1
$\begingroup$

I'm trying to teach myself a bit of time series analysis. I came across this equation that looks like it may be solvable, but I can't be sure. I've just gotten through the first few chapters of Hamiltons Time-Series Analysis, so I apologize if this question is a bit premature.

So its the standard $Ax = b$ equation. We know $x$ and $b$ vectors, and all but the first row of matrix $A$. The rest of matrix $A$ looks like the identity matrix shifted down a row. So if $A$ was $3\times 3$, row $1 = [x,~y,~z]$; row $2 = [1,~0,~0]$; row $3 = [0,~1,~0]$.

Since $xx^T$ wouldn't have an inverse, I can't move it to the other side. So I'm not sure if I would be able to find a solution to this, it just looks so close. I think in the book they are assuming these $x,y,z$ (phi's) are known at this point. But it would be much more useful to find a solution/approximation for these.

If anyone is familiar with time series or ARMA/ARIMA models, do you know how these phi values are approximated? I know phi can be approximated as $1 - \frac{d}{2}$ ($d = $ Durbin Watson statistic) for a first order model. But how are phi values approximated for higher order models?

$\endgroup$
1
$\begingroup$

When you multiply $A$ with $x$ to get $b$ then the first row of $A$ will only influence the first element of $b$. There is therefore not a unique solution for $A$ when dealing with a $n\times n$ matrix (with $n>1$) as we have $n$ unknowns, but only one equation constraining them.

For your $3\times 3$ example we get, letting $a_1,a_2,a_3$ denote the unknowns of row $1$,

$$Ax = \pmatrix{a_1 & a_2 & a_3\\ \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots }\pmatrix{x_1\\x_2\\x_3} = \pmatrix{a_1x_1+a_2x_2+a_3x_3\\\ldots\\\ldots} = b = \pmatrix{b_1\\b_2\\b_3}$$ so we need $a_1x_2+a_2x_2 + a_3x_3 = b_1$. This is just one equation with $3$ unknowns $\implies$ infinitely many solutions for $\{a_1,a_2,a_3\}$.

To construct a solution let the first row of $A$ be the vector $\vec{a}=\{a_1,a_2,\ldots,a_n\}$ and pick any vector $\vec{c}$ satisfying $\vec{c}\cdot \vec{x} \not = 0$ then

$$\vec{a} = \frac{b_1}{\vec{c}\cdot \vec{x}}\vec{c}$$

is a solution.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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