Solved – Roots of characteristic equation reciprocal to roots of its inverse

Question

This is probably just as much an algebra question as a time series question (no homework, self study).

I'm looking to understand why there is a reciprocal relationship between the roots of the characteristic equation and the roots of the inverse characteristic equation (of a difference equation in a time series context).

More specifically, the n-th order autoregressive difference equation

$$
y_t = a_0 + \sum_{i=1}^{n} a_i y_{t-i} + \epsilon_t
$$

can be written with the lag operator $L y_i = y_{i-1}$ as

$$
(1 – a_1 L – \dots – a_n L^n) y_t = a_0 + \epsilon_t\quad.
$$

The roots $\alpha_i$ of the characteristic equation (obtained by substituting trial solution $A \alpha^t$ into the homogeneous part of the difference equation)

$$
\alpha^n – a_1 \alpha^{n-1} – \dots – a_n = 0
$$

are the reciprocals $L_i$ of the inverse characteristic equation

$$
1 – a_1 L – \dots – a_n L^n = 0
$$

obtained from determining the roots of the lag polynomial above. (This is stated in W. Enders, "Applied Econometric Time Series", 4th edition, p. 42.)

What's a quick way to derive this result?

Answer 1

As whuber pointed out in the comments, this is almost trivial:

Dividing the characteristic equation

$$ \alpha^n - a_1 \alpha^{n-1} - \dots - a_n = 0 $$

by $\alpha^n$ yields

$$ 1 - a_1 \alpha^{-1} - a_2 \alpha^{-2} - \dots - a_n \alpha^{-n} = 0\quad. $$

Substituting $L = \alpha^{-1}$ gives the inverse characteristic equation

$$ 1 - a_1 L - a_2 L^2 - \dots - a_n L^n = 0\quad. $$

Therefore, the roots $L_i$ of the inverse characteristic equation are the reciprocals of the roots $\alpha_i$ of the characteristic equation,

$$ L_i = \frac{1}{\alpha_i}\quad. $$

(And as a consequence, the solution of a homogeneous autoregressive difference equation is stable if all roots of its characteristic polynomials are inside the unit circle, or equivalently, the roots of its inverse characteristic polynomial are outside of the unit circle.)