Question
Let$$x_{\mathrm{f}}^{\prime \prime}=\frac{1}{2 m+1} \sum_{j=-\mathrm{m}}^{m} x_{t+j}$$be a simple, centered $(2 m+1)$-point moving average. Show that$$x_{t+1}^{*}=x_{t}^{*} \frac{x_{t+m+1}-x_{t-m}}{2 m+1}$$How might this result be used in the efficient computation of series of centered moving averages?
Step 1
Step 1: First, we recall the general equation for the moving average: $$ x_{\mathrm{f}}^{\prime \prime}=\frac{1}{2 m+1} \sum_{j=-\mathrm{m}}^{m} x_{t+j} $$ This equation represents a simple, centered $(2 m+1)$-point moving average. Show more…
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