delta \ge 0$. Then $1-\rho _2 + (\rho _2 - \rho _1) T_{i,k} = 1 - \rho _1 + \delta ( T_{i,k}-1) > 0$.

It is shown that $\mathbf{F }_i \mathbf{R }_i = \mathbf{R }_i\mathbf{F }_i= \mathbf{I }_{T,i}$. Let $d_{i,k} = T_{i,k} a_{i,k}$, $\mathbf{d } = (d_{i,1}, \ldots , d_{i,K})^T$ and note that $\mathbf{D }^T_i \mathbf{D }_i = \text {diag}(T_{i,1}, \ldots ,T_{i,K})$, $\mathbf{D }^T_i \mathbf{J }_{T_i} = \text {diag}(T_{i,1}, \ldots , T_{i,K})\mathbf{J }_{K, T_i}$, $(1-\rho _2) \mathbf{a }^T_i + (\rho _2-\rho _1)\mathbf{d }^T_i = \mathbf{1 }^T_{K}$ and $\mathbf{D }_i \mathbf{J }_{K, T_i} = \mathbf{J }_{T_i}$. Then

$$\begin{aligned} \mathbf{F }_i \mathbf{R }_i&= \left\{ \frac{1}{1-\rho _2} \mathbf{I }_{T_i} - \mathbf{D }_i \left[ \frac{\rho _2 - \rho _1}{1-\rho _2} \text {diag}(\mathbf{a }_i) + \frac{\rho _1}{1+\rho _1 \sum _k d_{i,k}} \mathbf{a }_i\mathbf{a }^T_i\right] \mathbf{D }^T_i \right\} \\&= \mathbf{I }_{T_i} + \frac{(\rho _2 - \rho _1)}{1-\rho _2} \mathbf{D }_i \mathbf{D }^T_i + \frac{\rho _1 }{1-\rho _2} \mathbf{J }_{T_i}\\&\quad - \left\{ (\rho _2 - \rho _1) \mathbf{D }_i \text {diag}(\mathbf{a }_i) \mathbf{D }^T_i + \frac{ (\rho _2 - \rho _1)^2}{1-\rho _2} \mathbf{D }_i \text {diag}(\mathbf{a }_i) \mathbf{D }^T_i \mathbf{D }_i \mathbf{D }^T_i \right. \\&\left. \quad + \frac{\rho _1 (\rho _2 - \rho _1)}{1-\rho _2} \mathbf{D }_i \text {diag}(\mathbf{a }_i) \mathbf{D }^T_i \mathbf{J }_{T_i} + \frac{\rho _1(1-\rho _2 )}{1+\rho _1 \sum _k d_{i,k}}\mathbf{D }_i \mathbf{a }_i\mathbf{a }^T_i \mathbf{D }^T_i \right. \\&\left. \quad + \frac{\rho _1(\rho _2 - \rho _1)}{1+\rho _1 \sum _k d_{i,k}} \mathbf{D }_i \mathbf{a }_i\mathbf{a }^T_i \mathbf{D }^T_i \mathbf{D }_i \mathbf{D }^T_i + \frac{\rho ^2_1}{1+\rho _1 \sum _k d_{i,k}} \mathbf{D }_i \mathbf{a }_i\mathbf{a }^T_i \mathbf{D }^T_i \mathbf{J }_{T_i} \right\} . \end{aligned}$$

Further, using the results from above,

$$\begin{aligned} (\rho _2 - \rho _1) \mathbf{D }_i \text {diag}(\mathbf{a }_i) \mathbf{D }^T_i + \frac{ (\rho _2 - \rho _1)^2}{1-\rho _2} \mathbf{D }_i \text {diag}(\mathbf{a }_i) \mathbf{D }^T_i \mathbf{D }_i \mathbf{D }^T_i = \frac{(\rho _2 - \rho _1)}{1-\rho _2} \mathbf{D }_i \mathbf{D }^T_i \end{aligned}$$

