英国代考|  6YYD0017 Econometrics经济学

If $p=N / 2$, we ignore the term in $\beta \sin \omega t$, which is zero for all $t$, and find
\begin{aligned} &\hat{\mu}=\bar{x}, \ &\widehat{\alpha}=\sum(-1)^{t} x_{t} / N . \end{aligned}
The model in Equation (7.1) is essentially the one used before about 1900 to search for hidden periodicities, but this model has now gone out of fashion. However, it can still be useful if there is reason to suspect that a time series does contain a deterministic periodic component at a known frequency and it is desired to isolate this component (e.g. Bloomfield, 2000, Chapters 2,3 ).
Readers who are familiar with the analysis of variance (ANOVA) technique will be able to work out that the total corrected sum of squared deviations, namely,
$$\sum_{t=1}^{N}\left(x_{t}-\bar{x}\right)^{2}$$

can be partitioned into two components, namely, the residual sum of squares and the sum of squares ‘explained’ by the periodic component at frequency $\omega_{p}$. This latter component is given by
$$\sum_{t=1}^{N}\left(\widehat{\alpha} \cos \omega_{p} t+\hat{\beta} \sin \omega_{p} t\right)^{2}$$
which, after some algebra (Exercise $7.2$ ), can be shown to be
$$\begin{array}{cc} \left(\widehat{\alpha}^{2}+\widehat{\beta}^{2}\right) N / 2 & p \neq N / 2 \ \widehat{\alpha}^{2} N & p=N / 2 \end{array}$$

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