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Stephan Kolassa
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Time series (yearly figures) below and I use R to fit it withan ARIMA model to a time series (yearly granularity):

library(forecast)

beer <- c(150,241,361,403,504,684,706,862,879,806,840,846,1024,1196,1239,1237,1281,1342)

ts_beer = ts(beer, start = c(1980), frequency = 1)

dif.ts_beer <- diff(ts_beer)

acf(dif.ts_beer)
pacf(dif.ts_beer)

enter image description here

Based on the ACF and PACF, I make itfit an ARIMA(4,0,4) model.

dif.ts_beer.fit <- arima(dif.Gas, order = c(4,0,4))

dif.ts_beer.fit

It looks okOK. But whenthen I run auto.arimaauto.arima:

auto.arima(dif.ts_beer)

It gives:

Series: dif.ts_beer 
ARIMA(0,0,0) with non-zero mean 

Coefficients:
         mean
      70.1176
s.e.  17.0359

sigma^2 estimated as 5242:  log likelihood=-96.4
AIC=196.81   AICc=197.67   BIC=198.48

So the manual ARIMA(4,0,4) is not a good choice for this case? If so, what parameters shall be given to the ARIMA(p,d,q) model should I use?

Thank you.

Time series (yearly figures) below and I use R to fit it with ARIMA

library(forecast)

beer <- c(150,241,361,403,504,684,706,862,879,806,840,846,1024,1196,1239,1237,1281,1342)

ts_beer = ts(beer, start = c(1980), frequency = 1)

dif.ts_beer <- diff(ts_beer)

acf(dif.ts_beer)
pacf(dif.ts_beer)

enter image description here

Based on the ACF and PACF, I make it ARIMA(4,0,4)

dif.ts_beer.fit <- arima(dif.Gas, order = c(4,0,4))

dif.ts_beer.fit

It looks ok. But when I run auto.arima:

auto.arima(dif.ts_beer)

It gives:

Series: dif.ts_beer 
ARIMA(0,0,0) with non-zero mean 

Coefficients:
         mean
      70.1176
s.e.  17.0359

sigma^2 estimated as 5242:  log likelihood=-96.4
AIC=196.81   AICc=197.67   BIC=198.48

So the manual ARIMA(4,0,4) is not a good choice for this case? If so, what parameters shall be given to the ARIMA()?

Thank you.

I use R to fit an ARIMA model to a time series (yearly granularity):

library(forecast)

beer <- c(150,241,361,403,504,684,706,862,879,806,840,846,1024,1196,1239,1237,1281,1342)

ts_beer = ts(beer, start = c(1980), frequency = 1)

dif.ts_beer <- diff(ts_beer)

acf(dif.ts_beer)
pacf(dif.ts_beer)

enter image description here

Based on the ACF and PACF, I fit an ARIMA(4,0,4) model.

dif.ts_beer.fit <- arima(dif.Gas, order = c(4,0,4))

dif.ts_beer.fit

It looks OK. But then I run auto.arima:

auto.arima(dif.ts_beer)

It gives:

Series: dif.ts_beer 
ARIMA(0,0,0) with non-zero mean 

Coefficients:
         mean
      70.1176
s.e.  17.0359

sigma^2 estimated as 5242:  log likelihood=-96.4
AIC=196.81   AICc=197.67   BIC=198.48

So the manual ARIMA(4,0,4) is not a good choice for this case? If so, what ARIMA(p,d,q) model should I use?

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Richard Hardy
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Time series for Selecting ARIMA orders based on ACF-PACF vs. auto.arima

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Mark K
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Time series for ARIMA

Time series (yearly figures) below and I use R to fit it with ARIMA

library(forecast)

beer <- c(150,241,361,403,504,684,706,862,879,806,840,846,1024,1196,1239,1237,1281,1342)

ts_beer = ts(beer, start = c(1980), frequency = 1)

dif.ts_beer <- diff(ts_beer)

acf(dif.ts_beer)
pacf(dif.ts_beer)

enter image description here

Based on the ACF and PACF, I make it ARIMA(4,0,4)

dif.ts_beer.fit <- arima(dif.Gas, order = c(4,0,4))

dif.ts_beer.fit

It looks ok. But when I run auto.arima:

auto.arima(dif.ts_beer)

It gives:

Series: dif.ts_beer 
ARIMA(0,0,0) with non-zero mean 

Coefficients:
         mean
      70.1176
s.e.  17.0359

sigma^2 estimated as 5242:  log likelihood=-96.4
AIC=196.81   AICc=197.67   BIC=198.48

So the manual ARIMA(4,0,4) is not a good choice for this case? If so, what parameters shall be given to the ARIMA()?

Thank you.