and

$$\begin{aligned}&\frac{\rho _1 (\rho _2 - \rho _1)}{1-\rho _2} \mathbf{D }_i \text {diag}(\mathbf{a }_i) \mathbf{D }^T_i \mathbf{J }_{T_i} + \frac{\rho _1(1-\rho _2 )}{1+\rho _1 \sum _k d_{i,k}}\mathbf{D }_i \mathbf{a }_i\mathbf{a }^T_i \mathbf{D }^T_i \\&\qquad + \frac{\rho _1(\rho _2 - \rho _1)}{1+\rho _1 \sum _k d_{i,k}} \mathbf{D }_i \mathbf{a }_i\mathbf{a }^T_i \mathbf{D }^T_i \mathbf{D }_i \mathbf{D }^T_i + \frac{\rho ^2_1}{1+\rho _1 \sum _k d_{i,k}} \mathbf{D }_i \mathbf{a }_i\mathbf{a }^T_i \mathbf{D }^T_i \mathbf{J }_{T_i} \\&\quad =\rho _1 \mathbf{D }_i \left( \mathbf{a }_i + \frac{(\rho _2 - \rho _1)}{1-\rho _2} \mathbf{d }_i \right) \mathbf{1 }^T_{T_i} = \frac{\rho _1}{1-\rho _2} \mathbf{J }_{T_i} . \end{aligned}$$

Thus, $\mathbf{F }_i\mathbf{R }_i = \mathbf{I }_{T,i}$. $\mathbf{R }_i \mathbf{F }_i = \mathbf{I }_{T,i}$ follows from the symmetry of $\mathbf{R }_i$. $\square$

Further note that

$$\begin{aligned} \mathbf{D }^T_i \mathbf{R }^{-1}_i&= \frac{1}{1-\rho _2} \mathbf{D }^T_i - \left[ \frac{\rho _2 - \rho _1}{1-\rho _2} \text {diag}(\mathbf{d }_i) + \frac{\rho _1}{1+\rho _1 \sum _k d_{i,k}} \mathbf{d }_i\mathbf{a }^T_i\right] \mathbf{D }^T_i \\&= \left[ \frac{1}{1-\rho _2} \mathbf{I }_K - \frac{\rho _2 - \rho _1}{1-\rho _2} \text {diag}(\mathbf{d }_i) - \frac{\rho _1}{1+\rho _1 \sum _k d_{i,k}} \mathbf{d }_i\mathbf{a }^T_i\right] \mathbf{D }^T_i \\&= \left[ \text {diag}(\mathbf{a }_i) -\frac{\rho _1}{1+\rho _1 \sum _k d_{i,k}} \mathbf{d }_i\mathbf{a }^T_i\right] \mathbf{D }^T_i. \end{aligned}$$

If $\mathbf{Z }$ is regular, then its inverse exists and simplification of (7) and (8) leads to (10) and (11). On the other hand, we may write (5) as

$$\begin{aligned} \hat{\varvec{\theta }}^{(0)}&= \left( \sum _i \mathbf{X }^T_i \mathbf{W }_i^{-1} \mathbf{X }_i\right) ^{-1} \left( \sum _i \mathbf{X }^T_i \mathbf{W }_i^{-1} \mathbf{y }_i \right) \\&= \mathbf{Z }^{-1} \mathbf{V }^{1/2} \left( \sum _i \mathbf{D }^T_i \mathbf{R }_i^{-1} \mathbf{D }_i\right) ^{-1} \left( \sum _i \mathbf{D }^T_i \mathbf{R }_i^{-1} \widetilde{\mathbf{V }}^{-1/2}_i \mathbf{y }_i \right) . \end{aligned}$$

Taking the mean of $\hat{\varvec{\theta }}^{(0)}$ with respect to $\mathbf{y }$ conditional on $\mathbf{Z }$, noting that $E(\bar{\mathbf{y }}_i | \mathbf{Z }, \mathbf{r }_i) = E ( \bar{\mathbf{y }}_i | \mathbf{Z } )$ for all i, which follows from assuming that $\Pr (\mathbf{r } | \mathbf{Z }) = \Pr (\mathbf{r }| \mathbf{Z } , \tilde{\mathbf{y }})$ for all possible $\tilde{\mathbf{y }}$ and $\mathbf{r }$, and $E(\bar{\mathbf{y }}_i | \mathbf{Z }) = \mathbf{Z }\varvec{\theta }_0$ for all i [see model (1)],

