Newsgroup sci.stat.math 12150

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Ilias Kastanas wrote:
> 
> In article <329D8210.41C67EA6@sees.bangor.ac.uk>,
> Ian   wrote:
> >Ilias Kastanas wrote:
> >>
> >> In article <3297B7A3.2DE7@postoffice.worldnet.att.net>,
> >> kenneth paul collins   wrote:
> >> >Ilias Kastanas wrote:
> >> >>
> >> >> In article <329393FD.10CE@orci.com>, Bob Massey   wrote:
> >> >> >kenneth paul collins wrote:
> >> >> >>
> >> >> >> .....       . All such attempts can be disproven by presenting the
> >> >> >> system with something that "breaks" the syntax. (This is also my main
> >> >> >> objection to Goedel's "Incompleteness".)
> >> >>
> >> >>         Side note: I didn't follow this exchange, but "breaking the syntax"
> >> >>    is irrelevant to Goedel Incompleteness.
> >> >
> >> >By "syntax" I was referring to the "rules" of the "proof". I stand on what I
> >> >posted. ken collins
> >>
> >>         The "rules" of "proof" have been shown to be _the_ rules of proof;
> >>    that is the Completeness theorem.  If you see something wrong there, maybe
> >>    you could state what.  As it is, breaking the rules is pointless and
> >>    self-defeating... and irrelevant to the Incompleteness theorem.
> >>
> >>                                                         Ilias
> >
> >I'm intreged (spelling?) as to how the rules of proof were shown to be
> >_the_ rules of proof. Surely any such proof would have to be
> >self-referential, or rely on axioms.
> 
>         It is in fact remarkable.  The logical axioms and Modus Ponens
>    are straightforward and almost simplistic; and yet they suffice.  For
>    every semantic implication, "in every structure where P holds, Q also
>    holds" there is a formal deduction of Q from P using those rules.  It
>    is the Completeness Theorem.
> 
>                                                         Ilias
I am confused as to exactly what you mean by the completeness theorem.
You don't seem to have said anything here which invalidates my comments
on self-referentiality or reliance on axioms. 
Many mathematicians are sceptical about the law of the excluded middle.
You can construct a self consistent logic without it and using it leads
to some very odd conclusions (or so I'm told).
Ian

In , Joseph Strout  writes:
>The Canonical Timeline
>Assume that N transition times from Phase i to Phase i+1 are normally
>distributed, with mean m and standard deviation s.  Then in the canonical
>case, the first transition occurs when (N-1)/N of the distribution occurs
>later.  In general, the ith transition occurs at time t where
>
>	Integral of f(x,m,s) from t to infinity = (N - i) / N
>
>where f(x,m,s) is the frequency of x in a normal distribution with mean m
>and standard deviation s.
I don't think this is quite what you want since this forces the Nth transition 
to never occur (the normal integral from minus infinity to any finite value is
never equal to 1).  The values you describe are the (N-i)/N th quantiles of
the normal distribution.  Perhaps some slight modification of the choice of
quantiles for your canonical distribution would work for you.  Using the 
expected values of the N order statistics instead of quantiles may also prove
useful.  
Both quantiles and order statistics should be discussed in most intermediate
or advanced statistics texts, and related papers should be easy to find.

In , Joseph Strout  writes:
>The Canonical Timeline
>Assume that N transition times from Phase i to Phase i+1 are normally
>distributed, with mean m and standard deviation s.  Then in the canonical
>case, the first transition occurs when (N-1)/N of the distribution occurs
>later.  In general, the ith transition occurs at time t where
>
>	Integral of f(x,m,s) from t to infinity = (N - i) / N
>
>where f(x,m,s) is the frequency of x in a normal distribution with mean m
>and standard deviation s.
I don't think this is quite what you want since this forces the Nth transition 
to never occur (the normal integral from minus infinity to any finite value is
never equal to 1).  The values you describe are the (N-i)/N th quantiles of
the normal distribution.  Perhaps some slight modification of the choice of
quantiles for your canonical distribution would work for you.  Using the 
expected values of the N order statistics instead of quantiles may also prove
useful.  
Both quantiles and order statistics should be discussed in most intermediate
or advanced statistics texts, and related papers should be easy to find.

