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
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-ph
2 matches
Mail list logo
