Robust Linear Models¶

[1]:

%matplotlib inline

[2]:

import matplotlib.pyplot as plt
import numpy as np
import statsmodels.api as sm

Estimation¶

Load data:

[3]:

data = sm.datasets.stackloss.load()
data.exog = sm.add_constant(data.exog)

Huber’s T norm with the (default) median absolute deviation scaling

[4]:

huber_t = sm.RLM(data.endog, data.exog, M=sm.robust.norms.HuberT())
hub_results = huber_t.fit()
print(hub_results.params)
print(hub_results.bse)
print(
    hub_results.summary(
        yname="y", xname=["var_%d" % i for i in range(len(hub_results.params))]
    )
)

const       -41.026498
AIRFLOW       0.829384
WATERTEMP     0.926066
ACIDCONC     -0.127847
dtype: float64
const        9.791899
AIRFLOW      0.111005
WATERTEMP    0.302930
ACIDCONC     0.128650
dtype: float64
                    Robust linear Model Regression Results
==============================================================================
Dep. Variable:                      y   No. Observations:                   21
Model:                            RLM   Df Residuals:                       17
Method:                          IRLS   Df Model:                            3
Norm:                          HuberT
Scale Est.:                       mad
Cov Type:                          H1
Date:                Mon, 29 Nov 2021
Time:                        22:05:33
No. Iterations:                    19
==============================================================================
                 coef    std err          z      P>|z|      [0.025      0.975]
------------------------------------------------------------------------------
var_0        -41.0265      9.792     -4.190      0.000     -60.218     -21.835
var_1          0.8294      0.111      7.472      0.000       0.612       1.047
var_2          0.9261      0.303      3.057      0.002       0.332       1.520
var_3         -0.1278      0.129     -0.994      0.320      -0.380       0.124
==============================================================================

If the model instance has been used for another fit with different fit parameters, then the fit options might not be the correct ones anymore .

Huber’s T norm with ‘H2’ covariance matrix

[5]:

hub_results2 = huber_t.fit(cov="H2")
print(hub_results2.params)
print(hub_results2.bse)

const       -41.026498
AIRFLOW       0.829384
WATERTEMP     0.926066
ACIDCONC     -0.127847
dtype: float64
const        9.089504
AIRFLOW      0.119460
WATERTEMP    0.322355
ACIDCONC     0.117963
dtype: float64

Andrew’s Wave norm with Huber’s Proposal 2 scaling and ‘H3’ covariance matrix

[6]:

andrew_mod = sm.RLM(data.endog, data.exog, M=sm.robust.norms.AndrewWave())
andrew_results = andrew_mod.fit(scale_est=sm.robust.scale.HuberScale(), cov="H3")
print("Parameters: ", andrew_results.params)

Parameters:  const       -40.881796
AIRFLOW       0.792761
WATERTEMP     1.048576
ACIDCONC     -0.133609
dtype: float64

See help(sm.RLM.fit) for more options and module sm.robust.scale for scale options

Comparing OLS and RLM¶

Artificial data with outliers:

[7]:

nsample = 50
x1 = np.linspace(0, 20, nsample)
X = np.column_stack((x1, (x1 - 5) ** 2))
X = sm.add_constant(X)
sig = 0.3  # smaller error variance makes OLS<->RLM contrast bigger
beta = [5, 0.5, -0.0]
y_true2 = np.dot(X, beta)
y2 = y_true2 + sig * 1.0 * np.random.normal(size=nsample)
y2[[39, 41, 43, 45, 48]] -= 5  # add some outliers (10% of nsample)

Example 1: quadratic function with linear truth¶

Note that the quadratic term in OLS regression will capture outlier effects.

[8]:

res = sm.OLS(y2, X).fit()
print(res.params)
print(res.bse)
print(res.predict())

[ 5.05333008  0.53655535 -0.01514808]
[0.44265788 0.0683404  0.00604707]
[ 4.67462818  4.95293561  5.2261958   5.49440872  5.75757439  6.01569281
  6.26876397  6.51678787  6.75976452  6.99769392  7.23057605  7.45841094
  7.68119856  7.89893893  8.11163205  8.31927791  8.52187651  8.71942786
  8.91193195  9.09938879  9.28179837  9.45916069  9.63147576  9.79874357
  9.96096413 10.11813743 10.27026348 10.41734227 10.55937381 10.69635809
 10.82829511 10.95518488 11.07702739 11.19382265 11.30557065 11.4122714
 11.51392489 11.61053112 11.7020901  11.78860183 11.87006629 11.94648351
 12.01785346 12.08417616 12.14545161 12.2016798  12.25286073 12.29899441
 12.34008083 12.37612   ]

Estimate RLM:

[9]:

resrlm = sm.RLM(y2, X).fit()
print(resrlm.params)
print(resrlm.bse)

[ 4.98243613  0.52421267 -0.00502269]
[0.1200262  0.01853043 0.00163966]

Draw a plot to compare OLS estimates to the robust estimates:

[10]:

fig = plt.figure(figsize=(12, 8))
ax = fig.add_subplot(111)
ax.plot(x1, y2, "o", label="data")
ax.plot(x1, y_true2, "b-", label="True")
pred_ols = res.get_prediction()
iv_l = pred_ols.summary_frame()["obs_ci_lower"]
iv_u = pred_ols.summary_frame()["obs_ci_upper"]

ax.plot(x1, res.fittedvalues, "r-", label="OLS")
ax.plot(x1, iv_u, "r--")
ax.plot(x1, iv_l, "r--")
ax.plot(x1, resrlm.fittedvalues, "g.-", label="RLM")
ax.legend(loc="best")

[10]:

<matplotlib.legend.Legend at 0x7fd0918a5df0>

../../../_images/examples_notebooks_generated_robust_models_0_18_1.png

Example 2: linear function with linear truth¶

Fit a new OLS model using only the linear term and the constant:

[11]:

X2 = X[:, [0, 1]]
res2 = sm.OLS(y2, X2).fit()
print(res2.params)
print(res2.bse)

[5.66389028 0.38507459]
[0.38928827 0.03354265]