"Less Wright"  in
<01bbe808$c3a6ede0$2d292299@lessw10> tells a funny story and asks a
probability question. Two students are asked to name one of the four tires
on a car. What is the probability they will name the same one.
In this formulation the answer is obviously 0.25. Your 1/16 answer would
be true if the professor knew the correct tire and both students had to
guess (for example, suppose they had shown him a borrowed garage bill but
neither remember which tire was listed on the bill).
Actually the students may be able to do better than 0.25. For example,
consider the following question: you are to meet someone in New York City
at noon tomorrow but you have no place selected and no way to contact the
other person. What is your probability of selecting the same place?
According to the logic above the answer is almost zero since there are a
very large number of places in New York. However, upon reflection, it's
clear that only an idiot would go to, say, 1135 37th Street in Brooklyn.
If you went to, say, the clock in Grand Central Station, the Observation
Deck of the Empire State Building, or the 59th Street entrance to Central
Park; you might have a 1% or 2% chance of meeting the other person. If you
know something about them (for example, they love "An Affair to Remember"
and "Sleepless in Seattle") you have an even better chance.
My guess if that if you asked 100 people the tire question (an interesting
experiment), 35 would say "right front." Smart people who thought about it
would probably say the same thing in greater numbers.
Aaron C. Brown
New York, NY

Babak Fakhamzadeh  wrote:
>Hi,
>
>Can anyone tell me an easy conversion scheme from  variables, picked from a uniform distribution, to variables picked from a normal one?
>
>Thanx
>
>Babak
I can never quite recall the formula, but you can use the Box-Muller (
sp? ) method which converts a pair of uniform random variables into a
pair of normal random variables.  It is listed in the book Numerical
Recipies in C.
george c

In article <32B03222.41C67EA6@sees.bangor.ac.uk> Ian  writes:
>Ilias Kastanas wrote:
>> 
>> In article <329D8210.41C67EA6@sees.bangor.ac.uk>,
>> Ian   wrote:
>> >I'm intreged (spelling?) as to how the rules of proof were shown to be
>> >_the_ rules of proof. Surely any such proof would have to be
>> >self-referential, or rely on axioms.
>> 
>>         It is in fact remarkable.  The logical axioms and Modus Ponens
>>    are straightforward and almost simplistic; and yet they suffice.  For
>>    every semantic implication, "in every structure where P holds, Q also
>>    holds" there is a formal deduction of Q from P using those rules.  It
>>    is the Completeness Theorem.
>> 
>>                                                         Ilias
>
>
>I am confused as to exactly what you mean by the completeness theorem.
>You don't seem to have said anything here which invalidates my comments
>on self-referentiality or reliance on axioms. 
Actually, he did; he said that you hadn't done enough reading.
What you're looking for is Godel's Completeness Theorem.  Basically,
it demonstrates that, given a set of axioms (as a *variable*, in this
context), if a sentence is true in all models satisfying the axioms,
then it's derivable via 1-order logic (or in other words, true in all
models implies provable).
The tricky bit (clever chap, Kurt) is that by quantizing over axioms,
and because most of the work is done by the semantics, he can demonstrate
that it doesn't matter what axioms you pick.
	Patrick

A function f(t) is said to be of exponential order 
if there are positive constants c and a, and some
number T, such that |f(t)| < c*exp(a*t) for t > T.
(As usual, * denotes multiplication.)
I don't know how this relates to "rates of convergence".
If the question is what is the definition of exponential
order for a series, I'm afraid I don't know the answer,
unless it means it converges faster than the Taylor series
for an exponential function.
dpd

