226 lines
7.1 KiB
Python
226 lines
7.1 KiB
Python
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"""
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Created on Thu Aug 30 12:26:38 2012
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Author: Josef Perktold
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function jc = c_sja(n,p)
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% PURPOSE: find critical values for Johansen maximum eigenvalue statistic
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% ------------------------------------------------------------
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% USAGE: jc = c_sja(n,p)
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% where: n = dimension of the VAR system
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% p = order of time polynomial in the null-hypothesis
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% p = -1, no deterministic part
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% p = 0, for constant term
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% p = 1, for constant plus time-trend
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% p > 1 returns no critical values
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% ------------------------------------------------------------
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% RETURNS: a (3x1) vector of percentiles for the maximum eigenvalue
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% statistic for: [90% 95% 99%]
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% ------------------------------------------------------------
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% NOTES: for n > 12, the function returns a (3x1) vector of zeros.
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% The values returned by the function were generated using
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% a method described in MacKinnon (1996), using his FORTRAN
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% program johdist.f
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% ------------------------------------------------------------
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% SEE ALSO: johansen()
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% ------------------------------------------------------------
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% References: MacKinnon, Haug, Michelis (1996) 'Numerical distribution
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% functions of likelihood ratio tests for cointegration',
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% Queen's University Institute for Economic Research Discussion paper.
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% -------------------------------------------------------
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% written by:
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% James P. LeSage, Dept of Economics
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% University of Toledo
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% 2801 W. Bancroft St,
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% Toledo, OH 43606
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% jlesage@spatial-econometrics.com
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"""
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import numpy as np
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ss_ejcp0 = '''\
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2.9762 4.1296 6.9406
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9.4748 11.2246 15.0923
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15.7175 17.7961 22.2519
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21.8370 24.1592 29.0609
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27.9160 30.4428 35.7359
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33.9271 36.6301 42.2333
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39.9085 42.7679 48.6606
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45.8930 48.8795 55.0335
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51.8528 54.9629 61.3449
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57.7954 61.0404 67.6415
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63.7248 67.0756 73.8856
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69.6513 73.0946 80.0937'''
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ss_ejcp1 = '''\
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2.7055 3.8415 6.6349
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12.2971 14.2639 18.5200
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18.8928 21.1314 25.8650
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25.1236 27.5858 32.7172
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31.2379 33.8777 39.3693
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37.2786 40.0763 45.8662
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43.2947 46.2299 52.3069
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49.2855 52.3622 58.6634
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55.2412 58.4332 64.9960
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61.2041 64.5040 71.2525
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67.1307 70.5392 77.4877
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73.0563 76.5734 83.7105'''
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ss_ejcp2 = '''\
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2.7055 3.8415 6.6349
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15.0006 17.1481 21.7465
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21.8731 24.2522 29.2631
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28.2398 30.8151 36.1930
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34.4202 37.1646 42.8612
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40.5244 43.4183 49.4095
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46.5583 49.5875 55.8171
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52.5858 55.7302 62.1741
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58.5316 61.8051 68.5030
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64.5292 67.9040 74.7434
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70.4630 73.9355 81.0678
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76.4081 79.9878 87.2395'''
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ejcp0 = np.array(ss_ejcp0.split(),float).reshape(-1,3)
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ejcp1 = np.array(ss_ejcp1.split(),float).reshape(-1,3)
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ejcp2 = np.array(ss_ejcp2.split(),float).reshape(-1,3)
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def c_sja(n, p):
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if ((p > 1) or (p < -1)):
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jc = np.full(3, np.nan)
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elif ((n > 12) or (n < 1)):
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jc = np.full(3, np.nan)
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elif p == -1:
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jc = ejcp0[n-1,:]
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elif p == 0:
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jc = ejcp1[n-1,:]
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elif p == 1:
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jc = ejcp2[n-1,:]
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return jc
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'''
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function jc = c_sjt(n,p)
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% PURPOSE: find critical values for Johansen trace statistic
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% ------------------------------------------------------------
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% USAGE: jc = c_sjt(n,p)
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% where: n = dimension of the VAR system
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% NOTE: routine does not work for n > 12
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% p = order of time polynomial in the null-hypothesis
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% p = -1, no deterministic part
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% p = 0, for constant term
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% p = 1, for constant plus time-trend
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% p > 1 returns no critical values
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% ------------------------------------------------------------
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% RETURNS: a (3x1) vector of percentiles for the trace
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% statistic for [90% 95% 99%]
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% ------------------------------------------------------------
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% NOTES: for n > 12, the function returns a (3x1) vector of zeros.
