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$vec(\boldsymbol{\beta}\boldsymbol{\sigma}_{n-1}^2\boldsymbol{e}_j'\boldsymbol{\sigma}_{n-1}^2\boldsymbol{\alpha}_1^\prime)=((\boldsymbol{\alpha}_1\boldsymbol{e}_j')\otimes\boldsymbol{\beta})vec(\boldsymbol{\sigma}_{n-1}^2\boldsymbol{\sigma}_{n-1}^{2^\prime})$ where $\boldsymbol{\beta}$ is a $k \times k$ matrix and $\boldsymbol{e}_j, \boldsymbol{\sigma}_{n-1}^2, \boldsymbol{\alpha}_1$ are $k \times 1 $ vectors.

Can anyone help how does this follow? I tried using $vec(\boldsymbol{A}\boldsymbol{B}\boldsymbol{C})=(\boldsymbol{C}'\otimes\boldsymbol{A})vec(\boldsymbol{B})$ but had no success

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  • $\begingroup$ is $A'$ the transpose of $A$? $\endgroup$ – Alvin Lepik Jul 8 '18 at 7:34
  • $\begingroup$ yes the prime defines transpose @AlvinLepik $\endgroup$ – Anna Jul 8 '18 at 7:34
  • $\begingroup$ @AlvinLepik Sorry I had a typo, $\sigma_{n-1}^2$ is the variance which is a $k\times 1$ vector $\endgroup$ – Anna Jul 8 '18 at 7:44
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If $e_j$ is what I think it is, $j$-th coordinate is $1$ and the rest are zeros, then we could swap: $$e_j' \sigma ^2_{n-1} = (\sigma ^2_{n-1})'e_j $$ Use the identity $\mbox{vec}(ABC) = (C'\otimes A)\mbox{vec}(B)$ with $$A = \beta \qquad B = \sigma ^2_{n-1}(\sigma ^2_{n-1})'\qquad C = e_j\alpha _1'$$

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