$MA(q)$ : Show $\sqrt{n}\hat{\rho}(q + l)\overset{d}{\to} N(0, 1 + 2\sum _{j=1}^q\rho ^2\left(j\right)), l\ge 1 $

Question

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

Answer 1

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.