In <01bbe808$c3a6ede0$2d292299@lessw10>, "Less Wright"  writes:
>1)Question - worth 100 points:
>Which tire ?
>
>The bet:
> what is the probability of both students guessing the correct
>tire, given that they hadn't agreed beforehand on which tire had gone flat
>in their concocted story?  Everyone but me said 25%
Sorry, but you lose.  The correct answer is 25%.  Since the professor cannot
know which tire the two should choose (particularly since there was no flat),
all the students need to do is choose the same tire.  The probability that they
both choose the left front tire, for example, is indeed 1/16 following your
reasoning.  Since there are four possible ways they could agree, each having
probability 1/16, the final answer is 1/4 or 25%.  (Of course, the whole
discussion assumes that they each are equally likely to choose any given tire.)

Hi,
Daniel Nordlund wrote:
> >          float x1, x2, w, y1, y2;
> >
> >          do {
> >                  x1 = 2.0 * ranf() - 1.0;
> >                  x2 = 2.0 * ranf() - 1.0;
> >                  w = x1 * x1 + x2 * x2;
> >          } while ( w >= 1.0 );
> >
> >          w = sqrt( (-2.0 * log( w ) ) / w );
> >          y1 = x1 * w;
> >          y2 = x2 * w;
> >
This code blows up if w = 0. This can happen if ranf() can be zero.
Numerical Recipes in C V.2 page 289 has for the while line: 
while ( w >= 1.0 || w == 0.0 );
Also, for processors with fast trig functions or for applications that
cannot throw away values, this same page has the square-to-circle trig
mapping equations. These obviate the while loop and one gets 2-to-2
uniform->normal.
Cheers!
-Will
-- 
Dr. William C. Snyder   will@icess.ucsb.edu
Imaging Scientist       http://www.icess.ucsb.edu/~will/will.html  
Institute for Computational Earth Systems Science
University of California Santa Barbara

In article <01bbe808$c3a6ede0$2d292299@lessw10>,
Less Wright  wrote:
>Tonight on the way home from dinner I told some of my programming
>colleagues a funny story I got in email, but after the laughs it turned
>into a discussion of probability theory, followed by a bet on who's answer
>was right...please enjoy the short story below if you haven't already heard
>it, but if you can help settle our bet based on the resulting problem it
>brought up, I would appreciate it!
>
>The story as I recall it: (if you've already heard, please skip to 'the
>bet')...
>Two college students, supposedly from UVA, were doing quite well in their
>chemistry class and ended up having several free days during finals before
>their last exam in chemistry.  Given that they only needed to get a D on
>their chem final to pass the course, they decided to enjoy the free days
>before the exam by partying at a neighboring college...in their ensuing
>drunkeness, they ended up oversleeping on the day they were supposed to
>return to take their 'easy' chem exam...technically, this meant they got a
>0 and were now going to fail the course.  They drove back to campus, and on
>the way concocted a story about how they had returned on time, but were
>seriously delayed due to a flat tire, and the ensuing towing and repair
>time.  
>They told their professor the story, and begged for a make up exam - he
>consented, provided that they could not leave the classroom during the exam
>for any reason until they were done with their exam, and said they could
>take the exam the next day...they stayed up all night reviewing all of
>their chemistry knowledge.  They met the professor the next day, and he
>then placed them into seperate rooms on each side of the hall, gave them
>their exams, shut the door, and he then sat in a desk in the middle of the
>hallway.  The students then opened the exam book to see:
>1)Question - worth 100 points:
>Which tire ?
>
>The bet:
>Hopefully you enjoyed the story, but now for the statistical bet which
>evolved - what is the probability of both students guessing the correct
>tire, given that they hadn't agreed beforehand on which tire had gone flat
>in their concocted story?  Everyone but me said 25%, because their are four
>tires and only one correct choice.  I disagreed, because my vague
>recollection of my probability course says that the real  fact is that you
>are evaluating the probability of *two* people choosing the same tire, each
>of which has a 25% choice, so your odds of them both randomly picking the
>same tire should be less than 25%....i.e. potential outcomes are 16
>different combinations of tires (i.e. Student A chooses Right Front,
>Student B could choose RF, LF, RB, LB.  A could also choose Left Front, and
>B could again choose RF,LF,RB,LB...)  So is it still a 25% chance they both
>pick the same tire, or is it 1/16 probability or ??  Actually though, now
>that I have written this out, I am starting to think I am wrong - b/c A
>could pick any given tire, and B could then pick any tire with a 25%
>probability of B's pick matching A...anyway, if you know the definitive
>answer and could help us resolve this gentlemen's bet, I would greatly
>appreciate it....
>
>Thanks in Advance,
>Less
On "standard" independence assumptions, the answer is .25.
--
-----------------------------------------------------------------------------
David Nichols             Senior Support Statistician              SPSS, Inc.
Phone: (312) 329-3684     Internet:  nichols@spss.com     Fax: (312) 329-3668
-----------------------------------------------------------------------------

