Note

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Create a deconditioned random vector¶

In this example we are going to build the random vector $\inputRV$ defined as the deconditioned distribution of:

$\inputRV|\vect{\Theta}$

with respect to parameter random vector $\vect{\Theta}$ following the distribution $\cL_{\vect{\Theta}}$ .

This example creates a DeconditionedRandomVector. Remember that a DeconditionedRandomVector (and more generally a RandomVector) can only be sampled.

There is no restriction on the random vector $\vect{\Theta}$ . It can be created from any multivariate distribution or as the output of a function $f$ applied to an input random vector $\vect{Y}$ : $\vect{\Theta} = f(\vect{Y})$ .

Note that in some restricted cases, it is possible to create the distribution of $\inputRV$ using the class DeconditionedDistribution. The DeconditionedDistribution offers a lot of methods attached to the distribution, in particular the computation of the PDF, CDF, the moments if any, $\dots$ . The advantage of the DeconditionedRandomVector relies on the fact that it is fast to build and can be built in all cases. But it only offers the sampling capacity.

We consider the following cases:

Case 1: the parameter random vector has continuous marginals,
Case 2: the parameter random vector has dependent continuous and discrete marginals,
Case 3: the parameter random vector has any dependent marginals.

import openturns as ot
import openturns.viewer as otv

Case 1: the parameter random vector has continuous marginals¶

We consider the case where $X$ is of dimension 1 and follows an exponential distribution defined by:

Variable	Distribution	Parameter
$X$	`Exponential` ( $\Lambda, \Gamma$ )	( $\Lambda, \Gamma$ ) are random
$\Lambda$	`Uniform` ( $a, b$ )	$(a,b) = (0, 1)$
$\Gamma$	`Uniform` ( $a, b$ )	$(a,b) = (1, 2)$
Copula ( $\Lambda, \Gamma$ )	`ClaytonCopula` ( $\theta$ )	$\theta = 2$

Create the parameter random vector $\vect{\Theta} = (\Lambda, \Gamma)$ :

lambdaDist = ot.Uniform(0.0, 1.0)
gammaDist = ot.Uniform(1.0, 2.0)
thetaDist = ot.JointDistribution([lambdaDist, gammaDist], ot.ClaytonCopula(2))
thetaRV = ot.RandomVector(thetaDist)

Create the $\inputRV|\vect{\Theta}$ distribution: as the parameters have no importance, we create the default distribution.

XgivenThetaDist = ot.Exponential()

Create the $\inputRV$ random vector.

X_RV = ot.DeconditionedRandomVector(XgivenThetaDist, thetaRV)

Draw a sample

X_RV.getSample(5)

	X0
0	5.081867
1	1.987522
2	4.769532
3	4.779249
4	1.950028

If we generate a large sample of the random vector, we can fit its distribution with non-parametric techniques such as a kernel smoothing.

sample_large = X_RV.getSample(10000)
dist_KS = ot.KernelSmoothing().build(sample_large)
g_PDF = dist_KS.drawPDF()
g_PDF.setTitle("Case 1: PDF of X by kernel smoothing")
g_PDF.setXTitle("x")
g_PDF.setLegendPosition("")
view = otv.View(g_PDF)

Case 2: the parameter random vector has dependent continuous and discrete marginals¶

We consider the case where $X$ is of dimension 1 and follows an exponential distribution defined by:

Variable	Distribution	Parameter
$X$	`Exponential` ( $\Lambda, \Gamma$ )	( $\Lambda, \Gamma$ ) are random
$\Lambda$	1 + `Poisson` ( $\ell$ )	$\ell = 1$
$\Gamma$	`Uniform` ( $a, b$ )	$(a,b) = (1, 2)$
Copula ( $\Lambda, \Gamma$ )	`ClaytonCopula` ( $\theta$ )	$\theta = 2$

Create the parameter random vector $\vect{\Theta} = (\Lambda, \Gamma)$ . We shift the Poisson distribution to get positive values for $\Lambda$ .

lambdaDist = 1 + ot.Poisson(1)
gammaDist = ot.Uniform(1.0, 2.0)
thetaDist = ot.JointDistribution([lambdaDist, gammaDist], ot.ClaytonCopula(2))
thetaRV = ot.RandomVector(thetaDist)

Create the $\inputRV|\vect{\Theta}$ random vector.

XgivenThetaDist = ot.Exponential()
X_RV = ot.DeconditionedRandomVector(XgivenThetaDist, thetaRV)

If we generate a large sample of the random vector, we can fit its distribution with non-parametric techniques such as a kernel smoothing.

sample_large = X_RV.getSample(10000)
dist_KS = ot.KernelSmoothing().build(sample_large)
g_PDF = dist_KS.drawPDF()
g_PDF.setTitle("Case 2: PDF of X by kernel smoothing")
g_PDF.setXTitle("x")
g_PDF.setLegendPosition("")
view = otv.View(g_PDF)

Case 3: the parameter random vector has any dependent marginals¶

We consider the case where $X$ is of dimension 1 and follows an exponential distribution defined by:

Variable Distribution	Parameter
$X$	`Exponential` ( $\Lambda, \Gamma$ )	( $\Lambda, \Gamma$ ) are random
$\Lambda$	`Mixture`	see below
$\Gamma$	`Uniform` ( $a, b$ )	$(a,b) = (1, 2)$
Copula ( $\Lambda, \Gamma$ )	`ClaytonCopula` ( $\theta$ )	$\theta = 2$

where the mixture is built from the Exponential ( $\ell$ ) with $\ell = 1$ and the Geometric ( $p$ ) with $p = 0.1$ , with equal weights. In this case, the distribution of $\Lambda$ is not discrete nor continuous.

Create the parameter random vector $\vect{\Theta} = (\Lambda, \Gamma)$ :

lambdaDist = ot.Mixture([ot.Exponential(1.0), ot.Geometric(0.1)])
gammaDist = ot.Uniform(1.0, 2.0)
thetaDist = ot.JointDistribution([lambdaDist, gammaDist], ot.ClaytonCopula(2))
thetaRV = ot.RandomVector(thetaDist)

Create the $\inputRV|\vect{\Theta}$ random vector.

XgivenThetaDist = ot.Exponential()
X_RV = ot.DeconditionedRandomVector(XgivenThetaDist, thetaRV)

If we generate a large sample of the random vector, we can fit its distribution with non-parametric techniques such as a kernel smoothing.

sample_large = X_RV.getSample(10000)
dist_KS = ot.KernelSmoothing().build(sample_large)
g_PDF = dist_KS.drawPDF()
g_PDF.setTitle("Case 3: PDF of X by kernel smoothing")
g_PDF.setXTitle("x")
g_PDF.setLegendPosition("")
view = otv.View(g_PDF)

view.ShowAll()

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Create a deconditioned random vector¶

Case 1: the parameter random vector has continuous marginals¶

Case 2: the parameter random vector has dependent continuous and discrete marginals¶

Case 3: the parameter random vector has any dependent marginals¶