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Virgil
Posts:
7,011
Registered:
1/6/11


Re: Probabilities not in [0,1]?
Posted:
Mar 12, 2013 6:38 PM


In article <cef6abc9a61746499454ca2d2e22aa90@z3g2000vbg.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 10 Mrz., 19:36, "Peter Percival" <peterxperci...@hotmail.com> > wrote: > > Is there a theory of probability in which probabilities do not lie in the > > real interval [0,1]? > > Yes. > > http://www.sciencedirect.com/science/article/pii/0370157386901109 > > http://finden.nationallizenzen.de/Author/Home?author=M%C3%BCckenheim%2C%20W. > > Regards, WM
Not when using the standard definition of probability.
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WM has frequently claimed that a mapping from the set of all infinite binary sequences to the set of paths of a CIBT is a linear mapping. In order to show that such a mapping is a linear mapping, WM must first show that the set of all binary sequences is a vector space and that the set of paths of a CIBT is also a vector space, which he has not done and apparently cannot do, and then show that his mapping satisfies the linearity requirement that f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of a field of scalars and x and y are f(x) and f(y) are vectors in suitable linear spaces.
By the way, WM, what are a, b, ax, by and ax+by when x and y are binary sequences?
If a = 1/3 and x is binary sequence, what is ax ? and if f(x) is a path in a CIBT, what is af(x)?
Until these and a few other issues are settled, WM will still have failed to justify his claim of a LINEAR mapping from the set (but not yet proved to be vector space) of binary sequences to the set (but not yet proved to be vector space) of paths ln a CIBT.
Just another of WM's many wild claims of what goes on in his WMytheology that he cannot back up. 



