Newsgroup sci.stat.math 10943

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Herman Rubin wrote:
> 
> In article <50hch4$icj@www.umsmed.edu>, Warren   wrote:
> 
> >>There is no central limit.  However, there is a limit theorem that
> >>that is central to mathematical statistics.
> 
> >Should we follow the mathematicians model and call it the "Fundamental
> >Theorem of Statistics" to avoid confusion?
> 
> 1.  It is not a theorem of statistics; it is a theorem of probability.
> 
> 2.  What is fundamental about it?
Come on, the fact that almost any normalized sum of random variables
converges
weakly to a standard normal distribution? That's a pretty fundamental
property.
I also happen to think that it's pretty extraordinary.
--------------------------------------
Any philosophy that can be put in a nutshell belongs there.
                -- Sydney J. Harris
Joseph Edward Nemec                    
Operations Research Center	         
Room E40-149
Massachusetts Institute of Technology
Cambridge, MA 02139
nemecj@mit.edu
http://web.mit.edu/manuf-sys/www/nemec/

aacbrown@aol.com (AaCBrown) wrote:
>John Collins  in <5143t1$itc@acs1.star.net> writes:
>
>> In blackjack there are thirteen card values and multiple
>> decks. I think this is the correct distribution. Does anyone
>> have an expression?
>
>Actually in Blackjack 10, J, Q, and K all count the same so there are only
>10 different values. Call the numbers of remaining cards n1, n2, . . .,
>n10 for each of the 10 values. Let N = n1+n2+. . .+n10.
>
>The probability of any specific situation, say blackjack for both dealer
>and player at the same time, can be computed as:
>
>N! * n1 * (n1-1) * n10 * (n10 - 1) * 2 / (N-4)!*4!*3
>
>
>Aaron C. Brown
>New York, NY
Thank you for your comments. I like to retain the 13 card types. Yes there 
are 4 with a value 10 and the ace with 2 values. The crux of the question 
is about a distribution with more than 2 results, in this case 13, and a 
dependent draw probability, like the hypergeometric which is a similar 
version of the binominal distribution.
John

In article <323DF76F.9832E9E@mit.edu>,
Joseph Edward Nemec   wrote:
>I am looking for a copy of
>
>"Convergence of Probability Measures"
>
>by Billingsley, 1968.
>
>Anyone know where I can get a copy? I tried the publisher, but it's
>been three months...
>
>--------------------------------------
>Any philosophy that can be put in a nutshell belongs there.
>                -- Sydney J. Harris
>
>Joseph Edward Nemec                    
>Operations Research Center	         
>Room E40-149
>Massachusetts Institute of Technology
>Cambridge, MA 02139
>
>nemecj@mit.edu
>http://web.mit.edu/manuf-sys/www/nemec/
Try to Seminary Co-op Bookstore in Hyde Park here in Chicago. It's right
across the street from where Patrick Billingsley works (or has, and I
think still does), and they generally seem to have a copy on hand. They
can be reached at www.semcoop.com.
--
-----------------------------------------------------------------------------
David Nichols             Senior Support Statistician              SPSS, Inc.
Phone: (312) 329-3684     Internet:  nichols@spss.com     Fax: (312) 329-3668
-----------------------------------------------------------------------------

This is a conditional prob. question.3 criminals named A,B and C were to be
executed.Two days before the execution one of them is pardoned,picked
randomly.The name of the man pardoned is  secret.Criminal A asks the warden
if B will be executed.The warden answers yes.Criminal A thinks his P of
being pardoned has risen to 1/2.Warden thinks that A's P of being pardoned
has not changed,it remains 1/3.Who is right and why?If possible pls give me
the sample space.
Ari

Hi, when might one like to do a factor analysis on a covariance
which has NOT had means subtracted (xx' instead of (x-u)(x-u)')?
For example, in our data, we are interested in correlations but also
in similarity between the means of the series' being analized.
I recall that SAS has a NOINT option which leaves the intercept out of
the analysis, not sure if that's the same thing...
Thanks for any insight,
Ralph Gonzalez

gonzalez@crab.rutgers.edu (Ralph Gonzalez) in
<51memq$2b9@crab.rutgers.edu> asks:
>  when might one like to do a factor analysis on a
> covariance which has NOT had means subtracted
> (xx' instead of (x-u)(x-u)')?
One case is when you know the true means are all zero. A related case is
when zero has a natural meaning and you are more interested in variation
around zero than variation around the mean. An example of the latter might
be a study of security returns.
\
Aaron C. Brown
New York, NY

