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In Terrence C. Mills book Im currently reading The Foundation of Modern Time Series Analysis on while discussing Person's detrending methods, he mentions the following result.

... Concider the deviation of $x_t$ from a three year moving average centered on $$x_t:x_t-(x_{t+1}+x_t+x_{t-1})/3$$. This is equivalent to: $$-\frac{1}{3}(x_{t+1}-2x_t+x_{t-1})=-\frac{1}{3}\Delta^2x_{t+1}$$

so the correlation between the two series will be identical to the correlation between the deviations of the three-year moving averages...

pages:41-42

Im not sure how the above equality is derived. How does this work mathematically?

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    $\begingroup$ I just want to make doubly sure: the thing you want demonstrated is that this $-\frac{1}{3}(x_{t+1}-2x_t+x_{t-1})$ equals this $-\frac{1}{3}\Delta^2x_{t+1}$? (I presume you know the definition of $\Delta$ being used there?) $\endgroup$ – Glen_b Nov 12 '17 at 5:27
  • $\begingroup$ @Glen_b yeah I want to see how this equality exists. I know $\Delta$ refers to the number of times the data was differenced. $\endgroup$ – EconJohn Nov 12 '17 at 16:01
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\begin{eqnarray} \Delta^2(x_{t+1})&=&\Delta(\Delta(x_{t+1}))\\ &=&\Delta(x_{t+1}-x_t)\\ &=&\Delta(x_{t+1})-\Delta(x_t)\qquad\text{(by linearity of }\Delta \text{ operator)}\\ &=&(x_{t+1}-x_t)-(x_{t}-x_{t-1})\\ &=&x_{t+1}-2x_{t}+x_{t-1} \end{eqnarray}

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  • $\begingroup$ Wow, a value centered by a moving average is just a second difference over the number of variables we wish to take the average of. Thanks! $\endgroup$ – EconJohn Nov 12 '17 at 20:37

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