## 1 Correction to: AStA Advances in Statistical Analysis https://doi.org/10.1007/s10182-022-00437-9

In this article Eqs. 9, 11 and 12 have displayed a piece of code, "[2pt]", which is not part of the equations. They now read as follows:

\begin{aligned} \begin{aligned} {\varvec{\mu }}_{j,t}=\begin{pmatrix} {\varvec{\mu }}_{j,t}^{(o)}\\ {\varvec{\mu }}_{j,t}^{({\bar{o}})} \end{pmatrix}\\ \mathbf {\Sigma }_j=\begin{pmatrix}\mathbf {\Sigma }_j^{(o,o)}&{}\mathbf {\Sigma }_j^{(o,{\bar{o}})}\\ \mathbf {\Sigma }_j^{({\bar{o}},o)}&{}\mathbf {\Sigma }_j^{({\bar{o}},{\bar{o}})}\end{pmatrix} \end{aligned} \end{aligned}
(9)
\begin{aligned} \begin{aligned} \widehat{{\mathbf {B}}}_j=({\mathbb {X}}'{\mathbb {W}}{\mathbb {X}})^{-1}{\mathbb {X}}'{\mathbb {W}}{\mathbb {Y}}\\ \widehat{{\varvec{\mu }}}_{j,t}=\widehat{{\mathbf {B}}}_j'\left( 1,t,t^2,\ldots ,t^d\right) '\\ \widehat{\mathbf {\Sigma }}_j=\frac{\sum _{i=1}^n \pi _{i,j}\sum_{t=1}^T({\mathbf{y}}_{i,t}-\widehat{{\varvec{\mu}}}_{j,t})({\mathbf{y}}_{i,t}-\widehat{{\varvec{\mu}}}_{j,t})'}{\sum _{i=1}^n \pi _{i,j}} \end{aligned} \end{aligned}
(11)
\begin{aligned} \begin{aligned} \underset{(nT)\times (d+1)}{{\mathbb {X}}}=\begin{pmatrix} {\varvec{x}}_1\\ \ldots \\ {\varvec{x}}_i\\ \ldots \\ {\varvec{x}}_n \end{pmatrix}\hspace{0.9cm} \underset{T\times (d+1)}{{\varvec{x}}_i}=\begin{pmatrix} 1&{}1&{}1&{}\ldots &{}1\\ 1&{}2&{}2^2&{}\ldots &{}2^d\\ \ldots &{}\ldots &{}\ldots &{}\ldots &{}\ldots \\ 1&{}t&{}t^2&{}\ldots &{}t^d\\ \ldots &{}\ldots &{}\ldots &{}\ldots &{}\ldots \\ 1&{}T&{}T^2&{}\ldots &{}T^d\\ \end{pmatrix}\\ \underset{(nT)\times K}{{\mathbb {Y}}}=\begin{pmatrix} {\varvec{y}}_1\\ \ldots \\ {\varvec{y}}_i\\ \ldots \\ {\varvec{y}}_n\\ \end{pmatrix} \hspace{.9cm} \underset{nT\times nT}{{\mathbb {W}}}=\begin{pmatrix} {\varvec{1}}_{T\times T}\pi _{1,j}&{}0&{}\ldots &{}0\\ \ldots &{}\ldots &{}\ldots &{}\ldots \\ 0&{}{\varvec{1}}_{T\times T}\pi _{i,j}&{}\ldots &{}0\\ \ldots &{}\ldots &{}\ldots &{}\ldots \\ 0&{}0&{}\ldots &{}{\varvec{1}}_{T\times T}\pi _{n,j}\\ \end{pmatrix} \end{aligned} \end{aligned}
(12)

This does not influence the results of this work.

The original article has been corrected.