At
ftp://ftp.stat.ucla.edu/pub/apps/msdog/hedeker/
you will find various packages for mixed level linear modeling. We
already had MIXOR (for ordinal multi-category outcomes, either
probit or logit) and MIXREG (for numerical outcomes, but with
autocorrelated error structures). We now also have MIXGSUR, for
mixed level survival analysis. All three programs have manuals,
binaries for DOS, Mac (PPC), and Solaris 2.x, and examples.
Programs and manuals are written by Don Hedeker (School of
Public Health and Prevention Research Center, University of
Illinois at Chicago, hedeker@uic.edu). The Solaris and Mac(PPC)
binaries have been compiled by Jan de Leeuw (deleeuw@stat.ucla.edu).
-- 
Jan de Leeuw; UCLA Department of Statistics; UCLA Statistical Consulting
US mail: 8118 Math Sciences, 405 Hilgard Ave, Los Angeles, CA 90095-1554
phone (310)-825-9550;  fax (310)-206-5658;  email: deleeuw@stat.ucla.edu  
                 www: http://www.stat.ucla.edu/~deleeuw

This problem arises from signal detection theory, and concerns
composite hypothesis testing.
The observer receives a single sample r.
r = s + n
s is a sample from {s0, s1, s2, s3, s4, s5}, i.e. one of several signal
levels is given per trial.
n is a sample from N(0, sigma^2) where sigma is known
H0: r = n  (i.e. s=0)
H1: r = s + n (i.e. s>0); observer says "signal"
What is the likelihood ratio?
What is p("signal"|s)?
What happens as the signal set shrinks?  (In the limit, to {0,s1}).
Thanks very much for any help. (No this is not homework!)
Bill Simpson

Hi,
I would worry a little about the assumptions for ANOVA for these data.  
Looks to me like the equality of variances is questionable.  You could 
try a transformation [maybe log] to stabalize the variance a little.  
Plot out the residuals from the ANOVA against the "group" variables...if 
you don't see any noticable trends, then the ANOVA on the original data 
may be adequate.  
If the transformation doesn't help, you might try a nonparametric 
analysis.  Maybe Kruskal-Wallis would apply.
Of course, ANOVA on the raw data is much easier to interpret most of the 
time.  But you do need to be a little concerned about the unequal 
variances.
Looking at your data, the smallest mean has the smallest variance.  Is 
this a control of some kind?
Best luck :)

I was recently given a homework problem in which I was told after I 
turned it in that it was an infinite series problem but I can't figure 
out what type of series it is.  The problem is as follows:
I will spin the roulette wheel until the ball lands on green.  The prob 
of green in any one spin is 2/38.  What is the prob that the first green 
observed is observed on an odd numbered spin?
I came up with 2/38*[(36/38)^2 + (36/38)^4 + ....].  If correct, I don't 
know how to evaluate the last part.  If it included even exponents (all 
integers) it would just be a power function, but that's not the case 
here.  Any help is appreciated.
Mike

Ari wrote:
> 
> This is a conditional prob. question.3 criminals named A,B and C were to be
> executed.Two days before the execution one of them is pardoned,picked
> randomly.The name of the man pardoned is  secret.Criminal A asks the warden
> if B will be executed.The warden answers yes.Criminal A thinks his P of
> being pardoned has risen to 1/2.Warden thinks that A's P of being pardoned
> has not changed,it remains 1/3.Who is right and why?If possible pls give me
> the sample space.
> 
> Ari
 Ari-
 I think A is in luck with probability 1/2: The sample space can be viewed
 as the set of three points {a,b,c} where a is the event
 "A is the one pardoned", b is the event "B is the one pardoned" and c
 is the event "C is the one pardoned". Then Pr(a|not b) = Pr(a|a or c) =
 Pr(a and (a or c))/Pr(a or c) = Pr(a)/Pr(a or c)= (1/3)/(2/3) = 1/2.
 In case you're wondering, this is subtly different from the notorious
 Monty Hall problem in which a subject with Pr of 1/3 of being chosen for
 something asks for the name of *anyone* else who was not chosen. That
 information does not change his probability. 
 Ellen

