Which model should I prefer for time series forecasting?

Question

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

1. Autoregressive model

   > ar(t)
   
   Call:
       ar(x = t)
   
       Order selected 0  sigma^2 estimated as  0.9222 
   

2. 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
   

3. 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?

Answer 1

The Actual , Fit and Forecast 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.

Answer 2

I am tempted to say use autobox, it will pick a model for you and may find a better one than the ones you are looking at. I am sure IrishStat would be thinking that and be glad I said it. But to answer your question: Your series is relatively long with around 100 observations. All the fit statistics suggest that the first model is much better than the second. The first model appears to be a constant. if there is any sort of pattern in the data it would be easier to see it in a plot of the data rather than a table of numbers. If a plot shows that it seems to randomly vary around 1.26 and the acf looks likeit should for a white noise series the first model may be the best. Otherwise look for a different alternative than the second model which is clearly a poor model. Now using the first model the best forecast is the current mean which is the intercept value 1.2623. Not a very satisfactory result if you wanted to take advantage of the power of time series analysis. It really says that the recent past is no moreimportant in forecasting the next time point than the distant past and new observations only help in refining your estimate of the mean.