The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Probabilities not in [0,1]?
Replies: 8   Last Post: Mar 12, 2013 10:56 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 2,720
Registered: 2/15/09
Re: Probabilities not in [0,1]?
Posted: Mar 12, 2013 10:55 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Mar 12, 3:38 pm, Virgil <> wrote:
> In article
> <>,
>  WM <> wrote:

> > On 10 Mrz., 19:36, "Peter Percival" <>
> > wrote:

> > > Is there a theory of probability in which probabilities do not lie in the
> > > real interval [0,1]?

> > Yes.
> >
> >
> > Regards, WM
> Not when using the standard definition of probability.
> ***********************************************************************
> 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.
> --

Well, identity from the real value of the expansion to the real value
of the path is a continuous mapping, for the topology of the reals, as
the Cantor space and CIBT are each R_[0,1] with dual representation of
reals as expansions or paths.

Defining addition on the paths or expansions is simple enough with a
convention for the representation. Obviously those sets aren't closed
under that operation, or scalar multiplication that follows from the
definition of multiplication of real numbers.

Notions of topologies or spaces, of the Cantor space and CIBT, may
follow that aren't simply those of R_[0,1]. Look to union and for
simplicity in binary. Make something, dammit.

A breadth-first ordering of the CIBT or BT sees the same results as a
sweep of the Cantor space (not necessarily as with regards to Cantor
dust) with EF: the antidiagonal argument doesn't apply. Similarly,
countably many rays through ordinal points are dense in the paths of
the CIBT, similarly to that the rationals are dense in the reals, and
the irrationals are only their complement, in the reals, and the
rational paths have terminating or repeating ends.


Ross Finlayson

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.