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I have an original identity, that has been transformed into a different form as follows:

$z_t=a+\Gamma F_t +v_t$

$F_t=\mu +AF_{t-1}+u_t$ $\;\; u_t \sim N(0,Q)$

$v_t=B v_{t-1}+ \xi_t$ $\;\;\;\;\;\;\;\;\;\; \xi_t \sim N(0,R)$


$z_t=(n*1)$ vector
$\Gamma=(n*p)$ vector
$F=(p*1)$ vector
$v_t=(n*1)$ vector
$u_t=(p*1)$ vector
$\xi_t=(n*1)$ vector

This is then transformed into this form

$z_t=\Gamma^* F_t^* +v_t^* $ $\;\;\;\;\;\;\;\;\;\; u_t \sim N(0,R^*)$

$F_t^*=A^*F_{t-1}^* +u_t^*$ $\;\;\;\;\;\; \xi_t \sim N(0,Q^*)$


$ \Gamma^*_t = \left( \begin{array} {c} \Gamma \\a \\ I_n \end{array} \right) $

$ F_t^* = \left( \begin{array} {c} F_t \\c_t \\v_t\end{array} \right) $

$\def\iddots{ {\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}} A^* = \left( \begin{array} {c,|c|,c} A\;\;\;\mu &\cdots& 0 \\ \hline \vdots\;\;\; \iddots &1&\vdots \\ \hline 0 \;\;\;\cdots&\cdots &B \end{array} \right) $

$ u_t^* = \left( \begin{array} {c} u_t \\v_t \\xi_t\end{array} \right) $

$ Q^* = \left( \begin{array} {c,|c|,c} Q &\cdots& 0 \\ \hline \vdots & \epsilon&\vdots \\ \hline 0 & \cdots & R \end{array} \right) $

where $R^*_t=\epsilon I_n$

The part I don't understand is $A^*F^*_{t-1}$ how do I handle the fact that the first column contains two elements?

It seems that you should take the first element of $A^*$ i.e. the (1*2) submatrix $(A\;\;\; \mu)$ and mutiply it by the first two elements of $F^*_{t-1}$ but what happens to the element [1,2] of $A^*$ should it also be multiplied by c in $F_{t-1}^*$ looking at the next two rows of $A^*$ this is what should happen but there appears to be an element of double counting? Can someone tell me if I am doing this correctly and what is the correct term for a matrix like this (I realize it's partitioned but it also has one column with two sub-columns what is this part called?

Lastly what do the $\cdots, \vdots $ etc represent? Do they all signify that those elements with dots in should be considered zero? Or something else?

Kind Regards


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up vote 1 down vote accepted

It looks like

$$\def\iddots{ {\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}} A^* = \left( \begin{array} {c,|c|,c} A\;\;\;&\mu& 0 \\ \hline \vdots\;\;\; \iddots &1&\vdots \\ \hline 0 \;\;\;\cdots&\cdots &B \end{array} \right) $$

where all dots are zeros. $c_t$ should be 1. But there are some typos in your entry for $N(0,R^*)$ and $N(0,Q^*)$.

I encourage you to figure out exactly what the sizes of each block are (and no, $[A \mu]$ is not a 1x2 matrix!)

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Hi thanks for looking the matrix you give makes sense but I have taken the matrix from here:… see p24 are they doing something different? – Bazman Jun 4 '14 at 16:12
Seems like a simple typo. It can't be $[A \ \mu]$ because you're multiplying $\Gamma$ with it. – PA6OTA Jun 4 '14 at 16:29
Maybe but its not just the first element in the first row of A^* the first elements of the 2nd and third rows clearly have two elements too? Surely that's too many typos? – Bazman Jun 4 '14 at 17:11
You were of course correct! – Bazman Jun 6 '14 at 15:29

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