In article <3239A73A.1404@lamrc.com>, philippe.boyer@lamrc.com wrote:
> Does anyone can help me understanding these definitions ?
> 
> What are the formulas to convert the uniformity of a population (sigma)
> into CP or CPK ?
> 
> If you do know a book or better a WWW site where I can get detailed
> definitions, examples on how to use these indices, I would apreciate a
> lot.
> 
> Thanks in advance,
> 
> philippe.boyer@lamrc.com
A good reference is Samuel Kotz  Process Capability Indices
But you could also see any good reference on statistical process control
-- 
Stefan Steiner
Dept. of Statistics and Actuarial Science
University of Waterloo
Waterloo, Ont. Canada

Lina Santos wrote:
> 
> Hello, i am having a trouble with this test, can someone help me with
> this please :)?
> 
>  Here it goes :~
> 
>   We selected at random 10 students from one of the 4 biggers
> Universities on North America. We examinate their intelectual capacities.
> This 40 students realized one mathematical test and we registered the
> means and variances ( sample ones ) to each one of the 4 universities :
> 
>                  University          mean        variance
>                  ----------------------------------------
>                       1              105.7         30.3
>                       2              102.2         54.4
>                       3               93.5         25.0
>                       4              110.8         36.4
> 
>   test the hipothesis of the knowlegde levels are equal in the 4
> Univeristies...
> 
>                           Thanx in Advance
 This is another idea which may seem unorthodox in this context but seems
 legitimate: For each of the four groups we have a mean mi. mi is an estimate
 of the population mean and has estimated variance vi where
 vi is the variance in the last column divided by 10, ie, 3.03 for the
 first group.
 Let ui = mi/sqrt(vi).   Then let T1 = the sum for i = 1 to 4 of ui^2.
 T2 = (the sum for i = 1 to 4 of ui)^2/4. Under the null hypothesis
 that the 4 means are the same, T1-T2 is chi square with 3 df. A 
 reference for this is Statistical Methods for Ratios and Proportions
 by J. Fleiss.
 Good luck.
 Ellen

I have been reading Keppel's Design and Analysis text, and
he describes an iterative process using charts (from Pearson
and Hartley's Biometrika tables) for estimating sample size
for a given power and omega squared (for ANOVA analysis).
Are there equations available for these charts so that I can
make these calculations without the charts? I also checked
Cohen's book, and he gives tables for the power
calculations, but again no equations.
Thanks, Gary
--
Gary Muhonen
gary@iotasys.com    http://www.iotasys.com
gary@unr.edu

David Nichols (nichols@spss.com) wrote:
: In article <51292e$ih@news.injersey.com>,
: Richard S. Kain  wrote:
: >Stephen Barry (100665.1475@CompuServe.COM) wrote:
: >: Can anyone explain (or more precisely provide formulae) for how
: >: APR's are calculated; paticularly from nominal or effective rates.
: >
: >: Many thanks...
: >
: >The APR is the interest rate at the end of a year when it is compounded.
: >Compounding may be daily, monthly (usually) quarterly. It is higher than 
: >the simple annual interest rate. The highest is daily compounding, then 
: >monthly then quarterly ....
: >Compounding means interest on the interest.
: >You can construct your own compounding table when you understand how it 
: >is actually done.
: >
: >The above is on bank interest, investment interest.
: >
: >On a loan which starts with a low-ball rate and then converts to a higher 
: >rate, my assumption is that the APR is higher than the beginning rate and 
: >less than the longer term rate. But closer to the long=term rate the 
: >longer the loan.
: There is also now in common use the annual percentage yield (APY), which
: is quoted as the amount of interest earned on $100 left to compound for
: one year. APY is higher than APR if compounding is done more frequently
: than annually. I believe that financial institutions offerings savings
: vehicles in the United States are now required to state the APY, which
: is the appropriate number for comparisons between offers (perhaps this
: is only in Illinois).
: --
: -----------------------------------------------------------------------------
: David Nichols             Senior Support Statistician              SPSS, Inc.
: Phone: (312) 329-3684     Internet:  nichols@spss.com     Fax: (312) 329-3668
: -----------------------------------------------------------------------------
My apologies. Shortly after posting my definition above of APR I was 
reminded by a bank ad that I had described APY. Yield rather rthan rate.
Yes I know the difference but my aging brain got confused.
APR is the simple annual rate and APY is the effective rate after 
compounding at the end of a year.

