Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
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!)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.