fitHRegL1: L1 Fit of Four Parameter Harmonic Regression
In CliffordLai/harper: Harmonic Regression and Periodicity Test

Description Usage Arguments Details Value Author(s) Examples

Estimates mu, A, B and lambda in the harmonic regression, y(t)=mu+A*cos(2*pi*lambda*t)+B*sin(2*pi*lambda*t)+e(t), where e(t) is assumed IID mean zero and constant variance. The absolute sum of errors is computed for each lambda on a grid of values and lambdaHat corresponds to the smallest. An efficient linear programming method is used to fit the model to equi-spaced frequencies in (0,0.5). The fitting algorithm minimizes the sum of the absolute residuals. The frequency is estimated by the model that has the best fit on the grid. The accuracy is controlled K, usually K=500, is sufficient.

1 2	fitHRegL1(z, t = 1:n, K = 500, ncpu = 1, lambdaRange = c(1/n, 0.45))

`z`	series.
`t`	Time points.
`K`	number of subintervals.
`ncpu`	number of compute nodes for use with parallel.
`lambdaRange`	range of frequencies inside (0,0.5), see Details.

The three parameter harmonic regression model is fitted by least absolute deviation (LAD) for each frequency in the grid which is a subset of the equi-spaced point {j/(2K+1), j=1,…,K} possibly excluding low and high frequencies as specified by the lambdaRange argument. Frequencies lower than the first Fourier frequency, 1/n, where n is the length time series, n=length(z) are usually due to trend while high frequencies are often not of interest and the numerical algorithm for fitting the 3-parameter harmonic regression is sometimes inaccurate at high. Note that LAD is MLE when the errors are IID double exponential. The LAD fitting uses the function rq.fit.br from the package quantreg.

list with the following components

coefficients const, A, B, lambda
residuals
periodogram
z
t
K
model
lambdaRange

Yuanhao Lai and A.I. McLeod

z<-c(0.42, 0.89, 1.44, 1.98, 2.21, 2.04, 0.82, 0.62, 0.56, 0.8, 1.33)
fitHRegL1(z)
#
#on multicore pcs, more the package parallel may be used for the
#grid computation but unless n is very large this is not recommended.
## Not run:  #adjust ncpu
system.time(ans1 <- fitHRegL1(z)) #0.06 sec on my computer
system.time(ans2 <- fitHRegL1(z, ncpu=8)) #4.11 sec
identical(ans1, ans2)

## End(Not run)