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Harmonic Regression Model

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Harmonic Regression Model Statistics 910, #6 1 Harmonic Regression Overview 1. Example: periodic data 2. Regression at Fourier frequencies 3. Discrete Fourier transform (periodogram) 4. Examples of the DFT Example: Periodic Data Magnitude of variable star This integer time series is reported to be the magnitude of a variable star observed on 600 successive nights (Whittaker and Robinson, 1924, varstar.dat). Periodic model Classical example of \signal in noise," with a hidden (though not very hidden in this case) periodic signal. Model for periodic data with a single sinusoid is (Example 2.7) xt = A cos(2π!t + φ) + wt with amplitude A, frequency !, and phase φ. Don't confuse wt (white noise) for ! (frequency, the reciprocal of the period in t units). Caution: Aliasing and identifiability The frequency must be restricted to the range −1=2 < ! ≤ 1=2 in order for the model to be identified (or a translation of this range). For discrete-time, equally-spaced data, frequencies outside of this range are aliased into this range. For exam- ple, suppose that λ = 1 − δ; 0 < δ < 1=2, then for t = 1; 2;::: cos(2πλt) = cos(2π(1 − δ)t + φ) = cos(2πt + (−2πδt + φ)) = cos(2πt) cos(2πδt − φ) + sin(2πt) sin(2πδt + φ) = cos(2πδt − φ) : Thus, a sampled sinusoid with frequency ! > 1=2 appears as a sinusoid with frequency in the interval (−1=2; 1=2]. If the data is sampled at Statistics 910, #6 2 a spacing of ∆ time units, then the highest frequency observable in discrete data is 2=∆, which is known as the folding frequency or Nyquist frequency. Notation alert! Many books (including the ones I learned from) write the sinusoids using the angular frequency notation that embeds the 2π into the frequency as λ = 2π!. Thus, the sinusoidal model would look like xt = A cos(λt + φ) + wt Question How to estimate unknown parameters? the amplitude A, the frequency ! and the phase φ. Estimation by regression Convert the expression of the model into one in which the unknown parameters can be linearly estimated. (This is a common, and very useful theme.) Once the model is expressed with unknown that offer linear estimators, we can estimate them using regression and OLS. The conversion uses the first of the following two \double-angle" formulas: cos(a + b) = cos(a) cos(b) − sin(a) sin(b) even sin(a + b) = sin(a) cos(b) + cos(a) sin(b) odd Write it as the linear regression (sans intercept) xt = A cos(φ) cos(2π!t) + −A sin(φ) sin(2π!t) + wt | {z } | {z } β1 β2 2 2 2 Notice that β1 + β2 = A . The sum of the squared regression coef- ficients is the squared amplitude of the hidden sinusoid. This sum is also, as we'll see, proportional to the regression sum-of-squares asso- ciated with this pair of variables. Frequency The estimation is linear if ! is known. What's a good estimator for !? Hint: count cycles. The estimation of ! is non-standard: we can get an estimator within op(1=n) rather than the usual root-n rate. There's also a nice plot. If the period is M = 1=!, then plot xt on t mod M. This is useful as it introduces a nonparametric estimator for a general periodic function. Statistics 910, #6 3 What if the background is not white noise? Least squares provides good estimates (i.e., efficient) of β1 and β2 if wt in the regression yt = β1 cos 2π!t + β2 sin 2π!t + wt forms a stationary process. The expla- nation goes back to our comparison of OLS and GLS: OLS is fully efficient if the regression variable is an eigenvector of the covariance matrix of the process. Sinusoids at the Fourier frequences (next sec- tion) are just that { for any stationary process (asymptotically). Regression at the Fourier Frequencies Idea Rather than count peaks to guess the period or frequency (as in the variable star), fit regressions at many frequencies to find hidden sinu- soids (simulated data). Use the estimated amplitude at these frequen- cies to locate hidden periodic components. It is particularly easy to estimate the amplitude at a grid of evenly spaced frequencies from 0 to 1/2. Fourier frequencies determine the grid of frequencies. A frequency 0 ≤ ! ≤ 1=2 is known as a Fourier frequency if the associated sinusoid completes an integer number of cycles in the observed length of data, j ! = ; j = 0; 1; 2; : : : ; n=2; j n assuming n is an even integer. Because of aliasing, the set of j's stop at n=2. We also don't consider negative frequencies since cosine is an even function and sine is odd; these are redundant (aliased). Orthogonality Cosines and sines at the Fourier frequencies generate an orthogonal (though not orthonormal) set of regressors. Consider the n × n matrix (assuming n is even) given by 0 . 