Time series ARMA model estimation results different from textbook
2 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
I tried to estimate an ARMA(4,3) model using armax function in the system identification toolbox. I used the same data set, but the result was totally different from the textbook Pandit and Wu,1990, Time Series and System Analysis with Applications. The data I used is the sunspot data in Table A2 in the textbook (page487). I double checked to make sure the data is exactly the same as in the textbook.
The estimated ARMA(4,3) model in the textbook in page 170 is:
A(q)=1-0.42(+-1)q^-1-0.28(+-0.75)q^-2+0.08(+-0.49)q^-3+0.33(+-0.5)q^-4.
C(q)=1+0.89(+-1)q^-1+0.31(+-0.76)q^-2-0.14(+-0.23)q^-3.
However, the estimation result of armax is:
A(q)=1-1.572(+-0.3967)q^-1+0.1358(+-0.9669)q^-2+1.046(+-0.8845)q^-3-0.6024(+-0.3122)q^-4.
C(q)=1-0.3253(+-0.4095)q^-1-0.756(+-0.5163)q^-2 +0.08137(+-0.168)q^-3.
My code is as follows. My question is which result is correct?
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Read and plot raw data (rawdata has 176 points in one column)
rawdata=[80.9 83.4 47.7 47.8 30.7 12.2 9.6 ...
10.2 32.4 47.6 54 62.9 85.9 61.2 45.1 36.4 ...
20.9 11.4 37.8 69.8 106.1 100.8 81.6 66.5 ...
34.8 30.6 7 19.8 92.5 154.4 125.9 84.8 ...
68.1 38.5 22.8 10.2 24.1 82.9 132 130.9 ...
118.1 89.9 66.6 60 46.9 41 21.3 16 ...
6.4 4.1 6.8 14.5 34 45 43.1 47.5 ...
42.2 28.1 10.1 8.1 2.5 0 1.4 ...
5 12.2 13.9 35.4 45.8 41.1 30.4 23.9 ...
15.7 6.6 4 1.8 8.5 16.6 36.3 49.7 ...
62.5 67 71 47.8 27.5 8.5 13.2 56.9 ...
121.5 138.3 103.2 85.8 63.2 36.8 24.2 10.7 ...
15 40.1 61.5 98.5 124.3 95.9 66.5 64.5 ...
54.2 39 20.6 6.7 4.3 22.8 54.8 93.8 ...
95.7 77.2 59.1 44 47 30.5 16.3 7.3 ...
37.3 73.9 139.1 111.2 101.7 66.3 44.7 17.1 ...
11.3 12.3 3.4 6 32.3 54.3 59.7 63.7 ...
63.5 52.2 25.4 13.1 6.8 6.3 7.1 35.6 ...
73 84.9 78 64 41.8 26.2 26.7 12.1 9.5 ...
2.7 5 24.4 42 63.5 53.8 62 48.5 43.9 ...
18.6 5.7 3.6 1.4 9.6 47.4 57.1 103.9 ...
80.6 63.6 37.6 26.1 14.2 5.8 16.7]';
meanvalue=mean(rawdata);
%Fit an ARMA model
order=[4 3];
data=iddata(rawdata-meanvalue,[],1);
model=armax(data,order);
%Show the fitted model
present(model);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
2 Commenti
Walter Roberson
il 14 Nov 2011
I really have no idea, but the 1.572 coefficient is highly suggestive of Pi/2 and thus hints at a possible difference in whether the values are being interpreted in some absolute unit (e.g., Hz) or in some relative unit such as "cycles per radian"
Wayne King
il 14 Nov 2011
I don't have this book, but are you sure you did not leave out some important detail about how they preprocessed the data. Are you able to post a copy of the relevant pages?
Risposta accettata
Wayne King
il 21 Nov 2011
Hi Yong, I don't know what to say, but an ARMA(4,3) model fitted in R with arima is not close to what the book lists either. Perhaps there is a problem with the book? Did they use MATLAB, R, none of the above?
0 Commenti
Più risposte (4)
Gurudatha Pai
il 22 Nov 2011
Assuming that the there is no typo in the data (I dint check), I tried your code through system id GUI of Matlab system id toolbox. It give the same values as you have typed here (obviously, right!) for 'zero' initialization and gives (very) different answers for other initializations. So, I am not sure about the books answer! I am not familiar with this book. I am just wondering what the "initial values" are supposed to be! Is this something on any of the recursive system id algorithms which are using initial values of the parameters? I don't see that you have used these initial values anywhere. You might want to use these initial values and see what happens to the result. I don't know if system id toolbox allows you to use non-zero initial values.
If I get a chance, I will play more with your data and may be I will have a better answer later! Finally, I would take it for granted that any computation tool (Matlab, R or Fortan for that matter) should give visibly same answer (unless some numerical or theoretical exception is happening).
0 Commenti
Vibhav Gaur
il 24 Apr 2023
The book provides a hint as to why they may be different (at the end of the sunspot model fitting example (page 169 in my book):
"It should be noted ... that the actual results obtained by the computer programs listed at the end may be somewhat different due to a variety of reasons such as subjective choices of convergence tolerances, initial values, method of computing 's etc."
I wonder if checking their FORTRAN code at the end and making sure all settings are the same will result in estimation of similar (if not the same) parameters.
0 Commenti
Vedere anche
Categorie
Scopri di più su Time Series Analysis in Help Center e File Exchange
Prodotti
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!