Re: principal components reconstruction

On May 28, 6:11 pm, "Happy " <happygalpea...@xxxxxxxxx> wrote:
Hi,

I'm using PCA

(the contents of the parentheses indicate dimensions;
not component indices)

X(n,N) = T(n,n)*P(n,N),
P(n,N) = T'(n,n)*X(n,N),

to reduce the dimensions of my data

X(n,N) ~ T(n,r)*P(r,N), r < n,
P(r,N) ~ T'(n,r)*X(n,N),

before doing regression analysis (PRINCOMP).

Y(m,N) = Wp(m,r+1)*Pa(r+1,N) + Ey(r),
Pa(r+1,N) = [ones(1,N); P(r,N)],

and random Ey(r)is assumed to be zero mean Gaussian
distributed.

Once I have obtained the regression coefficients using
the lower dimensional data from PCA,

Wp(m,r+1) = Y(m,N)/Pa(r+1,N),

(contains uncertainty because of Ey(r)),

I can reconstruct the coefficients back to
the original dimensions by using the same matrix of
eigenvectors found in PCA.

Yhat(m,N) = Wp(m,r+1)*Pa(r+1,N)]
= Wp(m,r+1)*[ones(1,N);T'(n,r)*X(n,N)]

My question is, it is valid to reconstruct the standard
error obtained from regression in the lower dimensions back
to the original dimensions using the same matrix of
eigenvectors? Or do i need to use other methods like monte
carlo or bootstrap?

Do you mean "Given the standard error of Wp(m,r+1), is it
possible to obtain an exact analytic expression for the
standard error of Wx(m,n+1) where

Y(m,N) = Wx(m,n+1)*Xa(n+1,N) + Ey(n),
Xa(n+1,N) = [ones(1,N); X(n,N)],
Wx(m,n+1) = Y(m,N)/Xa(n+1,N)?


I think the answer is not exactly because Ey(r) ~= Ey(n).

Nevertheless, making the assumption Ey(r) ~ Ey(n) yields

Wx(m,n+1)*Xa(n+1,N) ~ Wp(m,r+1)*Pa(r+1,N)

and

Wx(m,n+1) ~ Wp(m,r+1)*Pa(r+1,N)/Xa(n+1,N)


The usual assumption is that X,T and P are error free so that
any uncertainty in W is caused by Ey. Consequently, the problem
is reduced to

Wx(m,n+1) = Wp(m,r+1)*A(r+1,n+1)

where A is constant.

The rest is up to you.

Hope this helps.

Greg





.

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