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Assume the 2x1 random vector, $E_t$, has a multivariate normal distribution with a mean vector of 0 and covariance matrix of $Q_t$. Note here that $E_t$ is a time series variable and the covariance matrix is differs across time

$$E_t = [e_{1,t}, e_{2,t}]^T\sim MVN(0, Q_t), \quad t=0,..,T$$

Now let $E$ be the matrix of all the time series observations stacked on top of each other:

$$E = \begin{pmatrix} e_{0,0} & e_{1, 0}\\ e_{0,1} & e_{1, 1}\\... & ...\\e_{0,T} & e_{1,T}\end{pmatrix} $$

Lastly, let $\hat{E}$ be the vectorization of $E$:

$$\hat{E} = vec(E) = \begin{pmatrix} e_{0,0} & e_{0,1} & ... & e_{0,T}& e_{1,0} & e_{1,1} & ... & e_{1,T}\end{pmatrix}^T$$

My question is how to specify the covariance function of the vectorized $\hat{E}$

That is $$vec(E) \sim MVN(0, ???)$$

What I've tried:

It's clear that $E$ has a Matrix Normal distribution with some row and column covariance matrices $U$ and $V$, which implies that $\hat{E} \sim MVN(0, V\bigotimes U)$ but I don't see how to make this work since the column-wise covariance matrix is time dependent. If $Q_t = Q$ for all $t$ and the observations are IID than we would have $$\hat{E}\sim MVN(0, Q\bigotimes I)$$ but in this case $Q$ is time varying and there is potentially dependence across time

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In general, you can't write the covariance matrix as a Kronecker product form, i.e. $E$ does not necessarily follow a matrix normal distribution, even if its vectorization is multivariate normal.

To see why, note that $$\hat{E}=\left(E^{0\top},E^{1\top}\right)^{\top}=\begin{bmatrix}E^{0}\\E^{1}\end{bmatrix},$$ where $E^{k}=\left(e_{k,0},e_{k,1},\ldots,e_{k,T}\right)$ for $k=0,1$. Therefore, the covariance matrix should be $$\mathbb{E}\left[\hat{E}\hat{E}^{\top}\right]=\begin{bmatrix}\mathbb{E}\left[E^{0}E^{0\top}\right]&\mathbb{E}\left[E^{0}E^{1\top}\right]\\\mathbb{E}\left[E^{1}E^{0\top}\right]&\mathbb{E}\left[E^{1}E^{1\top}\right]\end{bmatrix}.$$ You can see that the $t,t$ diagonal components in all four block matrices constitute $Q_{t}$, and you still need more information to recover the off-diagonal components.

An easier way to represent the covariance matrix is to consider vectorization of $E^{\top}$ instead of $E$ it self. Let $\tilde{E}=\mathrm{vec}\left(E^{\top}\right)$, then $$\tilde{E}=\left(E_{0}^{\top},E_{1}^{\top},\ldots,E_{T}^{\top}\right)^{\top}=\begin{bmatrix}E_{0}\\E_{1}\\\vdots\\E_{T}\end{bmatrix},$$ the covariance matrix is $$\mathbb{E}\left[\tilde{E}\tilde{E}^{\top}\right]=\begin{bmatrix}\mathbb{E}\left[E_{0}E_{0}^{\top}\right]&\mathbb{E}\left[E_{0}E_{1}^{\top}\right]&\cdots&\mathbb{E}\left[E_{0}E_{T}^{\top}\right]\\\mathbb{E}\left[E_{1}E_{0}^{\top}\right]&\mathbb{E}\left[E_{1}E_{1}^{\top}\right]&\cdots&\mathbb{E}\left[E_{1}E_{T}^{\top}\right]\\\vdots&\vdots&\ddots&\vdots\\\mathbb{E}\left[E_{T}E_{0}^{\top}\right]&\mathbb{E}\left[E_{T}E_{1}^{\top}\right]&\cdots&\mathbb{E}\left[E_{T}E_{T}^{\top}\right]\end{bmatrix}=\begin{bmatrix}Q_{0,0}&Q_{0,1}&\cdots&Q_{0,T}\\Q_{1,0}&Q_{1,1}&\cdots&Q_{1,T}\\\vdots&\vdots&\ddots&\vdots\\Q_{T,0}&Q_{T,1}&\cdots&Q_{T,T}\end{bmatrix},$$ where $Q_{t,s}=\mathbb{E}\left[E_{t}E_{s}^{\top}\right]$ for $t,s=0,1,\ldots,T$.

Remark 0. $Q_{t,t}=Q_{t}$ as defined in the question.
Remark 1. $Q_{t,s}=Q_{s,t}^{\top}$.
Remark 2. To make sure $E$ is matrix normal distributed, a sufficient condition is $\left\{E_{t}\right\}_{t=0}^{T}$ being stationary and $\mathrm{corr}\left(e_{0,t},e_{0,s}\right)=\mathrm{corr}\left(e_{1,t},e_{1,s}\right)$, in which case $\mathbb{E}\left[\tilde{E}\tilde{E}^{\top}\right]=R\otimes Q$ with $R$ being correlation matrix.

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