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, 01 Sep 2023
Time:                        11:12:48
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())
[ 5.12904695  0.51939631 -0.01304549]
[0.46485848 0.07176788 0.00635035]
[ 4.80290969  5.06598174  5.3247071   5.57908578  5.82911778  6.07480308
  6.31614171  6.55313364  6.78577889  7.01407746  7.23802934  7.45763453
  7.67289304  7.88380486  8.09036999  8.29258844  8.49046021  8.68398529
  8.87316368  9.05799539  9.23848041  9.41461875  9.5864104   9.75385536
  9.91695364 10.07570524 10.23011015 10.38016837 10.5258799  10.66724476
 10.80426292 10.9369344  11.06525919 11.1892373  11.30886873 11.42415346
 11.53509151 11.64168288 11.74392756 11.84182555 11.93537686 12.02458148
 12.10943942 12.18995067 12.26611524 12.33793312 12.40540431 12.46852882
 12.52730664 12.58173778]

Estimate RLM:

[9]:
resrlm = sm.RLM(y2, X).fit()
print(resrlm.params)
print(resrlm.bse)
[ 5.06153136e+00  5.05498374e-01 -2.78952691e-03]
[0.15744795 0.02430784 0.00215087]

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 0x7f6eb8f21c60>
../../../_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)
[5.65486008 0.3889414 ]
[0.40084994 0.03453885]

Estimate RLM:

[12]:
resrlm2 = sm.RLM(y2, X2).fit()
print(resrlm2.params)
print(resrlm2.bse)
[5.17628139 0.48009342]
[0.12507884 0.0107773 ]

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: Sep 01, 2023