Source code for statsmodels.sandbox.stats.diagnostic

# -*- coding: utf-8 -*-
"""Various Statistical Tests


Author: josef-pktd
License: BSD-3

Notes
-----
Almost fully verified against R or Gretl, not all options are the same.
In many cases of Lagrange multiplier tests both the LM test and the F test is
returned. In some but not all cases, R has the option to choose the test
statistic. Some alternative test statistic results have not been verified.


TODO
* refactor to store intermediate results
* how easy is it to attach a test that is a class to a result instance,
  for example CompareCox as a method compare_cox(self, other) ?
* StatTestMC has been moved and should be deleted

missing:

* pvalues for breaks_hansen
* additional options, compare with R, check where ddof is appropriate
* new tests:
  - breaks_ap, more recent breaks tests
  - specification tests against nonparametric alternatives


"""
from __future__ import print_function
from statsmodels.compat.python import iteritems, lrange, map, long
import numpy as np
from scipy import stats
from statsmodels.regression.linear_model import OLS
from statsmodels.tools.tools import add_constant
from statsmodels.tsa.stattools import acf, adfuller
from statsmodels.tsa.tsatools import lagmat
from statsmodels.compat.numpy import np_matrix_rank

#get the old signature back so the examples work
[docs]def unitroot_adf(x, maxlag=None, trendorder=0, autolag='AIC', store=False): return adfuller(x, maxlag=maxlag, regression=trendorder, autolag=autolag, store=store, regresults=False)
#TODO: I like the bunch pattern for this too. class ResultsStore(object): def __str__(self): return self._str
[docs]class CompareCox(object): '''Cox Test for non-nested models Parameters ---------- results_x : Result instance result instance of first model results_z : Result instance result instance of second model attach : bool Formulas from Greene, section 8.3.4 translated to code produces correct results for Example 8.3, Greene '''
[docs] def run(self, results_x, results_z, attach=True): '''run Cox test for non-nested models Parameters ---------- results_x : Result instance result instance of first model results_z : Result instance result instance of second model attach : bool If true, then the intermediate results are attached to the instance. Returns ------- tstat : float t statistic for the test that including the fitted values of the first model in the second model has no effect. pvalue : float two-sided pvalue for the t statistic Notes ----- Tests of non-nested hypothesis might not provide unambiguous answers. The test should be performed in both directions and it is possible that both or neither test rejects. see ??? for more information. References ---------- ??? ''' if not np.allclose(results_x.model.endog, results_z.model.endog): raise ValueError('endogenous variables in models are not the same') nobs = results_x.model.endog.shape[0] x = results_x.model.exog z = results_z.model.exog sigma2_x = results_x.ssr/nobs sigma2_z = results_z.ssr/nobs yhat_x = results_x.fittedvalues yhat_z = results_z.fittedvalues res_dx = OLS(yhat_x, z).fit() err_zx = res_dx.resid res_xzx = OLS(err_zx, x).fit() err_xzx = res_xzx.resid sigma2_zx = sigma2_x + np.dot(err_zx.T, err_zx)/nobs c01 = nobs/2. * (np.log(sigma2_z) - np.log(sigma2_zx)) v01 = sigma2_x * np.dot(err_xzx.T, err_xzx) / sigma2_zx**2 q = c01 / np.sqrt(v01) pval = 2*stats.norm.sf(np.abs(q)) if attach: self.res_dx = res_dx self.res_xzx = res_xzx self.c01 = c01 self.v01 = v01 self.q = q self.pvalue = pval self.dist = stats.norm return q, pval
def __call__(self, results_x, results_z): return self.run(results_x, results_z, attach=False)
compare_cox = CompareCox() compare_cox.__doc__ = CompareCox.__doc__
[docs]class CompareJ(object): '''J-Test for comparing non-nested models Parameters ---------- results_x : Result instance result instance of first model results_z : Result instance result instance of second model attach : bool From description in Greene, section 8.3.3 produces correct results for Example 8.3, Greene - not checked yet #currently an exception, but I don't have clean reload in python session check what results should be attached '''
[docs] def run(self, results_x, results_z, attach=True): '''run J-test for non-nested models Parameters ---------- results_x : Result instance result instance of first model results_z : Result instance result instance of second model attach : bool If true, then the intermediate results are attached to the instance. Returns ------- tstat : float t statistic for the test that including the fitted values of the first model in the second model has no effect. pvalue : float two-sided pvalue for the t statistic Notes ----- Tests of non-nested hypothesis might not provide unambiguous answers. The test should be performed in both directions and it is possible that both or neither test rejects. see ??? for more information. References ---------- ??? ''' if not np.allclose(results_x.model.endog, results_z.model.endog): raise ValueError('endogenous variables in models are not the same') nobs = results_x.model.endog.shape[0] y = results_x.model.endog x = results_x.model.exog z = results_z.model.exog #sigma2_x = results_x.ssr/nobs #sigma2_z = results_z.ssr/nobs yhat_x = results_x.fittedvalues #yhat_z = results_z.fittedvalues res_zx = OLS(y, np.column_stack((yhat_x, z))).fit() self.res_zx = res_zx #for testing tstat = res_zx.tvalues[0] pval = res_zx.pvalues[0] if attach: self.res_zx = res_zx self.dist = stats.t(res_zx.df_resid) self.teststat = tstat self.pvalue = pval return tstat, pval
def __call__(self, results_x, results_z): return self.run(results_x, results_z, attach=False)
