A linear process $x_{t}$ is the weighted sum of white noise variates $(w_{t})_{t}$, i.e. $$x_{t}=\mu+\sum\limits_{k \in \mathbb Z}\psi_{k}w_{t-k}$$ such that $$ \sum\limits_{j \in \mathbb Z}\lvert \psi_{j}\rvert<\infty$$

Show that for autocovariance function $\gamma(h)$, the following is satisfied $\sum\limits_{j \in \mathbb Z}\lvert \gamma(j)\rvert < \infty(*)$.

I have proven that $\gamma(h)=\sigma_{w}^{2}\sum\limits_{k \in \mathbb Z}\psi_{k+h}\psi_{k}$.

My attempt at $(*)$:

$\sum\limits_{j \in \mathbb Z}\lvert \gamma(j)\rvert=\sum\limits_{j \in \mathbb Z}\lvert\sigma_{w}^{2}\sum\limits_{k \in \mathbb Z}\psi_{k+j}\psi_{k}\rvert\leq \sigma_{w}^{2}\sum\limits_{j \in \mathbb Z} \sum\limits_{k \in \mathbb Z}\lvert \psi_{j+k}\psi_{k}\rvert$ and then using Cauchy-Schwarz,

$\sigma_{w}^{2}\sum\limits_{j \in \mathbb Z} \sum\limits_{k \in \mathbb Z}\lvert \psi_{j+k}\psi_{k}\rvert\leq \sigma_{w}^{2}\sum\limits_{j \in \mathbb Z}\left((\sum\limits_{l\in \mathbb Z}\lvert\psi_{j+l}\rvert^{2})^{\frac{1}{2}}\cdot (\sum\limits_{k\in \mathbb Z}\lvert\psi_{k}\rvert^{2})^{\frac{1}{2}}\right)$

I assume that the additional assumption that $\sum\limits_{k\in \mathbb Z}\lvert\psi_{k}\rvert^{2}<\infty$ is actually needed. Even if this is true, I still reach a deadend since

$$ \sigma_{w}^{2}\sum\limits_{j \in \mathbb Z}\left((\sum\limits_{l\in \mathbb Z}\lvert\psi_{j+l}\rvert^{2})^{\frac{1}{2}}\cdot (\sum\limits_{k\in \mathbb Z}\lvert\psi_{k}\rvert^{2})^{\frac{1}{2}}\right)=\sigma_{w}^{2}(\sum\limits_{k\in \mathbb Z}\lvert\psi_{k}\rvert^{2})^{\frac{1}{2}})\cdot\sum\limits_{j \in \mathbb Z}\left((\sum\limits_{l\in \mathbb Z}\lvert\psi_{j+l}\rvert^{2})^{\frac{1}{2}}\right).$$

I do not see anyway of progressing. Any ideas, or additional assumptions needed?

  • $\begingroup$ Hi: The proof is on page 2 of this: stat.tamu.edu/~suhasini/teaching673/chapter3.pdf $\endgroup$
    – mlofton
    Jun 19, 2020 at 21:44
  • $\begingroup$ I disagree with this proof, we obtain the bound $\sum\limits_{k\in \mathbb Z}\sum\limits_{j \in \mathbb Z}\lvert \psi_{j}\rvert \cdot \lvert \psi_{j-k}\rvert $ but how do we know that this is indeed finite? All we know is that $\sum\limits_{k \in \mathbb Z}\lvert \psi_{k}\rvert <\infty$ which is not sufficient. $\endgroup$
    – MinaThuma
    Jun 19, 2020 at 21:58
  • $\begingroup$ Thuma: if you have two expressions that are finite, doesn't their multiplication have to be finite ? I'm just asking and not claiming you're wrong. I could be mis-understanding something. $\endgroup$
    – mlofton
    Jun 20, 2020 at 3:44

1 Answer 1


Using the autocovariance form that you have already derived, and assuming that $\sigma_w^2 < \infty$, you have:

$$\begin{aligned} \sum_{z \in \mathbb{Z}} |\delta(z)| &= \sum_{z \in \mathbb{Z}} \Bigg| \sigma_w^2 \sum_{k \in \mathbb{Z}} \psi_{k+z} \psi_k \Bigg| \\[6pt] &\leqslant \sigma_w^2 \sum_{z \in \mathbb{Z}} \sum_{k \in \mathbb{Z}} |\psi_{k+z}| |\psi_k| \\[6pt] &= \sigma_w^2 \sum_{k \in \mathbb{Z}} \sum_{z \in \mathbb{Z}} |\psi_{k+z}| |\psi_k| \\[6pt] &= \sigma_w^2 \sum_{k \in \mathbb{Z}} |\psi_k| \sum_{z \in \mathbb{Z}} |\psi_{k+z}| \\[6pt] &= \sigma_w^2 \sum_{k \in \mathbb{Z}} |\psi_k| \sum_{r \in \mathbb{Z}} |\psi_{r}| \\[6pt] &= \sigma_w^2 \Bigg( \sum_{k \in \mathbb{Z}} |\psi_k| \Bigg)^2 \\[6pt] &< \infty. \\[6pt] \end{aligned}$$

(Note the interchange of infinite sums in the third line; this step is allowed by the fact that all terms in the sum are non-negative.)

  • $\begingroup$ Thanks Ben for spelling it out. It's much clearer now. When you say the interchange of infinite sums in the third line are you talking about the third to last line where you take the double summation and split it up into two seperate summations of each $\psi$ term ? $\endgroup$
    – mlofton
    Jun 20, 2020 at 13:22
  • $\begingroup$ No, I'm talking about the transition from the second to third line. Notice that the order of the two sums has been swapped (so that the sum over $k \in \mathbb{Z}$ now occurs before the sum over $z \in \mathbb{Z}$). Since infinite sums are limits of partial sums, swapping the order of infinite sums is an instance of interchange of limits, so it is something you can't always do, and it requires a justification. $\endgroup$
    – Ben
    Jun 20, 2020 at 23:14
  • $\begingroup$ gotcha. thanks. but the changing of the double sum into two seperate single sums of the individual $\psi's$ doesn't need any justification because the order shouldn't matter. thanks for confirmation. $\endgroup$
    – mlofton
    Jun 21, 2020 at 2:05
  • $\begingroup$ Kinda. The step from the fourth to the fifth line occurs because you use a change-of-variables $r=k+z$ and you then have an inner sum that does not depend on $k$. You can then split into the product of two seperate sums because the inner sum does not depend on the index for the outer sum. $\endgroup$
    – Ben
    Jun 21, 2020 at 5:25
  • $\begingroup$ Gotcha. I understand you on the chance of variable thing. Thanks. $\endgroup$
    – mlofton
    Jun 22, 2020 at 1:40

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