Estimate a multivariate distribution

In this example we are going to estimate a joint distribution from a multivariate sample by fitting marginals and finding a set of copulas.

While the estimation of marginals is quite straightforward, the estimation of the dependency structure takes several steps:

• find the dependent components

• estimate a copula on each dependent bloc

• assemble the estimated copulas

import openturns as ot
import math as m

ot.Log.Show(ot.Log.NONE)
ot.RandomGenerator.SetSeed(0)

generate some multivariate data to estimate, with correlation

cop1 = ot.AliMikhailHaqCopula(0.6)
cop2 = ot.ClaytonCopula(2.5)
copula = ot.ComposedCopula([cop1, cop2])
marginals = [
    ot.Uniform(5.0, 6.0),
    ot.Arcsine(),
    ot.Normal(-40.0, 3.0),
    ot.Triangular(100.0, 150.0, 300.0),
]
distribution = ot.ComposedDistribution(marginals, copula)
sample = distribution.getSample(10000).getMarginal([0, 2, 3, 1])

estimate marginals

dimension = sample.getDimension()
marginalFactories = []
for factory in ot.DistributionFactory.GetContinuousUniVariateFactories():
    if str(factory).startswith("Histogram"):
        # ~ non-parametric
        continue
    marginalFactories.append(factory)
estimated_marginals = [
    ot.FittingTest.BestModelBIC(sample.getMarginal(i), marginalFactories)[0]
    for i in range(dimension)
]
estimated_marginals

[class=Uniform name=Uniform dimension=1 a=5.00008 b=6, class=Normal name=Normal dimension=1 mean=class=Point name=Unnamed dimension=1 values=[-39.9843] sigma=class=Point name=Unnamed dimension=1 values=[3.05427] correlationMatrix=class=CorrelationMatrix dimension=1 implementation=class=MatrixImplementation name=Unnamed rows=1 columns=1 values=[1], class=Triangular name=Triangular dimension=1 a=100.476 m=150.62 b=298.489, class=Beta name=Beta dimension=1 alpha=0.500965 beta=0.499485 a=-1.0002 b=1.0002]

Find connected components of a graph defined from its adjacency matrix

def find_neighbours(head, covariance, to_visit, visited):
    visited[head] = 1
    to_visit.remove(head)
    current_component = [head]
    for i in to_visit:
        # If i is connected to head and has not yet been visited
        if covariance[head, i] > 0:
            # Add i to the current component
            component = find_neighbours(i, covariance, to_visit, visited)
            current_component += component
    return current_component


def connected_components(covariance):
    N = covariance.getDimension()
    to_visit = list(range(N))
    visited = [0] * N
    all_components = []
    for head in range(N):
        if visited[head] == 0:
            component = find_neighbours(head, covariance, to_visit, visited)
            all_components.append(sorted(component))
    return all_components

Estimate the copula

First find the dependent components : we compute the Spearman correlation

C = sample.computeSpearmanCorrelation()
print(C)

[[  1          -0.00167386  0.00312294  0.245006   ]
 [ -0.00167386  1           0.739083   -0.0138198  ]
 [  0.00312294  0.739083    1          -0.00164887 ]
 [  0.245006   -0.0138198  -0.00164887  1          ]]

We filter and consider only significantly non-zero correlations.

epsilon = 2.0 / m.sqrt(sample.getSize())
for j in range(dimension):
    for i in range(j):
        C[i, j] = 1.0 if abs(C[i, j]) > epsilon else 0.0
print(C)

[[ 1 0 0 1 ]
 [ 0 1 1 0 ]
 [ 0 1 1 0 ]
 [ 1 0 0 1 ]]

Note that we can apply the HypothesisTest.Spearman test. As the null hypothesis of the test is the independence, we must take the complementary of the binary measure as follow:

>>>   M = ot.SymmetricMatrix(dimension)
>>>   for i in range(dimension):
>>>       M[i,i] = 1
>>>       for j in range(i):
>>>           M[i, j] = 1 - ot.HypothesisTest.Spearman(sample[:,i], sample[:,j]).getBinaryQualityMeasure()

Now we find the independent blocs:

blocs = connected_components(C)
blocs

[[0, 3], [1, 2]]

For each dependent block, we estimate the most accurate parameteric copula.

To do this, we first need to transform the sample in such a way as to keep the copula intact but make all marginal samples follow the uniform distribution on [0,1].

copula_sample = ot.Sample(sample.getSize(), sample.getDimension())
copula_sample.setDescription(sample.getDescription())
for index in range(sample.getDimension()):
    copula_sample[:, index] = estimated_marginals[index].computeCDF(sample[:, index])

copulaFactories = []
for factory in ot.DistributionFactory.GetContinuousMultiVariateFactories():
    if not factory.build().isCopula():
        continue
    if factory.getImplementation().getClassName() == "BernsteinCopulaFactory":
        continue
    copulaFactories.append(factory)
estimated_copulas = [
    ot.FittingTest.BestModelBIC(copula_sample.getMarginal(bloc), copulaFactories)[0]
    for bloc in blocs
]
estimated_copulas

[class=AliMikhailHaqCopula name=AliMikhailHaqCopula dimension=2 theta=0.612473, class=ClaytonCopula name=ClaytonCopula dimension=2 theta=2.47924]

Finally we assemble the copula

estimated_copula_perm = ot.ComposedCopula(estimated_copulas)

Take care of the order of each bloc vs the order of original components !

permutation = []
for bloc in blocs:
    permutation.extend(bloc)
inverse_permutation = [-1] * dimension
for i in range(dimension):
    inverse_permutation[permutation[i]] = i
estimated_copula = estimated_copula_perm.getMarginal(inverse_permutation)
estimated_copula

MarginalDistribution

• name=MarginalDistribution
• dimension=4
• weight=1
• range=[0, 1] [0, 1] [0, 1] [0, 1]
• description=[X0,X2,X3,X1]
• isParallel=true
• isCopula=true

We build joint distribution from marginal distributions and dependency structure:

estimated_distribution = ot.ComposedDistribution(estimated_marginals, estimated_copula)
estimated_distribution

ComposedDistribution

• name=ComposedDistribution
• dimension: 4
• description=[X0,X2,X3,X1]
• copula: MarginalDistribution(distribution=ComposedCopula(AliMikhailHaqCopula(theta = 0.612473), ClaytonCopula(theta = 2.47924)), indices=[0,2,3,1])

Index	Variable	Distribution
0	X0	Uniform(a = 5.00008, b = 6)
1	X2	Normal(mu = -39.9843, sigma = 3.05427)
2	X3	Triangular(a = 100.476, m = 150.62, b = 298.489)
3	X1	Beta(alpha = 0.500965, beta = 0.499485, a = -1.0002, b = 1.0002)