compare_j = CompareJ() compare_j.__doc__ = CompareJ.__doc__
[docs]def acorr_ljungbox(x, lags=None, boxpierce=False): """ Ljung-Box test for no autocorrelation Parameters ---------- x : array_like, 1d data series, regression residuals when used as diagnostic test lags : None, int or array_like If lags is an integer then this is taken to be the largest lag that is included, the test result is reported for all smaller lag length. If lags is a list or array, then all lags are included up to the largest lag in the list, however only the tests for the lags in the list are reported. If lags is None, then the default maxlag is 'min((nobs // 2 - 2), 40)' boxpierce : {False, True} If true, then additional to the results of the Ljung-Box test also the Box-Pierce test results are returned Returns ------- lbvalue : float or array test statistic pvalue : float or array p-value based on chi-square distribution bpvalue : (optionsal), float or array test statistic for Box-Pierce test bppvalue : (optional), float or array p-value based for Box-Pierce test on chi-square distribution Notes ----- Ljung-Box and Box-Pierce statistic differ in their scaling of the autocorrelation function. Ljung-Box test is reported to have better small sample properties. TODO: could be extended to work with more than one series 1d or nd ? axis ? ravel ? needs more testing *Verification* Looks correctly sized in Monte Carlo studies. not yet compared to verified values Examples -------- see example script References ---------- Greene Wikipedia """ x = np.asarray(x) nobs = x.shape[0] if lags is None: lags = np.arange(1, min((nobs // 2 - 2), 40) + 1) elif isinstance(lags, (int, long)): lags = np.arange(1, lags + 1) lags = np.asarray(lags) maxlag = max(lags) # normalize by nobs not (nobs-nlags) # SS: unbiased=False is default now acfx = acf(x, nlags=maxlag, fft=False) acf2norm = acfx[1:maxlag+1]**2 / (nobs - np.arange(1,maxlag+1)) qljungbox = nobs * (nobs+2) * np.cumsum(acf2norm)[lags-1] pval = stats.chi2.sf(qljungbox, lags) if not boxpierce: return qljungbox, pval else: qboxpierce = nobs * np.cumsum(acfx[1:maxlag+1]**2)[lags-1] pvalbp = stats.chi2.sf(qboxpierce, lags) return qljungbox, pval, qboxpierce, pvalbp
def acorr_lm(x, maxlag=None, autolag='AIC', store=False, regresults=False): '''Lagrange Multiplier tests for autocorrelation This is a generic Lagrange Multiplier test for autocorrelation. I don't have a reference for it, but it returns Engle's ARCH test if x is the squared residual array. A variation on it with additional exogenous variables is the Breusch-Godfrey autocorrelation test. Parameters ---------- resid : ndarray, (nobs,) residuals from an estimation, or time series maxlag : int highest lag to use autolag : None or string If None, then a fixed number of lags given by maxlag is used. store : bool If true then the intermediate results are also returned Returns ------- lm : float Lagrange multiplier test statistic lmpval : float p-value for Lagrange multiplier test fval : float fstatistic for F test, alternative version of the same test based on F test for the parameter restriction fpval : float pvalue for F test resstore : instance (optional) a class instance that holds intermediate results. Only returned if store=True See Also -------- het_arch acorr_breusch_godfrey acorr_ljung_box ''' if regresults: store = True x = np.asarray(x) nobs = x.shape[0] if maxlag is None: #for adf from Greene referencing Schwert 1989 maxlag = int(np.ceil(12. * np.power(nobs/100., 1/4.)))#nobs//4 #TODO: check default, or do AIC/BIC xdiff = np.diff(x) # xdall = lagmat(x[:,None], maxlag, trim='both') nobs = xdall.shape[0] xdall = np.c_[np.ones((nobs,1)), xdall] xshort = x[-nobs:] if store: resstore = ResultsStore() if autolag: #search for lag length with highest information criteria #Note: I use the same number of observations to have comparable IC results = {} for mlag in range(1, maxlag+1): results[mlag] = OLS(xshort, xdall[:,:mlag+1]).fit() if autolag.lower() == 'aic': bestic, icbestlag = min((v.aic,k) for k,v in iteritems(results)) elif autolag.lower() == 'bic': icbest, icbestlag = min((v.bic,k) for k,v in iteritems(results)) else: raise ValueError("autolag can only be None, 'AIC' or 'BIC'") #rerun ols with best ic xdall = lagmat(x[:,None], icbestlag, trim='both') nobs = xdall.shape[0] xdall = np.c_[np.ones((nobs,1)), xdall] xshort = x[-nobs:] usedlag = icbestlag if regresults: resstore.results = results else: usedlag = maxlag resols = OLS(xshort, xdall[:,:usedlag+1]).fit() fval = resols.fvalue fpval = resols.f_pvalue lm = nobs * resols.rsquared lmpval = stats.chi2.sf(lm, usedlag) # Note: degrees of freedom for LM test is nvars minus constant = usedlags #return fval, fpval, lm, lmpval if store: resstore.resols = resols resstore.usedlag = usedlag return lm, lmpval, fval, fpval, resstore else: return lm, lmpval, fval, fpval
[docs]def het_arch(resid, maxlag=None, autolag=None, store=False, regresults=False, ddof=0): '''Engle's Test for Autoregressive Conditional Heteroscedasticity (ARCH) Parameters ---------- resid : ndarray residuals from an estimation, or time series maxlag : int highest lag to use autolag : None or string If None, then a fixed number of lags given by maxlag is used. store : bool If true then the intermediate results are also returned ddof : int Not Implemented Yet If the residuals are from a regression, or ARMA estimation, then there are recommendations to correct the degrees of freedom by the number of parameters that have been estimated, for example ddof=p+a for an ARMA(p,q) (need reference, based on discussion on R finance mailinglist) Returns ------- lm : float Lagrange multiplier test statistic lmpval : float p-value for Lagrange multiplier test fval : float fstatistic for F test, alternative version of the same test based on F test for the parameter restriction fpval : float pvalue for F test resstore : instance (optional) a class instance that holds intermediate results. Only returned if store=True Notes ----- verified agains R:FinTS::ArchTest ''' return acorr_lm(resid**2, maxlag=maxlag, autolag=autolag, store=store, regresults=regresults)
