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:                Wed, 19 Feb 2025
Time:                        12:39:33
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())
[ 5.1576005   0.50315179 -0.01162381]
[0.44593294 0.06884603 0.00609181]
[ 4.8670053   5.117881    5.36488371  5.60801343  5.84727016  6.08265391
  6.31416466  6.54180243  6.7655672   6.98545899  7.20147779  7.4136236
  7.62189642  7.82629625  8.0268231   8.22347695  8.41625781  8.60516569
  8.79020058  8.97136248  9.14865139  9.32206731  9.49161024  9.65728018
  9.81907714  9.9770011  10.13105208 10.28123006 10.42753506 10.56996707
 10.70852609 10.84321212 10.97402517 11.10096522 11.22403228 11.34322636
 11.45854745 11.56999554 11.67757065 11.78127277 11.88110191 11.97705805
 12.0691412  12.15735137 12.24168854 12.32215273 12.39874393 12.47146213
 12.54030735 12.60527959]

Estimate RLM:

[9]:
resrlm = sm.RLM(y2, X).fit()
print(resrlm.params)
print(resrlm.bse)
[ 5.09849015e+00  4.87916022e-01 -1.54050930e-03]
[0.15434397 0.02382863 0.00210847]

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 0x7f09c5bb78b0>
../../../_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)
[5.62611113 0.38691371]
[0.38234977 0.0329448 ]

Estimate RLM:

[12]:
resrlm2 = sm.RLM(y2, X2).fit()
print(resrlm2.params)
print(resrlm2.bse)
[5.14262045 0.47543968]
[0.12342497 0.01063479]

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

Last update: Feb 19, 2025