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You have a matrix containing $T$ observations of each of $K$ random variables \begin{align} U = \begin{bmatrix} u_{11} & \dots & u_{1T} \\ \vdots & \ddots & \vdots \\ u_{K1} & \dots & u_{KT} \end{bmatrix}. \end{align} The covariance matrix of the $u_{kt}$'s is given by $\Sigma_u := E(u_t u_t')$ for $t = 1, \dots, T$ and where $ u_t = (u_{1t}, \dots, u_{Kt})'$ is a column vector by convention.

Now, we introduce the $vec$ operator which stacks columns of a matrix one over the next, i.e. \begin{align} vec(U) = \begin{bmatrix} u_{11} \\ \vdots \\ u_{1T} \\ \vdots \\ u_{K1} \\ \vdots \\ u_{KT} \end{bmatrix} = \begin{bmatrix} u_{1}' \\ \vdots \\ u_{K}' \end{bmatrix} \end{align} again using as convention that $u_k = ( u_{k1}, \dots, u_{kT} )'$ are column vectors for $k = 1, \dots, K$. It turns out that the covariance matrix of $vec(U)$ is \begin{align} E\left( vec(U) vec(U)' \right) &= I_T \otimes \Sigma_u \end{align} which is to say, a block diagonal $TK \times TK$ matrix with entries $E( u_t u_t' )$. While I can see the intuition, I don't know how to formally establish this result. Any help would be appreciated.

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  • $\begingroup$ You're assuming mean zero vectors right? $\endgroup$ – Firebug Apr 13 at 22:27
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    $\begingroup$ @Firebug Yes. I forgot to mention that detail, but they would be error terms in a regression, so they are assumed to be centered about 0. $\endgroup$ – Stéphane Apr 14 at 18:45
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I think the easiest way is just index chasing, not anything elegant.

I'll assume the $u$ are all mean zero; if not, the identity does not hold, and the left-hand side isn't the covariance matrix.

A. $E[u_{it}u_{js}]=\sigma_{ij}$ if $t=s$ and zero otherwise.

B. Now write $S$ for the matrix on the right-hand side, and $s_{itjs}$ for the entry corresponding to $u_{it}$ and $u_{js}$. Clearly $E[u_{it}u_{js}]=s_{itjs}$; that's just what A said.

C. The entries of the right-hand side matrix (call it $S$) are also either $\sigma_{ij}$ or 0 for some $(i,j)$. In fact, $T$ of them are $\sigma_{ij}$ for each $(i,j)$, and $T(T-1)K^2$ of them are zero

D. So we just need to work out if the indices match up. The $i$ index varies fastest, so the top left block of the left-hand side has $u_{i1,j1}$, then the next block to the right has $u_{i1,j2}$ and the first block on the second row has $u_{i2,j1}$ and so on. That is, the $(it,jt)$ entries -- the diagonal blocks -- have entries $\sigma_{ij}$ and the $(it,js)$ entries for $s\neq t$ are zero. That's exactly $I_T\otimes \Sigma_u$.

More formally, if we write $(p,q)$ for the indices of the big matrices, then $p=i+(t-1)K$ and $p=j+(s-1)K$, and that works for both sides.

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  • $\begingroup$ This makes sense. I was simply expecting a proof involving a trick with, say, an identity matrix that I couldn't somehow see. $\endgroup$ – Stéphane Apr 14 at 18:47
  • $\begingroup$ I almost forgot to come back to this question, but I talked with my thesis supervisor and it does seem there isn't an elegant way to do it. For some of those formulas, he told me tracking indexes is all we have -- and, usually, he just gets a feel for it using small vectors and matrices... So, thanks for the effort. Hopefully, it will help others too. $\endgroup$ – Stéphane Apr 26 at 15:50

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