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How to decide the stationary about time series.

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Hong Zhang
Hong Zhang 2018 年 4 月 17 日
コメント済み: Hong Zhang 2020 年 5 月 3 日
1)When I use the matlab build-in function adftest ,the results is >> [H p]=adftest(x); H = 0 p = 0.2677 This mean that the data is not stationary.
2)if I do 1 oder difference the results is [H p]=adftest(diff(TVSQ(:,1),2)) H = 1 p = 1.0000e-03 This mean that the data is stationary.
3)But when I use the software tools 'eviews', the result shows that the data is stationary. The results is as follows:
I am very confuse how to decide the stationarity.
my row data: [71.2955 60.3591 50.2819 43.5664 41.1082 44.0974 45.8533 48.3995 48.2788 53.7030 58.8999 61.2222 61.5451 61.0036 61.5284 62.5502 62.9354 63.6123 63.9352 64.9714 63.9818 65.0413 64.5228 47.2740 40.6282 37.6279 35.2846 34.2732 34.0366 35.8862 46.3137 50.7998 52.0396 54.9744 56.9747 56.6138 57.7891 58.0364 57.7064 56.3850 54.4739 53.1415 53.4890 53.2135 53.0051 54.0178 56.6830 57.5005 58.1247 59.3933 60.2135 60.8124 61.8094 63.0197 62.4239 49.7056 44.7863 41.2664 39.0328 36.6799 35.8447 34.9218 34.0544 33.9111 32.8596 32.5181 32.0295 42.8910 52.3912 53.1898 54.2287 54.4167 54.8854 56.1082 55.9900 56.0964 56.2029 55.6622 55.3722 55.4419 55.9324 56.3929 56.6637 57.0997 57.3540 58.1640 58.5248 57.2931 55.4843 55.7493 55.6987 54.8490 54.8738 53.3408 51.6901 52.4476 56.0069 57.8444 58.3679 59.5909 60.6797 55.2465 49.0235 48.1973 47.3786 45.5947 45.3466 46.1310 46.2881 46.4721 46.3603 47.0461 47.7488 47.4928 47.5015 46.9833 47.9807 50.0121 53.1143 54.9539 55.9725 56.2395 56.4108 56.6412 57.8002 58.4386 57.8929 57.4919 57.8076 57.8022 58.0150 58.0724 58.2363 60.2029 61.7465 61.7847 61.8456 62.2273 61.9209 60.1059 58.4930 57.2365 55.3890 53.7136 54.0211 52.9078 52.0987 51.6844 50.9243 50.0750 49.6969 47.2838 45.7404 44.3817 44.0440 43.2900 43.4942 43.4829 43.6377 42.9912 43.1449 43.3120 50.5656 53.9648 56.6141 58.5364 59.5783 55.4172 40.0105 34.7344 32.2913 31.6840 31.9129 32.6177 32.7513 32.2156 32.4589 33.7396 35.0430 36.1355 36.2683 36.0295 35.9575 38.0034 42.2240 48.6775 53.5205 55.5958 57.2173 55.8627 54.5309 54.6592 54.9133 55.6472 56.1204 56.1610 56.5122 55.1607 39.6605 33.6982 31.4149 30.6356 30.2366 29.6821 29.6150 29.4808 33.7856 36.7863 37.9275 39.5438 39.6645 36.8690 37.5553 47.3699 52.7951 55.3160 55.4035 48.1759 40.5816 37.7998 36.8194 36.1271 35.1887 34.1268 35.0291 35.1119 35.8907 39.9687 41.2146 43.1077 43.2588 41.6554 42.1174 43.5568 49.5629 55.4203 58.4069 61.0009 61.3185 62.2714 63.1061 63.8822 64.1916 62.7986 45.7381 37.6424 35.2271 34.4352 33.6902 33.3955 47.3577 55.0197 56.8228 58.0238 58.7672 59.1700 59.0949 59.4507 59.7406 59.0563 57.2616 55.1346 54.6623 54.7818 55.8213 57.0512 59.0010 57.9271 56.8775 56.5069 56.6923 57.1512 56.8557 57.7096 57.8691 59.1573 59.2433 59.8385 57.7784 53.1444 51.4961 51.1329 50.1868 49.5523 45.6965 44.5725 44.5524 45.0179 44.6705 43.5227 42.8591 41.6612 40.1141 37.8156 36.7472 35.6783 35.2476 34.0893 39.2460 47.8766 ];
Could somebody encouter the same problem and give me some intruction.Thanks a lot.

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採用された回答

Hristo Zhivomirov
Hristo Zhivomirov 2020 年 4 月 29 日
編集済み: Hristo Zhivomirov 2020 年 4 月 29 日
Hi Hong,
If we consider your data as a time series i.e., a time domain signal, my test clearly indicates that the signal is non-stationary. One also could check that by visual inspection of the oscillogram of the signal - there is some linear trend and clearly some heteroscedasity. Please, try my functon Stationarity Estimation of a Signal with Matlab in order to ensure yourself.
All best,
Hristo Zhivomirov

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Hong Zhang
Hong Zhang 2020 年 5 月 3 日
Dear Hristo,
Thanks for your concern. I will use your method if I need it.
Thanks.
Best,
Hong

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その他の回答 (1 件)

Hang Qian
Hang Qian 2018 年 4 月 17 日
Hi Hong,
As the name suggests, ADF is Dickey-Fuller test augmented by lagged regressors. The default value of MATLAB function 'adftest' is zero lag, so we may want to add more lags. Also, different models have an impact on the test results, I would use 'ARD' in this case.
[h,pValue,stat]=adftest(x,'lags',1,'model','ARD')
MATLAB gives sensible results, comparable to those given by Eviews:
h = 1
pValue = 1.0000e-03
stat = -5.8455
Hope that helps,
Regards,
Hang Qian

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Hong Zhang
Hong Zhang 2018 年 4 月 18 日
Dear Hang,
Thank you for your kindly reply. I am confused how can I choose the right model,such as AR,ARD and TS in 'adftest'. In other words, I cannot make it clear that wethear the data has a intercept,trend and intercept or none.
Best Wishes,
Hong

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