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:                Thu, 16 Apr 2026
Time:                        11:53:13
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())
[ 4.93218646  0.52099024 -0.01169552]
[0.43705981 0.06747614 0.0059706 ]
[ 4.63979852  4.89823596  5.15277653  5.40342021  5.65016701  5.89301692
  6.13196995  6.36702611  6.59818538  6.82544776  7.04881327  7.26828189
  7.48385363  7.69552849  7.90330647  8.10718756  8.30717177  8.5032591
  8.69544955  8.88374311  9.0681398   9.2486396   9.42524252  9.59794855
  9.76675771  9.93166998 10.09268537 10.24980388 10.4030255  10.55235025
 10.69777811 10.83930909 10.97694318 11.1106804  11.24052073 11.36646418
 11.48851075 11.60666044 11.72091324 11.83126916 11.9377282  12.04029036
 12.13895563 12.23372403 12.32459554 12.41157017 12.49464791 12.57382878
 12.64911276 12.72049986]

Estimate RLM:

[9]:
resrlm = sm.RLM(y2, X).fit()
print(resrlm.params)
print(resrlm.bse)
[ 4.85061903e+00  5.11439382e-01 -2.28855133e-03]
[0.12889962 0.01990036 0.00176087]

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 0x7fd3fbe29db0>
../../../_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)
[5.40358743 0.40403507]
[0.3754672  0.03235177]

Estimate RLM:

[12]:
resrlm2 = sm.RLM(y2, X2).fit()
print(resrlm2.params)
print(resrlm2.bse)
[4.90828079 0.49369184]
[0.09935433 0.00856077]

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

[13]:
pred_ols = res2.get_prediction()
iv_l = pred_ols.summary_frame()["obs_ci_lower"]
iv_u = pred_ols.summary_frame()["obs_ci_upper"]

fig, ax = plt.subplots(figsize=(8, 6))
ax.plot(x1, y2, "o", label="data")
ax.plot(x1, y_true2, "b-", label="True")
ax.plot(x1, res2.fittedvalues, "r-", label="OLS")
ax.plot(x1, iv_u, "r--")
ax.plot(x1, iv_l, "r--")
ax.plot(x1, resrlm2.fittedvalues, "g.-", label="RLM")
legend = ax.legend(loc="best")
../../../_images/examples_notebooks_generated_robust_models_0_24_0.png