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[math-fun] Relativity Challenge
by Thane Plambeck 29 Sep '05

29 Sep '05

Please pass on by if you're not interested in cranks. > Scientists and researchers have scrutinized Einstein's work and have been unable to definitively find anything wrong with it... until now! > Best of all, you don't have to have a PhD to understand and verify Einstein's mistake for yourself. > All that's required is an understanding of basic Algebra, access to Einstein's papers, and information to guide the way.... > http://www.relativitychallenge.com/index.htm -- Thane Plambeck http://www.plambeck.org/ehome.htm

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Re: [math-fun] another kind of median (trying again)
by dasimov＠earthlink.net 29 Sep '05

29 Sep '05

Jim writes: << >But: For very small samples, the median doesn't generally convey much information. >For a large sample, however, this "bar-graph" median can be unduly influenced >by outlying data -- much as an average can. Really? It seems to me that the bar-graph median is very close to the "true" median. In one case, your probability measure on R is a combination of point-masses; in the other case, it's a combination of uniform distributions on disjoint intervals. But it's only near the median that it makes any difference which you're using. >> I agree that if Jim's idea is translated into the corresponding thing for a continuous probability density, it's exactly thr usual median of a continuous distribution. I don't think it's the same for a discrete distribution. (There's a big difference between a discrete distribution and one that's a combination of uniform continuous distributions on intervals.) For example, given just one datum for each n, 1 <= n <= 100, one gets the bar-graph median at x = 50.5. But now say you just replace the single datum of 100 by a datum of 1000. The vertical line bisecting the area under the bar graph is at x = 999.9505 . --Dan

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RE: [math-fun] Relativity Challenge
by dasimov＠earthlink.net 29 Sep '05

29 Sep '05

Thane wrote: << Please pass on by if you're not interested in cranks. <<< Scientists and researchers have scrutinized Einstein's work and have been unable to definitively find anything wrong with it... until now! Best of all, you don't have to have a PhD to understand and verify Einstein's mistake for yourself. All that's required is an understanding of basic Algebra, access to Einstein's papers, and information to guide the way.... >>> http://www.relativitychallenge.com/index.htm >> Actually, Einstein's first paper (more of an appendix) on E = mc^2 -- which I was just looking at yesterday -- contains several errors (as I learned from the footnotes), though these may all be chalked up to typos. Most likely there are similar errors in his other early papers. (These errors in the 1905 E = mc^2 were all found & corrected in print soon after.) The guy behind "Relativity Challenge" seems to have merely caught some of these early typos. --Dan

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[math-fun] Re: Median
by Thomas Colthurst 29 Sep '05

29 Sep '05

I have a general philosophy that descriptive statistics such as the median are only interesting when viewed through the lens of inferential statistics. Thus, averages are important because they are the maximum likelihood point estimates of the center of a Gaussian distribution, and medians are important only so far as the maximum likelihood point estimate of the center of the Laplace distribution pi^(n/2) n Gamma(n) ------------------- e^( - | S^(-1) ( x - mu ) | ) (n/2)! det S is important. In one dimension (n=1), any mu satisfying sum_i | x_i - mu | / ( x_i - mu ) = 0 <=> sum_i sign( x_i - mu ) = 0 will maximize the Laplace distribution's likelihood. For the "hard" cases such as the median of {1, 2} which have a continuum of maximum likelihood point estimates, the use of the "median average" 1.5 is easy to motivate: it is the loss minimizing point estimate for any symmetric cost function. Since this is math-fun and we are talking about medians, here's a question I've had for a while. Say you have n real numbers, x_1 ... x_n. If we define sum_j x_j / | u_i - x_j | u_{i+1} = --------------------------- ( j = 1 .. n ), sum_j 1 / | u_i - x_j | and assuming that u_i is never equal to any x_j, then does the sequence { u_i } always converge to a median of the x_j ? -Thomas C

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[math-fun] Median
by Eugene Salamin 29 Sep '05

29 Sep '05

This idea of fudging the median by introducing some kind of averaging bothers me in a more fundamental way. If a distribution Y exists over a set X, then for the median to be defined, it suffices that X be an ordered set. There is no necessity that X possess an addition, although of course Y must have an addition so that you know when it's split in half. For example, if I define the order: mouse < cat < bear, and I have a mouse and 2 cats, then my median pet is a cat. But if I adopt a bear, what would the median become? Perhaps the pair (cat, bear); that's much more sensible than (cat+bear)/2. Even for distributions over the real line, medians and quantiles are often preferred to averages and standard deviations, since the former always exist while the latter need not. Example: the Cauchy distribution 1/(1+x^2). Suppose one has a distribution p(x), and one wishes to describe it by a single number, a central value, c, of x. Under the criterion of minimizing the mean square error int(p(x) (x-c)^2 dx), assuming the integral exists, we know that c is the mean. But here is something perhaps not as familiar. If instead we wish to minimize the mean absolute error, int(p(x) |x-c| dx), again assuming that the integral exists, then c is the median. Gene __________________________________ Yahoo! Mail - PC Magazine Editors' Choice 2005 http://mail.yahoo.com

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RE: [math-fun] Multiplicative Magic Squares
by Christian Boyer 29 Sep '05

29 Sep '05

Sorry Michael, I have not completely read one of your previous messages, you mentioned already the same square: > (p,q,pq; pp,qq,r) [4,4,1] 6480 -----Message d'origine----- De : Christian Boyer [mailto:cboyer@club-internet.fr] Envoyé : lundi 26 septembre 2005 16:14 À : 'math-fun' Objet : RE: [math-fun] Multiplicative Magic Squares >From: "Michael Kleber" <michael.kleber(a)gmail.com> > >So abcdABCD makes a pretty small fraction of all multiplicative magic >squares, as the magic product (and the exponents on each prime in its >factorization) get large. Here is an interesting family of 4x4 multiplicative squares, using only three variables a, b, c: 1 abbb bc aaa ac aab ab bb aaab c abb b bbb a aa abc Magic product = a^4 * b^4 * c, for the 4 rows, 4 columns, 2 diagonals and 2 broken diagonals. And also for other sets of four cells (the corners for example, the 4 numbers of each quarter, ...). Apply a=2 and b=3, and you will get one of the various possible examples with magic product 6480. 6480 is the second smallest product after 5040, as listed in my previous email. Christian.

