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:                Sat, 09 Jul 2022
Time:                        10:15:30
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())

[ 5.14380904  0.49992059 -0.01238451]
[0.44551594 0.06878165 0.00608611]
[ 4.83419632  5.08673133  5.33513989  5.579422    5.81957766  6.05560687
  6.28750963  6.51528594  6.7389358   6.95845921  7.17385616  7.38512667
  7.59227073  7.79528834  7.99417949  8.1889442   8.37958246  8.56609426
  8.74847962  8.92673852  9.10087098  9.27087699  9.43675654  9.59850964
  9.7561363   9.9096365  10.05901025 10.20425756 10.34537841 10.48237281
 10.61524077 10.74398227 10.86859732 10.98908592 11.10544807 11.21768377
 11.32579302 11.42977582 11.52963217 11.62536207 11.71696552 11.80444252
 11.88779307 11.96701717 12.04211481 12.11308601 12.17993076 12.24264906
 12.3012409  12.3557063 ]

Estimate RLM:

[9]:

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

[ 5.07467625e+00  4.82490712e-01 -1.12882951e-03]
[0.10945501 0.01689838 0.00149525]

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 0x7f7cda5de020>

../../../_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)

[5.64298058 0.3760755 ]
[0.3838729  0.03307604]