Dilip V. Sarwate <sarwate@uiuc.edu> wrote in message news:<giURa.2858$o7.37232@vixen.cso.uiuc.edu>...
> 
> Come to think of it, could someone explain *why* I have to
> pay for electricity?  After all, I return one electron to
> the power company to replace each and every electron it
> sends to me.  Not the same electron, of course, but another
> one that is identical in all respects...

Because they do all the work of giving you the electron and taking it
back from you. They come to your doorstep to collect that electron you
are returning :)

Dilip V. Sarwate wrote:
..
>  Thus, heuristically speaking, a reasonable model for
> a "uniform distribution" on the natural numbers is one that
> assigns probability 1/n to the set A_n, the multiples of n,
> for each choice of n.  As Andor points out, this does not
> actually assign equal probability to each singleton event {n}.

Try as I might, I could not show that this set function which you define
above is a probability measure at all. On the contrary, I seem to be able to
prove now (with outside help and one shaky point), that such a distribution
is impossible.

Perhaps I can quickly outline this proof (by contradiction: assume the
distribution exists):
As I have shown, the sigma algebra generated by the sets A_n is the power
set of NN (the natural numbers). I have shown this by explicitely
constructing the sets {n} (for all n) from the sets A_n with countable
unions and complements.

It is clear that not all of the sets {n} can have zero probability (because
of sigma-additivity on disjoint unions). Let P[{n}] =: b_n. I now want to
calculate the distribution on the sets {n}. From the definition of the
distribution on A_n, we know that
P[A_1] = P[ U{n}_{n=1}^infty ] = 1. So we have the equation:

Sum(b_n)_{n=1}^infty = 1

Similarly, from the definition P[A_k] = 1/k we get the equations

Sum(b_{k n})_{n=1}^infty = 1/k

This is a set of countably may linear equations, in matrix notation:

( 1 1 1 1 1 1 1 1 1 ... )   (b_1)     (1/1)
( 0 1 0 1 0 1 0 1 0 ... )   (b_2)     (1/2)
( 0 0 1 0 0 1 0 0 1 ... ) * (b_3)  =  (1/3)
( 0 0 0 1 0 0 0 1 0 ... )   (b_4)     (1/4)
( .......               )   (...)      ...

Now we can look at any number k and its multiples, and get the equations:

( 1 1 1 1 1 1 1 1 1 ... )   (b_k )     (1/1)/k
( 0 1 0 1 0 1 0 1 0 ... )   (b_2k)     (1/2)/k
( 0 0 1 0 0 1 0 0 1 ... ) * (b_3k)  =  (1/3)/k
( 0 0 0 1 0 0 0 1 0 ... )   (b_4k)     (1/4)/k
( .......               )   (... )      ...

In the finite case, if the matrix is regular, this implies that

b_k  = (b_1) / k             (**)

And here comes the shaky point: I know didley about solving a set of
inifinitely many linear equations. I also don't know if this matrix is
"regular" (whatever that means for infinitely big matrices). But if we
assume that (**) is also implied in the infinite case, then we get that

b_1 = c
b_2 = c / 2
b_3 = c / 3
...

So the Sum(b_n) diverges for c =/= 0, we cannot satisfy the first equations.
Therefore this distribution does not exist.

Regards,
Andor

My apologies to Andor for mis-understanding what he wrote.



Since it is not possible to assign equal nonzero probability
to each of the natural numbers, it is natural to ask what
alternative might be used to model one's desire that each
integer is "equally likely" to be chosen.  So, let us consider
a uniform distribution on the first N integers, say
{1, 2, .. , 30} so that each has probability 1/30 of being
chosen.  Then, what is the probability that the outcome of
the experiment is a multiple of n?  The answer is exactly 1/n
for n = 1, 2, 3, 5, 6, 10, 15  and **approximately** 1/n for
other values of n.  If we choose N = 120, we get exact results
1/n for more choices of n, and better approximations for the
rest.  Thus, heuristically speaking, a reasonable model for
a "uniform distribution" on the natural numbers is one that
assigns probability 1/n to the set A_n, the multiples of n,
for each choice of n.  As Andor points out, this does not
actually assign equal probability to each singleton event {n}.

