Please solve step by step

4.11 Let the observed series * be composed of a periodic signal and noise so it can be written as X, = 81 cos(2xwx!) + 82 sin(2mux!) + Wr where w, is a white noise process with variance ow. The frequency wx is assumed to be known and of the form & in this problem. Suppose we consider estimating B1, By and ow by least squares, or equivalently, by maximum likelihood if the w, are assumed to be Gaussian. (a) Prove, for a fixed wx, the minimum squared error is attained by = 2n-1/2 |de(x) where the cosine and sine transforms (4.31) and (4.32) appear on the right-hand side. (b) Prove that the error sum of squares can be written as SSE = > x, - 21(WE) so that the value of ox that minimizes squared error is the same as the value that maximizes the periodogram /,(wx) estimator (4.28). (c) Under the Gaussian assumption and fixed up , show that the F-test of no regression leads to an F-statistic that is a monotone function of 1(wx ).I(w, ) = |d(w/)[ = n' > >(x -f)(xs - the Exley (!-8) 1 3 1 - E While Exlaugh (4.30) h= (m-1) for j * 0, where we have put h = t-s, with y(h) as given in (1.36).* In view of (4.30), the periodogram, I(w,), is the sample version of f(w) given in (4.17). That is, we may think of the periodogram as the sample spectral density of x. At first, (4.30) seems to be an obvious way to estimate a spectral density (4.17); i.e, simply put a hat on y(h) and sum as far as the sample size will allow. However, after further consideration, it turns out that this is not a very good estimator because it uses some bad estimates of y(). For example, there is only one pair of observations, (x], x) for estimating y(n - 1), and only two pairs (x], Xm-1), and (x2, X, ) that can be used to estimate y(n - 2), and so on. We will discuss this problem further as we progress, but an obvious improvement over (4.30) would be something like f(w) = [lism Y(h)e , where m is much smaller than n. It is sometimes useful to work with the real and imaginary parts of the DFT Individually. To this end, we define the following transforms. Definition 4.3 Given data x1, . .., An, we define the cosine transform de(wj) = n-1/25 x, COS(2mu,!) (4.31) and the sine transform *, sin(2nal,t) (4.32) where wj = j for j = 0, 1, . . .." - 1. We note that d(wy) = de(wj) - ids(wy) and hence I( w / ) = de(w/) + d;(w/). (4.33)Definition 4.1 Given data X1. .... In, we define the discrete Fourier transform (DFT) to be d(w/) = n-1/2 (4.26) for j = 0, 1. . ...n - 1, where the frequencies w, = j are called the Fourier or fundamental frequencies. If n is a highly composite integer (1.e., it has many factors), the DFT can be computed by the fast Fourier transform (FFT) introduced in Cooley and Tukey [44]. Also, different packages scale the FFT differently, so it is a good idea to consult the documentation. R computes the DFT defined in (4.26) without the factor , , but with an additional factor of easley that can be ignored because we will be interested in the squared modulus of the DFT. Sometimes it is helpful to exploit the inversion result for DFTs, which shows the linear transformation is one-to-one. For the inverse DFT we have, (4.27) 1=0 forf = 1. .... #. The following example shows how to calculate the DFT and its inverse in R for the data set {1, 2, 3, 4); note that R writes a complex number z = a + ib as a+bi. (dft = fft(1:4)/sqrt(4)) [1] 5+01 -1+1i -1+01 -1-1i (idft = fft(dft, inverse TRUE)/sqrt(4)) [1] 1+01 2401 3+01 4+0i (ReCidft)) # keep it real [1] 1 2 3 4 We now define the periodogram as the squared modulus of the DFT. Definition 4.2 Given data X1, ..., An, we define the periodogram to be I( w/) = |d(wil (4.28) for j = 0, 1, 2, . ..,n - 1. Note that /(0) = nx-, where & is the sample mean. Also, Er_, exp(-2mir;) =0 for j + 0,3 so we can write the DFT as (4.29)