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I was wondering if this equation is considered a Moving Average process of order 13? If so, does that mean that the coefficients at times t-2 to t-11 are 0? As they are clearly not present in this equation. (B is the Back shift operator and at is a white noise random variable)

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Yes, this is a 13-th order MA process, i.e. MA(13). The order of the process/filter is defined as the largest exponent of the back shift operator. It doesn’t matter if the coefficients of other lower order components are zero or not. For example, $x^2+1$ would still be a second order polynomial, even though we don’t have $x$ inside the expression.

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  • $\begingroup$ Thank you for the clarification gunes. I thought that was the correct idea, but I couldn't find anywhere in my textbooks to support that notion. $\endgroup$ – Corey Haverstick Jan 19 at 16:35

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