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I wanted to know given generated data from an ARMA(p,q) model, can I create data from an ARIMA (p,1,q) model?I tried in the web but unable to find anything .

For the ARMA parameters, choose μ,φ1,φ2,θ1, and θ2 so that the model is stationary and invertible. Based on above parameter 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)

But above code will ARMA process data but I want to generate ARIMA(2,2,2).Also how to check if mean(mu) is stationary and invertible.Like I use polyroot function to check for stationary/invertible

My apologies for long post.Would appreciate any input. Use R for explanation.

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1 Answer 1

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Suppose $X_t$ is an ARMA$(p,q)$ that I can simulate using arima.sim function.

You can create a process $Y_t$ which is ARIMA$(p,1,q)$ using this steps :

1- $Y_1=X_1$

2- for $t=1,\dots,n$, $Y_t = X_t+Y_{t+1}$

You can repeat this process to simulate an ARIMA$(p,2,q)$ base on $Y_t$ instead of $X_t$.

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  • $\begingroup$ Thank you but could you provide detailed explanation or point me to a resource as I am unable to understand above explanation.My apologies for inconvenience.How did you arrive at 1-Y1=X1?Your detailed response will be helpful in understanding the concept. $\endgroup$
    – Wolf Gupta
    Aug 5, 2021 at 23:50
  • $\begingroup$ As you just want a sample, the first value is not really important. If fact, I suppose that $Y_0=0$, which is the mean. Then, as $Y_t-Y_{t-1}$ should be an ARMA$(p,q)$, we can set $X_t= Y_t-Y_{t-1}$ and compute it recursively $\endgroup$ Aug 6, 2021 at 0:00
  • $\begingroup$ @Abdoul Haki : I think the expression in the second step should be $Y_{t} = X_{t} + Y_{t-1}$ $\endgroup$
    – mlofton
    Oct 17, 2023 at 3:10

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