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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?

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  • $\begingroup$ It is striking that the majority of the values in this time series occur multiple times (one as many as 11 times). How were these values measured? $\endgroup$ – whuber May 18 '12 at 20:28
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The Actual , Fit and Forecast enter image description here suggest a Mean Model where the forecast would be based upon the following equation enter image description here . 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.

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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.

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    $\begingroup$ Thank you for your acknowledging AUTOBOX . Your expert opinion means a lot to me. I will take his data and provide him with a step-by-step answer. $\endgroup$ – IrishStat May 18 '12 at 16:43
  • $\begingroup$ i don't know what is "Autobox",i search on net & take first hit http://www.autobox.com/cms/ but it gets error like "Database Error: Unable to connect to the database:Could not connect to MySQL" i want p(autoregressive order) & q(moving avg. order) value of above models for further use(for making distance matrix)in my project is it like that,if above model not fit,then it is not time series data, & how can i evaluates that given sequence of numbers is time series or not ? $\endgroup$ – Sagar Nikam May 19 '12 at 6:31
  • $\begingroup$ @Sagar I was able to find autobox.com/cms so I don't understand the error that you received . Please try again and if it continues contact sales@autobox.com Since the final model was a p=0 ; q=0 and no differencing and no need for seasonal pulses the conclusion would be that the data has an invariant mean and variance and is free of time series structure . One could conclude that this is not time series data. $\endgroup$ – IrishStat May 20 '12 at 15:30

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