modelling conditional distributions in chaospy

[gaoutham92]

Hi,

Can conditional distributions be modelled in Chaospy, does it support them. Or the only way to model dependency is by using copulas.

[jonathf]

You definetly can model conditional distributions in chaospy, but it is a big topic that deserves a lot more attention in the documentation than what is actually there. I really should get around to fixing that one of these days...

In the short term though, maybe I can help. What do you want to do?

[gaoutham92]

Hi,

Thanks for the reply. I want to model conditional lognormal distribution, conditioned upon Weibull marginal distribution. To be more precise, I have data for ocean waves - wave height and wave period. I want to fit a Weibull distribution based on wave height data and conditional lognormal distribution - conditioned upon Weibull based on wave period data and make a joint distribution.

So if you can make an example code snippet for generating conditional lognormal using any synthetic and less data samples(like 20 samples), it would be fine and helpful.

[jonathf]

In chaospy dependencies are declared through parameters. So when you say "conditional lognormal" that can either be declared as:

chaospy.LogNormal(mu=chaospy.Weibull())

or

chaospy.LogNormal(sigma=chaospy.Weibull())

The latter is illigal as the Weibull distribution overlaps with 0 which causes a pole in the distribution which the software can not handle. This can not be easily fixed.

If you mean the former, that should be possible, but I discovered a bug trying it out. Log-normal's mu parameter lacks a handler for dealing with dependencies properly. I will look into fixing that soon.

Short term, you can make a workaround like this:

dist1 = numpy.e**chaospy.Weibull(2)
dist2 = chaospy.LogNormal(mu=0)*dist1
joint = chaospy.J(chaospy.Log(dist1), dist2)
samples = joint.sample(20)

This should be equivalent declaring dependencies through the mu parameter because Log(e**dist) == dist and LogNormal(mu=0)*e^dist == LogNormal(mu=dist).

[gaoutham92]

Hi,

Thanks for your quick response, I will try it out.