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Suppose $a(L)$ and $b(L)$ are series of negative powers of the lag operator, that is $$a(L) = \sum_{j=1}^\infty a_jL^{-j},\quad\text{and}\quad b(L) = \sum_{j=1}^\infty b_jL^{-j}.$$ Also suppose that $||a(L)||<1$ with the operator norm, and suppose that $c$ is a scalar with $|c|<1$.

Which assumption must $b(L)$ satisfy to guarantee that $$\left\Vert \frac{b(L)+c}{1-a(L)} \right\Vert <1 ?$$

Obs: If $$\left\Vert \frac{b(L)+c}{1-a(L)} \right\Vert \leq \frac{|| b(L)+c||}{||1-a(L)||},$$ then it is easy to see that $$ ||a(L)||+||b(L)||+c<1$$ would be sufficient.

Obs2: By 'lag operator' I mean the shift operator with a unit shift, so that $||L||=1$.

mzp
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  • When you write $\frac{b(L) + c}{1 - a(L)}$, do you mean $[b(L) + c][1 - a(L)]^{-1}$? – Tom Chen Sep 25 '19 at 15:59
  • Is there a difference between the two in this case? – mzp Sep 25 '19 at 16:01
  • Since we are working with operators, not real numbers, I just want to make sure what you mean by division. Therefore, I assume you take $1/T$ as $T^{-1}$. In your example, we may have also interpreted $\frac{b(L)+c}{1 - a(L)}$ to be $[1 - a(L)]^{-1}[b(L)+c]$, but since the lag operator commutes, there is no ambiguity. – Tom Chen Sep 25 '19 at 16:05
  • But then we reach another problem: $|T^{-1}| \neq |T|^{-1}$. Indeed, $1 = |I| = |TT^{-1}| \le |T| |T^{-1}| \implies |T^{-1}| \ge |T|^{-1}$, so the inequality you present in Obs is not necessarily true. – Tom Chen Sep 25 '19 at 16:08
  • I see what you mean. This still doesn't rule it out though, right? To be sure, I am not claiming that the Obs is correct, just that if it is I would know the answer to the original question. – mzp Sep 25 '19 at 16:26
  • I guess one condition is this: $|[b(L) + c][1 - a(L)]^{-1}| \le (|b(L)| + c)|(1 - a(L))^{-1}| < 1 \implies |b(L)| < |(1 - a(L))^{-1}|^{-1} - c$ – Tom Chen Sep 25 '19 at 19:01
  • Combining your result with this answer, the condition in Obs can be shown to be sufficient. If you want to write this up as an answer I would be happy to give you the bounty! – mzp Sep 30 '19 at 11:23

1 Answers1

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Since the lag operator is commutative, we can safely read \begin{align*} \frac{b(L)+c}{1-a(L)} = [b(L)+c][1-a(L)]^{-1} \end{align*} Taking norms, \begin{align*} \|[b(L)+c][1-a(L)]^{-1}\| \le \|b(L)+c\|\cdot\|[1-a(L)]^{-1}\| \end{align*} Then, as @mzp pointed out from this post with $T = 1 - a(L)$, we have \begin{align*} \|[1-a(L)]^{-1}\| \le \frac{1}{1 -\|a(L)\|} \end{align*} And so \begin{align*} \|[b(L)+c][1-a(L)]^{-1}\| \le \frac{\|b(L)\|+c}{1 - \|a(L)\|} \end{align*} A sufficient condition is therefore $\frac{\|b(L)\|+c}{1 - \|a(L)\|} < 1$ or $\|b(L)\| < 1 - \|a(L)\| - c$, which is exactly the condition @mzp presents under Obs!

Tom Chen
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