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robert bristow-johnson wrote:
> jim wrote:
> > Good analogy. But, you mustn't say this:
> > 
> > a) It is impossible to keep track of goods on the shelf without resorting
> > to negative numbers.
> > b) Some days, I have negative number of goods on the shelf.
> > 
> > If you do say things analogous to that, its fair to say that you've lost
> > touch with reality.
> 
> 
> even if all tangible physical quantity could be or would normally be
> measured as positive values (which i'm not conceding is the case),
> there should always be the notion of the differences of like
> dimensioned physical quantities.  say pressure differences or
> difference of rotational position of some shaft or something (somewhat
> alluded to by Jerry below).  that's a reality that i am in touch with
> and it needs negative numbers.  so i think that "real" is a very
> appropriate adjective for negative numbers.

A mathematician, a physicist and a biologist stand before an elevator.
Nine people enter the elevator. After some time the elevator returns.
Ten people exit.

Biologist: "Oh my, people reproduced in there!"
Physicist:  "A measurement error of ten precent is negligible."
Mathematician: "Now if one person goes back inside, none are left in
the elevator."



Regards,
Andor

jim <sjedging@mwt.net> wrote in message news:<3F192CA4.3E492608@mwt.net>...

...
> 
> Good analogy. But, you mustn't say this:
> 
> a) It is impossible to keep track of goods on the shelf without resorting
> to negative numbers.
> b) Some days, I have negative number of goods on the shelf.
> 
> If you do say things analogous to that, its fair to say that you've lost
> touch with reality.


even if all tangible physical quantity could be or would normally be
measured as positive values (which i'm not conceding is the case),
there should always be the notion of the differences of like
dimensioned physical quantities.  say pressure differences or
difference of rotational position of some shaft or something (somewhat
alluded to by Jerry below).  that's a reality that i am in touch with
and it needs negative numbers.  so i think that "real" is a very
appropriate adjective for negative numbers.


Jerry Avins <jya@ieee.org> wrote in message news:<3F1C392F.B1AEBDC4@ieee.org>...

...
> 
> There is linear or rotary motion involved in playing a tape or record
> backward, so direction reversal makes sense. What about the loudspeaker
> cone? Does that vibrate backwards?
> 
> The examples above are a sort of time reversal. Does that really negate
> the frequencies? If so, how? 

not for the real signal.  for this conceptual device that we call the
"analytic signal" (which is x(t) + j*hilbert{x(t)}), it would negate
the frequencies.

(i'm in colorado springs, which BTW is a real political culture
adjustment from burlington VT, so my internet presence is sparse. 
just a note in case you don't hear from me for another week.)

r b-j

Dilip V. Sarwate wrote:
> an2or@mailcircuit.com (Andor) writes:
>
> >Yes. But there are also just as many number which are multiples of 3
> >as there are non-multiples of 3. However, your intuition tells you
> >that the probability of choosing a multiple of 3 from the natural
> >number should be 1/3. Again, this probability space is not
> >constructable via the axiomatic approach.
>
>
> Actually, this probability space **is** constructable via the
> axiomatic approach.  The field of events is defined to contain
> all sets of the form
>
> A_n = {all integer multiples of n}
>
> and any other sets that can be constructed from the A_n's by
> the operations of complementation, countable union and countable
> intersection.
>
> The probability assigned to A_n is 1/n, to the complement of
> A_n is 1 - 1/n.  What this assignment "lacks" is that it does
> not necessarily assign a probability to each individual outcome
> because a particular outcome may not be expressible in terms of
> complements, countable unions, and countable intersections of
> the A_n's, and is thus not a member of the field of events.  That
> is, P({n}) may have no meaning unless we can express the singleton
> event {n} in terms of complements, countable unions, and
> countable intersections of the A_n's

P[{n}] indeed has meaning for all n in NN, because {n} is an element of
sigma(A_1, A_2, ...):

B_n := {n} = A_n \ U(A_{n*k})_{k=2}^infty, n>=1,    (where U denotes
"union")

(which also shows that sigma(A_1, A_2, ...) is equal to the power set of
NN).

