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From: Huang on 14 Jan 2010 08:46
On Jan 14, 12:13 am, "preedmont" <nos...(a)spamless.com> wrote:
> "Huang" <huangxienc...(a)yahoo.com> wrote in message
>
> news:001d62db-a3e2-47da-b185-4b8d041b1e8d(a)p8g2000yqb.googlegroups.com...
>
> > Consider 3 trials of a random variable { X |  H, T } .
>
> > The result is no different than a single trial of the random variable
> > { X |  HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
>
> the first has 3 events, the second has just one event.
> the first is has 2 states, the second has 8.
>
> The results can be the same, does not imply the process is the same.
>
> The first can be a coin, the second is an 8 sided thingie.
>
> > How do we get back and forth from one to the other ?
>
> by moving eyes around from one to the other.
>
> > How to we make
> > the conversion ?
>
> trivial, you did it
>
> >How to perform such a transform ? Obviously these
> > things are equivalent, but how to do it without the hand wave ?
>
> they are not the same.


True. These things are not the same. They differ in process, but the
result is only trivially different. In each case you will get a
squence of 3 outcomes, and both processes produce an identical outcome
space. They are just worded differently.

It is true that the processes differ. But the difference is trivial.
It is very much like saying that

5 = 5 + 0 = 5 + 0 + 0 = 5 + 0 + 0 + 0 + 0 etc

These things are different, but the difference is trivial.

There should be a way to do this algebraically. Shoul be a way to show
symbolically that :

3 trials of a random variable
{ X | H, T }
is no different than a single trial of the random variable
{ X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .


I think that the way we can show this is by treating conservation as a
trivial operator. This operator would work equally well with discrete
and continuous random variables, because it is trivial it does not
care if it is multiplying or adding.











From: Huang on 14 Jan 2010 09:19
On Jan 14, 7:46 am, Huang <huangxienc...(a)yahoo.com> wrote:
> On Jan 14, 12:13 am, "preedmont" <nos...(a)spamless.com> wrote:
>
>
>
>
>
> > "Huang" <huangxienc...(a)yahoo.com> wrote in message
>
> >news:001d62db-a3e2-47da-b185-4b8d041b1e8d(a)p8g2000yqb.googlegroups.com...
>
> > > Consider 3 trials of a random variable { X |  H, T } .
>
> > > The result is no different than a single trial of the random variable
> > > { X |  HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
>
> > the first has 3 events, the second has just one event.
> > the first is has 2 states, the second has 8.
>
> > The results can be the same, does not imply the process is the same.
>
> > The first can be a coin, the second is an 8 sided thingie.
>
> > > How do we get back and forth from one to the other ?
>
> > by moving eyes around from one to the other.
>
> > > How to we make
> > > the conversion ?
>
> > trivial, you did it
>
> > >How to perform such a transform ? Obviously these
> > > things are equivalent, but how to do it without the hand wave ?
>
> > they are not the same.
>
> True. These things are not the same. They differ in process, but the
> result is only trivially different. In each case you will get a
> squence of 3 outcomes, and both processes produce an identical outcome
> space. They are just worded differently.
>
> It is true that the processes differ. But the difference is trivial.
> It is very much like saying that
>
> 5 = 5 + 0 = 5 + 0 + 0 = 5 + 0 + 0 + 0 + 0    etc
>
> These things are different, but the difference is trivial.
>
> There should be a way to do this algebraically. Shoul be a way to show
> symbolically that :
>
> 3 trials of a random variable
> { X |  H, T }
> is no different than a single trial of the random variable
> { X |  HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
>
> I think that the way we can show this is by treating conservation as a
> trivial operator. This operator would work equally well with discrete
> and continuous random variables, because it is trivial it does not
> care if it is multiplying or adding.- Hide quoted text -
>
> - Show quoted text -


Lets make a tentative symbol to represent the trivial operator,
something like this :

[@]

Then, for discrete random variables, from the example above, we might
write something like this

3 [@] { X | H, T } = { X | HHH, HHT, HTT, THH, THT, TTT, HTH,
TTH }

or possibly

{ X | H, T } [@] { X | H, T } [@] { X | H, T } = { X | HHH,
HHT, HTT, THH, THT, TTT, HTH, TTH }

and of course the operator [@] should behave the same whether we are
considering continuous or discrete random variables, I dont have a
continuous example at hand but it should work.

