M-Estimators for Robust Linear Modeling¶
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%matplotlib inline
from __future__ import print_function
from statsmodels.compat import lmap
import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
import statsmodels.api as sm
- An M-estimator minimizes the function
where $\rho$ is a symmetric function of the residuals
- The effect of $\rho$ is to reduce the influence of outliers
- $s$ is an estimate of scale.
- The robust estimates $\hat{\beta}$ are computed by the iteratively re-weighted least squares algorithm
- We have several choices available for the weighting functions to be used
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norms = sm.robust.norms
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def plot_weights(support, weights_func, xlabels, xticks):
fig = plt.figure(figsize=(12,8))
ax = fig.add_subplot(111)
ax.plot(support, weights_func(support))
ax.set_xticks(xticks)
ax.set_xticklabels(xlabels, fontsize=16)
ax.set_ylim(-.1, 1.1)
return ax
Andrew's Wave¶
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help(norms.AndrewWave.weights)
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a = 1.339
support = np.linspace(-np.pi*a, np.pi*a, 100)
andrew = norms.AndrewWave(a=a)
plot_weights(support, andrew.weights, ['$-\pi*a$', '0', '$\pi*a$'], [-np.pi*a, 0, np.pi*a]);
Hampel's 17A¶
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help(norms.Hampel.weights)
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c = 8
support = np.linspace(-3*c, 3*c, 1000)
hampel = norms.Hampel(a=2., b=4., c=c)
plot_weights(support, hampel.weights, ['3*c', '0', '3*c'], [-3*c, 0, 3*c]);
Huber's t¶
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help(norms.HuberT.weights)
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t = 1.345
support = np.linspace(-3*t, 3*t, 1000)
huber = norms.HuberT(t=t)
plot_weights(support, huber.weights, ['-3*t', '0', '3*t'], [-3*t, 0, 3*t]);
Least Squares¶
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help(norms.LeastSquares.weights)
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support = np.linspace(-3, 3, 1000)
lst_sq = norms.LeastSquares()
plot_weights(support, lst_sq.weights, ['-3', '0', '3'], [-3, 0, 3]);
Ramsay's Ea¶
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help(norms.RamsayE.weights)
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a = .3
support = np.linspace(-3*a, 3*a, 1000)
ramsay = norms.RamsayE(a=a)
plot_weights(support, ramsay.weights, ['-3*a', '0', '3*a'], [-3*a, 0, 3*a]);
Trimmed Mean¶
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help(norms.TrimmedMean.weights)
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c = 2
support = np.linspace(-3*c, 3*c, 1000)
trimmed = norms.TrimmedMean(c=c)
plot_weights(support, trimmed.weights, ['-3*c', '0', '3*c'], [-3*c, 0, 3*c]);
Tukey's Biweight¶
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help(norms.TukeyBiweight.weights)
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c = 4.685
support = np.linspace(-3*c, 3*c, 1000)
tukey = norms.TukeyBiweight(c=c)
plot_weights(support, tukey.weights, ['-3*c', '0', '3*c'], [-3*c, 0, 3*c]);
Scale Estimators¶
- Robust estimates of the location
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x = np.array([1, 2, 3, 4, 500])
- The mean is not a robust estimator of location
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x.mean()
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- The median, on the other hand, is a robust estimator with a breakdown point of 50%
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np.median(x)
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- Analagously for the scale
- The standard deviation is not robust
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x.std()
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Median Absolute Deviation
$$ median_i |X_i - median_j(X_j)|) $$Standardized Median Absolute Deviation is a consistent estimator for $\hat{\sigma}$
$$\hat{\sigma}=K \cdot MAD$$where $K$ depends on the distribution. For the normal distribution for example,
$$K = \Phi^{-1}(.75)$$In [22]:
stats.norm.ppf(.75)
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print(x)
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sm.robust.scale.mad(x)
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np.array([1,2,3,4,5.]).std()
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- The default for Robust Linear Models is MAD
- another popular choice is Huber's proposal 2
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np.random.seed(12345)
fat_tails = stats.t(6).rvs(40)
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kde = sm.nonparametric.KDEUnivariate(fat_tails)
kde.fit()
fig = plt.figure(figsize=(12,8))
ax = fig.add_subplot(111)
ax.plot(kde.support, kde.density);
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print(fat_tails.mean(), fat_tails.std())
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print(stats.norm.fit(fat_tails))
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print(stats.t.fit(fat_tails, f0=6))
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huber = sm.robust.scale.Huber()
loc, scale = huber(fat_tails)
print(loc, scale)
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sm.robust.mad(fat_tails)
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sm.robust.mad(fat_tails, c=stats.t(6).ppf(.75))
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sm.robust.scale.mad(fat_tails)
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Duncan's Occupational Prestige data - M-estimation for outliers¶
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from statsmodels.graphics.api import abline_plot
from statsmodels.formula.api import ols, rlm
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prestige = sm.datasets.get_rdataset("Duncan", "car", cache=True).data
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print(prestige.head(10))
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fig = plt.figure(figsize=(12,12))
ax1 = fig.add_subplot(211, xlabel='Income', ylabel='Prestige')
ax1.scatter(prestige.income, prestige.prestige)
xy_outlier = prestige.ix['minister'][['income','prestige']]
ax1.annotate('Minister', xy_outlier, xy_outlier+1, fontsize=16)
ax2 = fig.add_subplot(212, xlabel='Education',
ylabel='Prestige')
ax2.scatter(prestige.education, prestige.prestige);
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ols_model = ols('prestige ~ income + education', prestige).fit()
print(ols_model.summary())
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infl = ols_model.get_influence()
student = infl.summary_frame()['student_resid']
print(student)
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print(student.ix[np.abs(student) > 2])
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print(infl.summary_frame().ix['minister'])
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sidak = ols_model.outlier_test('sidak')
sidak.sort('unadj_p', inplace=True)
print(sidak)
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fdr = ols_model.outlier_test('fdr_bh')
fdr.sort('unadj_p', inplace=True)
print(fdr)
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rlm_model = rlm('prestige ~ income + education', prestige).fit()
print(rlm_model.summary())
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print(rlm_model.weights)
Hertzprung Russell data for Star Cluster CYG 0B1 - Leverage Points¶
- Data is on the luminosity and temperature of 47 stars in the direction of Cygnus.