The answer is:
2/38*[ 1+ (36/38)^2 + (36/38)^4 +....] = 2/38*[1/{ 1 -(36/38)^2 
}]

A good encryption algorithm can be used as the basis of a random
number generator.  I implemented three random number generators using
Robert Scott's NewDES encryption algorithm and tested them George
Marsaglia's Diehard collection of random number tests.  The random
number generators passed these tests (with one exception, explained
below).
The original purpose for running these tests was to test the quality
of the NewDES encryption.  However, the random number generators based
on NewDES proved to be about 40% faster than the mrand48 random number
generator provided with the HP/UX operating system, which makes them
useful for practical applications.
The three random number generators I tested were created by using NewDES
as follows:
 1)  The sequence of 64 bit numbers 0, 1, 2, ... was encrypted using the
     key "RandomNumberKey".  Numbers were represented using "big-endian"
     order, so B0 was the most significant octet of the 64 bit number and
     B7 was the least significant octet.
 2)  Zero was encrypted using the sequence of 120 bit numbers 0, 1, 2, ...
     as keys, again using "big-endian" representation for the numbers.
 3)  A buffer was initialized to zero and then repeatedly encrypted using
     the key "RandomNumberKey".  More formally,
        V_0 = E(0)
        V_(n + 1) = E(V_n)
     where E(x) is x encrypted with the key "RandomNumberKey"
These schemes all produced 64 bits of random data per encryption, which
was then split into two 32 bit random numbers.
I discovered two problems with Diehard.  First, running the overlapping
5-permutation test produced a floating point exception on some data
sets, including the output of random number generator 3 described above.
Second, all data sets I tried failed the minimum distance test.  I wrote
my own implementation of this test in Ada and got different results.
Studying the Fortran source for Diehard revealed that the code passes an
array of single precision floating point numbers to a routine expecting
an array of double precision floating point numbers.  Presumably this
had the effect of sorting pairs of floating point numbers on the machine
where Diehard was developed, but doesn't do this on the HP because the
format of floating point numbers is different.
				Kenneth Almquist

A Two Year Research Project with Bursary.
This project offers the opportunity to apply quantitative approaches to
modelling the diffusion and codification of knowledge and competence within
and between organisations.  Understanding of this process is key to the
effective management of research and development programmes.  The output of
the project will be a prototype simulation model with potential uses in
management training and business gaming.
The successful applicant will be expected to register for an MPhil degree
and have a background in operational research, mathematics or a related area.
Interested applicants should send a cv to Dr Nigel Meade, The Management
School, Imperial College, London SW7 2PG by post or EMail (nme@ic.ac.uk) by
7 October 1996.

A Two Year Research Project with Bursary.
This project offers the opportunity to apply quantitative approaches to
modelling the diffusion and codification of knowledge and competence within
and between organisations.  Understanding of this process is key to the
effective management of research and development programmes.  The output of
the project will be a prototype simulation model with potential uses in
management training and business gaming.
The successful applicant will be expected to register for an MPhil degree
and have a background in operational research, mathematics or a related area.
Interested applicants should send a cv to Dr Nigel Meade, The Management
School, Imperial College, London SW7 2PG by post or EMail (nme@ic.ac.uk) by
7 October 1996.

______________________________________________________________
Sorry for repeating my previous help request but I didn't get *
any help which I really need. So I hope for better luck this  *
time.                                                         * 
______________________________________________________________
I would like to simulate observations from a "multicenter clinical
trial" for didactical purposes. The analysis of the data will be
performed with SAS using the following program code:
        proc glm;
        model y = x treatment center x*center treatment*center;
        random center;
where y: response variable ("after treatment" value - x)
      x: covariate, "first visit" value
      treatment: treatment effect, fixed
      center: center effect (multicenter study...)
The factor center is considered random while treatment is fixed. I would
like to get significant center effects, but not-significant treatment by
center and covariate by center interaction. 
 * How do I simulate (?!) a fixed effect? 
 * Is it possible to control the significance (non-significance) of the
factors by choosing large/small standard deviation of the df? What about
the interaction effect?
Available software: SAS (Win), Mathematica and Minitab for Macintosh.
Any suggestion/help would be highly appreciated.
Thanks,
Peter Niman

