The basis expression of PCE for uniform distributions

Dear all,

I am trying to build an explicit expression based on the obtained PCE model. My random inputs are all standard uniform variables. However, when I built up the PCE expression as shown in point 2 of this post and tested several samples, I found that the output results were not the same as the results obtained from uq_evalModel.

Specifically, I was using the PCE.Basis.Indices, PCE.Coefficients in the PCE model, and standard Legendre polynomials to build explicit expressions. I am now wondering what basis function I should use. I understand that for standard uniform distributions, the orthogonal polynomials are the standard Legendre polynomials defined on [−1,1]. However, there is also Hilbertian basis in Table 1 of PCE manual. And when I build up the expression with the Hilbertian basis instead of Legendre polynomials, the test results will be the same as those from the function uq_evalModel.

Can you help to clarify this point for me? Thank you.

Best regards,
Yuanxi Wu

1 Like

Dear @YuanXi_Wu

Thanks for your questions: the classical Legendre polynomials are orthogonal, yet not orthonormal. As you can see here, their squared integral is:

\int_{-1}^1 P_n^2(x) \, \textrm{d}x =\frac{2}{2n+1}.

When weighting the integral with the probability distribution function of the uniform distribution over [-1, 1], i.e., \displaystyle{\frac{1}{2}}, one gets:

||P_n||^2 = \int_{-1}^1 P_n^2(x) \, \frac{1}{2} \, \textrm{d}x =\frac{1}{2n+1}.

The orthonormal Legendre polynomials are thus obtained by:

\tilde{P}_n(x) = \sqrt{2n+1} \, P_n(x),

so that:

||\tilde P_n||^2 = \int_{-1}^1 \tilde P_n^2(x) \, \frac{1}{2} \, \textrm{d}x =1,

which is implemented in UQLab.

I hope this clarifies the question.
Best regards

1 Like

Dear @bsudret

Thank you very much for your help. Your response is very clear and helps me to better understand the issue at hand.

So, in UQLab, I think the orthonormal polynomials are implemented as the Hilbertian basis in Table 1 of the PCE manual. Is that correct?

I appreciate your time and expertise in addressing my questions.

Best regards,
Yuanxi Wu