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dta = sm.datasets.get_rdataset("starsCYG", "robustbase", cache=True).data
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from matplotlib.patches import Ellipse
fig = plt.figure(figsize=(12,8))
ax = fig.add_subplot(111, xlabel='log(Temp)', ylabel='log(Light)', title='Hertzsprung-Russell Diagram of Star Cluster CYG OB1')
ax.scatter(*dta.values.T)
# highlight outliers
e = Ellipse((3.5, 6), .2, 1, alpha=.25, color='r')
ax.add_patch(e);
ax.annotate('Red giants', xy=(3.6, 6), xytext=(3.8, 6),
arrowprops=dict(facecolor='black', shrink=0.05, width=2),
horizontalalignment='left', verticalalignment='bottom',
clip_on=True, # clip to the axes bounding box
fontsize=16,
)
# annotate these with their index
for i,row in dta.ix[dta['log.Te'] < 3.8].iterrows():
ax.annotate(i, row, row + .01, fontsize=14)
xlim, ylim = ax.get_xlim(), ax.get_ylim()
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from IPython.display import Image
Image(filename='star_diagram.png')
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y = dta['log.light']
X = sm.add_constant(dta['log.Te'], prepend=True)
ols_model = sm.OLS(y, X).fit()
abline_plot(model_results=ols_model, ax=ax)
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rlm_mod = sm.RLM(y, X, sm.robust.norms.TrimmedMean(.5)).fit()
abline_plot(model_results=rlm_mod, ax=ax, color='red')
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- Why? Because M-estimators are not robust to leverage points.
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infl = ols_model.get_influence()
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h_bar = 2*(ols_model.df_model + 1 )/ols_model.nobs
hat_diag = infl.summary_frame()['hat_diag']
hat_diag.ix[hat_diag > h_bar]
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sidak2 = ols_model.outlier_test('sidak')
sidak2.sort('unadj_p', inplace=True)
print(sidak2)
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fdr2 = ols_model.outlier_test('fdr_bh')
fdr2.sort('unadj_p', inplace=True)
print(fdr2)
- Let's delete that line
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del ax.lines[-1]
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weights = np.ones(len(X))
weights[X[X['log.Te'] < 3.8].index.values - 1] = 0
wls_model = sm.WLS(y, X, weights=weights).fit()
abline_plot(model_results=wls_model, ax=ax, color='green')
- MM estimators are good for this type of problem, unfortunately, we don't yet have these yet.
- It's being worked on, but it gives a good excuse to look at the R cell magics in the notebook.
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yy = y.values[:,None]
xx = X['log.Te'].values[:,None]
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%load_ext rpy2.ipython
%R library(robustbase)
%Rpush yy xx
%R mod <- lmrob(yy ~ xx);
%R params <- mod$coefficients;
%Rpull params
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%R print(mod)
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print(params)
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abline_plot(intercept=params[0], slope=params[1], ax=ax, color='green')
Exercise: Breakdown points of M-estimator¶
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np.random.seed(12345)
nobs = 200
beta_true = np.array([3, 1, 2.5, 3, -4])
X = np.random.uniform(-20,20, size=(nobs, len(beta_true)-1))
# stack a constant in front
X = sm.add_constant(X, prepend=True) # np.c_[np.ones(nobs), X]
mc_iter = 500
contaminate = .25 # percentage of response variables to contaminate
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all_betas = []
for i in range(mc_iter):
y = np.dot(X, beta_true) + np.random.normal(size=200)
random_idx = np.random.randint(0, nobs, size=int(contaminate * nobs))
y[random_idx] = np.random.uniform(-750, 750)
beta_hat = sm.RLM(y, X).fit().params
all_betas.append(beta_hat)
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all_betas = np.asarray(all_betas)
se_loss = lambda x : np.linalg.norm(x, ord=2)**2
se_beta = lmap(se_loss, all_betas - beta_true)
Squared error loss¶
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np.array(se_beta).mean()
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all_betas.mean(0)
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beta_true
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se_loss(all_betas.mean(0) - beta_true)
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