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I am trying to show that for an $MA(q)$ process, $$\sqrt{n}\hat{\rho}(q + l)\overset{d}{\to} N(0, 1 + 2\sum _{j=1}^q\rho ^2\left(j\right)), \quad l\ge 1. $$

I'm having a hard time doing this. I'm not exactly sure how to start this off. I thought it might be similar to showing $$\sqrt{n}(\bar{X}-\mu)\overset{d}{\to} N(0, \sum _{h=-\infty}^\infty\gamma_X\left(h\right))$$ but it doesn't seem to work.

I tried to show $$\sum _{h=-\infty}^\infty\rho\left(h\right) = 1 + 2\sum _{j=1}^q\rho ^2\left(j\right)$$ but this doesn't seem to work and I think it was a bad idea, which resulted in me wasting 5 pages and 4 hours, and still no breakthroughs on how to solve the problem.

I think it might be a Bartlett's Formula problem, but I'm not sure how to apply it here.

Any help would be appreciated. This problem is on time series and the textbook being used in class can be found here.

I have an exam coming up and I really want to be able to understand this problem and how to solve it. Please help!

Thank you

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    $\begingroup$ @Zhanxiong the relation I wrote is exactly the practice problem given to solve. I think it requires use of the Bartlett's formula to prove it, but I'm not quite sure how to apply it $\endgroup$
    – eddie
    Commented Mar 26, 2023 at 1:43
  • $\begingroup$ Why estimate the autocorrelation at lags greater than q? $\endgroup$
    – Taylor
    Commented Mar 26, 2023 at 2:57
  • $\begingroup$ @Taylor I'm not quite sure, i thought that the idea was that up to q, the values are observed, then after q, they are future values. So when we estimate correlation at lags greater than q, we get relation for next values. I'm not sure if I'm thinking about it correctly $\endgroup$
    – eddie
    Commented Mar 26, 2023 at 3:08
  • $\begingroup$ @Zhanxiong I'm so confused. What exactly do I need to show/prove then? The material I've covered so far is chapters 1-3 from the textbook linked in the post. Am I not thinking of the right concepts then? Is it not asymptotic distribution? $\endgroup$
    – eddie
    Commented Mar 26, 2023 at 3:19
  • $\begingroup$ @eddie After a second thought, the problem still makes sense. I will add an answer below. $\endgroup$
    – Zhanxiong
    Commented Mar 26, 2023 at 3:25

1 Answer 1

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Because for MA($q$), the true autocorrelation $\rho(q + l) = 0$ for $l \geq 1$, the key of the proof is to apply the Bartlett's formula (as you correctly pointed out) to compute the asymptotic variance of $\sqrt{n}\hat{\rho}(q + l)$. The Bartlett's formula (Theorem 7.2.1, Time Series: Theory and Methods by Brockwell and Davis) gives the asymptotic covariance between $\sqrt{n}\hat{\rho}(i)$ and $\sqrt{n}\hat{\rho}(j)$: \begin{align} w_{ij} = \sum_{k = -\infty}^\infty\{&\rho(k + i)\rho(k + j) + \rho(k - i)\rho(k + j) + 2\rho(i)\rho(j)\rho^2(k) \\ & - 2\rho(i)\rho(k)\rho(k + j) - 2\rho(j)\rho(k)\rho(k + i)\}. \tag{1} \end{align}

Since the problem of interest only concerns with $\hat{\rho}(q + l)$, the asymptotic variance of $\sqrt{n}\hat{\rho}(q + l)$ can be obtained by setting $i = j = q + l =: r$ in $(1)$, which returns \begin{align} w_{rr} &= \sum_{k = -\infty}^\infty\{\rho^2(k + r) + \rho(k - r)\rho(k + r) + 2\rho^2(r)\rho^2(k) - 4\rho(r)\rho(k)\rho(k + r) \} \\ &= \sum_{k = -\infty}^\infty\{\rho^2(k + r) + \rho(k - r)\rho(k + r)\} \tag{$\dagger$} \\ \end{align} In $(\dagger)$, we use $\rho(r) = 0$ since $r > q$ and it is an MA($q$) process. So it remains to evaluate the two summations in $(\dagger)$.

The first summation is easy, again by noting $\rho(s) = 0$ for all $|s| > q$ in MA($q$), it follows that \begin{align} \sum_{k = -\infty}^\infty \rho^2(k + r) = \sum_{t = -\infty}^\infty\rho^2(t) = \sum_{t = -q}^q\rho^2(t) = 1 + 2\sum_{j = 1}^q\rho^2(j). \tag{$\dagger\dagger$} \end{align}

For the second summation, note that the summand $\rho(k - r)\rho(k + r) \equiv 0$ for all $k \in \mathbb{Z}$. This can be verified by considering $k - r < -q$ and $k + r > q$ separately (for each category, the product of $\rho(k - r)$ and $\rho(k + r)$ is zero -- again, use $\rho(s) = 0$ for all $|s| > q$ in MA($q$), and the union of these two categories covers $\mathbb{Z}$).

Combining $(\dagger)$ and $(\dagger\dagger)$ gives the desired result.

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  • $\begingroup$ I don't completely understand this part "Since the problem of interest only concerns with $\hat{\rho}(q + l)$, the asymptotic variance of $\sqrt{n}\hat{\rho}(q + l)$ can be obtained by setting $i = j = q + l =: r$ in $(1)$" Why do we set $i=j=q+l=r$? $\endgroup$
    – eddie
    Commented Mar 26, 2023 at 4:01
  • $\begingroup$ Can you read the reference I mentioned in the answer? Basically $(1)$ gives the covariance of two estimates $\hat{\rho}(i)$ and $\hat{\rho}(j)$, but your problem only asked the variance of $\hat{\rho}(r)$. It is just as simple as substitution. $\endgroup$
    – Zhanxiong
    Commented Mar 26, 2023 at 4:03
  • $\begingroup$ oh okay, I didn't know that I could apply substitution into the formula like that $\endgroup$
    – eddie
    Commented Mar 26, 2023 at 4:11
  • $\begingroup$ How do the last 2 summations here equal each other?\begin{align} \sum_{k = -\infty}^\infty \rho^2(k + r) = \sum_{t = -\infty}^\infty\rho^2(t) = \sum_{t = -q}^q\rho^2(t) = 1 + 2\sum_{j = 1}^q\rho^2(j). \tag{$\dagger\dagger$} \end{align} $\endgroup$
    – eddie
    Commented Mar 26, 2023 at 23:11
  • $\begingroup$ Use $\rho(t) = \rho(-t)$ and $\rho(0) = 1$. $\endgroup$
    – Zhanxiong
    Commented Mar 26, 2023 at 23:58

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