This problem was part of the preliminaries required to answer a question in an assignment set for a subject I'm doing: I've been scratching my head about it for two days now (having submitted the assignment already: the problem here is a very small part (<5%) of the assignment grade, and is the only thing I stuffed up). I'm reasonably certain that I learned this exact stuff 20 years ago (in Econometrics).

Problem setup:

Let $ r_t = \mu + \epsilon_t + \theta \epsilon_{t+1} ; \epsilon \text{ ~ i.i.d.} (0,{\sigma}^2 )$

And let $$ \mathbf{r} = \sum_{j=i}^k r_{t+j} $$

The eventual problem (calculating the variance ratio) required Var(r).

Now for an MA(1) process, $$ cov(r_{t+i}, r_{t+j}) = \begin{cases} (1+\theta^2)\cdot \sigma^2 \text { where } i=j; \\ \theta \cdot \sigma^2 \text { where } |i-j| = 1; \\ 0 \forall i,j \text { where } |i-j| > 1. \\ \end {cases} $$

(I realise that $cov(r_{t+i}, r_{t+j})$ is $var(r_{t+i})$ (and $var(r_{t+j})$ ) when i=j.)

The eventual answer is

$$ \begin{align} Var(\mathbf{r}) & = k \cdot var(r_t) + 2\cdot (k-1)\cdot cov(r_t, r_{t-1}) \\ & = k\cdot \sigma^2 + 2\cdot (k-1)\cdot \theta \cdot \sigma^2 \\ \end{align}$$

When I did it by hand (the most boring penwork I've done in two decades), I wound up with $ k\cdot var(r_t) + 2\cdot k \cdot cov (r_t, r_{t-1}) $ ; k instead of (k-1)... missed it by that much, Chief.

And finally, we get to my question: is there a 'shorthand' formula for the second moments of sums MA(q) variables? That is to say, if the question had instead used (say) an MA(2) process, would the answer have been

$$ Var(\mathbf{r}) = k \cdot var(r_t) + 2\cdot (k-1)\cdot cov(r_t, r_{t-1}) + \mathbf m\cdot (k-\mathbf n) \cdot cov(r_t, r_{t-2}) $$

       where *m* and *n* are known to the cognoscenti? 

An itch in the back of my skull says yes (and that perhaps m is always 2, and that n = 2 in this case), but for the life of me I have not been able to find a definitive answer. I've thought about the structure of a covariance matrix, but frankly my linear algebra is as flabby as my 48-year-old frame.

Hazarding a guess, I venture that for the sum of a sequence of k MA(q) variables, $$ var(\mathbf r) = k \cdot var(r_t) + 2\cdot \sum_{i=0}^{q}\left ((k-i)\cdot cov(r_t, r_{t-i})\right) $$

I've had a search through the archive, and while this question and this question are 'close', they aren't quite right.

[Edit: there was a typo in my guess at the variance of the sum of k MA(q) variables: I originally mistyped x instead of r in the $ cov(r_t, r_{t-i}) $ terms .]


Although there are possibly some typos (or things I misunderstand) in the question's notation, an $MA(q)$ process can be defined as (I omit the drift component since it doesn't affect the variance)

$$ r_t(q) = u_t + \theta_1u_{t-1} +...+ \theta_qu_{t-q}$$

with $u_t$ white noise, and the sum of the current and $k$ lags of this process as

$$R_t(q,k) = \sum_{i=0}^{k}r_{t-i}$$

I will give the easy formula, for $k=q$, and leave the boring penwork for the other cases to the OP. Writing $R_t(q,q)$ analytically we have

$$\begin{align} u_t + \theta_1&u_{t-1} +\theta_2u_{t-2}+...+ \theta_qu_{t-q} \\ + &u_{t-1} +\theta_3u_{t-2}+...+ \theta_{q-1}u_{t-q} +\theta_{q}u_{t-q-1} \\ &+\qquad \;\;\;\;\,u_{t-2}+...+ \theta_{q-2}u_{t-q} +\theta_{q-1}u_{t-q-1} + \theta_{q}u_{t-q-2}\\ +... \end{align} $$ The pattern is clear, and we obtain

$$R_t(q,q)= u_t + (1+\theta_{1})u_{t-1} +...+(1+\theta_{1} +...+\theta{q})u_{t-q}$$

$$+(\theta_1+...+\theta_q)u_{t-q-1} + (\theta_2+...+\theta_q)u_{t-q-2} +...+\theta_qu_{t-q-q}$$

$$\Rightarrow \text {Var}(R_t;q,q)= \sigma^2\left\{1+\sum_{i=1}^{q}\left[\left(\sum_{j=0}^{i}\theta_{j}\right)^2 + \left(\sum_{j=i}^{q}\theta_{j}\right)^2\right]\right\} \qquad \theta_0=1$$

  • 1
    $\begingroup$ Thanks for that, Alecos - what I was hoping to be able to do was express var(R) in terms of the autocovariance functions of the $r_t$, since for an MA(q), $$ cov(r_t, r_{t-j})=\sigma^2 \left [ \sum_{i=0}^{q-j} \theta_i \theta_{j+i} \right ]^2 $$ (for j = 1,...,q - 0 otherwise) - I have no doubt that it can be done, but it would involve decomposing the summation in your expression into combinations of the auto covariances. I think I'll have to do it by hand for a couple of q's and k's and see if I can see a pattern. Cheers GT $\endgroup$ – GT. Oct 13 '13 at 5:35

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