M-Estimators for Robust Linear Modeling

python

%matplotlib inline

python

import matplotlib.pyplot as plt
import numpy as np
import statsmodels.api as sm
from scipy import stats
from statsmodels.compat import lmap

An M-estimator minimizes the function

$$Q(e_i, \rho) = \sum_i~\rho \left (\frac{e_i}{s}\right )$$

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

python

norms = sm.robust.norms

python

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(-0.1, 1.1)
    return ax

Andrew's Wave

python

help(norms.AndrewWave.weights)

python

a = 1.339
support = np.linspace(-np.pi * a, np.pi * a, 100)
andrew = norms.AndrewWave(a=a)
plot_weights(
    support, andrew.weights, [r"$-\pi*a$", "0", r"$\pi*a$"], [-np.pi * a, 0, np.pi * a]
)

Hampel's 17A

python

help(norms.Hampel.weights)

python

c = 8
support = np.linspace(-3 * c, 3 * c, 1000)
hampel = norms.Hampel(a=2.0, b=4.0, c=c)
plot_weights(support, hampel.weights, ["3*c", "0", "3*c"], [-3 * c, 0, 3 * c])

Huber's t

python

help(norms.HuberT.weights)

python

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

python

help(norms.LeastSquares.weights)

python

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

python

help(norms.RamsayE.weights)

python

a = 0.3
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

python

help(norms.TrimmedMean.weights)

python

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

python

help(norms.TukeyBiweight.weights)

python

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

python

x = np.array([1, 2, 3, 4, 500])

The mean is not a robust estimator of location

python

x.mean()

The median, on the other hand, is a robust estimator with a breakdown point of 50%

python

np.median(x)

Analogously for the scale
The standard deviation is not robust

python

x.std()

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,

$$K = \Phi^{-1}(.75)$$

python

stats.norm.ppf(0.75)

python

print(x)

python

sm.robust.scale.mad(x)

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np.array([1, 2, 3, 4, 5.0]).std()

Another robust estimator of scale is the Interquartile Range (IQR)

$$\left(\hat{X}{0.75} - \hat{X}{0.25}\right),$$

where $\hat{X}_{p}$ is the sample p-th quantile and $K$ depends on the distribution.

The standardized IQR, given by $K \cdot \text{IQR}$ for $$K = \frac{1}{\Phi^{-1}(.75) - \Phi^{-1}(.25)} \approx 0.74,$$ is a consistent estimator of the standard deviation for normal data.

python

sm.robust.scale.iqr(x)

The IQR is less robust than the MAD in the sense that it has a lower breakdown point: it can withstand 25% outlying observations before being completely ruined, whereas the MAD can withstand 50% outlying observations. However, the IQR is better suited for asymmetric distributions.

Yet another robust estimator of scale is the $Q_n$ estimator, introduced in Rousseeuw & Croux (1993), 'Alternatives to the Median Absolute Deviation'. Then $Q_n$ estimator is given by $$ Q_n = K \left\lbrace \vert X_{i} - X_{j}\vert : i<j\right\rbrace_{(h)} $$ where $h\approx (1/4){{n}\choose{2}}$ and $K$ is a given constant. In words, the $Q_n$ estimator is the normalized $h$-th order statistic of the absolute differences of the data. The normalizing constant $K$ is usually chosen as 2.219144, to make the estimator consistent for the standard deviation in the case of normal data. The $Q_n$ estimator has a 50% breakdown point and a 82% asymptotic efficiency at the normal distribution, much higher than the 37% efficiency of the MAD.

python

sm.robust.scale.qn_scale(x)

The default for Robust Linear Models is MAD
another popular choice is Huber's proposal 2

python

np.random.seed(12345)
fat_tails = stats.t(6).rvs(40)

python

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)

python

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)

python

sm.robust.mad(fat_tails)

python

sm.robust.mad(fat_tails, c=stats.t(6).ppf(0.75))

python

sm.robust.scale.mad(fat_tails)

Duncan's Occupational Prestige data - M-estimation for outliers

python

from statsmodels.formula.api import ols, rlm
from statsmodels.graphics.api import abline_plot

python

prestige = sm.datasets.get_rdataset("Duncan", "carData", cache=True).data

python

print(prestige.head(10))

python

fig = plt.figure(figsize=(12, 12))
ax1 = fig.add_subplot(211, xlabel="Income", ylabel="Prestige")
ax1.scatter(prestige.income, prestige.prestige)
xy_outlier = prestige.loc["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)

python

ols_model = ols("prestige ~ income + education", prestige).fit()
print(ols_model.summary())

python

infl = ols_model.get_influence()
student = infl.summary_frame()["student_resid"]
print(student)

python

print(student.loc[np.abs(student) > 2])

python

print(infl.summary_frame().loc["minister"])

python

sidak = ols_model.outlier_test("sidak")
sidak.sort_values("unadj_p", inplace=True)
print(sidak)

python

fdr = ols_model.outlier_test("fdr_bh")
fdr.sort_values("unadj_p", inplace=True)
print(fdr)

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rlm_model = rlm("prestige ~ income + education", prestige).fit()
print(rlm_model.summary())

python

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.

python

dta = sm.datasets.get_rdataset("starsCYG", "robustbase", cache=True).data

python

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), 0.2, 1, alpha=0.25, 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.loc[dta["log.Te"] < 3.8].iterrows():
    ax.annotate(i, row, row + 0.01, fontsize=14)
xlim, ylim = ax.get_xlim(), ax.get_ylim()

python

from IPython.display import Image

Image(filename="star_diagram.png")

python

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)

python

rlm_mod = sm.RLM(y, X, sm.robust.norms.TrimmedMean(0.5)).fit()
abline_plot(model_results=rlm_mod, ax=ax, color="red")

Why? Because M-estimators are not robust to leverage points.

python

infl = ols_model.get_influence()

python

h_bar = 2 * (ols_model.df_model + 1) / ols_model.nobs
hat_diag = infl.summary_frame()["hat_diag"]
hat_diag.loc[hat_diag > h_bar]

python

sidak2 = ols_model.outlier_test("sidak")
sidak2.sort_values("unadj_p", inplace=True)
print(sidak2)

python

fdr2 = ols_model.outlier_test("fdr_bh")
fdr2.sort_values("unadj_p", inplace=True)
print(fdr2)

Let's delete that line

python

l = ax.lines[-1]
l.remove()
del l

python

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 do not 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.

python

yy = y.values[:, None]
xx = X["log.Te"].values[:, None]

Note: The R code and the results in this notebook has been converted to markdown so that R is not required to build the documents. The R results in the notebook were computed using R 3.5.1 and robustbase 0.93.

ipython

%load_ext rpy2.ipython

%R library(robustbase)
%Rpush yy xx
%R mod <- lmrob(yy ~ xx);
%R params <- mod$coefficients;
%Rpull params

ipython

%R print(mod)

Call:
lmrob(formula = yy ~ xx)
 \--> method = "MM"
Coefficients:
(Intercept)           xx  
     -4.969        2.253

python

params = [-4.969387980288108, 2.2531613477892365]  # Computed using R
print(params[0], params[1])

python

abline_plot(intercept=params[0], slope=params[1], ax=ax, color="red")

Exercise: Breakdown points of M-estimator

python

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 = 0.25  # percentage of response variables to contaminate

python

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)

python

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

python

np.array(se_beta).mean()

python

all_betas.mean(0)

python

beta_true

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se_loss(all_betas.mean(0) - beta_true)