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From the first plot, I have determined that there is a seasonal pattern of period $L = 12$. However, in my ACF plot (a), it appears that the period is $L = 6$.

Am I misinterpreting?

The data had a cubic trend, so I performed a third order difference in the ACF plot (b). The series is clearly non-stationary, but am I able to determine the a seasonal pattern?

I performed a Lag 12 difference of the third difference in the ACF plot (c), which tends to die down fairly quickly, so I'm assuming is stationary?

The ACF plot (d) illustrates the Lag 12 difference of the original data with differences = 3. However, given the sinusoidal pattern tells me the series is not stationary contradicting ACF Plot (c).

Am I understanding the plots correctly, or am I way off base. I included the PACF plots for reference.

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    $\begingroup$ Differencing is only warranted when series have unit roots. Deterministic trends do not warrant differencing. What you end up with is introducing unit roots in the moving average part of the model and increasing the error variance. This is known as overdifferencing and should be avoided. $\endgroup$ Commented Mar 16, 2021 at 13:27

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The peaks in the ACF plot (a) are at 1.0 and 2.0 periods, so 1 and 2 years. Six months correspond to the troughs at 0.5. Yes, the autocorrelations at a lag of six month are still positive, because there is a lot of autocorrelation in your data, but they are at a minimum. So I would say that your ACF confirms the 12 month seasonality that is obvious in your time series plot (and that we know should be there based on domain knowledge).

For visual inspection of seasonality, I would recomment seasonal plots and seasonal subseries plots. For ARIMA modeling, I would rather trust an information criteria based approach like the one implemented in forecast::auto.arima() than one based on interpreting (P)ACF plots - which is a little too much like reading animal entrails to me. The seasonality tests in auto.arima() should have no problems in picking up your seasonality here.

And I would be very careful about seeing a cubic trend and therefore taking third differences. The long-term time course may simply be autoregression on a larger scale, or (more likely) driven by macroeconomic variables that may or may not continue in certain ways. Reversing three differencings may give you very unstable forecasts, and remember that a third order polynomial will explode upwards or downwards sooner or later. Much better to trust the integration tests in auto.arima().

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  • $\begingroup$ Understood. I have fitted my initial plot, and have found a cubic model to be highly significant. So in the case of differencing, should I not be using a third order difference? Use a lower order difference? Not use differencing at all? $\endgroup$ Commented Mar 16, 2021 at 10:36
  • $\begingroup$ Significance only tells you about in-sample fit. If that is all you want, go ahead, but usually people want to forecast their time series, and then in-sample fit gets misleading very quickly. I would really recommend relying on auto.arima() as to differencing - it's very hard to beat. $\endgroup$ Commented Mar 16, 2021 at 10:43
  • $\begingroup$ I have completed this task, however It produces a notation I have not seen before. What is meant by ARIMA(0,1,2)(0,1,1)[12]? If this should be asked in the R section of stackoverflow, please let me know. $\endgroup$ Commented Mar 16, 2021 at 13:15
  • $\begingroup$ I think the answer should mention the glaring problem of overdifferencing, i.e. differencing data that does not contain unit roots but contains deterministic trends. You have addressed differencing but not in this way. $\endgroup$ Commented Mar 16, 2021 at 13:29
  • $\begingroup$ Here is the answer to my previous question: otexts.com/fpp2/seasonal-arima.html $\endgroup$ Commented Mar 16, 2021 at 13:34

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