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From: saswss@hotellng.unx.sas.com (Warren Sarle)
Subject: Re: Normalized RBF Net
Originator: saswss@hotellng.unx.sas.com
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Date: Fri, 31 Mar 1995 00:34:33 GMT
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References:  <3le5bq$fja@obelix.si.sintef.no>
Organization: SAS Institute Inc.
Keywords: Normalized RBF, Reg.analysis/Func.approx., linear dependencies?
Lines: 34


In article <3le5bq$fja@obelix.si.sintef.no>, Oystein.Amundrud@si.sintef.no (Oystein Amundrud) writes:
|> ...
|> Let's say I have determined the centers and weights and calculated the
|> "basis function matrix" B = [b_ij] where
|>      i = 1,..,N (N = number of observations/patterns)
|>      j = 1,..,M (M = number of centers/hidden layer nodes.
|>
|> The linear problem is to solve for the weights, w, in :
|>      y = B*w
|>
|> If I use the normalized version, I divide each element of B with the sum
|> of the current row
|>      newb_ij = b_ij/sum_j(b_ij)
|> This makes the sum of the rows in the newB matrix equal to one, that is,
|> I have linear dependencies among the rows.
|> Thus, to solve for the weights in y = newB*w, I throw away one of the
|> columns (since the last column doesn't give any additional information).
|>
|> My question is: Is this the correct way to perform Normalized RBFN-
|>                 calculation? If this is the case, the normalized
|>                 representation of the Net will have (M-1) terms
|>                 (one hidden node less than I started out with).

That will work. An alternative is to omit the intercept ("bias" in NN
terminology) from the linear model instead of one of the RBF columns;
that is preferable if you're doing stepwise regression to select a
subset of the RBF centers to use.

-- 

Warren S. Sarle       SAS Institute Inc.   The opinions expressed here
saswss@unx.sas.com    SAS Campus Drive     are mine and not necessarily
(919) 677-8000        Cary, NC 27513, USA  those of SAS Institute.
