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I've got the proof but I don't understand a small detail.

As you know for an MA process:

$X_n = \sum _{i=0} ^q \beta_i Z_{n-i}$

where $Z_n$ is WGN (pure Gaussian random process).

Then the ACF is:

$\gamma(k) = Cov(\sum _{i=0} ^q \beta_i Z_{n-i}, \sum _{j=0} ^q \beta_j Z_{n-j + k}) = \sum _{i=0} ^q \sum _{j=0} ^q \beta_i \beta_j Cov(Z_n-i, Z_{n-j+k}) = \sum _{i=0} ^q \sum _{j=0} ^q \beta_i \beta_j Cov(Z_n, Z_{n +i-j+k})$

But because {Z_{n+i}} is iid wrt i then:

$Cov(Z_n, Z_{n +i-j+k}) = 0$ for $k + i - j \neq 0$ and $Cov(Z_n, Z_{n +i-j+k}) = \sigma_z ^2$ for $k + i - j = 0$.


$\gamma(k) = \sigma_z ^2 \sum _{i=0} ^q \sum _{j=0} ^q \beta_i \beta_j$

But the book says this equals: $\sigma_z ^2 \sum_{i=0} ^{q-k} \beta_i \beta_{i+k}$ .

For some reason I can't see how. If the sums were to infinity then I would agree, but they are not.

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up vote 2 down vote accepted

Hint: How many $\mathrm{Cov}(Z_n,Z_{n+i-j+k})$ terms, with $0\leqslant i\leqslant q$ and $0\leqslant j\leqslant q$, are not zero?

You say $q^2$ and the book says $q-k+1$. So, let us fix $i$ with $0\leqslant i\leqslant q$. How many $j$ such that $\mathrm{Cov}(Z_n,Z_{n+i-j+k})\ne0$ and $0\leqslant j\leqslant q$?

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Yes, I can see that. I know that $Cov$ is non-zero only when $k+i-j = 0$ in other words $j = k+i$ and since $\begin{cases} 0 \le j \le q \\ 0 \le i \le q \end{cases}$ this leads to $ 0 \le k+i \le q$ which leads to $-k \le i \le q-k$. Union with $0 \le i \le q$ you get $0 \le i \le q-k$. It just seemed to me that it involved too much logic instead of mathematical manipulations and tricks. – s5s May 14 '12 at 6:13

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