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RE: [math-fun] another kind of median (trying again)
by dasimov＠earthlink.net 28 Sep '05

28 Sep '05

<< The picture was supposed to show a rectangle of width 1 and height 2 whose bottom is centered at x=1, and to the right of it, a rectangle of width 1 and height 1 whose bottom is centered at x=2. The base of the first rectangle goes from x=1/2 to x=3/2, and the base of the second rectangle goes from x=3/2 to x=5/2. The total area under the histogram is (1)(2)+(1)(1) = 3. The area to the left of the line x=5/4 is (5/4-1/2)(2) = 3/2, which is half of the total area. So x=5/4 is the "median". >> After getting an explanatory e-mail from John Robertson, I thought I posted a "now I get it" post to math-fun, but it apparently didn't get through. (Your original message was perfectly clear, and my attempt to reproduce your calculation neglected to take account of the +1/2 poisitioning of the bars, even though I implied that positioning myself.) But: For very small samples, the median doesn't generally convey much information. For a large sample, however, this "bar-graph" median can be unduly influenced by outlying data -- much as an average can. But the idea of the median, as I see it, is to characterize the middle of the sorted data without much influence on the size of data points not close to the middle. Or am I still misunderstanding something? --Dan

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Re: [math-fun] Sex, Lies and Government Conducted Surveys
by dasimov＠earthlink.net 28 Sep '05

28 Sep '05

Jim writes: << Here's another kind of median I could imagine statisticians using: Suppose our data are the numbers 1, 1, and 2. The bar graph is ___ | | | |___ | | | |___|___| 1 2 We could then choose the median to be the value m such that the line x=m bisects the area inside the histogram, which (if I'm not mistaken) is m=1.25. >> I'm proably not understanding the bar graph; it doesn't seem very clear to me even in an equal-width font. But I'd expect the bar graph for data 1, 1, 2 would be a bar of width 1 and height 2 around x = 1, adjacent to a bar of width 1 and height 1 around x = 2. Which would mean that the vertical line bisecting the area would lie at the solution to 2 - 2(1-x) = 2(1-x) + 1, giving x = 3/4. --Dan

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Re: [math-fun] Sex, Lies and Government Conducted Surveys
by dasimov＠earthlink.net 28 Sep '05

28 Sep '05

David Gale writes: << Mike Greenwald asks, how a median can be 3.8. Good question! . . . Anyhow, someone please help me out on those fractional medians >> In my fairly extensive interaction with statisticians + stat textbooks, all definitions I've seen of a median of a finite sample of N real-valued data are identical: If N odd, median = middle value; if N even, median = average of the 2 middle values. BUT -- there is a more sophisticated definition that may exist: If N is even, and the two middle values are distinct but at least one of them occurs with multiplicity > 1, one might prefer to use the "asimov median", defined as follows: Assume the data sorted by size: x(1) <= x(2) <= . . . <= x(N). Let J be the smallest interval of indices that is symmetrical about the real number N/2, such that J contains a) all repetitions of x((N+1)/2) (if N is odd), or b) all repetitions of both data x(N/2) and x(N/2 + 1) (if N is even). Now define the "weighted median" as the average of all values whose indices lie in J. Thus if N = 200 and the sorted data looks like this: sort #: . . . 94 95 96 97 98 99 100 101 102 103 104 105 106 . . . datum: . . . 2 3 3 3 4 4 4 4 4 4 4 4 5 . . . then one may wish to say the asimov median is the average of data whose indices k satisfy 100.5 - 5 <= k <= 100.5 + 5, i.e., 96 <= k <= 105, so we get a.m. = (3+3+4+4+4+4+4+4+4+4)/5 = 3.8 . --Dan

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CORRECTION: Re: [math-fun] Sex, Lies and Government Conducted Surveys
by dasimov＠earthlink.net 28 Sep '05

28 Sep '05

CORRECTIONS: I stupidly rushed the "asimov median" post into print with a number of errors. Here is the corrected definition -- sorry. Correct definition: ------------------------------------------------------------------------------------------------ Assume the data sorted by size: x(1) <= x(2) <= . . . <= x(N). Let J be the smallest [interval of indices] that is symmetrical about the real number (N+1)/2, such that J contains a) all repetitions of x((N+1)/2) if N is odd, or b) all repetitions of both data x(N/2) and x(N/2 + 1) if N is even. Now define the "asimov median" as the average of all values whose indices lie in J. Thus if N = 200 and the sorted data looks like this: sort #: . . . 94 95 96 97 98 99 100 101 102 103 104 105 106 . . . datum: . . . 2 3 3 3 4 4 4 4 4 4 4 4 5 . . . then one may wish to say the asimov median is the average of data whose indices k satisfy 100.5 - 5 <= k <= 100.5 + 5, i.e., 96 <= k <= 105, so we get a.m. = (3+3+4+4+4+4+4+4+4+4)/10 = 3.8 . --Dan

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