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
