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johansen_test.py
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'''
function result = johansen(x,p,k)
% PURPOSE: perform Johansen cointegration tests
% -------------------------------------------------------
% USAGE: result = johansen(x,p,k)
% where: x = input matrix of time-series in levels, (nobs x m)
% p = order of time polynomial in the null-hypothesis
% p = -1, no deterministic part
% p = 0, for constant term
% p = 1, for constant plus time-trend
% p > 1, for higher order polynomial
% k = number of lagged difference terms used when
% computing the estimator
% -------------------------------------------------------
% RETURNS: a results structure:
% result.eig = eigenvalues (m x 1)
% result.evec = eigenvectors (m x m), where first
% r columns are normalized coint vectors
% result.lr1 = likelihood ratio trace statistic for r=0 to m-1
% (m x 1) vector
% result.lr2 = maximum eigenvalue statistic for r=0 to m-1
% (m x 1) vector
% result.cvt = critical values for trace statistic
% (m x 3) vector [90% 95% 99%]
% result.cvm = critical values for max eigen value statistic
% (m x 3) vector [90% 95% 99%]
% result.ind = index of co-integrating variables ordered by
% size of the eigenvalues from large to small
% -------------------------------------------------------
% NOTE: c_sja(), c_sjt() provide critical values generated using
% a method of MacKinnon (1994, 1996).
% critical values are available for n<=12 and -1 <= p <= 1,
% zeros are returned for other cases.
% -------------------------------------------------------
% SEE ALSO: prt_coint, a function that prints results
% -------------------------------------------------------
'''
import numpy as np
from numpy import zeros, ones, flipud, log
from numpy.linalg import inv, eig, cholesky as chol
from statsmodels.regression.linear_model import OLS
tdiff = np.diff
class Holder(object):
pass
def rows(x):
return x.shape[0]
def trimr(x, front, end):
if end > 0:
return x[front:-end]
else:
return x[front:]
import statsmodels.tsa.tsatools as tsat
mlag = tsat.lagmat
def mlag_(x, maxlag):
'''return all lags up to maxlag
'''
return x[:-lag]
def lag(x, lag):
return x[:-lag]
def detrend(y, order):
if order == -1:
return y
return OLS(y, np.vander(np.linspace(-1, 1, len(y)), order + 1)).fit().resid
def resid(y, x):
r = y - np.dot(x, np.dot(np.linalg.pinv(x), y))
return r
def coint_johansen(x, p, k, print_on_console=True):
# % error checking on inputs
# if (nargin ~= 3)
# error('Wrong # of inputs to johansen')
# end
nobs, m = x.shape
# why this? f is detrend transformed series, p is detrend data
if (p > -1):
f = 0
else:
f = p
x = detrend(x, p)
dx = tdiff(x, 1, axis=0)
# dx = trimr(dx,1,0)
z = mlag(dx, k) # [k-1:]
# print z.shape
z = trimr(z, k, 0)
z = detrend(z, f)
# print dx.shape
dx = trimr(dx, k, 0)
dx = detrend(dx, f)
# r0t = dx - z*(z\dx)
r0t = resid(dx, z) # diff on lagged diffs
# lx = trimr(lag(x,k),k,0)
lx = lag(x, k)
lx = trimr(lx, 1, 0)
dx = detrend(lx, f)
# print 'rkt', dx.shape, z.shape
# rkt = dx - z*(z\dx)
rkt = resid(dx, z) # level on lagged diffs
skk = np.dot(rkt.T, rkt) / rows(rkt)
sk0 = np.dot(rkt.T, r0t) / rows(rkt)
s00 = np.dot(r0t.T, r0t) / rows(r0t)
sig = np.dot(sk0, np.dot(inv(s00), (sk0.T)))
tmp = inv(skk)
# du, au = eig(np.dot(tmp, sig))
au, du = eig(np.dot(tmp, sig)) # au is eval, du is evec
# orig = np.dot(tmp, sig)
# % Normalize the eigen vectors such that (du'skk*du) = I
temp = inv(chol(np.dot(du.T, np.dot(skk, du))))
dt = np.dot(du, temp)
# JP: the next part can be done much easier
# % NOTE: At this point, the eigenvectors are aligned by column. To
# % physically move the column elements using the MATLAB sort,
# % take the transpose to put the eigenvectors across the row
# dt = transpose(dt)
# % sort eigenvalues and vectors
# au, auind = np.sort(diag(au))
auind = np.argsort(au)
# a = flipud(au)
aind = flipud(auind)
a = au[aind]
# d = dt[aind,:]
d = dt[:, aind]