arte@panix.com (Arthur Ellen) in <58nk1e$grc@panix.com> asks:
> Can someone post an algorithm for the 4 moments
> in either basic or pascal with a brief explanation?
Moments can be defined either for a sample or a distribution. If it is a
distribution then you must perform a numerical integration, the technique
will depend on the functional form and properties.
For a sample it is easy. Suppose you have N points in an array X(I). Then
(in Basic):
M1 = 0
M2 = 0
M3 = 0
M4 = 0
For I = 1 to N
     M1 = M1 + X(I)
     M2 = M2 + X(I)^2
     M3 = M3 + X(I)^3
     M4 = M4 + X(I)^4
Next I
The next step depends on whether you want moments around the mean or
around zero (or some other number) and how you want to adjust the third
and fourth moments (the third moment is sometimes divided by the second
moment to the 1.5 power, the fourth moment is sometimes divided by the
second moment squared, and sometimes 3 is subtracted). Assuming you want
uncorrected moments around the mean:
First Moment = M1 / N
Second Moment = M2 / N - (M1 / N)^2
Third Moment = M3 / N - 3 (M1 / N) (M2 / N) + 2 (M1 / N)^3
Fourth Moment = M4 / N - 4 (M1 / N) (M3 / N) + 6 (M1 / N)^2 (M2 / N) - 3
(M1 / N)^4
Aaron C. Brown
New York, NY

I have an economics problem that is a variation of the St. Petersberg 
Paradox.
A regular, six-sided, fair die will be rolled repeatedly until a 2 comes 
up.  When a 2 comes up on the nth roll, you will be paid 6^n dollars.  
The paradox comes in because it was assumed for a long time that people's 
willingness to pay to be in such a gamble should be based on its 
expected monetary value.  However, a game such as this has an infinite 
expected value, leading people to change their views and assume that 
people base their attitudes toward gambles on the expected utility (or 
satisfaction) a gamble will bring them.(I'm sorry for the bad 
explanation, but I'd rather  get to the meat of the problem)
Anyway, here's the problem.  If a person has a utility function 
u(x) = ln(x), what is the expected utility of this gamble?
Basically you are going to replace the payoff x that you used when 
calculating the expected value of the gamble with ln(x) for calculating 
the expected utility of the gamble.  I've reduced it to this:
Expected utility = (1/6)*ln(6)*[n*sum(5/6)^n-1].  I'm having trouble even 
verifying that this series even converges, much less what to.  Any help 
is appreciated.
Mike LePore

Hi,
This is a beginners' SAS question.
I am a social scientist trying to 
finish my final project in a research class.
Is it possible to perform tests of simple effects
(as defined in APPLIED STATISTICS by HINKEL/WIERSMA/JURS)
in SAS ? I am using the following setup
PROC ANOVA DATA=PROJECT;
     CLASSES A B;
     MODEL S=A B A*B;
     MEANS A B A*B
     MEANS A B A*B/TUKEY BON;
     FORMAT A AA. B BB.;
     TITLE 'THE TWO-WAY FIXED-MODEL ANOVA';
The second means statement doesn't perform the simple
effects as I would have expected.
Please help !

