Let W = [U,U]. If you set v(W) = W[1] - W[2], then you find that E[v(W)]=0 for nonzero v. - ThreadSky

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ekernf01.bsky.social • 11 days ago

Let W = [U,U]. If you set v(W) = W[1] - W[2], then you find that E[v(W)]=0 for nonzero v.

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This is inconsistent with the usual meaning of "complete": buying a second copy of the Complete Guide to Playing the Lute should not render your lute library incomplete.

ekernf01.bsky.social•11 days ago

A similar funny thing happens when you distribute a complete statistic over two pieces: let W = [U + X, U - X] for unit normal X; set v(W)=W[1]+W[2].

ekernf01.bsky.social•11 days ago

Buying the Complete Guide to Playing the Lute as a two-volume set should not render your lute library incomplete.

ekernf01.bsky.social•11 days ago

If you want to knock me for buying a second copy, or for including a second set of liner notes, and you don't care that **I only bought the first chapter**, "parsimonious" is what you want.

ekernf01.bsky.social•11 days ago

"Complete" statistics contain nothing but the truth, but they do not contain the full truth. They would be better called "parsimonious" statistics. End rant.

neurostats.org•9 days ago

parsimonious doesn't reflect this property for me. A parsimonious representation can be entirely non-lossy in compression. But

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