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% The values returned by the function were generated using
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% a method described in MacKinnon (1996), using his FORTRAN
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% program johdist.f
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% ------------------------------------------------------------
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% SEE ALSO: johansen()
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% ------------------------------------------------------------
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% % References: MacKinnon, Haug, Michelis (1996) 'Numerical distribution
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% functions of likelihood ratio tests for cointegration',
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% Queen's University Institute for Economic Research Discussion paper.
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% -------------------------------------------------------
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% written by:
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% James P. LeSage, Dept of Economics
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% University of Toledo
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% 2801 W. Bancroft St,
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% Toledo, OH 43606
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% jlesage@spatial-econometrics.com
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% these are the values from Johansen's 1995 book
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% for comparison to the MacKinnon values
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%jcp0 = [ 2.98 4.14 7.02
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% 10.35 12.21 16.16
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% 21.58 24.08 29.19
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% 36.58 39.71 46.00
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% 55.54 59.24 66.71
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% 78.30 86.36 91.12
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% 104.93 109.93 119.58
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% 135.16 140.74 151.70
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% 169.30 175.47 187.82
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% 207.21 214.07 226.95
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% 248.77 256.23 270.47
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% 293.83 301.95 318.14];
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%
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'''
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ss_tjcp0 = '''\
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2.9762 4.1296 6.9406
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10.4741 12.3212 16.3640
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21.7781 24.2761 29.5147
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37.0339 40.1749 46.5716
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56.2839 60.0627 67.6367
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79.5329 83.9383 92.7136
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106.7351 111.7797 121.7375
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137.9954 143.6691 154.7977
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173.2292 179.5199 191.8122
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212.4721 219.4051 232.8291
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255.6732 263.2603 277.9962
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302.9054 311.1288 326.9716'''
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ss_tjcp1 = '''\
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2.7055 3.8415 6.6349
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13.4294 15.4943 19.9349
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27.0669 29.7961 35.4628
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44.4929 47.8545 54.6815
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65.8202 69.8189 77.8202
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91.1090 95.7542 104.9637
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120.3673 125.6185 135.9825
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153.6341 159.5290 171.0905
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190.8714 197.3772 210.0366
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232.1030 239.2468 253.2526
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277.3740 285.1402 300.2821
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326.5354 334.9795 351.2150'''
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ss_tjcp2 = '''\
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2.7055 3.8415 6.6349
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16.1619 18.3985 23.1485
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32.0645 35.0116 41.0815
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51.6492 55.2459 62.5202
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75.1027 79.3422 87.7748
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102.4674 107.3429 116.9829
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133.7852 139.2780 150.0778
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169.0618 175.1584 187.1891
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208.3582 215.1268 228.2226
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251.6293 259.0267 273.3838
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298.8836 306.8988 322.4264
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350.1125 358.7190 375.3203'''
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tjcp0 = np.array(ss_tjcp0.split(),float).reshape(-1,3)
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tjcp1 = np.array(ss_tjcp1.split(),float).reshape(-1,3)
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tjcp2 = np.array(ss_tjcp2.split(),float).reshape(-1,3)
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def c_sjt(n, p):
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if ((p > 1) or (p < -1)):
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jc = np.full(3, np.nan)
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elif ((n > 12) or (n < 1)):
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jc = np.full(3, np.nan)
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elif p == -1:
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jc = tjcp0[n-1,:]
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elif p == 0:
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jc = tjcp1[n-1,:]
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elif p == 1:
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jc = tjcp2[n-1,:]
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else:
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raise ValueError('invalid p')
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return jc
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if __name__ == '__main__':
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for p in range(-2, 3, 1):
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for n in range(12):
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print(n, p)
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print(c_sja(n, p))
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print(c_sjt(n, p))
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