Hi Matt, The ARMA model write the MA part with positive coefficients, therefore the formula that you write for the variance is wrong. Here an example where the answer is the same for both methods.
phi=0.75 theta=-0.4 coefvar=(1+theta^2+2*phi*theta)/(1-phi^2) coefvar result <- ARMAtoMA(ar=c(0.75), ma=(-0.4), lag.max=40) v=sum(result^2)+1 v HTH Marlene 2009/11/11 Pietrzykowski, Matthew (GE, Research) <pietr...@research.ge.com> > Hello R Users! > > I have a question about the output of ARMAtoMA when used to calculate > the variance of > a model. I have a mixed model of the form ARMA(1,1). The actual > model takes the form: > > X(t) = 0.75X(t-12) + a(t) - 0.4a(t-1) > > Given that gamma(0) takes the form [(1 + theta^2 - > 2*theta*phi)/(1-phi^2)]*sigma(a), I would > expect a process variance of 4.02*sigma(a) when I substitute 0.75 for > phi and -0.4 for theta. > > When I run ARMAtoMA, > > result <- ARMAtoMA(ar=c(0.75), ma=(-0.4), lag.max=40) > sum(result^2)+1 > > I get 1.28. If I input 0.4 instead of -0.4 in ARMAtoMA I get the > result I expected. Is there a sign > dependence in the R function I am overlooking? > > Thanks in advance. > > Matt > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
