请问为什么我做的单位根检验,差分之后的统计量越来越大,对原始数据进行对数变化再做单位根检验也是这样
还有我想做协整检验,怎么判断这个是同阶单整的呢
a.df=ur.df(a,lags=5,type='trend')
summary(a.df)
a1=diff(a)
a1.df=ur.df(a1,lags=5,type="trend")
summary(a1.df) 这个是我做的程序
结果如下:
Call:
lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
Residuals:
Min 1Q Median 3Q Max
-12.4233 -7.2875 -0.5495 2.8957 26.6279
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.86535 26.65402 -0.070 0.9465
z.lag.1 -0.31343 0.11011 -2.846 0.0293 *
tt 62.24479 21.06403 2.955 0.0254 *
z.diff.lag1 0.09653 0.29551 0.327 0.7550
z.diff.lag2 -0.43170 0.29438 -1.466 0.1929
z.diff.lag3 0.08881 0.30675 0.290 0.7819
z.diff.lag4 0.02555 0.29217 0.087 0.9332
z.diff.lag5 0.04607 0.31895 0.144 0.8899
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 15.23 on 6 degrees of freedom
Multiple R-squared: 0.9263, Adjusted R-squared: 0.8403
F-statistic: 10.77 on 7 and 6 DF, p-value: 0.005017
Value of test-statistic is: -2.8465 5.7202 4.3693
Critical values for test statistics:
1pct 5pct 10pct
tau3 -4.38 -3.60 -3.24
phi2 8.21 5.68 4.67
phi3 10.61 7.24 5.91
Call:
lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
Residuals:
1 2 3 4 5 6 7 8
-19.8223 -3.3538 10.2666 36.7955 -26.8049 0.3257 6.7007 6.2390
9 10 11 12 13
1.3203 3.2205 -15.4283 -2.3918 2.9328
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 60.2801 42.0703 1.433 0.211
z.lag.1 -0.6749 0.8214 -0.822 0.449
tt 4.4224 10.3259 0.428 0.686
z.diff.lag1 0.1171 0.8156 0.144 0.891
z.diff.lag2 -0.1455 0.8242 -0.177 0.867
z.diff.lag3 0.2013 0.7430 0.271 0.797
z.diff.lag4 0.2594 0.5971 0.434 0.682
z.diff.lag5 0.1408 0.4882 0.288 0.785
Residual standard error: 24.21 on 5 degrees of freedom
Multiple R-squared: 0.5352, Adjusted R-squared: -0.1155
F-statistic: 0.8225 on 7 and 5 DF, p-value: 0.608
Value of test-statistic is: -0.8216 1.3334 1.5478
Critical values for test statistics:
1pct 5pct 10pct
tau3 -4.38 -3.60 -3.24
phi2 8.21 5.68 4.67
phi3 10.61 7.24 5.91