help-octave
[Top][All Lists]
Advanced

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: AR model


From: Marco Atzeri
Subject: Re: AR model
Date: Sun, 5 Jul 2020 11:56:21 +0200
User-agent: Mozilla/5.0 (Windows NT 10.0; WOW64; rv:68.0) Gecko/20100101 Thunderbird/68.10.0

please reply to the mailing list

On 04.07.2020 22:25, estefaniame@gmail.com wrote:
Thank you for your answer.
The fact is that yt=b0+b1*y^-1+b2*y^-2 is also written as (1-b1-b2)*L being L the lag operator. This is why I wrote [1 -b1 -b2].

I want to find the roots of the corresponding polynomial for that AR model. Should then I put: [1 b1 b2]? I find examples for polynomials like x^2+x+1, for instance, but I need the way to express an AR model.
Could you help me? Thank you very much.

Kind regards,

Estefanía


I am not an expert in AR model,
but taking the definition at

https://en.wikipedia.org/wiki/Autoregressive_model

it seems not a simple polynomial operation.

The TSA package as some functions
https://octave.sourceforge.io/tsa/overview.html

that is referring to AR.
You may want it, of clarify better what you want to meet.

El sáb., 4 jul. 2020 22:12, Marco Atzeri <marco.atzeri@gmail.com <mailto:marco.atzeri@gmail.com>> escribió:

    On 04.07.2020 21:37, estefaniame@gmail.com
    <mailto:estefaniame@gmail.com> wrote:
     > Good evening
     >
     > I have a doubt on polynomials. If I have an AR model like
     > Yt=b0+b1*Yt-1+b2*Yt-2+et and I want to write the corresponding
     > polynomial, would it be [1 -b1 -b2]?
     >
     > Thank you very much.
     >
     > Kind regards,
     >
     > Estefanía
     >
     >

    polynomials have positive exponents.
    You seems to use negative ones b0+b1*Yt^-1+b2*Yt^-2

    so they are not polynomial

    https://octave.org/doc/v5.2.0/Polynomial-Manipulations.html

    Marco






reply via email to

[Prev in Thread] Current Thread [Next in Thread]