$$\begin{aligned} \begin{aligned}&E\left( \hat{\varvec{\theta }}^{(0)}|\mathbf{Z },\mathbf{r }\right) \\&\quad = \mathbf{Z }^{-1} \mathbf{V }^{1/2} \left( \sum _i \mathbf{D }^T_i \mathbf{R }_i^{-1} \mathbf{D }_i\right) ^{-1} \left( \sum _i \mathbf{D }^T_i \mathbf{R }_i^{-1} \widetilde{\mathbf{V }}^{-1/2}_i E\left[ \mathbf{y }_i | \mathbf{Z },T_{i,1},\ldots T_{i,K} \right] \right) \\&\quad = \mathbf{Z }^{-1} \mathbf{V }^{1/2} \left( \sum _i \mathbf{D }^T_i \mathbf{R }_i^{-1} \mathbf{D }_i\right) ^{-1} \left( \sum _i \mathbf{D }^T_i \mathbf{R }_i^{-1} \mathbf{D }_i \mathbf{V }^{-1/2} E\left[ \bar{\mathbf{y }}_i | \mathbf{Z } \right] \right) \\&\quad = \mathbf{Z }^{-1} E\left( \bar{\mathbf{y }}_i | \mathbf{Z }\right) \\&\quad = \varvec{\theta }_0. \end{aligned} \end{aligned}$$

(14)

Therefore, replacing $E(\bar{\mathbf{y }}_i | \mathbf{Z })$ by $\bar{\mathbf{y }} = \sum _i \bar{\mathbf{y }}_i / n$, we can estimate $\varvec{\theta }$ as given by (12) in Sect. 3. Note that $\hat{\varvec{\theta }}_{\text {m}}$ is unbiased. From $\text {Var}(\hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z }) = E(\text {Var}(\hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z },\mathbf{r })) + \text {Var}(E(\hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z },\mathbf{r }))$ and $\text {Var}(E(\hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z },\mathbf{r })) = \mathbf{0 }$, we have $\text {Var}(\hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z }) = E(\text {Var}(\hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z },\mathbf{r }))$. Thus, the variance can be written as

$$\begin{aligned} \begin{aligned} \text {Var}\left( \hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z }\right)&= E\left( \text {Var}\left( \hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z },\mathbf{r }\right) |\mathbf{Z }\right) \\&= \frac{1}{n^2}\, \mathbf{Z }^{-1} E\left( \sum _i \text {Var}\left( \bar{\mathbf{y }}_i|\mathbf{Z }, \mathbf{r }_i\right) |\mathbf{Z } \right) \mathbf{Z }^{-T} \\ \end{aligned} \end{aligned}$$

(15)

estimated by (13). From (14) and (15) it follows that $\hat{\varvec{\theta }}_{\text {m}}$ is consistent.

For a comparison of linear combinations of $\hat{\varvec{\theta }}_{\text {m}}$ with linear combinations of other unbiased estimators linear in $\bar{\mathbf{y }} = n^{-1} \sum _i \mathbf{y }_i$ and marginal with respect to $\mathbf{r }$, consider $\hat{\varvec{\vartheta }}=\mathbf{A }^T \bar{\mathbf{y }}$. The matrix $\mathbf{A }$ may account for dependencies within $\bar{\mathbf{y }}$. Unbiasedness of $\hat{\varvec{\vartheta }}$ implies $E(\hat{\varvec{\vartheta }} | \mathbf{Z }) = \mathbf{A }^T E(\bar{\mathbf{y }} | \mathbf{Z }) = \mathbf{A }^T \mathbf{Z } \varvec{\theta }_0 = \varvec{\theta }_0$ and thus $\mathbf{A }^{T} = \mathbf{Z }^{-1}$. Hence, there is no other unbiased estimator which is linear in $\bar{\mathbf{y }}$ than $\hat{\varvec{\theta }}_{\text {m}}$ that could be more efficient.

Another interesting point is a comparison of the variances of two estimators (7) and (12) if the “working” covariance matrix is correctly specified, i.e., $\mathbf{W }_i =\varvec{\Sigma }_i$ for all i. Defining random variables

$$\begin{aligned} \mathbf{M }&= n^{-1} \sum _i \left( \mathbf{D }^T_i \mathbf{D }_i\right) ^{-1} \mathbf{D }^T_i \left( \mathbf{y }_i - \mathbf{D }_i \mathbf{Z } \varvec{\theta }_0\right) \\ \widetilde{\mathbf{M }}&= \left( \sum _i \mathbf{D }^T_i \varvec{\Sigma }^{-1}_i \mathbf{D }_i \right) ^{-1} \sum _i \mathbf{D }^T_i \varvec{\Sigma }^{-1}_i \left( \mathbf{y }_i - \mathbf{D }_i \mathbf{Z } \varvec{\theta }_0\right) \end{aligned}$$