# %NOTE: The eigenvectors have been sorted by row based on auind and moved to array "d".
# % Put the eigenvectors back in column format after the sort by taking the
# % transpose of "d". Since the eigenvectors have been physically moved, there is
# % no need for aind at all. To preserve existing programming, aind is reset back to
# % 1, 2, 3, ....
# d = transpose(d)
# test = np.dot(transpose(d), np.dot(skk, d))
# %EXPLANATION: The MATLAB sort function sorts from low to high. The flip realigns
# %auind to go from the largest to the smallest eigenvalue (now aind). The original procedure
# %physically moved the rows of dt (to d) based on the alignment in aind and then used
# %aind as a column index to address the eigenvectors from high to low. This is a double
# %sort. If you wanted to extract the eigenvector corresponding to the largest eigenvalue by,
# %using aind as a reference, you would get the correct eigenvector, but with sorted
# %coefficients and, therefore, any follow-on calculation would seem to be in error.
# %If alternative programming methods are used to evaluate the eigenvalues, e.g. Frame method
# %followed by a root extraction on the characteristic equation, then the roots can be
# %quickly sorted. One by one, the corresponding eigenvectors can be generated. The resultant
# %array can be operated on using the Cholesky transformation, which enables a unit
# %diagonalization of skk. But nowhere along the way are the coefficients within the
# %eigenvector array ever changed. The final value of the "beta" array using either method
# %should be the same.
# % Compute the trace and max eigenvalue statistics */
lr1 = zeros(m)
lr2 = zeros(m)
cvm = zeros((m, 3))
cvt = zeros((m, 3))
iota = ones(m)
t, junk = rkt.shape
for i in range(0, m):
tmp = trimr(log(iota - a), i , 0)
lr1[i] = -t * np.sum(tmp, 0) # columnsum ?
# tmp = np.log(1-a)
# lr1[i] = -t * np.sum(tmp[i:])
lr2[i] = -t * log(1 - a[i])
cvm[i, :] = c_sja(m - i, p)
cvt[i, :] = c_sjt(m - i, p)
aind[i] = i
# end
result = Holder()
# % set up results structure
# estimation results, residuals
result.rkt = rkt
result.r0t = r0t
result.eig = a
result.evec = d # transposed compared to matlab ?
result.lr1 = lr1
result.lr2 = lr2
result.cvt = cvt
result.cvm = cvm
result.ind = aind
result.meth = 'johansen'
if print_on_console == True:
print ('--------------------------------------------------')
print ('--> Trace Statistics')
print ('variable statistic Crit-90% Crit-95% Crit-99%')
for i in range(len(result.lr1)):
print ('r =', i, '\t', round(result.lr1[i], 4), result.cvt[i, 0], result.cvt[i, 1], result.cvt[i, 2])
print ('--------------------------------------------------')
print ('--> Eigen Statistics')
print ('variable statistic Crit-90% Crit-95% Crit-99%')
for i in range(len(result.lr2)):
print ('r =', i, '\t', round(result.lr2[i], 4), result.cvm[i, 0], result.cvm[i, 1], result.cvm[i, 2])
print ('--------------------------------------------------')
print ('eigenvectors:\n', result.evec)
print ('--------------------------------------------------')
print ('eigenvalues:\n', result.eig)
print ('--------------------------------------------------')
return result
def c_sjt(n, p):
