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Topic: Predicting ARMA model: where does the delay come from?
Replies: 3   Last Post: Nov 9, 2011 7:24 AM

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Tikkuhirvi Tietavainen

Posts: 99
Registered: 4/22/08
Re: Predicting ARMA model: where does the delay come from?
Posted: Nov 9, 2011 7:24 AM
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"Aino" <aino.tietavainen@removeThis.helsinki.fi> wrote in message <j90j6b$l0v$1@newscl01ah.mathworks.com>...
> Rune Allnor <allnor@tele.ntnu.no> wrote in message <6234421e-b5bd-4ecd-b9b3-b9f0b9638587@h34g2000yqd.googlegroups.com>...
> > On 4 Nov, 12:00, "Aino" <aino.tietavai...@removeThis.helsinki.fi>
> > wrote:

> > > Hello all!
> > >
> > > I am trying to model my time series with ARMA model and then predict the signal with 'predict'. Here's an example:
> > >
> > > close all;clear all;clc;
> > > k=25;%prediction step
> > > mAR=10;%AR order
> > > mMA=10;%MA order
> > >
> > > %First signal:
> > > y=smooth(rand(1500,1),100);
> > > y=y(250:end-251);
> > > y=y-mean(y);
> > > data=iddata(y,[],0.01);
> > > m=armax(data,[mAR,mMA]);
> > > yp=predict(m,data,k);
> > > figure;plot(y);hold on;plot(yp.OutputData,'r')
> > >
> > > %Second signal:
> > > y=sin(1:0.01:10)';
> > > data=iddata(y,[],0.01);
> > > m=armax(data,[mAR,mMA]);
> > > yp=predict(m,data,k);
> > > figure;plot(y);hold on;plot(yp.OutputData,'r')
> > >
> > > The prediction signal (yp.OutputData) seems to have no delay with the second signal (sine wave), but compared to the first signal it is delayed about the prediction step. The signals I use look more like the first one, so this certainly will affect my prediction error, something like sqrt(sum((data.OutputData-yp.OutputData).^2)).
> > >
> > > Where does this delay come from and how can I get rid of it? Why does it only affect the first signal?
> > >
> > > Thank you,
> > > Aino

> >
> > I don't have the toolbox or functions you use in your
> > simulation, but I see that you in the second example
> > use a noiseless sinusoidal, whereas you in the first
> > example use a random signal.
> >
> > From a prediction POV that makes all the difference:
> >
> > 1) Sinusoidals are perfectly predictable at the outset.
> > 2) There is no noise to messthings up.
> >
> > So the second example is so perfect that it is totally
> > useless for anything other than a test of concept or
> > implementation.
> >
> > The results from your first example are far more
> > representative for how the method will work with
> > random signals.
> >
> > Rune

>
> Thank you for your quick reply.
>
> The sine wave is indeed almost perfectly predicted, and obviously the noisy signal cannot be so perfectly predicted, but what I cannot understand is that why is there a _delay_ in the prediction with the noisy signal? In other words, if I move the prediction signal by 25 steps and calculate the prediction error, sqrt(sum((data.OutputData(1:end-25+1)-yp.OutputData(25:end)).^2)), it is a lot smaller smaller than without moving it, sqrt(sum((data.OutputData(25:end)-yp.OutputData(25:end)).^2)). With the sine wave this is not true.
>
> Should I for example use the 'predict' with the time-reversed signal as well and calculate the average of the two predictions or should I just calculate the prediction error as above or what should I do?
>
> ~Aino


I still haven't figured this one out, but I have a little question about the model. What exactly is the loss function that the armax function gives out? How is it calculated?

Thanks,
Aino



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