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:                Sun, 20 Feb 2022
Time:                        20:03:58
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())

[ 5.01047523  0.52892395 -0.01286807]
[0.46798555 0.07225065 0.00639307]
[ 4.68877346  4.95503974  5.21701845  5.47470959  5.72811316  5.97722916
  6.22205758  6.46259844  6.69885173  6.93081744  7.15849559  7.38188616
  7.60098917  7.8158046   8.02633247  8.23257276  8.43452549  8.63219064
  8.82556822  9.01465823  9.19946067  9.37997555  9.55620285  9.72814258
  9.89579474 10.05915933 10.21823634 10.37302579 10.52352767 10.66974198
 10.81166872 10.94930788 11.08265948 11.2117235  11.33649996 11.45698885
 11.57319016 11.6851039  11.79273008 11.89606868 11.99511971 12.08988318
 12.18035907 12.26654739 12.34844814 12.42606132 12.49938693 12.56842497
 12.63317544 12.69363834]

Estimate RLM:

[9]:

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

[ 4.92861950e+00  5.12874344e-01 -1.84593051e-03]
[0.1267724  0.01957195 0.00173182]

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 0x7f9566ab91f0>

../../../_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)

[5.52913727 0.40024324]
[0.40288132 0.03471388]