Which one of these looks stationary?

Question

Step 1. To answer "Final Question"

( linked: "https://stats.stackexchange.com/questions/79246/the-final-question-order-of-differencing-to-achieve-stationary-and-interpreta")

Expecting to find correct order of difference to achieve stationary.

Some Result of test for stationary on monthly times series only.

Dickey fuller
KPSS test
Phillips–Perron test

If required i can provide test result for the any order difference.

  > urdfTest(Monthly, lags=1, type = c("nc","c","ct"), doplot = TRUE)

Title:
 Augmented Dickey-Fuller Unit Root Test

Test Results:

  Test regression none 

  Call:
  lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)

  Residuals:
      Min      1Q  Median      3Q     Max 
  -526683 -135297   24570  176578 1078796 

  Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
  z.lag.1    -0.08703    0.09829  -0.885 0.383219    
  z.diff.lag -0.61315    0.15711  -3.903 0.000521 ***
  ---
  Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

  Residual standard error: 329300 on 29 degrees of freedom
  Multiple R-squared:  0.4195,  Adjusted R-squared:  0.3794 
  F-statistic: 10.48 on 2 and 29 DF,  p-value: 0.0003761


  Value of test-statistic is: -0.8854 

  Critical values for test statistics: 
        1pct  5pct 10pct
  tau1 -2.62 -1.95 -1.61  


     > urkpssTest(Monthly,  type = c("mu","tau"), lag = c("short","long","nil"), use.lag = NULL, doplot = TRUE)

    Title:
     KPSS Unit Root Test

    Test Results:

      Test is of type: mu with 3 lags. 

      Value of test-statistic is: 0.0986 

      Critical value for a significance level of: 
                      10pct  5pct 2.5pct  1pct
      critical values 0.347 0.463  0.574 0.739



    > urppTest(Monthly,  type = c("Z-alpha","Z-tau"), model = c("constant","trend"), lags = c("short","long"),use.lag = NULL, doplot = TRUE)

    Title:
     Phillips-Perron Unit Root Test

    Test Results:

      Test regression with intercept 

      Residuals:
          Min      1Q  Median      3Q     Max 
      -397594 -251734   26252  144863  954417 

      Coefficients:
                    Estimate Std. Error t value Pr(>|t|)    
      (Intercept)  6.169e+05  1.183e+05   5.215 1.27e-05 ***
      y.l1        -8.733e-02  1.857e-01  -0.470    0.641    
      ---
      Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

      Residual standard error: 290200 on 30 degrees of freedom
      Multiple R-squared:  0.007322,    Adjusted R-squared:  -0.02577 
      F-statistic: 0.2213 on 1 and 30 DF,  p-value: 0.6415


      Value of test-statistic, type: Z-alpha  is: -39.5851 

               aux. Z statistics
      Z-tau-mu            5.2378

enter image description here

No sir, this one is of my personal interest. I never had studied time series before. — New Bie, Dec 15 '13 at 11:20

score 1 · Answer 1 · edited Apr 13 '17 at 12:44

1

All three series look stationary; the second & third look over-differenced (the strong negative auto-correlation at the first lag is a tell-tale sign). Recall that the sign of a non-stationary series is strong auto-correlations that take a long time to die out with increasing lag. Plotting residuals is also a good idea—say a quantile–quantile plot & a time series plot of standardized residuals—to check their distribution & spot outliers.

There are no clear signs of auto-correlation in your original time series, suggesting you needn't fit an ARMA model unless you have good reason to. That's not to say there is no auto-correlation, just that if there is it's weak to moderate & its nature can't be determined from such a short series (36 observations?). A white noise model about the mean might be appropriate, perhaps a Gaussian or perhaps a t-distribution with 5–10 degrees of freedom.

edited Apr 13 '17 at 12:44

Community

1

answered Dec 11 '13 at 09:36

Scortchi - Reinstate Monica

27,560
8
81
248

Yes: in this case some form of white noise looks to be a good candidate model. I don't know what you mean by an exponential model in this context. – Scortchi - Reinstate Monica Dec 11 '13 at 18:58
To have a residual plot i should be fitting a model first since my final objective is to create a relevant model. and since it is asked for i had fitted exponential smoothing via `Monthlyforcast – New Bie Dec 11 '13 at 19:00
How can i post the image of same ? – New Bie Dec 11 '13 at 19:04
what i know is an MA model is white noise model, please correct me i am wrong. so, should i try MA, required some more assistance in the content of same – New Bie Dec 11 '13 at 19:15
I meant just white noise about a constant mean: there's not really much sign of auto-correlation from your ACF plots, but the time-series looks perhaps a little spiky. – Scortchi - Reinstate Monica Dec 11 '13 at 19:16

score 1 · Answer 2 · answered Dec 11 '13 at 18:37

Non-stationarity is a symptom with possibly many causes. One cause is a shift in the mean at one or more points in time. Another possible cause is a change in parameters at one or more points in time. Another cause is a deterministic change in error variance at one or more points in time. Prof. Spyros Makridakis wrote an article http://www.insead.edu/facultyresearch/research/doc.cfm?did=46900 of the danger of using differencing to render a series stationary. While I don't fully agree with all of his propositions it certainly details the error of assuming differencing as the remedy for non-stationarity.

I will go about this doc and try to fetch as many information as i can. — New Bie, Dec 11 '13 at 19:10

Which one of these looks stationary?

2 Answers2