[docs]def acorr_breusch_godfrey(results, nlags=None, store=False): '''Breusch Godfrey Lagrange Multiplier tests for residual autocorrelation Parameters ---------- results : Result instance Estimation results for which the residuals are tested for serial correlation nlags : int Number of lags to include in the auxiliary regression. (nlags is highest lag) store : bool If store is true, then an additional class instance that contains intermediate results is returned. Returns ------- lm : float Lagrange multiplier test statistic lmpval : float p-value for Lagrange multiplier test fval : float fstatistic for F test, alternative version of the same test based on F test for the parameter restriction fpval : float pvalue for F test resstore : instance (optional) a class instance that holds intermediate results. Only returned if store=True Notes ----- BG adds lags of residual to exog in the design matrix for the auxiliary regression with residuals as endog, see Greene 12.7.1. References ---------- Greene Econometrics, 5th edition ''' x = np.asarray(results.resid) exog_old = results.model.exog nobs = x.shape[0] if nlags is None: #for adf from Greene referencing Schwert 1989 nlags = np.trunc(12. * np.power(nobs/100., 1/4.))#nobs//4 #TODO: check default, or do AIC/BIC nlags = int(nlags) x = np.concatenate((np.zeros(nlags), x)) #xdiff = np.diff(x) # xdall = lagmat(x[:,None], nlags, trim='both') nobs = xdall.shape[0] xdall = np.c_[np.ones((nobs,1)), xdall] xshort = x[-nobs:] exog = np.column_stack((exog_old, xdall)) k_vars = exog.shape[1] if store: resstore = ResultsStore() resols = OLS(xshort, exog).fit() ft = resols.f_test(np.eye(nlags, k_vars, k_vars - nlags)) fval = ft.fvalue fpval = ft.pvalue fval = np.squeeze(fval)[()] #TODO: fix this in ContrastResults fpval = np.squeeze(fpval)[()] lm = nobs * resols.rsquared lmpval = stats.chi2.sf(lm, nlags) # Note: degrees of freedom for LM test is nvars minus constant = usedlags #return fval, fpval, lm, lmpval if store: resstore.resols = resols resstore.usedlag = nlags return lm, lmpval, fval, fpval, resstore else: return lm, lmpval, fval, fpval
msg = "Use acorr_breusch_godfrey, acorr_breush_godfrey will be removed " \ "in 0.9 \n (Note: misspelling missing 'c')," acorr_breush_godfrey = np.deprecate(acorr_breusch_godfrey, 'acorr_breush_godfrey', 'acorr_breusch_godfrey', msg)
[docs]def het_breuschpagan(resid, exog_het): '''Breusch-Pagan Lagrange Multiplier test for heteroscedasticity The tests the hypothesis that the residual variance does not depend on the variables in x in the form :math: \sigma_i = \\sigma * f(\\alpha_0 + \\alpha z_i) Homoscedasticity implies that $\\alpha=0$ Parameters ---------- resid : array-like For the Breusch-Pagan test, this should be the residual of a regression. If an array is given in exog, then the residuals are calculated by the an OLS regression or resid on exog. In this case resid should contain the dependent variable. Exog can be the same as x. TODO: I dropped the exog option, should I add it back? exog_het : array_like This contains variables that might create data dependent heteroscedasticity. Returns ------- lm : float lagrange multiplier statistic lm_pvalue :float p-value of lagrange multiplier test fvalue : float f-statistic of the hypothesis that the error variance does not depend on x f_pvalue : float p-value for the f-statistic Notes ----- Assumes x contains constant (for counting dof and calculation of R^2). In the general description of LM test, Greene mentions that this test exaggerates the significance of results in small or moderately large samples. In this case the F-statistic is preferrable. *Verification* Chisquare test statistic is exactly (<1e-13) the same result as bptest in R-stats with defaults (studentize=True). Implementation This is calculated using the generic formula for LM test using $R^2$ (Greene, section 17.6) and not with the explicit formula (Greene, section 11.4.3). The degrees of freedom for the p-value assume x is full rank. References ---------- http://en.wikipedia.org/wiki/Breusch%E2%80%93Pagan_test Greene 5th edition Breusch, Pagan article ''' x = np.asarray(exog_het) y = np.asarray(resid)**2 nobs, nvars = x.shape resols = OLS(y, x).fit() fval = resols.fvalue fpval = resols.f_pvalue lm = nobs * resols.rsquared # Note: degrees of freedom for LM test is nvars minus constant return lm, stats.chi2.sf(lm, nvars-1), fval, fpval
het_breushpagan = np.deprecate(het_breuschpagan, 'het_breushpagan', 'het_breuschpagan', "Use het_breuschpagan, het_breushpagan will be " "removed in 0.9 \n(Note: misspelling missing 'c')")
