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, 24 Jul 2022
Time:                        16:51:07
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())

[ 5.15035135  0.51253272 -0.01256028]
[0.47761634 0.07373752 0.00652463]
[ 4.83634428  5.09471526  5.34890123  5.59890217  5.8447181   6.08634902
  6.32379491  6.55705579  6.78613165  7.01102249  7.23172832  7.44824913
  7.66058492  7.8687357   8.07270146  8.2724822   8.46807792  8.65948863
  8.84671432  9.02975499  9.20861064  9.38328128  9.5537669   9.72006751
  9.88218309 10.04011366 10.19385921 10.34341975 10.48879527 10.62998577
 10.76699125 10.89981172 11.02844716 11.1528976  11.27316301 11.38924341
 11.50113879 11.60884915 11.7123745  11.81171482 11.90687014 11.99784043
 12.08462571 12.16722597 12.24564121 12.31987144 12.38991664 12.45577684
 12.51745201 12.57494217]

Estimate RLM:

[9]:

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

[ 5.07961319e+00  4.97150832e-01 -1.53850538e-03]
[0.14933713 0.02305564 0.00204007]

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 0x7ffa89ba93f0>

../../../_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)

[5.65660765 0.38692989]
[0.40977825 0.03530815]