# PURPOSE: find critical values for Johansen trace statistic
# ------------------------------------------------------------
# USAGE: jc = c_sjt(n,p)
# where: n = dimension of the VAR system
# NOTE: routine doesn't work for n > 12
# p = order of time polynomial in the null-hypothesis
# p = -1, no deterministic part
# p = 0, for constant term
# p = 1, for constant plus time-trend
# p > 1 returns no critical values
# ------------------------------------------------------------
# RETURNS: a (3x1) vector of percentiles for the trace
# statistic for [90# 95# 99#]
# ------------------------------------------------------------
# NOTES: for n > 12, the function returns a (3x1) vector of zeros.
# The values returned by the function were generated using
# a method described in MacKinnon (1996), using his FORTRAN
# program johdist.f
# ------------------------------------------------------------
# SEE ALSO: johansen()
# ------------------------------------------------------------
# # References: MacKinnon, Haug, Michelis (1996) 'Numerical distribution
# functions of likelihood ratio tests for cointegration',
# Queen's University Institute for Economic Research Discussion paper.
# -------------------------------------------------------
# written by:
# James P. LeSage, Dept of Economics
# University of Toledo
# 2801 W. Bancroft St,
# Toledo, OH 43606
#
# Ported to Python by Javier Garcia
# these are the values from Johansen's 1995 book
# for comparison to the MacKinnon values
# jcp0 = [ 2.98 4.14 7.02
# 10.35 12.21 16.16
# 21.58 24.08 29.19
# 36.58 39.71 46.00
# 55.54 59.24 66.71
# 78.30 86.36 91.12
# 104.93 109.93 119.58
# 135.16 140.74 151.70
# 169.30 175.47 187.82
# 207.21 214.07 226.95
# 248.77 256.23 270.47
# 293.83 301.95 318.14];
jcp0 = ((2.9762, 4.1296, 6.9406),
(10.4741, 12.3212, 16.3640),
(21.7781, 24.2761, 29.5147),
(37.0339, 40.1749, 46.5716),
(56.2839, 60.0627, 67.6367),
(79.5329, 83.9383, 92.7136),
(106.7351, 111.7797, 121.7375),
(137.9954, 143.6691, 154.7977),
(173.2292, 179.5199, 191.8122),
(212.4721, 219.4051, 232.8291),
(255.6732, 263.2603, 277.9962),
(302.9054, 311.1288, 326.9716))
jcp1 = ((2.7055, 3.8415, 6.6349),
(13.4294, 15.4943, 19.9349),
(27.0669, 29.7961, 35.4628),
(44.4929, 47.8545, 54.6815),
(65.8202, 69.8189, 77.8202),
(91.1090, 95.7542, 104.9637),
(120.3673, 125.6185, 135.9825),
(153.6341, 159.5290, 171.0905),
(190.8714, 197.3772, 210.0366),
(232.1030, 239.2468, 253.2526),
(277.3740, 285.1402, 300.2821),
(326.5354, 334.9795, 351.2150))
jcp2 = ((2.7055, 3.8415, 6.6349),
(16.1619, 18.3985, 23.1485),
(32.0645, 35.0116, 41.0815),
(51.6492, 55.2459, 62.5202),
(75.1027, 79.3422, 87.7748),
(102.4674, 107.3429, 116.9829),
(133.7852, 139.2780, 150.0778),
(169.0618, 175.1584, 187.1891),
(208.3582, 215.1268, 228.2226),
(251.6293, 259.0267, 273.3838),
(298.8836, 306.8988, 322.4264),
(350.1125, 358.7190, 375.3203))
if (p > 1) or (p < -1):
jc = (0, 0, 0)
elif (n > 12) or (n < 1):
jc = (0, 0, 0)
elif p == -1:
jc = jcp0[n - 1]
elif p == 0:
jc = jcp1[n - 1]
elif p == 1:
jc = jcp2[n - 1]
return jc
def c_sja(n, p):
# PURPOSE: find critical values for Johansen maximum eigenvalue statistic
# ------------------------------------------------------------
# USAGE: jc = c_sja(n,p)
# where: n = dimension of the VAR system
# p = order of time polynomial in the null-hypothesis
# p = -1, no deterministic part
# p = 0, for constant term
# p = 1, for constant plus time-trend
# p > 1 returns no critical values
# ------------------------------------------------------------
# RETURNS: a (3x1) vector of percentiles for the maximum eigenvalue
# statistic for: [90# 95# 99#]
# ------------------------------------------------------------
# NOTES: for n > 12, the function returns a (3x1) vector of zeros.
# The values returned by the function were generated using
# a method described in MacKinnon (1996), using his FORTRAN
# program johdist.f
# ------------------------------------------------------------
# SEE ALSO: johansen()
# ------------------------------------------------------------
# References: MacKinnon, Haug, Michelis (1996) 'Numerical distribution
# functions of likelihood ratio tests for cointegration',
# Queen's University Institute for Economic Research Discussion paper.
# -------------------------------------------------------
# written by:
# James P. LeSage, Dept of Economics
# University of Toledo
# 2801 W. Bancroft St,
# Toledo, OH 43606
# Ported to Python by Javier Garcia
jcp0 = ((2.9762, 4.1296, 6.9406),
(9.4748, 11.2246, 15.0923),
(15.7175, 17.7961, 22.2519),
(21.8370, 24.1592, 29.0609),
(27.9160, 30.4428, 35.7359),
(33.9271, 36.6301, 42.2333),
(39.9085, 42.7679, 48.6606),
(45.8930, 48.8795, 55.0335),
(51.8528, 54.9629, 61.3449),
(57.7954, 61.0404, 67.6415),
(63.7248, 67.0756, 73.8856),
(69.6513, 73.0946, 80.0937))
jcp1 = ((2.7055, 3.8415, 6.6349),
(12.2971, 14.2639, 18.5200),
(18.8928, 21.1314, 25.8650),
(25.1236, 27.5858, 32.7172),
(31.2379, 33.8777, 39.3693),
(37.2786, 40.0763, 45.8662),
(43.2947, 46.2299, 52.3069),
(49.2855, 52.3622, 58.6634),
(55.2412, 58.4332, 64.9960),
(61.2041, 64.5040, 71.2525),
(67.1307, 70.5392, 77.4877),
(73.0563, 76.5734, 83.7105))
jcp2 = ((2.7055, 3.8415, 6.6349),
(15.0006, 17.1481, 21.7465),
(21.8731, 24.2522, 29.2631),
(28.2398, 30.8151, 36.1930),
(34.4202, 37.1646, 42.8612),
(40.5244, 43.4183, 49.4095),
(46.5583, 49.5875, 55.8171),
(52.5858, 55.7302, 62.1741),
(58.5316, 61.8051, 68.5030),
(64.5292, 67.9040, 74.7434),
(70.4630, 73.9355, 81.0678),
(76.4081, 79.9878, 87.2395))
if (p > 1) or (p < -1):
jc = (0, 0, 0)
elif (n > 12) or (n < 1):
jc = (0, 0, 0)
elif p == -1:
jc = jcp0[n - 1]
elif p == 0:
jc = jcp1[n - 1]
elif p == 1:
jc = jcp2[n - 1]
return jc