[docs]def het_white(resid, exog, retres=False): '''White's Lagrange Multiplier Test for Heteroscedasticity Parameters ---------- resid : array_like residuals, square of it is used as endogenous variable exog : array_like possible explanatory variables for variance, squares and interaction terms are included in the auxilliary regression. resstore : instance (optional) a class instance that holds intermediate results. Only returned if store=True Returns ------- lm : float lagrange multiplier statistic lm_pvalue :float p-value of lagrange multiplier test fvalue : float f-statistic of the hypothesis that the error variance does not depend on x. This is an alternative test variant not the original LM test. f_pvalue : float p-value for the f-statistic Notes ----- assumes x contains constant (for counting dof) question: does f-statistic make sense? constant ? References ---------- Greene section 11.4.1 5th edition p. 222 now test statistic reproduces Greene 5th, example 11.3 ''' x = np.asarray(exog) y = np.asarray(resid) if x.ndim == 1: raise ValueError('x should have constant and at least one more variable') nobs, nvars0 = x.shape i0,i1 = np.triu_indices(nvars0) exog = x[:,i0]*x[:,i1] nobs, nvars = exog.shape assert nvars == nvars0*(nvars0-1)/2. + nvars0 resols = OLS(y**2, exog).fit() fval = resols.fvalue fpval = resols.f_pvalue lm = nobs * resols.rsquared # Note: degrees of freedom for LM test is nvars minus constant #degrees of freedom take possible reduced rank in exog into account #df_model checks the rank to determine df #extra calculation that can be removed: assert resols.df_model == np_matrix_rank(exog) - 1 lmpval = stats.chi2.sf(lm, resols.df_model) return lm, lmpval, fval, fpval
def _het_goldfeldquandt2_old(y, x, idx, split=None, retres=False): '''test whether variance is the same in 2 subsamples Parameters ---------- y : array_like endogenous variable x : array_like exogenous variable, regressors idx : integer column index of variable according to which observations are sorted for the split split : None or integer or float in intervall (0,1) index at which sample is split. If 0<split<1 then split is interpreted as fraction of the observations in the first sample retres : boolean if true, then an instance of a result class is returned, otherwise 2 numbers, fvalue and p-value, are returned Returns ------- (fval, pval) or res fval : float value of the F-statistic pval : float p-value of the hypothesis that the variance in one subsample is larger than in the other subsample res : instance of result class The class instance is just a storage for the intermediate and final results that are calculated Notes ----- TODO: add resultinstance - DONE maybe add drop-middle as option maybe allow for several breaks recommendation for users: use this function as pattern for more flexible split in tests, e.g. drop middle. can do Chow test for structural break in same way ran sanity check ''' x = np.asarray(x) y = np.asarray(y) nobs, nvars = x.shape if split is None: split = nobs//2 elif (0<split) and (split<1): split = int(nobs*split) xsortind = np.argsort(x[:,idx]) y = y[xsortind] x = x[xsortind,:] resols1 = OLS(y[:split], x[:split]).fit() resols2 = OLS(y[split:], x[split:]).fit() fval = resols1.mse_resid/resols2.mse_resid if fval>1: fpval = stats.f.sf(fval, resols1.df_resid, resols2.df_resid) ordering = 'larger' else: fval = 1./fval; fpval = stats.f.sf(fval, resols2.df_resid, resols1.df_resid) ordering = 'smaller' if retres: res = ResultsStore() res.__doc__ = 'Test Results for Goldfeld-Quandt test of heterogeneity' res.fval = fval res.fpval = fpval res.df_fval = (resols2.df_resid, resols1.df_resid) res.resols1 = resols1 res.resols2 = resols2 res.ordering = ordering res.split = split #res.__str__ res._str = '''The Goldfeld-Quandt test for null hypothesis that the variance in the second subsample is %s than in the first subsample: F-statistic =%8.4f and p-value =%8.4f''' % (ordering, fval, fpval) return res else: return fval, fpval
[docs]class HetGoldfeldQuandt(object): '''test whether variance is the same in 2 subsamples Parameters ---------- y : array_like endogenous variable x : array_like exogenous variable, regressors idx : integer column index of variable according to which observations are sorted for the split split : None or integer or float in intervall (0,1) index at which sample is split. If 0<split<1 then split is interpreted as fraction of the observations in the first sample drop : None, float or int If this is not None, then observation are dropped from the middle part of the sorted series. If 0<split<1 then split is interpreted as fraction of the number of observations to be dropped. Note: Currently, observations are dropped between split and split+drop, where split and drop are the indices (given by rounding if specified as fraction). The first sample is [0:split], the second sample is [split+drop:] alternative : string, 'increasing', 'decreasing' or 'two-sided' default is increasing. This specifies the alternative for the p-value calculation. Returns ------- (fval, pval) or res fval : float value of the F-statistic pval : float p-value of the hypothesis that the variance in one subsample is larger than in the other subsample res : instance of result class The class instance is just a storage for the intermediate and final results that are calculated Notes ----- The Null hypothesis is that the variance in the two sub-samples are the same. The alternative hypothesis, can be increasing, i.e. the variance in the second sample is larger than in the first, or decreasing or two-sided. Results are identical R, but the drop option is defined differently. (sorting by idx not tested yet) ''' #TODO: can do Chow test for structural break in same way
