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, 20 Feb 2022
Time:                        20:03:58
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())
[ 5.09174198  0.53380263 -0.01476704]
[0.47521152 0.07336624 0.00649178]
[ 4.72256596  4.99825807  5.26902989  5.53488141  5.79581263  6.05182355
  6.30291418  6.54908452  6.79033455  7.02666429  7.25807373  7.48456288
  7.70613172  7.92278027  8.13450853  8.34131649  8.54320415  8.74017151
  8.93221858  9.11934535  9.30155182  9.478838    9.65120388  9.81864946
  9.98117474 10.13877973 10.29146443 10.43922882 10.58207292 10.71999672
 10.85300023 10.98108343 11.10424634 11.22248896 11.33581128 11.4442133
 11.54769502 11.64625645 11.73989758 11.82861841 11.91241895 11.99129919
 12.06525913 12.13429877 12.19841812 12.25761717 12.31189593 12.36125439
 12.40569255 12.44521041]

Estimate RLM:

[9]:
resrlm = sm.RLM(y2, X).fit()
print(resrlm.params)
print(resrlm.bse)
[ 5.03316343e+00  5.11977408e-01 -2.78517276e-03]
[0.1388404  0.02143508 0.00189667]

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 0x7fe0ebfcc790>
../../../_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)
[5.68694414 0.38613222]
[0.41357701 0.03563547]

Estimate RLM:

[12]:
resrlm2 = sm.RLM(y2, X2).fit()
print(resrlm2.params)
print(resrlm2.bse)
[5.11255222 0.48812715]
[0.10698696 0.00921843]

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