(It would'nt fit in the margin)

One very important thing to note, if we write
3 [@] { X | H, T } ,
then we are combining numbers with things which are not numbers. This
is not neccesarily a criminal act at this point but definately an
important consideration.

It does make sense for many reasons that conservation would be
regarded as a genuine operator. Trivial, but genuine.

From: Uncle Al on 14 Jan 2010 12:32
Huang wrote:
>
> On Jan 13, 9:29 am, Uncle Al <Uncle...(a)hate.spam.net> wrote:
> > Huang wrote:
> >
> > > Consider 3 trials of a random variable { X | H, T } .
> >
> > > The result is no different than a single trial of the random variable
> > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
> >
> > > How do we transform from one situation to the other ? Is it enough to
> > > say they are "equivalent" ??
> >
> > How do you know both outcomes are accessible given a single
> > observation? Flipping a coin has three outcomes not two. The third
> > is unlikely but not impossible.
> >
> > --
> > Uncle Alhttp://www.mazepath.com/uncleal/
> > (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/qz4.htm
>
> Consider 3 trials of a random variable { X | H, T } .
>
> The result is no different than a single trial of the random variable
> { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
>
> How do we get back and forth from one to the other ? How to we make
> the conversion ? How to perform such a transform ?
>
> Here's a hint for you Auntie....it has to do with conservation. And if
> you say that conservation has no use in mathematics....I say you are
> wrong. You need to re-read what I said about conservation.....what it
> really is.
>
> That is your hint.
>
> Solve this riddle and I'll pay $1,500, a slightly used Canon 5D, and
> dinner with a hot chick ( void where prohibited and everywhere else ).

idiot
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
From: Ostap S. B. M. Bender Jr. on 14 Jan 2010 20:22
On Jan 14, 6:19 am, Huang <huangxienc...(a)yahoo.com> wrote:
> On Jan 14, 7:46 am, Huang <huangxienc...(a)yahoo.com> wrote:
>
>
>
> > On Jan 14, 12:13 am, "preedmont" <nos...(a)spamless.com> wrote:
>
> > > "Huang" <huangxienc...(a)yahoo.com> wrote in message
>
> > >news:001d62db-a3e2-47da-b185-4b8d041b1e8d(a)p8g2000yqb.googlegroups.com....
>
> > > > Consider 3 trials of a random variable { X |  H, T } .
>
> > > > The result is no different than a single trial of the random variable
> > > > { X |  HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
>
> > > the first has 3 events, the second has just one event.
> > > the first is has 2 states, the second has 8.
>
> > > The results can be the same, does not imply the process is the same.
>
> > > The first can be a coin, the second is an 8 sided thingie.
>
> > > > How do we get back and forth from one to the other ?
>
> > > by moving eyes around from one to the other.
>
> > > > How to we make
> > > > the conversion ?
>
> > > trivial, you did it
>
> > > >How to perform such a transform ? Obviously these
> > > > things are equivalent, but how to do it without the hand wave ?
>
> > > they are not the same.
>
> > True. These things are not the same. They differ in process, but the
> > result is only trivially different. In each case you will get a
> > squence of 3 outcomes, and both processes produce an identical outcome
> > space. They are just worded differently.
>
> > It is true that the processes differ. But the difference is trivial.
> > It is very much like saying that
>
> > 5 = 5 + 0 = 5 + 0 + 0 = 5 + 0 + 0 + 0 + 0    etc
>
> > These things are different, but the difference is trivial.
>
> > There should be a way to do this algebraically. Shoul be a way to show
> > symbolically that :
>
> > 3 trials of a random variable
> > { X |  H, T }
> > is no different than a single trial of the random variable
> > { X |  HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
>
> > I think that the way we can show this is by treating conservation as a
> > trivial operator. This operator would work equally well with discrete
> > and continuous random variables, because it is trivial it does not
> > care if it is multiplying or adding.- Hide quoted text -
>
> > - Show quoted text -
>
> Lets make a tentative symbol to represent the trivial operator,
> something like this :
>
> [@]
>
> Then, for discrete random variables, from the example above, we might
> write something like this
>
> 3  [@]  { X |  H, T } = { X |  HHH, HHT, HTT, THH, THT, TTT, HTH,
> TTH }
>
> or possibly
>
> { X |  H, T }  [@]  { X |  H, T }  [@]  { X |  H, T }  =  { X |  HHH,
> HHT, HTT, THH, THT, TTT, HTH, TTH }
>
> and of course the operator  [@]  should behave the same whether we are
> considering continuous or discrete random variables, I dont have a
> continuous example at hand but it should work.
>
> (It would'nt fit in the margin)
>
> One very important thing to note, if we write
> 3  [@]  { X |  H, T } ,
> then we are combining numbers with things which are not numbers. This
> is not neccesarily a criminal act at this point but definately an
> important consideration.
>
> It does make sense for many reasons that conservation would be
> regarded as a genuine operator. Trivial, but genuine.
>