[docs] def run(self, y, x, idx=None, split=None, drop=None, alternative='increasing', attach=True): '''see class docstring''' x = np.asarray(x) y = np.asarray(y)#**2 nobs, nvars = x.shape if split is None: split = nobs//2 elif (0<split) and (split<1): split = int(nobs*split) if drop is None: start2 = split elif (0<drop) and (drop<1): start2 = split + int(nobs*drop) else: start2 = split + drop if not idx is None: xsortind = np.argsort(x[:,idx]) y = y[xsortind] x = x[xsortind,:] resols1 = OLS(y[:split], x[:split]).fit() resols2 = OLS(y[start2:], x[start2:]).fit() fval = resols2.mse_resid/resols1.mse_resid #if fval>1: if alternative.lower() in ['i', 'inc', 'increasing']: fpval = stats.f.sf(fval, resols1.df_resid, resols2.df_resid) ordering = 'increasing' elif alternative.lower() in ['d', 'dec', 'decreasing']: fval = fval; fpval = stats.f.sf(1./fval, resols2.df_resid, resols1.df_resid) ordering = 'decreasing' elif alternative.lower() in ['2', '2-sided', 'two-sided']: fpval_sm = stats.f.cdf(fval, resols2.df_resid, resols1.df_resid) fpval_la = stats.f.sf(fval, resols2.df_resid, resols1.df_resid) fpval = 2*min(fpval_sm, fpval_la) ordering = 'two-sided' else: raise ValueError('invalid alternative') if attach: res = self res.__doc__ = 'Test Results for Goldfeld-Quandt test of heterogeneity' res.fval = fval res.fpval = fpval res.df_fval = (resols2.df_resid, resols1.df_resid) res.resols1 = resols1 res.resols2 = resols2 res.ordering = ordering res.split = split #res.__str__ #TODO: check if string works res._str = '''The Goldfeld-Quandt test for null hypothesis that the variance in the second subsample is %s than in the first subsample: F-statistic =%8.4f and p-value =%8.4f''' % (ordering, fval, fpval) return fval, fpval, ordering
#return self def __str__(self): try: return self._str except AttributeError: return repr(self) #TODO: missing the alternative option in call def __call__(self, y, x, idx=None, split=None, drop=None, alternative='increasing'): return self.run(y, x, idx=idx, split=split, drop=drop, attach=False, alternative=alternative)
het_goldfeldquandt = HetGoldfeldQuandt() het_goldfeldquandt.__doc__ = het_goldfeldquandt.run.__doc__
[docs]def linear_harvey_collier(res): '''Harvey Collier test for linearity The Null hypothesis is that the regression is correctly modeled as linear. Parameters ---------- res : Result instance Returns ------- tvalue : float test statistic, based on ttest_1sample pvalue : float pvalue of the test Notes ----- TODO: add sort_by option This test is a t-test that the mean of the recursive ols residuals is zero. Calculating the recursive residuals might take some time for large samples. ''' #I think this has different ddof than #B.H. Baltagi, Econometrics, 2011, chapter 8 #but it matches Gretl and R:lmtest, pvalue at decimal=13 rr = recursive_olsresiduals(res, skip=3, alpha=0.95) from scipy import stats return stats.ttest_1samp(rr[3][3:], 0)
[docs]def linear_rainbow(res, frac = 0.5): '''Rainbow test for linearity The Null hypothesis is that the regression is correctly modelled as linear. The alternative for which the power might be large are convex, check Parameters ---------- res : Result instance Returns ------- fstat : float test statistic based of F test pvalue : float pvalue of the test ''' nobs = res.nobs endog = res.model.endog exog = res.model.exog lowidx = np.ceil(0.5 * (1 - frac) * nobs).astype(int) uppidx = np.floor(lowidx + frac * nobs).astype(int) mi_sl = slice(lowidx, uppidx) res_mi = OLS(endog[mi_sl], exog[mi_sl]).fit() nobs_mi = res_mi.model.endog.shape[0] ss_mi = res_mi.ssr ss = res.ssr fstat = (ss - ss_mi) / (nobs-nobs_mi) / ss_mi * res_mi.df_resid from scipy import stats pval = stats.f.sf(fstat, nobs - nobs_mi, res_mi.df_resid) return fstat, pval
[docs]def linear_lm(resid, exog, func=None): '''Lagrange multiplier test for linearity against functional alternative limitations: Assumes currently that the first column is integer. Currently it doesn't check whether the transformed variables contain NaNs, for example log of negative number. Parameters ---------- resid : ndarray residuals of a regression exog : ndarray exogenous variables for which linearity is tested func : callable If func is None, then squares are used. func needs to take an array of exog and return an array of transformed variables. Returns ------- lm : float Lagrange multiplier test statistic lm_pval : float p-value of Lagrange multiplier tes ftest : ContrastResult instance the results from the F test variant of this test Notes ----- written to match Gretl's linearity test. The test runs an auxilliary regression of the residuals on the combined original and transformed regressors. The Null hypothesis is that the linear specification is correct. ''' from scipy import stats if func is None: func = lambda x: np.power(x, 2) exog_aux = np.column_stack((exog, func(exog[:,1:]))) nobs, k_vars = exog.shape ls = OLS(resid, exog_aux).fit() ftest = ls.f_test(np.eye(k_vars - 1, k_vars * 2 - 1, k_vars)) lm = nobs * ls.rsquared lm_pval = stats.chi2.sf(lm, k_vars - 1) return lm, lm_pval, ftest
def _neweywestcov(resid, x): ''' Did not run yet from regstats2 :: if idx(29) % HAC (Newey West) L = round(4*(nobs/100)^(2/9)); % L = nobs^.25; % as an alternative hhat = repmat(residuals',p,1).*X'; xuux = hhat*hhat'; for l = 1:L; za = hhat(:,(l+1):nobs)*hhat(:,1:nobs-l)'; w = 1 - l/(L+1); xuux = xuux + w*(za+za'); end d = struct; d.covb = xtxi*xuux*xtxi; ''' nobs = resid.shape[0] #TODO: check this can only be 1d nlags = int(round(4*(nobs/100.)