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We know that ARIMA(p,1,q) can be generated as ct=yt-y(t-1) where ct is ARMA and yt is ARIMA while yt-1 is lag of ARIMA?I want to generate ARIMA from above equation which will be equal to yt=ct/(1-B) where B is lag operator.

How to generate ARIMA in R from above equation?

Your input will be appreciated.

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1 Answer 1

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I think this is already answered here.

To summarize, depending on how many times your series has to be differentiated to become stationary (in your case that should be 1), you can generate your ARIMA with:

library(forecast)
model <- Arima(timeseries,order=c(p-1,1,q))

to doublecheck you can compare

model_ARMA <- Arima(timeseries,order=c(p,0,q))
model_ARIMA <- Arima(timeseries,order=c(p-1,1,q))

both should yield same results.
Maybe as a more generalized form:

model_ARMA <- Arima(timeseries,order=c(p,0,q))
model_ARIMA <- Arima(timeseries,order=c(p-n,n,q)) 

I hope I understood your question correctly.
best regards

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  • $\begingroup$ I wanted to know Using parameter μ,φ1,φ2,θ1, and θ2 . Generate three time series for the ARIMA(2,2,2) for each of the following values of σa: 0.8, 0.1, and 0.01. I attempted the question like this: #Check whether AR root are stationary rootsar=polyroot(c(1,0.15,-0.20)) Mod(rootar) #Check whether MA root are invertibe rootsma=polyroot(c(1,0.05,-0.25)) Mod(rootsma) #Generating ARMA for different value of sigma(assuming n=1008) arma=arima.sim(n=1004,list(ar=c(0.15,-0.20),ma=c(0.05,-0.25)),mean=1,s=0.8) Using above equation could how I generate ARIMA(2,2,2)? $\endgroup$
    – Wolf Gupta
    Aug 10, 2021 at 17:51

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