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:                Fri, 04 Aug 2023
Time:                        17:42:09
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())
[ 4.9104817   0.53042093 -0.01177281]
[0.48157592 0.07434882 0.00657872]
[ 4.61616154  4.87875083  5.13741749  5.39216151  5.6429829   5.88988165
  6.13285777  6.37191125  6.6070421   6.83825032  7.0655359   7.28889885
  7.50833916  7.72385684  7.93545188  8.14312429  8.34687407  8.54670121
  8.74260572  8.93458759  9.12264683  9.30678343  9.4869974   9.66328873
  9.83565743 10.0041035  10.16862693 10.32922773 10.48590589 10.63866142
 10.78749432 10.93240458 11.0733922  11.2104572  11.34359955 11.47281928
 11.59811636 11.71949082 11.83694264 11.95047182 12.06007838 12.16576229
 12.26752358 12.36536222 12.45927824 12.54927162 12.63534236 12.71749047
 12.79571595 12.87001879]

Estimate RLM:

[9]:
resrlm = sm.RLM(y2, X).fit()
print(resrlm.params)
print(resrlm.bse)
[ 4.83206544e+00  5.17766555e-01 -9.51488034e-04]
[0.12703035 0.01961177 0.00173534]

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 0x7f9b4d865e90>
../../../_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)
[5.38499787 0.41269286]
[0.41111921 0.03542369]

Estimate RLM:

[12]:
resrlm2 = sm.RLM(y2, X2).fit()
print(resrlm2.params)
print(resrlm2.bse)
[4.86052521 0.50981062]
[0.10340455 0.00890975]

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