**(2/9.))) hhat = resid * x.T xuux = np.dot(hhat, hhat.T) for lag in range(nlags): za = np.dot(hhat[:,lag:nobs], hhat[:,:nobs-lag].T) w = 1 - lag/(nobs + 1.) xuux = xuux + np.dot(w, za+za.T) xtxi = np.linalg.inv(np.dot(x.T, x)) #QR instead? covbNW = np.dot(xtxi, np.dot(xuux, xtxi)) return covbNW def _recursive_olsresiduals2(olsresults, skip): '''this is my original version based on Greene and references keep for now for comparison and benchmarking ''' y = olsresults.model.endog x = olsresults.model.exog nobs, nvars = x.shape rparams = np.nan * np.zeros((nobs,nvars)) rresid = np.nan * np.zeros((nobs)) rypred = np.nan * np.zeros((nobs)) rvarraw = np.nan * np.zeros((nobs)) #XTX = np.zeros((nvars,nvars)) #XTY = np.zeros((nvars)) x0 = x[:skip] y0 = y[:skip] XTX = np.dot(x0.T, x0) XTY = np.dot(x0.T, y0) #xi * y #np.dot(xi, y) beta = np.linalg.solve(XTX, XTY) rparams[skip-1] = beta yipred = np.dot(x[skip-1], beta) rypred[skip-1] = yipred rresid[skip-1] = y[skip-1] - yipred rvarraw[skip-1] = 1+np.dot(x[skip-1],np.dot(np.linalg.inv(XTX),x[skip-1])) for i in range(skip,nobs): xi = x[i:i+1,:] yi = y[i] xxT = np.dot(xi.T, xi) #xi is 2d 1 row xy = (xi*yi).ravel() # XTY is 1d #np.dot(xi, yi) #np.dot(xi, y) print(xy.shape, XTY.shape) print(XTX) print(XTY) beta = np.linalg.solve(XTX, XTY) rparams[i-1] = beta #this is beta based on info up to t-1 yipred = np.dot(xi, beta) rypred[i] = yipred rresid[i] = yi - yipred rvarraw[i] = 1 + np.dot(xi,np.dot(np.linalg.inv(XTX),xi.T)) XTX += xxT XTY += xy i = nobs beta = np.linalg.solve(XTX, XTY) rparams[i-1] = beta rresid_scaled = rresid/np.sqrt(rvarraw) #this is N(0,sigma2) distributed nrr = nobs-skip sigma2 = rresid_scaled[skip-1:].var(ddof=1) rresid_standardized = rresid_scaled/np.sqrt(sigma2) #N(0,1) distributed rcusum = rresid_standardized[skip-1:].cumsum() #confidence interval points in Greene p136 looks strange? #this assumes sum of independent standard normal #rcusumci = np.sqrt(np.arange(skip,nobs+1))*np.array([[-1.],[+1.]])*stats.norm.sf(0.025) a = 1.143 #for alpha=0.99 =0.948 for alpha=0.95 #following taken from Ploberger, crit = a*np.sqrt(nrr) rcusumci = (a*np.sqrt(nrr) + a*np.arange(0,nobs-skip)/np.sqrt(nrr)) \ * np.array([[-1.],[+1.]]) return (rresid, rparams, rypred, rresid_standardized, rresid_scaled, rcusum, rcusumci)
[docs]def recursive_olsresiduals(olsresults, skip=None, lamda=0.0, alpha=0.95): '''calculate recursive ols with residuals and cusum test statistic Parameters ---------- olsresults : instance of RegressionResults uses only endog and exog skip : int or None number of observations to use for initial OLS, if None then skip is set equal to the number of regressors (columns in exog) lamda : float weight for Ridge correction to initial (X'X)^{-1} alpha : {0.95, 0.99} confidence level of test, currently only two values supported, used for confidence interval in cusum graph Returns ------- rresid : array recursive ols residuals rparams : array recursive ols parameter estimates rypred : array recursive prediction of endogenous variable rresid_standardized : array recursive residuals standardized so that N(0,sigma2) distributed, where sigma2 is the error variance rresid_scaled : array recursive residuals normalize so that N(0,1) distributed rcusum : array cumulative residuals for cusum test rcusumci : array confidence interval for cusum test, currently hard coded for alpha=0.95 Notes ----- It produces same recursive residuals as other version. This version updates the inverse of the X'X matrix and does not require matrix inversion during updating. looks efficient but no timing Confidence interval in Greene and Brown, Durbin and Evans is the same as in Ploberger after a little bit of algebra. References ---------- jplv to check formulas, follows Harvey BigJudge 5.5.2b for formula for inverse(X'X) updating Greene section 7.5.2 Brown, R. L., J. Durbin, and J. M. Evans. “Techniques for Testing the Constancy of Regression Relationships over Time.” Journal of the Royal Statistical Society. Series B (Methodological) 37, no. 2 (1975): 149-192. ''' y = olsresults.model.endog x = olsresults.model.exog nobs, nvars = x.shape if skip is None: skip = nvars rparams = np.nan * np.zeros((nobs,nvars)) rresid = np.nan * np.zeros((nobs)) rypred = np.nan * np.zeros((nobs)) rvarraw = np.nan * np.zeros((nobs)) #intialize with skip observations x0 = x[:skip] y0 = y[:skip] #add Ridge to start (not in jplv XTXi = np.linalg.inv(np.dot(x0.T, x0)+lamda*np.eye(nvars)) XTY = np.dot(x0.T, y0) #xi * y #np.dot(xi, y) #beta = np.linalg.solve(XTX, XTY) beta = np.dot(XTXi, XTY) #print('beta', beta rparams[skip-1] = beta yipred = np.dot(x[skip-1], beta) rypred[skip-1] = yipred rresid[skip-1] = y[skip-1] - yipred rvarraw[skip-1] = 1 + np.dot(x[skip-1],np.dot(XTXi, x[skip-1])) for i in range(skip,nobs): xi = x[i:i+1,:] yi = y[i] #xxT = np.dot(xi.T, xi) #xi is 2d 1 row xy = (xi*yi).ravel() # XTY is 1d #np.dot(xi, yi) #np.dot(xi, y) #print(xy.shape, XTY.shape #print(XTX #print(XTY # get prediction error with previous beta yipred = np.dot(xi, beta) rypred[i] = yipred residi = yi - yipred rresid[i] = residi #update beta and inverse(X'X) tmp = np.dot(XTXi, xi.T) ft = 1 + np.dot(xi, tmp) XTXi = XTXi - np.dot(tmp,tmp.T) / ft #BigJudge equ 5.5.15 #print('beta', beta beta = beta + (tmp*residi / ft).ravel() #BigJudge equ 5.5.14 # #version for testing # XTY += xy # beta = np.dot(XTXi, XTY) # print((tmp*yipred / ft).shape # print('tmp.shape, ft.shape, beta.shape', tmp.shape, ft.shape, beta.shape rparams[i] = beta rvarraw[i] = ft i = nobs #beta = np.linalg.solve(XTX, XTY) #rparams[i] = beta rresid_scaled = rresid/np.sqrt(rvarraw) #this is N(0,sigma2) distributed nrr = nobs-skip #sigma2 = rresid_scaled[skip-1:].var(ddof=1) #var or sum of squares ? #Greene has var, jplv and Ploberger have sum of squares (Ass.:mean=0) #Gretl uses: by reverse engineering matching their numbers sigma2 = rresid_scaled[skip:].var(ddof=1) rresid_standardized = rresid_scaled/np.sqrt(sigma2) #N(0,1) distributed rcusum = rresid_standardized[skip-1:].cumsum() #confidence interval points in Greene p136 looks strange. Cleared up #this assumes sum of independent standard normal, which does not take into #account that we make many tests at the same time #rcusumci = np.sqrt(np.arange(skip,nobs+1))*np.array([[-1.],[+1.]])*stats.norm.sf(0.025) if alpha == 0.95: a = 0.948 #for alpha=0.95 elif alpha == 0.99: a = 1.143 #for alpha=0.99 elif alpha == 0.90: a = 0.850 else: raise ValueError('alpha can only be 0.9, 0.95 or 0.99') #following taken from Ploberger, crit = a*np.sqrt(nrr) rcusumci = (a*np.sqrt(nrr) + 2*a*np.arange(0,nobs-skip)/np.sqrt(nrr)) \ * np.array([[-1.],[+1.]]) return (rresid, rparams, rypred, rresid_standardized, rresid_scaled, rcusum, rcusumci)
