According to arima(p,0,q) model if we have n data and our total parameter is p+q then it is said that degree of freedom is n-(p+q). Could you mathematically demonstrate it? No sufficient information on internet and books.



The degrees of freedom in a model is the number of obervations minus the number of parameters to be estimated. So, there is nothing to be proven because it's a definition.

  • $\begingroup$ But this situation can be explained mathematically in multilinear regression model using idempotent matrix. Are you sure that there is no proof that ARIMA(p,0,q) model has n-p-q degree of freedom? $\endgroup$ – mertcan Apr 1 '19 at 10:09
  • $\begingroup$ I do not only want to memorize arıma model degree of freedom is n-p-q. Could you help me understand via mathematical proof? $\endgroup$ – mertcan Apr 1 '19 at 21:40
  • $\begingroup$ Hi:IIRC, the idempotentcy proof in multiple regression proves the unbiasedness of the estimate of sigma squared in multiple regression. But that's still defining the dof as number of samples - number of parameters estimated. In the arima case, the degrees of freedom is not used to prove unbiasedness so I'm not clear on what you're asking. $\endgroup$ – mlofton Apr 2 '19 at 14:05
  • $\begingroup$ I believe there must be analytical proof that ARIMA(P,0,Q) has n-p-q degree of freedom. Otherwise it has no meaning actually. I just would like to proofs. Could you help me? $\endgroup$ – mertcan Apr 2 '19 at 15:09
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    $\begingroup$ I do not think this is the definition of degrees of freedom. The notion of degrees of freedom is more complicated than that. There are models/techniques where this does not hold, e.g. k nearest neighbours (see James et al. "Introduction to Statistical Learning" and/or nearest neighbors degrees of freedom). If I remember correctly, ridge regression would be another example. @mertcan $\endgroup$ – Richard Hardy Jun 15 '19 at 6:41

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