and noting that the difference

$$\begin{aligned} \text {Var}\left( \hat{\varvec{\theta }}_{\text {m}}|\mathbf{Z },\mathbf{r }\right) - \text {Var}\left( \hat{\varvec{\theta }}|\mathbf{Z },\mathbf{r }\right) = \mathbf{Z }^{-1} E\left( \left( \mathbf{M } - \widetilde{\mathbf{M }}\right) \left( \mathbf{M } - \widetilde{\mathbf{M }}\right) ^T|\mathbf{Z },\mathbf{r }\right) \mathbf{Z }^{-T} \end{aligned}$$

is positive semi-definite, implies that (7) is asymptotically at least as efficient as (12).

B Estimators for Variances and Correlations

The variances as well as the correlations in $\mathbf{W }_i$ are unknown. To calculate $\hat{\varvec{\theta }}$ and $\widehat{\text {Var}}(\hat{\varvec{\theta }})$ we therefore have to estimate $\sigma ^2_{k}$, $k=1,\ldots ,K$ , $\rho _1$ and $\rho _2$. This can be done by solving moment conditions as well, i.e., solving

$$\begin{aligned} \varvec{0}&= \sum _{i=1}^n \mathbf{G }_i \mathbf{K }_i \text {vec}\left( \mathbf{S }_i - \mathbf{W }_i\right) , \quad \text {where} \quad \mathbf{W }_i = \widetilde{\mathbf{V }}^{1/2}_i \mathbf{R }_i \widetilde{\mathbf{V }}^{1/2}_i , \\ \mathbf{K }_i&= \frac{\partial \, \mathbf{W }_i}{\partial \, \varvec{\xi }} , \quad \varvec{\xi } = \left( \sigma ^2_1,\ldots ,\sigma ^2_K, \sigma _{1,2} ,\ldots , \sigma _{K-1,K}\right) ^T, \end{aligned}$$

$\sigma _{k,k'}$ are the covariances of observations in the k and $k'$th cells, $\mathbf{G }_i = \text {bdiag}(\mathbf{I }_K,\mathbf{Q }_i)$

$$\begin{aligned} \mathbf{Q }_i = \frac{\partial \, \varvec{\xi }}{\partial \, \begin{pmatrix} \rho _1 \\ \rho _2 \end{pmatrix}} \, , \quad \mathbf{S }_i = \left( \mathbf{y }_i - \mathbf{X }_i \hat{\varvec{\theta }}\right) \left( \mathbf{y }_i - \mathbf{X }_i \hat{\varvec{\theta }}\right) ^T \end{aligned}$$

and $\text {vec}(\mathbf{A })$ stacks the m rows of the $(m \times p)$ matrix $\mathbf{A }$ as $(p \times 1)$ column vectors over each other to form an $(mp \times 1)$ column vector (Swaminathan, 1976). For the derivations we use the calculus of McDonald (1976), McDonald & Swaminathan (1973) and Swaminathan (1976). These estimating equations are based on the assumption that the differences $((y_{i,k,t}-\mathbf{z }^T_k \hat{\varvec{\theta }})(y_{i,k',t'}-\mathbf{z }^T_k \hat{\varvec{\theta }}) - \hat{\zeta })$, where $\hat{\zeta }$ is one of $\hat{\sigma }^2_{k}$, $\hat{\rho }_1$ and $\hat{\rho }_2$, are linearly independent. If this is not true, then the estimators for variances and correlations are not (asymptotically) efficient. However this is not necessary for $\hat{\varvec{\theta }}$ to be (asymptotically) efficient (Newey & McFadden, 1994). Solving these estimating equations gives the explicit solutions for $\sigma ^2_k$, $\rho _1$ and $\rho _2$ presented in Sect. 3.

C Counter Example

One of the necessary and sufficient conditions in Theorem 2 of Zyskind (1967) under which the simple linear least squares estimator is also a best linear unbiased estimator is that $\varvec{\Sigma }\mathbf{P }=\mathbf{P }\varvec{\Sigma }$ is symmetric, where $\varvec{\Sigma }=\text {bdiag}(\varvec{\Sigma }_1,\ldots ,\varvec{\Sigma }_n)$, $\mathbf{P }= \mathbf{X }(\mathbf{X }^T \mathbf{X })^{-1} \mathbf{X }^T$ and $\mathbf{X } = (\mathbf{X }^T_1,\ldots ,\mathbf{X }^T_n)^T$.

