Sparse PCE, q-norm: 1.00, standard total-degree trucation

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1

Hi, for sparse PCE, if q-norm is 1.00, hyperbolic trucation corresponds to standard total-degree trucation, then why is the PCE still sparse using q-norm: 1.00?

2

Hi @AmosJoseph,

it would be easier to answer your question if an example is provided. Maybe you could paste here the results you obtained that confirms your assumption.

Best
Gian

3

Number of input variables: 9
Maximal degree: 3
q-norm: 1.00
Size of full basis: 220
Size of sparse basis: 65

Hi, this is the result, please check.

4

Dear @AmosJoseph

As you correctly noted, the q-norm truncation with q=1.0 is equivalent to the truncation of the total degree. Thus, the PCE model is still sparse in the sense that the size of the basis is reduced. However, truncation by total degree with only 3 input variables is a weak truncation, as you can also see from the numbers you gave in your second post. So I can’t see any contradiction between what you wrote and the results you obtained. Maybe you specify your question more precisely.

Best regards
Styfen

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5

Dear @AmosJoseph

Thanks for your question. I think there is a misunderstanding about sparsity. When you choose a max total degree p and a value of q for the q-norm, you define a candidate basis. If you were to use ordinary least squares (OLS) as a solver, you would compute all the coefficients.

If in contrast, you choose a sparse solver such as LARS, only relevant coefficients of the candidate basis are computed, thus the numbers you got (your second post).
With M=9 inputs, p=3 and q=1, the “full” basis (candidate terms) is of size P = \binom{9+3}{3} = 12\cdot11\cdot10/6 = 220. The sparse solver only computes 65 relevant terms, the other coefficients are set to zero.

Hope it helps!
Best regards
Bruno

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