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In Brockwell and Davis's book (Time Series Theory and Methods 2nd Edition), provide the following problem:

Show that if $\{X_t, t=0, \pm1, \dots\}$ is weak stationary and $|\theta| < 1$ then for each $n$, $\sum_{j=1}^{m} \theta^j X_{n+1-j}$ converges in mean square as $m \rightarrow \infty $.

I tried to show Cauchy convergence in the mean square sense as follows:

$$\mathbb{E}\left(\sum_{j=1}^{m} \theta^j X_{n+1-j} - \sum_{j=1}^{k} \theta^j X_{n+1-j}\right)^2 \quad \text { for } m > k\\ = \mathbb{E}\left(\sum_{j=k+1}^{m} \theta^j X_{n+1-j}\right)^2\\ = \mathbb{E}\left(\sum_{i=k+1}^{m} \sum_{j=k+1}^{m} \theta^i \theta^j X_{n+1-i} X_{n+1-j}\right)$$

But I am a bit stuck from here. Anyone to help?

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Write $\sigma^2$ for the marginal variance. $$\left|E[X_{n+1-i}X_{n+1-j}]\right|\leq \sigma^2$$ so $$\sum_{i=k+1}^m\sum_{j=k+1}^m \theta^i\theta^j\left|E[X_{n+1-i}X_{n+1-j}]\right|\leq\sigma^2\sum_{i=k+1}^m\sum_{j=k+1}^m \theta^i\theta^j=\sigma^2\sum_{i=k+1}^m\sum_{j=k+1}^m \theta^{i+j}$$

We have a square. Think of one index along the diagonal and the other perpendicular to it, so take $l=i+j-2k$ counting along the diagonal. There is one term with $l=0$, two with $l=1$, and so on (eventually decreasing, but we don't need that)

$$\sum_{i=k+1}^m\sum_{j=k+1}^m \theta^{i+j}< \theta^{2k}\sum_{l=0}^{\infty} l\theta^l$$

Now, $\sum_{l=0}^{\infty} l\theta^l$ is finite by the ratio test, so $$\sum_{i=k+1}^m\sum_{j=k+1}^m \theta^{i+j}=O(\theta^{2k})$$ and is small whenever $k$ is large and $m>k$.

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  • $\begingroup$ How is that first equality true if $\mu \ne 0$? $\endgroup$
    – Darby Bond
    Oct 28 '20 at 18:11

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