Now you have a probability space on NN. But it does not give uniform
distribution for the sets B_n, which we are interested in. One could think
that P[ B_n ] = 0 for all n, but that cannot be true under the distribution
that you specify because

NN = A_1 = U (B_n)_{n=1}^infty  (where the B_n are disjoint).

Taking the probability on both sides we get:

1 = P[ A_1 ] = P[U (B_n)_{n=1}^infty] = Sum(P[B_n])_{n=1}^infty,

so not all P[B_n] can be zero. OTH, P[B_n] cannot be constant for all n,
because of the convergence of the sum on the right.

That's what I meant by not constructable uniform distribution for drawing
natural numbers.

Regards,
Andor

an2or@mailcircuit.com (Andor) writes:

>Yes. But there are also just as many number which are multiples of 3
>as there are non-multiples of 3. However, your intuition tells you
>that the probability of choosing a multiple of 3 from the natural
>number should be 1/3. Again, this probability space is not
>constructable via the axiomatic approach.


Actually, this probability space **is** constructable via the
axiomatic approach.  The field of events is defined to contain
all sets of the form

	A_n = {all integer multiples of n}

and any other sets that can be constructed from the A_n's by
the operations of complementation, countable union and countable
intersection.

The probability assigned to A_n is 1/n, to the complement of
A_n is 1 - 1/n.  What this assignment "lacks" is that it does
not necessarily assign a probability to each individual outcome
because a particular outcome may not be expressible in terms of
complements, countable unions, and countable intersections of
the A_n's, and is thus not a member of the field of events.  That
is, P({n}) may have no meaning unless we can express the singleton
event {n} in terms of complements, countable unions, and
countable intersections of the A_n's

Note that the intersection of A_n and A_m is A_{LCM(n,m)}
and that A_n and A_m are independent events if m and n are
relatively prime integers (that is, have no factors in common
other than 1; m and n themselves need not be prime, e.g. 15
and 16).

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/ D \ I / L \ I / P \   / S \ A / R \ W / A \ T / E \
     `-'     `-'     `-'     `-'     `-'     `-'

Andor wrote:

> If you have a discrete distribution function, a question like P[X = x]
> makes very good sense. You can also mix absolute continuous and

Yep, but is this case you are in a different
condition, I guess.

bye,

-- 

piergiorgio

Piergiorgio Sartor wrote:
> > For example, when you choose a number from the interval [0,1] with
> > uniform distribution, the probability of it being rational is zero.
> > Which doesn't mean that it can't happen!
> 
> Actually that's why they prefer P[x <= y]
> instead of P[x = y]...

If you have a discrete distribution function, a question like P[X = x]
makes very good sense. You can also mix absolute continuous and
discrete distributions (think of a density function with a Dirac delta
included somewhere).

 
> > The other problem you get is that you cannot create a uniform
> > distribution on an infinite countable set. For example, you cannot say
> > what the probability is of choosing an even number from natural
> > numbers (even though you feel that it ought to be 1/2). That is
> > because of the additivity of probabilites (measures) on disjoint sets.
> 
> Well, that's also because even, odd and all natural
> are _exactly_ the same amount of numbers...

Yes. But there are also just as many number which are multiples of 3
as there are non-multiples of 3. However, your intuition tells you
that the probability of choosing a multiple of 3 from the natural
number should be 1/3. Again, this probability space is not
constructable via the axiomatic approach.

Andor wrote:

> For example, when you choose a number from the interval [0,1] with
> uniform distribution, the probability of it being rational is zero.
> Which doesn't mean that it can't happen!

Actually that's why they prefer P[x <= y]
instead of P[x = y]...

> The other problem you get is that you cannot create a uniform
> distribution on an infinite countable set. For example, you cannot say
> what the probability is of choosing an even number from natural
> numbers (even though you feel that it ought to be 1/2). That is
> because of the additivity of probabilites (measures) on disjoint sets.

Well, that's also because even, odd and all natural
are _exactly_ the same amount of numbers...

bye,

-- 
   Piergiorgio Sartor