Let $y_{t}$ be a scalar stationary stochastic process with its Wold representation

$y_{t} = \mu_{t} + \sum_{k = 0}^{\infty} \phi_{k}\epsilon_{t-k}$.

I understand that the theorem says that $\sum_{k=0}^{\infty} \phi_{k}^2$ is finite. Although, we if assume that $Var(\epsilon_{t}) = \sigma^2, \forall t$, can we state that $\sum_{k=0}^{\infty} |\phi_{k}|$ is also finite?

I don't think so, but I would like some clarification. For example, if we let $\mu_{t} = 0, \forall t$ and $\phi_{k} = 1/k$ for every $k$, then the above conjecture fails. What do you guys think? Is this a valid counterexample? Thanks in advance!

  • $\begingroup$ I don't see how $\operatorname{Var}(\epsilon_t) = \sigma^2$ has anything to with whether $\sum_{k=0}^{\infty} |\phi_{k}|$ or $\sum_{k=0}^{\infty} |\phi_{k}|^2$ are finite. $\endgroup$ – Matthew Gunn Dec 14 '17 at 19:24

Yes, that's a valid counterexample.

It might be fun/useful to think about this more broadly. An equivalent question to yours might be, "Are there sequences in $ℓ_2$, the space of square summable sequences, that aren't in $ℓ_1$, the space of absolutely convergent sequences?

Those sums you have are basically the p-norm.

The p-norm and its generalizations:

  • In a finite, $n$ dimensional vector space, the the p-norm is: $$ \|\phi \|_p = \left( \sum_{k=1}^n |\phi_k|^p \right) ^\frac{1}{p}$$
  • In a countably infinite $ℓ_p$ space, the p-norm is: $$ \|\phi \|_p = \left( \sum_{k=1}^\infty |\phi_k|^p \right) ^\frac{1}{p}$$
  • In $L_p$ spaces, the generalization for the p-norm is: $$ \|f\|_p = \left( \int_S |f|^p d \mu\right)^\frac{1}{p}$$

As described in this answer, $f(x) = \frac{1}{x}$ for $x >0$ and $f(0)=0$ is an example of a function that is in $L_2$ but not $L_1$.

  • $\begingroup$ Thanks! I have already some stuff on Lp spaces but I didn't see the link! That kind of stuff you learn in different course along life and fail to connect dots. Very welcome answer! $\endgroup$ – Raul Guarini Dec 14 '17 at 19:30

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.