1 . ::: . B C X = B cos(2π!0t) cos(2π!1t) sin(2π!1t) ::: cos(2π!n=2−1t) sin(2π!n=2−1t) cos(2π!n=2t) C @ . A . ::: . Note that the first column is all 1's (!0 = 0) and the last column 1 alternates ±1 (!n=2 = 2 ). (There's no sine term at !0 or !n=2 since Statistics 910, #6 4 these are both identically zero; sin 0 = sin π = 0. It follows that 0 n 0 0 ::: 0 1 B 0 n=2 0 ::: 0 C B C 0 B . C X X = B 0 0 .. 0 0 C B C B C @ 0 ::: 0 n=2 0 A 0 ::: 0 0 n The non-constant columns are orthogonal to the constant (i.e.sum to zero) since n X X cos(2π!kt) = sin(2π!kt) = 0 t=1 at the Fourier frequencies. (Imagine summing the sine and cosines around the unit circle in the plane.) The cross-products between columns reduce to these since, for instance, n X 1 X cos(2π!kt) sin(2π!jt) = 2 sin(2π!k+jt) − sin(2π!k−jt) t=1 = 0; For diagonal terms, notice that (from the double-angle expressions) 1 cos(a) cos(b) = 2 (cos(a + b) + cos(a − b)) 1 sin(a) sin(b) = 2 (cos(a − b) − cos(a + b)) allow us handle the sums of squares as follows: X 2 1 X cos(2π!jt) = 2 cos(4π!jt) + 1 = n=2; j 6= 0; n=2 : Harmonic regression If we use Fourier frequencies in our harmonic re- gression, the regression coefficients are easily found since \X0X" is almost diagonal. Specifically, (with b2s on the cosine terms and b1s on the sines) X X t b0 = xt=n bn=2 = xt(−1) =n t t 2 X 2 X b = x cos 2π! t b = x sin 2π! t: 2j n t j 1j n t j t t Statistics 910, #6 5 Note that there is no sine component at 0 or n=2 since sin 0 = sin π = 0. 2 2 To find the hidden sinusoid, we now can plot b2j + b1j versus the \frequency" j. There should be a peak for frequencies near that of the hidden sinusoid. Estimated amplitude The sum of squares captured by a specific sine/cosine pair is (j 6= 0; n=2) n n (b2 + b2 ) = A2 : 2 1j 2j 2 j That is, the amplitude of the fitted sinusoid at frequency !j determines the variance explained by this term in a regression model. Since we have fit a regression of n parameters to n observations x1; : : : ; xn, the model fits perfectly. Thus the variation in the fitted values is ex- actly that of the original data, and we obtain the following decompo- sition of the variance: X n X X2 = n(A2 + A2 ) + A2; t 0 n=2 2 j t j where the weights on A0 and An=2 differ since there is no sine term at these frequencies. Thus 0 Vector space The data X = (x1; : : : ; xn) is a vector in n-dimensional space. The usual basis for this space is the set of vectors 0 ej = (0 ::: 0 1j 0 0 ::: 0) : P Thus we can write X = t Xtet. The harmonic model uses a different orthogonal basis, namely the sines and cosines associated with the Fourier frequencies. The \saturated" harmonic regression n=2−1 t X Xt = b0 + bn=2(−1) + (b2j cos(2π!jt) + b1j sin(2π!jt)) j=1 represents the vector X in this new basis. The coordinates of X in this basis are the coefficients b1j and b2j. Statistics 910, #6 6 Discrete Fourier transform Time origin When dealing with the Fourier transform, it is common to index time from t = 0 to t = n − 1. Definition Changing to complex variables via Euler's form exp(iθ) = cos θ + i sin θ (1) leads to the discrete Fourier transform. The discrete Fourier transform (DFT) of the real-valued n sequence y0; : : : ; yn−1 is defined as n−1 1 X d(! ) = D = y exp(−i2π! t); j = 0; 1; 2; : : : ; n − 1: j j n t j t=0 The DFT is the set of harmonic regression coefficients, written using n complex variables. For j = 0; 1;:::; 2 , n−1 1 X D = y exp(−i2π! t) j n t j t=0 n−1 1 X = y (cos 2π! t − i sin 2π! t) n t j j t=0 1 = (b − ib ) 2 2j 1j p Note: SS define the DFT using a divisor n rather than n; R defines the DFT with no divisor. Matrix form The DFT amounts to a change of basis transformation.
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