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.