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Mea culpa!  I sure hope nothing but the program blew up as a 
result of my error.  As Will Snyder points out, it would be 
prudent to trap for w=0, and he provides a correct solution. 
However, the problem does not occur when the random number 
function, ranf(), returns a 0; the problem occurs if both x1 
and x2 evaluate to 0.  This will happen if ranf() returns 0.5 
twice in a row so that 2*ranf()=1, at least within the 
precision of the machine.
As current computing power has increased, it is probably less 
necessary to avoid trig functions, and Will's comments below 
are very appropriate.
Dan
William Snyder wrote:
> 
> Hi,
> 
> Daniel Nordlund wrote:
> 
> > >          float x1, x2, w, y1, y2;
> > >
> > >          do {
> > >                  x1 = 2.0 * ranf() - 1.0;
> > >                  x2 = 2.0 * ranf() - 1.0;
> > >                  w = x1 * x1 + x2 * x2;
> > >          } while ( w >= 1.0 );
> > >
> > >          w = sqrt( (-2.0 * log( w ) ) / w );
> > >          y1 = x1 * w;
> > >          y2 = x2 * w;
> > >
> 
> This code blows up if w = 0. This can happen if ranf() can be zero.
> Numerical Recipes in C V.2 page 289 has for the while line:
> while ( w >= 1.0 || w == 0.0 );
> 
> Also, for processors with fast trig functions or for applications that
> cannot throw away values, this same page has the square-to-circle trig
> mapping equations. These obviate the while loop and one gets 2-to-2
> uniform->normal.
> 
> Cheers!
> 
> -Will
> 
> --
> Dr. William C. Snyder   will@icess.ucsb.edu
> Imaging Scientist       http://www.icess.ucsb.edu/~will/will.html
> Institute for Computational Earth Systems Science
> University of California Santa Barbara

nichols@spss.com (David Nichols) wrote:
>In article <01bbe808$c3a6ede0$2d292299@lessw10>,
>Less Wright  wrote:
>>Tonight on the way home from dinner I told some of my programming
>>colleagues a funny story I got in email, but after the laughs it turned
>>into a discussion of probability theory, followed by a bet on who's answer
>>was right...please enjoy the short story below if you haven't already heard
>>it, but if you can help settle our bet based on the resulting problem it
>>brought up, I would appreciate it!
>>
>>The story as I recall it: (if you've already heard, please skip to 'the
>>bet')...
>>Two college students, supposedly from UVA, were doing quite well in their
BW saving snip inserted here ....
>>probability of B's pick matching A...anyway, if you know the definitive
>>answer and could help us resolve this gentlemen's bet, I would greatly
>>appreciate it....
>>
>>Thanks in Advance,
>>Less
>
>On "standard" independence assumptions, the answer is .25.
>
>--
>-----------------------------------------------------------------------------
>David Nichols             Senior Support Statistician              SPSS, Inc.
>Phone: (312) 329-3684     Internet:  nichols@spss.com     Fax: (312) 329-3668
>-----------------------------------------------------------------------------
I agree with the 0.25.  The confusion between the 1/4 and the 1/16
answers may be the fact that there are 4 correct answers out of the 16
combinations ( usually there would be one correct combination ).  The
students don't have to get the 'correct' tire ... just the same one.
I think the correct answer is 4/16.
george c

Ihave recently been asked to include the geometric mean and geometric standard deviation
in a stat table I am creating.
I have lightly researched the definition of geometric mean.
Why would someone want to see the geometric mean and compare
it to the arithmetic mean (and same for standard deviation).
What does this statistic tell you about the datset?
Oh, and I am having a devil of a time finding the equation used to calculated the geometric Standard
Deviation.
Thank you in advance.
Rock Neveau
Denver, CO