3
$\begingroup$

In a system of first order difference equations, I get the following solution

$\begin{bmatrix} x_{1,t} \\ x_{2,t}\end{bmatrix} = \begin{bmatrix} \mathcal{A} \\ 1 \end{bmatrix} \lambda_1^t + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \lambda_{2}^t$

For some $\mathcal{A} \neq 0$.

I know I need one eigenvalue to be less than one and the to be other greater than one in absolute terms for the system to be saddle path stable, which is what I want.

My question is about the fact that I get a zero in the second eigenvector. Does this have any particular meaning?

For example (please correct me if I am wrong), if $|\lambda_1| < 1$ and $|\lambda_2| > 1$, $x_{2,t}$ will stabilise and $x_{1,t}$ will go to zero as $t\rightarrow 0$.

But if the converse is true, i.e. $|\lambda_1| > 1$ and $|\lambda_2| < 1$, then although $x_{2,t}$ continues to be stable, $x_{1,t}$ will now explode.

This would not happen if that zero wasn't there, correct? I would always have both $x_{1,t}$ and $x_{2,t}$ saddle path stable as long as one of the eigenvalues was greater than one and the other less than one in absolute terms. Is that correct?

PS: how do I call that format of solution I wrote on top? Is this the Jordan form?

$\endgroup$

1 Answer 1

2
+25
$\begingroup$

The fact that your second eigenvector has a zero on top just means that the second basis vector of your space (in your initial basis) is that eigenvector. Since you provided no context, it is hard to say what that second basis vector represents, and if there is anything intrinsically interesting about it. In any case if you transform the problem by choosing a different basis initially, then this phenomenon would not occur, while the problem is otherwise equivalent.

Your equation implies $x_{2,t}=\lambda_1^t+\lambda_2^t$; if $\lambda_1,\lambda_2$ are on opposite sides of $1$ but both distinct from it, then as $t\to\infty$ the value of $x_{2,t}$ will rapidly tend to $\lambda_i^t$ where $\lambda_i>1$. This means it does not stabilise, but goes to infinity. If you are talking about stabilising, then maybe you are implicitly rescaling vectors? For the first variable you get $x_{1,t}=\mathcal A\lambda_1^t$ which does what it suggests.

$\endgroup$
1
  • $\begingroup$ Thanks. I am still trying to get my head around this whole thing but I think I am getting there. I think I understand now what I want to understand but I will leave my question for a few more days hoping to get more insight on my problem. By the way, do you know that answer to my question on the last line (the PS part)? $\endgroup$
    – Vivi
    Jun 5, 2013 at 1:06

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.