Wikipedia:Reference desk/Archives/Mathematics/2010 February 18

= February 18 =

Standard Deviation and Semi-Interquartile Range
Hello. How can I prove that a set of data cannot have a standard deviation smaller than its semi-interquartile range? Thanks in advance. --Mayfare (talk) 10:47, 18 February 2010 (UTC)


 * What you want is the Law of total variance. To get the minimum standard deviation you want as small a variance as possible on both sides of the medians giving the interquartile range. Which means you need to bunch each side to a single point so you have values only at -x 0 and +y, this quickly works out to needing values only at -x 0 and +x for the minimum standard deviation compared to the semi interquartile range. Dmcq (talk) 11:34, 18 February 2010 (UTC)


 * I followed the link to see if it confirmed my interpretation of the assertion that the question implies, but am still unsure. Is it true that, for any data set, the standard deviation is equal to or greater than half of the interquartile range? --NorwegianBluetalk 19:32, 19 February 2010 (UTC)


 * Actually it is false. Try it out with just over a quarter of the values at -1, the same amount at +1 and just under half the total at 0. The interquartile range is 2. The semi-interquartile range is 1. The standard deviation is sqrt((-1)2*1/4+0*1/2+12*1/4) = 1/sqrt(2) which is less than 1. Dmcq (talk)

Alternation Operator
Hello. I'm currently reading a book on manifolds and I've gotten confused by a really simple exercise, or I'm missing something. The book defines an action for permutations of k symbols on k-linear maps via pF(v1,...vk) = F(vp(1),...vp(k)). It, then, defines Alt f = the sum of all terms sgn(p)(pf), where p ranges over the symmetric group. The exercise is to consider Alt f in the case that f is 3-linear. I arrived at the following: 123, 312, and 231 are even permutations and 213, 321, 132 are odd; thus, we should have Alt f(v1, v2, v3) = f(v1 + v3 + v2 - v2 - v3 - v1, v2 + v1 + v3 -...etc = f(0, 0, 0) = 0. Clearly, this isn't acurate since the wedge of a 3-linear f and a constant c should be cf, which isn't 0. What did I screw up? Thanks for any help:) 67.165.56.56 (talk) 16:10, 18 February 2010 (UTC)

It was indeed a stupid error, for whatever reason I was assuming f(a,b,c) + f(x,y,z) = f(a+x,b+y,c+z)...obviously not the case...:) 67.165.56.56 (talk) 16:32, 18 February 2010 (UTC)