Unbiased estimation of variance of autocorrelated time series

Question

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

$$Y_{t}=\mu+\epsilon_{t}-\sum_{k=1}^{11}\theta_{k}\epsilon_{t+k}$$

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

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

correct?

Answer 1

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.

Edit

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 $\rho_Y(h) = (12-|h|)/12$.