Jive Dadson wrote:
> 
> Suppose X and Y are two Gaussian distributions with std-deviation 1,
> and that mean(Y)-mean(X) = t. Then the probability that a sample drawn
> from X will be smaller than a sample drawn from Y is,
> 
>                oo
>      g(t) = integral  N(x)*(1-CN(x-t)) dx
>             x = -oo
> 
> where N is the standard normal function, and CN is the cumulative
> distribution of the standard normal (assuming I didn't get a sign reversed
> somewhere).
> 
> Is it the case that g(t) = CN(.5*sqrt(2)*t) ?
> 
>     -- J.
J.-
I think g(t) = CN(t/sqrt(2)). X-Y+t is normal with mean 0 and variance 2.
P(X-Y < 0) = P(X-Y+t < t) = P((X-Y+t)/sqrt(2) < t/sqrt(2)) = CN(t/sqrt(2)).
Ellen

Ari wrote:
> 
> This is a conditional prob. question.3 criminals named A,B and C were to be
> executed.Two days before the execution one of them is pardoned,picked
> randomly.The name of the man pardoned is  secret.Criminal A asks the warden
> if B will be executed.The warden answers yes.Criminal A thinks his P of
> being pardoned has risen to 1/2.Warden thinks that A's P of being pardoned
> has not changed,it remains 1/3.Who is right and why?If possible pls give me
> the sample space.
> 
> Ari
The warden is wrong. A can now estimate his chances at 1/2.
A simple way to determine whether A's chances have changed is to
ask whether the answer he got carried any information. Decidedly it did.
To the criminal, (for all the difference it makes ... is he applying
for life-insurance?), his chances have improved, because there was
a fatally worse answer he could have heard: "No."
    P(A|B) = PB(A&B;)/PB(B) = (1/3)/(2/3) = 1/2
where "PB" stands for "prior belief" - the _a priori_ probability
(before A learned that B was a goner).
This seems much too easy. Are you sure you got the problem right?
    J.

In Stephen Jay Gould's latest book, _Full House_ (Harmony
Books, 1996), he says [p. 32]:
	"... a famous statistician once showed a precise
	correlation between arrests for public drunkenness and
	the number of Baptist preachers in nineteenth-century
	America."
Can anyone identify the statistician and provide a reference?
E-mail appreciated.
Jeffrey Shallit, Computer Science, University of Waterloo,
Waterloo, Ontario  N2L 3G1 Canada shallit@graceland.uwaterloo.ca
URL = http://math.uwaterloo.ca/~shallit/

ka@socrates.hr.att.com (Kenneth Almquist) wrote:
>
>The three random number generators I tested were created by using NewDES
>as follows:
> 1)  The sequence of 64 bit numbers 0, 1, 2, ... was encrypted using the
>     key "RandomNumberKey".  Numbers were represented using "big-endian"
>     order, so B0 was the most significant octet of the 64 bit number and
>     B7 was the least significant octet.
> 2)  Zero was encrypted using the sequence of 120 bit numbers 0, 1, 2, ...
>     as keys, again using "big-endian" representation for the numbers.
> 3)  A buffer was initialized to zero and then repeatedly encrypted using
>     the key "RandomNumberKey".  More formally,
>        V_0 = E(0)
>        V_(n + 1) = E(V_n)
>     where E(x) is x encrypted with the key "RandomNumberKey"
     The first two methods at least guarantee long periods, but the
third method is not advisable for random number generation.  Even if
NEWDES behaves like a randomly-chosen permuation acting on the set of
64-bit numbers, the expected value of the length of the cycle
containing any given 64-bit number is *much* smaller than 2^^64.
(Because of the "birthday problem").  If you happen to be unlucky
enough to pick a starting 64-bit number that is contained in a short
cycle under the permutation defined by encryption, then the
distribution of the random numbers so obtained will be poor indeed.
My vote is for encrypting 0,1,2,3, under a fixed key, as described in
your first method.
   -Bob Scott