# How does the the variance of a moving average process vary with its order?

Let $\ MA(q)$ be a moving average process of order $\ q$, and $\hat{y} = \mu + a_t$ be a constant forecast model whose level is represented by $\mu$ and the random term at time $\ t$ by $\ a_t$.

Now, suppose $\mu \equiv \hat{a_1}$ so one can write $\hat{y} = \ a_1 + a_t$.

If $\hat{a}_1$ is calculated by a $\ MA(q)$, what would the variance of $\hat{y}$ be?

What if, then, $\hat{a}_1$ is calculated by a $\ MA(2q)$ or a $\ MA(q^2)$ - how would that affect the variance of $\hat{y}$?

My goal is understand how the variance of my forecasted values will vary as I vary$\ q$.

Bob Stine, from UPenn, made some actual calculations on page 9 of this document, but the way he calculates the variance for 3 steps ahead is something I'm yet to understand. More precisely, I don't get why he keeps getting rid of his $a_n$ terms.

Thank you.

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Edit: I'll try to show some of my thoughts on this matter here:

On page 86, Montgomery, Johnson,and Gardiner (1990) write that:

$\large var(M_T) = \large var[\frac{1}{N}\sum_{k=0}^{N-1} x_{T-k}] = \large \frac{\sigma^2_\epsilon}{N}$

where $\ M_T$ is the $\ MA$ process, $\ N$ is the number of the most recent observations taken into account, and $\sigma^2_\epsilon$ is the variance of the noise (random component).

However, if I write the $\ MA$ process as $\phi_p. z_t = \theta_q . a_t$, for $\phi_p$ equals 1, I have that my time series $z_t$ is equal to $\theta_q . a_t$, where $\theta_q$ is the lag operator of order $q$ and $a_t$ is my random component.

Thus, I can write $z_t$ as $z_t = (1 + B + B^2 + ... + B^q).a_t$, and then $z_t = a_t + a_{t-1} + a_{t-2} + ... + a_{t-q+1}$.