Wikipedia:Reference desk/Archives/Mathematics/2010 November 22

= November 22 =

1/3 Anglo-Saxon
At the pub the other day the allegation was made that someone was 1/3rd Anglo-Saxon. This led to a discussion of whether this was possible or not. Is there a proof (possibly using mathematical induction) that it is possible or that it is impossible? Joaq99 (talk) 00:30, 22 November 2010 (UTC)
 * This depends, of course, on how one is defining fractions of ethnicity. If one takes the fairly standard approach of going back to ancestors of clear and unambiguous ethnicity, and assigning fractions based on that, then the only fractions possible are the dyadic rationals. 1/3 is not dyadic, so it's not possible. Algebraist 00:33, 22 November 2010 (UTC)
 * Assuming that ancestry can be traced as far back as one likes and that the Anglo-Saxons existed as long ago as one looks (neither of which assumptions is true, of course), one can have a 1/3 Anglo-Saxon ancestry in a limit. Otherwise, the best "1/3" can be is an approximation.&mdash;msh210 &#x2120; 19:57, 22 November 2010 (UTC)
 * Added a title heading for the question. &#x2013; b_jonas 11:39, 22 November 2010 (UTC)
 * Gilgamesh is 1/3 part mortal and 2/3 part divine. &#x2013; b_jonas 11:44, 22 November 2010 (UTC)

But what if one has only three grandparents? Suppose you're the product of an incestuous mating of half-siblings. Michael Hardy (talk) 23:53, 22 November 2010 (UTC)
 * Then one of those grandparents is your grandparent twice-over, so what I take to be the standard approach would make that grandparent responsible for half of your ethnicity and your other grandparents responsible for a quarter each. Algebraist 00:00, 23 November 2010 (UTC)
 * OK..... Michael Hardy (talk) 00:04, 23 November 2010 (UTC)

Someone at the pub spoke 1/3rd Anglo-Saxon. Perhaps he or she spoke English? Bo Jacoby (talk) 00:15, 23 November 2010 (UTC).

IID Random variables with uniformly distributed difference
Hello everyone,

What would be the best way to show there do not exist independent identically distributed random variables X, Y such that X-Y ~ U([-1,1])? I am comfortable with what I believe would be all the relevant probability/measure theoretical concepts, and I expect there's probably a very clever concise way of doing this problem which takes almost no time at all - is there such a solution? I can't think of an easy way to get at it, despite the fact it certainly looks like there should be one.

Thankyou,

Otherlobby17 (talk) 06:52, 22 November 2010 (UTC)


 * Compute explicitly the distribution of X-Y. That depends on four parameters: the upper and lower bounds of X and Y.  &#x2013; b_jonas 11:38, 22 November 2010 (UTC)
 * ... and the actual distribution of X and Y, which the OP didn't say should be uniform. -- Meni Rosenfeld (talk) 12:38, 22 November 2010 (UTC)
 * Sorry, you're right. I answered the wrong question.  &#x2013; b_jonas 20:23, 23 November 2010 (UTC)


 * Consider the cumulants of X and of &minus;Y. Compute the cumulants of X&minus;Y. Compare to the cumulants of the continuous uniform distribution. Bo Jacoby (talk) 12:20, 22 November 2010 (UTC).
 * You don't know what the distributions of X and Y are so how do you get their upper and lower bounds or the cumulants? It's easy to see that if X-Y ~ U([-1,1]) then the distributions of X and Y must be contained in and interval of length 1. If the distributions are continuous and densities bounded then it's not hard to do by playing with inequalities. What happens if, for example X and Y are distributed like the x-coordinate of a random point on a circle of radius 1/2 (or sphere, etc.)?--RDBury (talk) 18:21, 22 November 2010 (UTC)
 * Bo's point was to show that for any choice of cumulants of X (and Y), you won't get the uniform cumulants for their difference. If X and Y were required to be uniform, b_jonas' method would be easy - for any choice of bounds, the difference won't be uniform.
 * Taking the distribution of the x-coordinate on a circle, which is $$f(t)\propto\frac{1}{\sqrt{1-t^2}}$$, doesn't work. -- Meni Rosenfeld (talk) 19:30, 22 November 2010 (UTC)
 * That was extremely helpful, thankyou all very much! :) Otherlobby17 (talk) 01:16, 25 November 2010 (UTC)

log
today I participated in a math competition and one of the problems was express log 2 and log 5 in terms of k if k=log 5 * log 2 (log here denoting the common log). I put the obvious answer k/log 5 and k/log 2 but I got marked wrong! What is the correct answer? 24.92.78.167 (talk) 21:38, 22 November 2010 (UTC)
 * I suspect the problem was so phrased that you're not allowed to use log 2 and log 5 explicitly in the answer. Michael Hardy (talk) 23:56, 22 November 2010 (UTC)

OK, so log 5 + log 2 = log 10 = 1, and hence
 * $$ k = (\log 5)(\log 2) = u(1 - u), \, $$

where
 * $$ u = \log 2\text{ or }u = \log 5. \, $$

Then
 * $$ u - u^2 = k \, $$
 * $$ u^2 - u + k = 0. \, $$

This is a quadratic equation in u. So
 * $$ u = \frac{1 \pm \sqrt{1 - 4k}}{2}. $$

The smaller of the two solutions is log 2 and the larger is log 5. Michael Hardy (talk) 00:02, 23 November 2010 (UTC)
 * Should be $$ u = \frac{1 \pm \sqrt{1 - 4k}}{2}. $$ -- Meni Rosenfeld (talk) 09:12, 23 November 2010 (UTC)
 * I've changed that now---thanks. Michael Hardy (talk) 21:31, 23 November 2010 (UTC)
 * Putting Michael Hardy's method another way, if we set $$\alpha = \log(2)$$ and $$\beta = \log(5)$$, then $$\alpha + \beta = \log(2) + \log(5) = \log(2*5) = \log(10) = 1$$, and $$\alpha\beta = k$$ (by definition), so $$\alpha$$ and $$\beta$$ are the roots of $$x^2 - x + k = 0$$, and the rest follows from the quadratic formula. (Is this "theory of equations" stuff still taught these days?) AndrewWTaylor (talk) 10:03, 23 November 2010 (UTC)