[docs]def breaks_hansen(olsresults): '''test for model stability, breaks in parameters for ols, Hansen 1992 Parameters ---------- olsresults : instance of RegressionResults uses only endog and exog Returns ------- teststat : float Hansen's test statistic crit : structured array critical values at alpha=0.95 for different nvars pvalue Not yet ft, s : arrays temporary return for debugging, will be removed Notes ----- looks good in example, maybe not very powerful for small changes in parameters According to Greene, distribution of test statistics depends on nvar but not on nobs. Test statistic is verified against R:strucchange References ---------- Greene section 7.5.1, notation follows Greene ''' y = olsresults.model.endog x = olsresults.model.exog resid = olsresults.resid nobs, nvars = x.shape resid2 = resid**2 ft = np.c_[x*resid[:,None], (resid2 - resid2.mean())] s = ft.cumsum(0) assert (np.abs(s[-1]) < 1e10).all() #can be optimized away F = nobs*(ft[:,:,None]*ft[:,None,:]).sum(0) S = (s[:,:,None]*s[:,None,:]).sum(0) H = np.trace(np.dot(np.linalg.inv(F), S)) crit95 = np.array([(2,1.9),(6,3.75),(15,3.75),(19,4.52)], dtype = [('nobs',int), ('crit', float)]) #TODO: get critical values from Bruce Hansens' 1992 paper return H, crit95, ft, s
[docs]def breaks_cusumolsresid(olsresidual, ddof=0): '''cusum test for parameter stability based on ols residuals Parameters ---------- olsresiduals : ndarray array of residuals from an OLS estimation ddof : int number of parameters in the OLS estimation, used as degrees of freedom correction for error variance. Returns ------- sup_b : float test statistic, maximum of absolute value of scaled cumulative OLS residuals pval : float Probability of observing the data under the null hypothesis of no structural change, based on asymptotic distribution which is a Brownian Bridge crit: list tabulated critical values, for alpha = 1%, 5% and 10% Notes ----- tested agains R:strucchange Not clear: Assumption 2 in Ploberger, Kramer assumes that exog x have asymptotically zero mean, x.mean(0) = [1, 0, 0, ..., 0] Is this really necessary? I don't see how it can affect the test statistic under the null. It does make a difference under the alternative. Also, the asymptotic distribution of test statistic depends on this. From examples it looks like there is little power for standard cusum if exog (other than constant) have mean zero. References ---------- Ploberger, Werner, and Walter Kramer. “The Cusum Test with Ols Residuals.” Econometrica 60, no. 2 (March 1992): 271-285. ''' resid = olsresidual.ravel() nobs = len(resid) nobssigma2 = (resid**2).sum() if ddof > 0: #print('ddof', ddof, 1. / (nobs - ddof) * nobs nobssigma2 = nobssigma2 / (nobs - ddof) * nobs #B is asymptotically a Brownian Bridge B = resid.cumsum()/np.sqrt(nobssigma2) # use T*sigma directly sup_b = np.abs(B).max() #asymptotically distributed as standard Brownian Bridge crit = [(1,1.63), (5, 1.36), (10, 1.22)] #Note stats.kstwobign.isf(0.1) is distribution of sup.abs of Brownian Bridge #>>> stats.kstwobign.isf([0.01,0.05,0.1]) #array([ 1.62762361, 1.35809864, 1.22384787]) pval = stats.kstwobign.sf(sup_b) return sup_b, pval, crit
#def breaks_cusum(recolsresid): # '''renormalized cusum test for parameter stability based on recursive residuals # # # still incorrect: in PK, the normalization for sigma is by T not T-K # also the test statistic is asymptotically a Wiener Process, Brownian motion # not Brownian Bridge # for testing: result reject should be identical as in standard cusum version # # References # ---------- # Ploberger, Werner, and Walter Kramer. “The Cusum Test with Ols Residuals.” # Econometrica 60, no. 2 (March 1992): 271-285. # # ''' # resid = recolsresid.ravel() # nobssigma2 = (resid**2).sum() # #B is asymptotically a Brownian Bridge # B = resid.cumsum()/np.sqrt(nobssigma2) # use T*sigma directly # nobs = len(resid) # denom = 1. + 2. * np.arange(nobs)/(nobs-1.) #not sure about limits # sup_b = np.abs(B/denom).max() # #asymptotically distributed as standard Brownian Bridge # crit = [(1,1.63), (5, 1.36), (10, 1.22)] # #Note stats.kstwobign.isf(0.1) is distribution of sup.abs of Brownian Bridge # #>>> stats.kstwobign.isf([0.01,0.05,0.1]) # #array([ 1.62762361, 1.35809864, 1.22384787]) # pval = stats.kstwobign.sf(sup_b) # return sup_b, pval, crit def breaks_AP(endog, exog, skip): '''supLM, expLM and aveLM by Andrews, and Andrews,Ploberger p-values by B Hansen just idea for computation of sequence of tests with given change point (Chow tests) run recursive ols both forward and backward, match the two so they form a split of the data, calculate sum of squares for residuals and get test statistic for each breakpoint between skip and nobs-skip need to put recursive ols (residuals) into separate function alternative: B Hansen loops over breakpoints only once and updates x'x and xe'xe update: Andrews is based on GMM estimation not OLS, LM test statistic is easy to compute because it only requires full sample GMM estimate (p.837) with GMM the test has much wider applicability than just OLS for testing loop over single breakpoint Chow test function ''' pass #delete when