Test independence

from __future__ import print_function
import openturns as ot
import openturns.viewer as viewer
from matplotlib import pylab as plt
ot.Log.Show(ot.Log.NONE)

Sample independence test

In this paragraph we perform tests to assess whether two 1-d samples are independent or not.

The following tests are available :

  • the ChiSquared test: it tests if both scalar samples (discrete ones only) are independent. If n_{ij} is the number of values of the sample i=(1,2) in the modality 1 \leq j \leq m, \displaystyle n_{i.} = \sum_{j=1}^m n_{ij} \displaystyle n_{.j} = \sum_{i=1}^2 n_{ij}, and the ChiSquared test evaluates the decision variable:

D^2 = \sum_{i=1}^2 \sum_{j=1}^m \frac{( n_{ij} - \frac{n_{i.} n_{.j}}{n} )^2}{\frac{n_{i.} n_{.j}}{n}}

which tends towards the \chi^2(m-1) distribution. The hypothesis of independence is rejected if D^2 is too high (depending on the p-value threshold).

  • the Pearson test: it tests if there exists a linear relation between two scalar samples which form a gaussian vector (which is equivalent to have a linear correlation coefficient not equal to zero). If both samples are \underline{x} = (x_i)_{1 \leq i \leq n} and \underline{y} = (y_i)_{1 \leq i \leq n}, and \bar{x} = \displaystyle \frac{1}{n}\sum_{i=1}^n x_i and \bar{y} = \displaystyle \frac{1}{n}\sum_{i=1}^n y_i, the Pearson test evaluates the decision variable:

D = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^n (x_i - \bar{x})^2\sum_{i=1}^n (y_i - \bar{y})^2}}

The variable D tends towards a \chi^2(n-2), under the hypothesis of normality of both samples. The hypothesis of a linear coefficient equal to 0 is rejected (which is equivalent to the independence of the samples) if D is too high (depending on the p-value threshold).

  • the Spearman test: it tests if there exists a monotonous relation between two scalar samples. If both samples are \underline{x} = (x_i)_{1 \leq i \leq n} and \underline{y}= (y_i)_{1 \leq i \leq n},, the Spearman test evaluates the decision variable:

D = 1-\frac{6\sum_{i=1}^n (r_i - s_i)^2}{n(n^2-1)}

where r_i = rank(x_i) and s_i = rank(y_i). D is such that \sqrt{n-1}D tends towards the standard normal distribution.

The continuous case

We create two different continuous samples :

sample1 = ot.Normal().getSample(100)
sample2 = ot.Normal().getSample(100)

We first use the Pearson test and store the result :

resultPearson = ot.HypothesisTest.Pearson(sample1, sample2, 0.10)

We can then display the result of the test as a yes/no answer with the getBinaryQualityMeasure. We can retrieve the p-value and the threshold with the getPValue and getThreshold methods.

print('Component is normal?', resultPearson.getBinaryQualityMeasure(),
      'p-value=%.6g' % resultPearson.getPValue(),
      'threshold=%.6g' % resultPearson.getThreshold())

Out:

Component is normal? False p-value=0.0451584 threshold=0.1

We can also use the Spearman test :

resultSpearman = ot.HypothesisTest.Spearman(sample1, sample2, 0.10)
print('Component is normal?', resultSpearman.getBinaryQualityMeasure(),
      'p-value=%.6g' % resultSpearman.getPValue(),
      'threshold=%.6g' % resultSpearman.getThreshold())

Out:

Component is normal? False p-value=0.0603411 threshold=0.1

The discrete case

Testing is also possible for discrete distribution. Let us create discrete two different samples :

sample1 = ot.Poisson(0.2).getSample(100)
sample2 = ot.Poisson(0.2).getSample(100)

We use the Chi2 test to check independence and store the result :

resultChi2 = ot.HypothesisTest.ChiSquared(sample1, sample2, 0.10)

and display the results :

print('Component is normal?', resultChi2.getBinaryQualityMeasure(),
      'p-value=%.6g' % resultChi2.getPValue(),
      'threshold=%.6g' % resultChi2.getThreshold())

Out:

Component is normal? True p-value=0.20552 threshold=0.1

Test samples independence using regression

Independence testing with regression is also an option in OpenTURNS. It consists in detecting a linear relation between two scalar samples.

We generate a sample of dimension 3 with component 0 correlated to component 2 :

marginals = [ot.Normal()] * 3
S = ot.CorrelationMatrix(3)
S[0, 2] = 0.9
copula = ot.NormalCopula(S)
distribution = ot.ComposedDistribution(marginals, copula)
sample = distribution.getSample(30)

Next, we split it in two samples : firstSample of dimension=2, secondSample of dimension=1.

firstSample = sample[:, :2]
secondSample = sample[:, 2]

We test independence of each component of firstSample against the secondSample :

test_results = ot.LinearModelTest.FullRegression(firstSample, secondSample)
for i in range(len(test_results)):
    print('Component', i, 'is independent?', test_results[i].getBinaryQualityMeasure(),
          'p-value=%.6g' % test_results[i].getPValue(),
          'threshold=%.6g' % test_results[i].getThreshold())

Out:

Component 0 is independent? True p-value=0.646138 threshold=0.05
Component 1 is independent? False p-value=1.30057e-10 threshold=0.05
Component 2 is independent? True p-value=0.342379 threshold=0.05

Total running time of the script: ( 0 minutes 0.009 seconds)

Gallery generated by Sphinx-Gallery