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:                Thu, 03 Aug 2023
Time:                        21:28:34
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())

[ 5.03027791  0.52652647 -0.01369165]
[0.42919039 0.0662612  0.00586309]
[ 4.68798666  4.95649872  5.22044879  5.47983689  5.734663    5.98492712
  6.23062927  6.47176943  6.70834761  6.94036381  7.16781803  7.39071026
  7.60904051  7.82280878  8.03201506  8.23665936  8.43674168  8.63226202
  8.82322037  9.00961675  9.19145114  9.36872354  9.54143397  9.70958241
  9.87316887 10.03219335 10.18665584 10.33655635 10.48189488 10.62267143
 10.75888599 10.89053857 11.01762917 11.14015779 11.25812442 11.37152908
 11.48037174 11.58465243 11.68437113 11.77952786 11.87012259 11.95615535
 12.03762612 12.11453491 12.18688172 12.25466655 12.31788939 12.37655025
 12.43064913 12.48018603]

Estimate RLM:

[9]:

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

[ 4.97283906  0.51653512 -0.005167  ]
[0.13877351 0.02142476 0.00189576]

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 0x7f3fcaa03a90>

../../../_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)

[5.58213524 0.38960997]
[0.37452175 0.03227031]