testing is finished class StatTestMC(object): """class to run Monte Carlo study on a statistical test''' TODO print(summary, for quantiles and for histogram draft in trying out script log this has been copied to tools/mctools.py, with improvements """ def __init__(self, dgp, statistic): self.dgp = dgp #staticmethod(dgp) #no self self.statistic = statistic # staticmethod(statistic) #no self def run(self, nrepl, statindices=None, dgpargs=[], statsargs=[]): '''run the actual Monte Carlo and save results ''' self.nrepl = nrepl self.statindices = statindices self.dgpargs = dgpargs self.statsargs = statsargs dgp = self.dgp statfun = self.statistic # name ? #single return statistic if statindices is None: self.nreturn = nreturns = 1 mcres = np.zeros(nrepl) for ii in range(nrepl-1): x = dgp(*dgpargs) #(1e-4+np.random.randn(nobs)).cumsum() mcres[ii] = statfun(x, *statsargs) #unitroot_adf(x, 2,trendorder=0, autolag=None) #more than one return statistic else: self.nreturn = nreturns = len(statindices) self.mcres = mcres = np.zeros((nrepl, nreturns)) for ii in range(nrepl-1): x = dgp(*dgpargs) #(1e-4+np.random.randn(nobs)).cumsum() ret = statfun(x, *statsargs) mcres[ii] = [ret[i] for i in statindices] self.mcres = mcres def histogram(self, idx=None, critval=None): '''calculate histogram values does not do any plotting ''' if self.mcres.ndim == 2: if idx is not None: mcres = self.mcres[:,idx] else: raise ValueError('currently only 1 statistic at a time') else: mcres = self.mcres if critval is None: histo = np.histogram(mcres, bins=10) else: if not critval[0] == -np.inf: bins=np.r_[-np.inf, critval, np.inf] if not critval[0] == -np.inf: bins=np.r_[bins, np.inf] histo = np.histogram(mcres, bins=np.r_[-np.inf, critval, np.inf]) self.histo = histo self.cumhisto = np.cumsum(histo[0])*1./self.nrepl self.cumhistoreversed = np.cumsum(histo[0][::-1])[::-1]*1./self.nrepl return histo, self.cumhisto, self.cumhistoreversed def quantiles(self, idx=None, frac=[0.01, 0.025, 0.05, 0.1, 0.975]): '''calculate quantiles of Monte Carlo results ''' if self.mcres.ndim == 2: if not idx is None: mcres = self.mcres[:,idx] else: raise ValueError('currently only 1 statistic at a time') else: mcres = self.mcres self.frac = frac = np.asarray(frac) self.mcressort = mcressort = np.sort(self.mcres) return frac, mcressort[(self.nrepl*frac).astype(int)] if __name__ == '__main__': examples = ['adf'] if 'adf' in examples: x = np.random.randn(20) print(acorr_ljungbox(x,4)) print(unitroot_adf(x)) nrepl = 100 nobs = 100 mcres = np.zeros(nrepl) for ii in range(nrepl-1): x = (1e-4+np.random.randn(nobs)).cumsum() mcres[ii] = unitroot_adf(x, 2,trendorder=0, autolag=None)[0] print((mcres<-2.57).sum()) print(np.histogram(mcres)) mcressort = np.sort(mcres) for ratio in [0.01, 0.025, 0.05, 0.1]: print(ratio, mcressort[int(nrepl*ratio)]) print('critical values in Green table 20.5') print('sample size = 100') print('with constant') print('0.01: -19.8, 0.025: -16.3, 0.05: -13.7, 0.01: -11.0, 0.975: 0.47') print('0.01: -3.50, 0.025: -3.17, 0.05: -2.90, 0.01: -2.58, 0.975: 0.26') crvdg = dict([map(float,s.split(':')) for s in ('0.01: -19.8, 0.025: -16.3, 0.05: -13.7, 0.01: -11.0, 0.975: 0.47'.split(','))]) crvd = dict([map(float,s.split(':')) for s in ('0.01: -3.50, 0.025: -3.17, 0.05: -2.90, 0.01: -2.58, 0.975: 0.26'.split(','))]) ''' >>> crvd {0.050000000000000003: -13.699999999999999, 0.97499999999999998: 0.46999999999999997, 0.025000000000000001: -16.300000000000001, 0.01: -11.0} >>> sorted(crvd.values()) [-16.300000000000001, -13.699999999999999, -11.0, 0.46999999999999997] ''' #for trend = 0 crit_5lags0p05 =-4.41519 + (-14.0406)/nobs + (-12.575)/nobs**2 print(crit_5lags0p05) adfstat, _,_,resstore = unitroot_adf(x, 2,trendorder=0, autolag=None, store=1) print((mcres>crit_5lags0p05).sum()) print(resstore.resols.model.exog[-5:]) print(x[-5:]) print(np.histogram(mcres, bins=[-np.inf, -3.5, -3.17, -2.9 , -2.58, 0.26, np.inf])) print(mcressort[(nrepl*(np.array([0.01, 0.025, 0.05, 0.1, 0.975]))).astype(int)]) def randwalksim(nobs=100, drift=0.0): return (drift+np.random.randn(nobs)).cumsum() def normalnoisesim(nobs=500, loc=0.0): return (loc+np.random.randn(nobs)) def adf20(x): return unitroot_adf(x, 2,trendorder=0, autolag=None)[:2] print('\nResults with MC class') mc1 = StatTestMC(randwalksim, adf20) mc1.run(1000, statindices=[0,1]) print(mc1.histogram(0, critval=[-3.5, -3.17, -2.9 , -2.58, 0.26])) print(mc1.quantiles(0)) print('\nLjung Box') def lb4(x): s,p = acorr_ljungbox(x, lags=4) return s[-1], p[-1] def lb4(x): s,p = acorr_ljungbox(x, lags=1) return s[0], p[0] print('Results with MC class') mc1 = StatTestMC(normalnoisesim, lb4) mc1.run(1000, statindices=[0,1]) print(mc1.histogram(1, critval=[0.01, 0.025, 0.05, 0.1, 0.975])) print(mc1.quantiles(1)) print(mc1.quantiles(0)) print(mc1.histogram(0)) nobs = 100 x = np.ones((nobs,2)) x[:,1] = np.arange(nobs)/20. y = x.sum(1) + 1.01*(1+1.5*(x[:,1]>10))*np.random.rand(nobs) print(het_goldfeldquandt(y,x, 1)) y = x.sum(1) + 1.01*(1+0.5*(x[:,1]>10))*np.random.rand(nobs) print(het_goldfeldquandt(y,x, 1)) y = x.sum(1) + 1.01*(1-0.5*(x[:,1]>10))*np.random.rand(nobs) print(het_goldfeldquandt(y,x, 1)) print(het_breuschpagan(y,x)) print(het_white(y,x)) f, fp, fo = het_goldfeldquandt(y,x, 1) print(f, fp) resgq = het_goldfeldquandt(y,x, 1, retres=True) print(resgq) #this is just a syntax check: print(_neweywestcov(y, x)) resols1 = OLS(y, x).fit() print(_neweywestcov(resols1.resid, x)) print(resols1.cov_params()) print(resols1.HC0_se) print(resols1.cov_HC0) y = x.sum(1) + 10.*(1-0.5*(x[:,1]>10))*np.random.rand(nobs) print(HetGoldfeldQuandt().run(y,x, 1, alternative='dec'))