Let's say I have $n$ measurements (let's assume monthly measurements) of some underlying random variable $X_{t}$ for $t\in\{1,\ldots,n\}$. Now, let's say I then construct a new measure

$$Y_{t}=\tfrac{1}{12}\big(X_{t}+X_{t+1}+\cdots+X_{t+11}\big)$$ which is the forward 12-month average of the underlying random variable. Now, due to the introduction of the new measure, we have reduced much of the original volatility in the time series $X_{t}$. So when we estimate the variance of this new time series $Y_{t}$, the estimate will be biased downward.

We would like to get an unbiased estimator of the variance that reflects the volatility of $X_{t}$. Now, according to this, we can calculate our unbiased estimate of the variance using the provided equation. It also states in a caveat that we need the analytical expression for the ACF function and it cannot be estimated from the biased data.

This leads me to my question, is it correct that my process $Y_{t}$ is a $\text{MA}(11)$ process


and that, as a result, I can calculate the analytical ACF for my data given that I know the parameters of my process are all equal. Is this intuition correct?

Additionally, is the analytical ACF of this $\text{MA}(12)$ process

$$\begin{align} \rho(\tau)=\frac{\gamma(\tau)}{\gamma(0)}&=\frac{\sigma^{2}\sum_{j=0}^{12-|\tau|}\theta_{j}\theta_{j+|\tau|}}{\sigma^{2}\sum_{j=0}^{12}\theta_{j}\theta_{j}}\\ &=\frac{\sum_{j=0}^{12-|\tau|}\theta^{2}}{\sum_{j=0}^{12}\theta^{2}}\\ &=\frac{\sum_{j=0}^{12-|\tau|}}{\sum_{j=0}^{12}}\\ &=\frac{13-|\tau|}{13},\quad\quad\quad |\tau|\le 12 \end{align}$$


  • $\begingroup$ If you look at your right hand side carefully, you will note that the subscript of $\epsilon_k$ is independent of $t$, which is not what you want! $\endgroup$ – jbowman Jan 24 '18 at 0:04
  • $\begingroup$ @jbowman Thanks, I'm still not sure my notation is exactly correct. Does my intuition for solving the problem seem correct? $\endgroup$ – Ed P Jan 24 '18 at 0:10
  • $\begingroup$ Given that it's a forward average, I think the subscript $t+k$ would be correct. But yes, given that minor caveat, your intuition is (also) correct. $\endgroup$ – jbowman Jan 24 '18 at 0:15

If your data are independent with the same mean and different variances then you can write $X_t = \mu + b_t \epsilon_t$ and

\begin{align*} Y_t &= \frac{1}{12}\sum_{k=0}^{11} X_{t+k}\\ &= \frac{1}{12}\sum_{k=0}^{11} \mu + b_{t+k} \epsilon_{t+k} \\ &= \mu + \sum_{k=0}^{11} \theta_k \epsilon_{t+k} \end{align*} where $\theta_k = b_{t+k}/12$.

If your data was from a causal linear process to begin with, then you can write your $X_t$ as a moving average process (who knows what the order is, though), and your $Y_t$ will always be a moving average process. However, it will not necessarily be of order 12. If you start with a linear process, and you filter it (which is what you're doing), you will always have a linear process as a result.


To answer your new question, I will assume your data are iid and I will shift the time index for $Y$ forward, which means $$ Y_t = \mu + \theta\epsilon_t + \sum_{i=1}^{11} \theta\epsilon_{t-i} = \mu + W_t + \sum_{i=1}^{11} W_{t-i}, $$ which is a non-invertible MA(11) process.

\begin{align*} \gamma_Y(h) &= \text{Cov}(Y_{t+h},Y_t)\\ &= \text{Cov}\left(\sum_{i=0}^{11} W_{t+h-i} ,\sum_{j=0}^{11} W_{t-j}\right)\\ &= \sum_{i=0}^{11}\sum_{j=0}^{11}\gamma_W(j+h-i)\\ &= \theta^2(12-|h|) \end{align*} which means $\rho_Y(h) = (12-|h|)/12$.

  • $\begingroup$ My goal was to calculate the analytical ACF for the $Y_{t}$ series. Are you saying I can't actually do that because I don't know the order $q$? $\endgroup$ – Ed P Jan 24 '18 at 2:58
  • $\begingroup$ @StatsPlease no it's doable, but you need to be clear about what your assumptions about $X_t$ are $\endgroup$ – Taylor Jan 24 '18 at 2:59
  • $\begingroup$ So under the assumption that $X_{t}$ are iid I can say that the order $q=12$? $\endgroup$ – Ed P Jan 24 '18 at 3:01
  • $\begingroup$ @StatsPlease yes, if they are independent AND IDENTICAL, then it simplifies the situation a little bit. In that case, $b_t$ are all the same, and you have an MA(12) process where all the $\theta$ are the same. Be aware that there is a convention, though, that people will write down an MA(1) process with a leading coefficient of $1$ (e.g. $Y_t = \epsilon_t + \theta_1 \epsilon_{t-1}$), so you might need to make an adjustment. $\endgroup$ – Taylor Jan 24 '18 at 3:03

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