In the following, we will show that this does not hold for the design and the covariance matrix given in Sect. 2. Let $\mathbf{D } = (\mathbf{D }^T_1, \ldots ,\mathbf{D }^T_n)^T$. Then $\mathbf{X } = \mathbf{D } \mathbf{Z }$ and $\mathbf{P } = \mathbf{D }\mathbf{Z } (\mathbf{Z }^T \mathbf{D }^T \mathbf{D } \mathbf{Z })^{-1} \mathbf{Z }^T \mathbf{D }^T$. However, $\mathbf{D }^T \mathbf{D } = \sum _i \mathbf{D }^T_i \mathbf{D }_i = \sum _i \text {diag}(T_{i,1}, \ldots , T_{i,K})$ and therefore $\mathbf{P } = \mathbf{D } (\sum _i \mathbf{D }^T_i \mathbf{D }_i)^{-1} \mathbf{D }^T$, since $\mathbf{Z }$ is regular. $\mathbf{P }$ is a partitioned matrix with matrices $\mathbf{P }_{i,j}=\text {bdiag}(T^{-1}_1 \mathbf{J }_{T_{i,1},T_{j,1}}, \ldots , T^{-1}_K \mathbf{J }_{T_{i,K},T_{j,K}})$, where $i,j=1,\ldots ,n$.

On the other hand, $\varvec{\Sigma } = \text {bdiag}( \varvec{\Sigma }_1 , \ldots , \varvec{\Sigma }_n )$, where $\varvec{\Sigma }_i = \widetilde{\mathbf{V }}^{1/2}_i \mathbf{R }_i \widetilde{\mathbf{V }}^{1/2}_i$, see Sect. 2. Thus, $\varvec{\Sigma }\mathbf{P }$ is a partitioned matrix as well, with matrices

$$\begin{aligned} \mathbf{A }_{i,j}&= \varvec{\Sigma }_i \mathbf{P }_{i,j} \\&= (1-\rho _2) \text {bdiag}\left( \sigma ^2_1/T_1 \mathbf{J }_{T_{i,1},T_{j,1}},\ldots , \sigma ^2_K/T_K \mathbf{J }_{T_{i,K},T_{j,K}} \right) \\&\quad + (\rho _2 - \rho _1) \text {bdiag}\left( \sigma ^2_1 T_{i,1} /T_1 \mathbf{J }_{T_{i,1},T_{j,1}},\ldots , \sigma ^2_K T_{i,K}/T_K \mathbf{J }_{T_{i,K},T_{j,K}} \right) \\&\quad + \rho _1 \begin{pmatrix} \frac{ \sigma ^2_1 T_{i,1}}{T_1 } \mathbf{J }_{T_{i,1},T_{j,1}} &{} \cdots &{} \frac{ \sigma _1 \sigma _K T_{i,K}}{T_K } \mathbf{J }_{T_{i,1},T_{j,K}} \\ \vdots &{} \ddots &{} \vdots \\ \frac{ \sigma _K \sigma _1 T_{i,1}}{T_1 } \mathbf{J }_{T_{i,K},T_{j,1}} &{} \cdots &{} \frac{\sigma ^2_K T_{i,K} }{T_K } \mathbf{J }_{T_{i,K},T_{j,K}} \end{pmatrix}, \end{aligned}$$

which are again partitioned matrices with entries

$$\begin{aligned} \mathbf{B }_{k,l}^{i,j}&= \left\{ \begin{array}{ll} \frac{\sigma ^2_k \left( 1+\rho _2\left( T_{i,k} -1\right) \right) }{T_k} \, \mathbf{J }_{T_{i,k},T_{j,k}} &{} \quad \text {if } k=l \\ \frac{\sigma _k \sigma _l \rho _1 T_{i,l}}{T_l} \, \mathbf{J }_{T_{i,k},T_{j,l}} &{} \quad \text {if } k\ne l \end{array} \right. . \end{aligned}$$

For symmetry of $\varvec{\Sigma }\mathbf{P }$ to hold, $\mathbf{A }_{i,j}=\mathbf{A }_{j,i}^T$ and thus $\mathbf{B }_{k,l}^{i,j} = (\mathbf{B }_{l,k}^{j,i})^T$ must hold for all i, j, k, l. Although this obviously holds if $T_{i,k}=T_{j,k}$ for all i, j, k, l, it does not hold if the number of observations varies.