Solved – Which model should I prefer for time series forecasting

rtime series

I have time series as

0.4385487 0.7024281 0.9381081 0.8235792 0.7779642 1.1670665 1.1958634 1.1958634 0.8235792 0.8530141 0.8802216 1.1958634 1.1235897 1.3542734 1.3245534 0.9381081 1.1670665 1.1958634 0.8802216 1.3542734 1.1670665 4.9167998 0.9651803 0.8221709 1.1070461 1.2006974 1.3542734 0.9651803 0.9381081 0.9651803 0.8854192 1.3245534 1.1235897 1.2006974 1.1958634 0.4385487 1.3245534 4.9167998 1.2277843 0.8530141 1.0018480 0.3588158 0.8530141 0.8867365 1.3542734 1.1958634 1.1958634 0.9651803 0.8802216 0.8235792 4.9167998 1.1958634 0.9651803 0.8854192 0.8854192 1.2006974 0.8867365 0.9381081 0.8235792 0.9651803 0.4385487 0.9936722 0.8821301 1.3542734 1.1235897 1.6132899 1.3245534 1.3542734 0.8132233 0.8530141 1.1958634 1.2279813 0.8354292 1.3578511 1.1070461 0.8530141 0.9670581 1.1958634 0.7779642 1.2006974 1.1958634 0.8235792 1.3245534 0.5119648 2.3386331 0.8890464 0.8867365 4.9167998 1.2006974 1.2006974 0.6715839 4.9167998 0.7747481 4.9167998 0.8867365 1.2277843 0.8890464 1.2277843 0.8890464 1.0541099 0.8821301

I am using package "itsmr"-autofit(),"forecast"-auto.arima(),"package"–functions

Autoregressive model

> ar(t)

Call:
    ar(x = t)

    Order selected 0  sigma^2 estimated as  0.9222

ARMA model

> autofit(t)
    $phi
    [1] 0

    $theta
    [1] 0

    $sigma2
    [1] 0.9130698

    $aicc
    [1] 279.4807

    $se.phi
    [1] 0

    $se.theta
    [1] 0

ARIMA model

    > auto.arima(t)
    Series: t 
    ARIMA(0,0,0) with non-zero mean 

    Coefficients:
          intercept
             1.2623
    s.e.     0.0951

    sigma^2 estimated as 0.9131:  log likelihood=-138.72
    AIC=281.44   AICc=281.56   BIC=286.67

The auto.arima function automatically differences time series: we don't have to worry about transformation.

> auto.arima(AirPassengers)
Series: AirPassengers 
ARIMA(0,1,1)(0,1,0)[12]                    

Coefficients:
          ma1
      -0.3184
s.e.   0.0877

sigma^2 estimated as 137.3:  log likelihood=-508.32
AIC=1020.64   AICc=1020.73   BIC=1026.39`

Which model should I select to get p,q values & for forecasting purpose?

Best Answer

The Actual , Fit and Forecast enter image description here suggest a Mean Model where the forecast would be based upon the following equation . Note that there are 12 anomalous data points yielding a robust mean of 1.0378 . The visual definitely supports the unusual data points. Good time series analysis detecting the underlying signal ( a mean model ) while also detecting any exceptional data points rendering a cleansed/robust mean of 1.0378 as compared to a simple mean of 1.26 . Now that the anomalous points have been identified , one needs to ask what they have in common as a possible explanatory variable. Additionally the ACF of the errors from this model indicate randomness. Furthermore there is no evidence that the expected value is systematic with the error variance or error standard deviation suggesting that a power transform is not warranted. Finally there is no evidence of a structural shift in the robust mean over time suggesting that the parameter(s) of the model are invariant.

Related Solutions

Solved – time series forecasting using auto.arima and exponential smoothing

Seasonality is probably not very strong. Different algorithms will give different results, unless seasonality is glaringly obvious.
The best measure is always to compare forecast accuracy on a holdout set: hold back the last $n$ observations, fit your models to all other observations, forecast into the last $n$ time periods with both models, then compare forecast accuracy using your error measure of choice (see 5 below).
Yes, this is a common complaint. I don't think there is an easy way to get the in-sample fit. But you can get the residuals: auto.arima(WWWusage)$residuals. Best to look into the code of auto.arima() to see whether you need to add or subtract them from the original series to get the fit. I'd say you have to subtract ("actuals=model+residuals"), but better check.
I recommend a good forecasting textbook. This is a very good start. Otherwise, read through the help pages.
The appropriate error measure will depend on your personal loss function. Is your pain symmetric, and will it increase more strongly with larger errors? Then use MSE. Is your pain proportional to absolute errors? Then use MAE. Best to look at multiple error measures.

One tip: averaging forecasts will usually improve accuracy. Consider taking the average of your two models' forecasts per future time bucket.
auto.arima() apparently fits no drift, even if you allow it.

Solved – When is the AIC a good model selection criterion for forecasting and when is it not

I am not completely satisfied with my answer, but here goes.

To an extent, you are comparing apples and oranges. Your two calls to Arima() use method="ML", whereas your auto.arima() uses the default, which is method="CSS-ML". Then again, refitting everything with the default does not make a real difference.
Minimizing the AIC is asymptotically equivalent to minimizing the one-step ahead squared prediction error. (I don't have a reference at hand, sorry.) Note that this is an asymptotic result in a suitable statistical sense. It's quite possible for a handpicked model to outperform AIC on a limited length time series. And on a single one, at that.
Finally, as you write in a comment, the AirPassengers dataset exhibits strong multiplicative seasonality. ARIMA does not model multiplicative seasonality or trend; it can only deal with additive effects. Your overparameterized model gets the multiplicative trend and seasonality right, but it may also forecast this in a series that does not exhibit such effects. There are reasons why such large models are typically not considered.

To model multiplicative effects, allow auto.arima() to use Box-Cox transformations:
```
> (foo <- auto.arima(train,lambda="auto"))
Series: train 
ARIMA(0,1,1)(0,1,1)[12] 
Box Cox transformation: lambda= -0.3096628 

Coefficients:
          ma1     sma1
      -0.3936  -0.5713
s.e.   0.1035   0.0863

> accuracy(forecast(foo,h=24,biasadj=TRUE),test)
                     ME      RMSE       MAE        MPE     MAPE      MASE       ACF1 Theil's U
Training set -0.7186038  8.915531  6.691014 -0.2079082 2.753580 0.2341638 0.04889565        NA
Test set     28.5600533 31.711896 28.884516  6.2710488 6.348486 1.0108644 0.17279165 0.6372069
```
I cut out the AIC, because that is not comparable to the AIC on nontransformed data. Note that we end up much closer to your large model in terms of the test RMSE, but the model is much more interpretable, and I personally would trust it a lot more than an ARIMA(15,1,15)(4,1,4)[12] one. Incidentally, searching through more possible ARIMA models yields the exact same model:
```
> (bar <- auto.arima(train,max.p=15,max.q=15,max.P=4,max.Q=4,
+ lambda="auto",stepwise=FALSE,approximation=FALSE))
Series: train 
ARIMA(0,1,1)(0,1,1)[12] 
Box Cox transformation: lambda= -0.3096628 

Coefficients:
          ma1     sma1
      -0.3936  -0.5713
s.e.   0.1035   0.0863
```

Best Answer

Related Solutions

Solved – time series forecasting using auto.arima and exponential smoothing

Solved – When is the AIC a good model selection criterion for forecasting and when is it not

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