# Ergodicity for the mean of a linear process without finite second moment

Suppose that $\{X_k:k\in\mathbb Z\}$ is a linear process, i.e. a sequence of random variables such that $$X_k=\sum_{j=0}^\infty\psi_j\varepsilon_{k-j}$$ for each $k\in\mathbb Z$, where $\{\psi_j:j\ge0\}$ is a real sequence such that $\sum_{j=0}^\infty|\psi_j|<\infty$ and $\{\varepsilon_k:k\in\mathbb Z\}$ are independent and identically distributed random variables with $\operatorname E|\varepsilon_0|<\infty$.

Is the linear process $\{X_k:k\in\mathbb Z\}$ ergodic for the mean, i.e. does the average $n^{-1}\sum_{k=1}^nX_k$ converge almost surely (or in probability) to $\operatorname EX_0$ as $n\to\infty$?

If we assume that $\operatorname E\varepsilon_0^2<\infty$, then $\sum_{j=0}^\infty|\psi_j|<\infty$ implies the ergodicity for the mean (see p. 52 of Time Series Analysis by James D. Hamilton), but I'm interested in the case when we only assume that $\operatorname E|\varepsilon_0|<\infty$.

Any help is much appreciated!

Denote by $\mu$ the expected value of $\varepsilon_0$ and let

$$S_{n,j} := \frac{1}{n} \sum_{k=1}^n \varepsilon_{k-j} \quad , \quad \bar{X}_n = \frac{1}{n} \sum_{k=1}^n X_k$$

We have, by definition of $X_k$ :

$$\bar{X}_n = \sum_{j=0}^{\infty} \psi_j S_{n,j} ,$$

and thus :

$$|\bar{X}_n - \mathbb{E}X_0| \leq \sum_{j=0}^{\infty} |\psi_j| |S_{n,j} - \mu|.$$

Because $S_{n,j}$ satisfies the assumptions of the strong law of large numbers, we have

$$\mathbb{E}|S_{n,j} - \mu| \to 0 \text{ as } n \to \infty$$

and this is true for any fixed $j$.

On the other hand, we have

$$\mathbb{E}|S_{n,j} - \mu| \leq 2 \mathbb{E}|\varepsilon_0|$$

By the discrete dominated convergence theorem we have :

$$\mathbb{E}|\bar{X}_n - \mathbb{E}X_0| \to 0$$

thus giving the convergence in probability of the whole sequence and a.s. convergence up to a subsequence.

On a side note, this post seems interesting, but the missing factor of $n$ in the sum may be a trouble.

• Writing $S_n$ instead of $S_{n,j}$ was a slight abuse of notation, but it does depend on $j$, so one cannot factor it out. The answer has been edited accordingly. – Hachino Jan 14 '15 at 14:26
• Indeed. As stated, this is an avatar of the dominated convergence theorem for series (hence the "discrete"). – Hachino Jan 14 '15 at 14:41
• We have that for each $j\ge0$ there exists $\Omega_j\subset\Omega$ such that $\Pr(\Omega_j)=1$ and $|S_{n,j}(\omega) - \mu|\to0$ as $n\to\infty$ for each $\omega\in\Omega_j$. But does it follow that there exists $\Omega_\infty\subset\Omega$ such that $\Pr(\Omega_\infty)=1$ and $|\bar X_n(\omega) - \mathbb{E}X_0|\to0$ as $n\to\infty$ for each $\omega\in\Omega_\infty$? Could you provide some details? – Cm7F7Bb Jan 14 '15 at 18:52
• For each $j \geq 0$, the set $N_j := \Omega \setminus \Omega_j$ is negligible. Thus, $N := \bigcup_{j=0}^{\infty} N_j$ is also negligible (basic measure theory). But outside $N$ (that is, we set $\Omega_{\infty} := \Omega \setminus N$), we have pointwise convergence everywhere of $|S_{n,j}(\omega) - \mu|$ to $0$. – Hachino Jan 15 '15 at 7:43
• Thank you very much for the answer and your comments! I'm still not sure how we can justify the interchange of the limits (I deleted my previous comments about the interchange of limits). If we want to use the dominated convergence theorem for series, we need to find a sequence of random variables $\{d_j:j\ge0\}$ such that $|\psi_j||S_{n,j}-\mu|\le d_j$ for each $n\ge1$ and $\sum_{j=0}^\infty d_j$ converges almost surely, right? How can we find such a sequence? Or is there another way to justify the interchange of the limits? – Cm7F7Bb Jan 15 '15 at 9:27