Null Hypothesis: LOG_AUSTRIA has a unit root
Exogenous: Constant
Lag Length: 1 (Automatic - based on SIC, maxlag=29)
t-Statistic Prob.*
Augmented Dickey-Fuller test statistic 1.765602 0.9997
Test critical values: 1% level -3.432059
5% level -2.862181
10% level -2.567155
*MacKinnon (1996) one-sided p-values.
Augmented Dickey-Fuller Test Equation
Dependent Variable: D(LOG_AUSTRIA)
Method: Least Squares
Date: 07/17/13 Time: 18:55
Sample (adjusted): 3 3450
Included observations: 3448 after adjustments
Variable Coefficient Std. Error t-Statistic Prob.
LOG_AUSTRIA(-1) 0.000659 0.000373 1.765602 0.0776
D(LOG_AUSTRIA(-1)) 0.053321 0.017030 3.131112 0.0018
C -0.003183 0.002040 -1.560515 0.1187
R-squared 0.003923 Mean dependent var 0.000428
Adjusted R-squared 0.003345 S.D. dependent var 0.010612
S.E. of regression 0.010595 Akaike info criterion -6.256058
Sum squared resid 0.386693 Schwarz criterion -6.250711
Log likelihood 10788.44 Hannan-Quinn criter. -6.254148
F-statistic 6.784764 Durbin-Watson stat 2.000233
Prob(F-statistic) 0.001146
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$\begingroup$ Welcome to the site, @user28136. I recognize that your title is informative, but it might be nice to add some context to your question in the body, rather than just paste output. What is it that's confusing you exactly? $\endgroup$– gung - Reinstate MonicaCommented Jul 17, 2013 at 16:23
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$\begingroup$ I have not expirience from Eviews programm.I want the interpretation particularly the t-statistic and Prob for my example (Log_austria) $\endgroup$– user28136Commented Jul 17, 2013 at 16:30
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2 Answers
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For the interpretation of Eviews output, just focus on top part. The lower one shows how the Eviews runs the regression. Eviews runs the regression in first difference form, so the null is coefficient on LOG_AUSTRIA(-1) is zero which means that there is an unit root.The alternate hypothesis is that it is less than zero, i.e., there is no unit root.
Your test statistic 1.765602 is greater than critical value -2.567155 at 10%. So, you accept the null that there is an unit root (see p=0.999). To reject the null at 10 %, p<=0.10 (test statistic should be less than -2.567155), to reject the null at 5% (test statistic should be less than-2.862181) p<=0.05, and to reject the null at 1%,(test statistic should be less than-3.432059) p<=0.01.
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$\begingroup$ thank you so much,your answer helps me to organize my results.And something else i always put absolute value in t-statistic to compare or not? $\endgroup$ Commented Jul 18, 2013 at 12:11
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$\begingroup$ this is one tailed test and so you can't use the absolute value for comparison. Also, note that you are comparing your calculated t-statistics with tabulated tau (and not t) statistics. Please check the answer as correct so that it will be helpful for future
Eviewsuser like you. $\endgroup$– MetricsCommented Jul 18, 2013 at 13:30 -
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$\begingroup$ Glad that you find my answer helpful. $\endgroup$– MetricsCommented Jul 18, 2013 at 18:07
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I am not sure if this is what you are asking, but one might conceive of an integrated process that looks something like this:
$$y_{t} = y_{t-1} + \varepsilon$$
Feel free to define the distribution of $\varepsilon$ according to your needs and whimsy, and you have got yourself a nice little nonstationary random walk process.
The "unit root" in the above case is in the invisible coefficient $\beta_{y_{t-1}}$, which equals $1$ or $-1$ ($|1|$ being the "unit" in the unit root). I think you may be asking about the intercept in a model like this:
$$y_{t} = \beta_{0} + y_{t-1} + \varepsilon$$
This model is very similar to the above, except for every unit of time, the time series changes by $\beta_{0}$. If you were to eliminate the random portion of this process, what you would be left with is a linear change in $y$ over unit time. So $\beta_{0}$ is the slope of a linear trend over time in a unit root process added to the nice random walk in the remaining portion of the model.
Of course, $\beta_{y_{t-1}}$ need not be invisible, and may be estimated from data, and may not equal exactly $1$ or $-1$. When $|\beta_{y_{t-1}}| < 1$, then one's time series is not technically "integrated," but rather has a long memory, or is near integrated (if the value is close to $1$ or $-1$). If $|\beta_{y_{t-1}}| > 1$, then your time series diverges/explodes. The linear trend implied by $\beta_{0}$ remains just that in such cases.
