Let $X_n = \sum\limits_{j=0}^{\infty} a_jZ_{t-j}$, where $Z_t$ is a (weak) white noise $(0,\sigma^2)$ and $a_j \in L^2$. Prove that ACF $\gamma_X(h)\longrightarrow 0$ as $h\longrightarrow +\infty$. So:

$\gamma_X(h)=\text{Cov}(X_{t+h},X_t)=\text{Cov}(\sum\limits_{i=0}^{\infty} a_iZ_{t+h-i}, \sum\limits_{j=0}^{\infty}a_jZ_{t-j})=\lim\limits_{n,m\to\infty} \text{Cov}(\sum\limits_{i=0}^{n} a_iZ_{t+h-i}, \sum\limits_{j=0}^{m}a_jZ_{t-j})\overset{\textbf{if true: why?}}{=} \lim\limits_{n,m\to\infty}\sum\limits_{i=0}^{n}\sum\limits_{j=0}^{m} \text{Cov}(a_iZ_{t+h-i}, a_jZ_{t-j})$

and if we have weak white noise, then only non-zero terms of $\text{Cov}$ above are these that satisfy $t+h-i=t-j \longleftrightarrow h=i-j$, but... how to proceed from that point?

  • $\begingroup$ One way is to observe that for any $\epsilon\gt 0,$ there exists $N$ such that for all $h\ge N,$ $a_h^2 + a_{h+1}^2 + \cdots \lt \epsilon.$ $\endgroup$
    – whuber
    Commented Mar 3, 2021 at 14:26
  • $\begingroup$ @whuber thanks, but is everything correct up to the point where I left this exercise? $\endgroup$
    – itsme
    Commented Mar 3, 2021 at 15:02
  • $\begingroup$ It looks fine to me: you approach this with appropriate rigor with the limits. But you have only begun the calculation: that last summation simplifies greatly because the $Z_{t+h-i}$ and $Z_{t-j}$ are independent--hence uncorrelated--when $i-h\ne j.$ $\endgroup$
    – whuber
    Commented Mar 3, 2021 at 15:04

1 Answer 1


In terms of the Kronecker delta $\delta(i,i)=1,$ $\delta(i,j)=0$ when $i\ne j,$ you are concerned about the double limit

$$\begin{aligned} \lim\limits_{n,m\to\infty} \operatorname{Cov}\left( a_iZ_{t+h-i}, \sum\limits_{j=0}^{m}a_jZ_{t-j}\right) &= \lim\limits_{n,m\to\infty}\sum\limits_{i=0}^{n} \sum\limits_{j=0}^{m}\operatorname{Cov}\left( a_iZ_{t+h-i}, a_jZ_{t-j}\right)\\ &= \lim\limits_{n,m\to\infty}\sum\limits_{i=0}^{n} \sum\limits_{j=0}^{m}a_ia_j\operatorname{Cov}\left( Z_{t+h-i}, Z_{t-j}\right)\\ &= \lim\limits_{n,m\to\infty}\sum\limits_{i=0}^{n} \sum\limits_{j=0}^{m}a_ia_j \delta(t+h-i, t-j)\sigma^2\\ &= \sigma^2\lim\limits_{n,m\to\infty}\sum\limits_{j=0}^{\min(n, m-h)} a_{j+h}a_j.\\ \end{aligned}$$

The Cauchy-Schwarz Inequality gives an upper bound for the size of this sum,

$$\begin{aligned} \left|\sum\limits_{j=0}^{\min(n, m-h)} a_{j+h}a_j\right| &\le \left(\sum\limits_{j=0}^{\min(n, m-h)}a^2_{j+h} \sum\limits_{j=0}^{\min(n, m-h)}a^2_j\right)^{1/2}\\ &\le \left(\sum\limits_{j=0}^{\infty}a^2_{j+h} \sum\limits_{j=0}^{\infty}a^2_j\right)^{1/2}\\ &=\left(\left[\sum\limits_{j=0}^{\infty}a^2_{j} - \sum\limits_{j=0}^{h-1}a^2_{j} \right]\sum\limits_{j=0}^{\infty}a^2_j\right)^{1/2}. \end{aligned}$$

"$(a)\in L^2$" means $\sum\limits_{j=0}^{\infty}a^2_{j}$ converges to a finite value $||a||_2^2.$ Thus, given any $\epsilon\gt 0$ there exists $h(\epsilon)$ such that $\sum\limits_{j=h(\epsilon)}^{\infty}a^2_{j} \le \epsilon^2.$ In these terms the upper bound is

$$\left(\left[\sum\limits_{j=0}^{\infty}a^2_{j} - \sum\limits_{j=0}^{h(\epsilon)-1}a^2_{j} \right]\sum\limits_{j=0}^{\infty}a^2_j\right)^{1/2} = \left(\sum\limits_{j=h(\epsilon)}^{\infty}a^2_{j} \sum\limits_{j=0}^{\infty}a^2_j\right)^{1/2}\le \left(\epsilon^2 ||a||_2^2\right)^{1/2} = \epsilon ||a||_2.$$

Consequently, $\left|\gamma_X(h)\right|$ is bounded above by $\epsilon\sigma^2||a||_2.$ The limit of $|\gamma_X(h)|$ must lie between $0$ and the least upper bound of all such values, also equal to $0,$ showing $\gamma_X(h)$ converges absolutely to $0,$ whence it converges to $0,$ QED.


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