Since this child play seems to never end, let me tell you that the
concept that you are looking for is called 'Cartesian Product'. It is
usually covered in the first month of any introductory course in
probability theory. You can read about it for example here:

http://media.wiley.com/product_data/excerpt/56/04712597/0471259756.pdf

1 Experiments and Probability

1.2 COMBINED EXPERIMENTS

Combined experiments play an important role in probability theory
applications. There are many ways we can combine experiments,
including cartesian products in which independent trials of the same
or different experiments can be described.

1.2.1 Cartesian Product of Two Experiments

Consider the case of having two separate experiments specified by the
following: i{ : ( 5 '1,^1,^1(.)) and S'2 : (S2, ^ 2 , ^2(.)). The
sample spaces Sx and S2 are usually different sets, for example,
results of a coin toss and results of a die roll, but they could be
the same sets representing separate trials of the same experiment as
in repeated coin-tossing experiments. We can define a new combined
experiment by using the cartesian product concept as S — $\®S'2, where
the new sample space S = S{ ® S2 is the cartesian product of the two
sample spaces expressible by the ordered pair of elements where the
first element is from S{ and the second is from S2.

1.2.2 Cartesian Product of n Experiments

Consider the case of having n separate experiments specified by Sk :
(S/c, 3Fk, &k(.)) for k = 1, 2 , . . . , n. Define a new combined
experiment S : (S, ^ , ^(.)) as a cartesian product: <f = #'t ® ^ 2® *
* *® $n where the new sample space S = Sx ® S2® • • -® Sn is the
cartesian product of the n spaces and expressible by the ordered ^-
tuples of elements whose first element is from Sx and the second from
S2 . . . , the nth from Sn. The $k are, in general, different, but in
many cases the experiment could be formed from independent trials of
the same experiment.



From: Huang on 14 Jan 2010 22:54
On Jan 14, 11:32 am, Uncle Al <Uncle...(a)hate.spam.net> wrote:
> Huang wrote:
>
> > On Jan 13, 9:29 am, Uncle Al <Uncle...(a)hate.spam.net> wrote:
> > > Huang wrote:
>
> > > > Consider 3 trials of a random variable { X |  H, T } .
>
> > > > The result is no different than a single trial of the random variable
> > > > { X |  HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
>
> > > > How do we transform from one situation to the other ? Is it enough to
> > > > say they are "equivalent" ??
>
> > > How do you know both outcomes are accessible given a single
> > > observation?  Flipping a coin has three outcomes not two.  The third
> > > is unlikely but not impossible.
>
> > > --
> > > Uncle Alhttp://www.mazepath.com/uncleal/
> > >  (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/qz4.htm
>
> > Consider 3 trials of a random variable { X |  H, T } .
>
> > The result is no different than a single trial of the random variable
> > { X |  HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } .
>
> > How do we get back and forth from one to the other ? How to we make
> > the conversion ? How to perform such a transform ?
>
> > Here's a hint for you Auntie....it has to do with conservation. And if
> > you say that conservation has no use in mathematics....I say you are
> > wrong. You need to re-read what I said about conservation.....what it
> > really is.
>
> > That is your hint.
>
> > Solve this riddle and I'll pay $1,500, a slightly used Canon 5D, and
> > dinner with a hot chick ( void where prohibited and everywhere else ).
>
> idiot



You turned down a date with this girl >>>

http://photo.net/photodb/photo?photo_id=10413431

and supposedly I am the idiot ???


That's intense.









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