[math-fun] another kind of median (trying again)
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". Jim
Jim, what is the Propp median if there are m zeros and m fives (and zero everything else)? Dan, if you're going to bring in averages at all then why not go all the way and use THE average? But maybe the CDC was using some sort of hybrid like the one you suggest. D At 09:14 PM 9/28/2005, you wrote:
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".
Jim
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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.
Jim, what is the Propp median if there are m zeros and m fives (and zero everything else)?
In that case it isn't well defined. Jim
So, if y(x) is the histogram, the median is the m such that integral(x<m) y(x) = integral(x>m) y(x) while the mean is the m such that integral(x<m) |x-m|*y(x) = integral(x>m) |x-m|*y(x) This suggests a whole family of averages (using various functions of (x-m) for the weighting), though what use they might have escapes me. --ms David Gale wrote:
Jim, what is the Propp median if there are m zeros and m fives (and zero everything else)? Dan, if you're going to bring in averages at all then why not go all the way and use THE average? But maybe the CDC was using some sort of hybrid like the one you suggest.
D
At 09:14 PM 9/28/2005, you wrote:
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".
Jim
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There is a whole class of "centers" which differ depending upon which "moment" (or combination of moments) are being considered. Of course, a symmetrical distribution will have the same centers, but highly asymmetrical distributions will have different centers (assuming that such a center even exists). The major problem with using classical means, medians, std's, etc., for distributions with heavy tails (e.g., Zipf's Law) is that they don't do a very good job of characterizing the distribution. You need an entirely new language to describe them. At 08:49 AM 9/29/2005, Mike Speciner wrote:
So, if y(x) is the histogram, the median is the m such that
integral(x<m) y(x) = integral(x>m) y(x)
while the mean is the m such that
integral(x<m) |x-m|*y(x) = integral(x>m) |x-m|*y(x)
This suggests a whole family of averages (using various functions of (x-m) for the weighting), though what use they might have escapes me.
--ms
David Gale wrote:
Jim, what is the Propp median if there are m zeros and m fives (and zero everything else)? Dan, if you're going to bring in averages at all then why not go all the way and use THE average? But maybe the CDC was using some sort of hybrid like the one you suggest. D At 09:14 PM 9/28/2005, you wrote:
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".
Jim
Instead of inventing new medians I suggest we take a closer look at the old one. As a prototypical model, consider a set S of n sticks or segments. These make up a PRE-ordered set with a relation >=,"greater than or equal". Namely given sticks s and s', we have s=s' if the sticks are congruent and s<s' if s is congruent to a subset of s'. This is straight out of Euclid's elements which never mentions the notion of length of a segment, and neither will I. What is a median for this set? NOTATION. if s is a stick then s+ ={s'in S; s'>=s} s- = {s' in S: s'<=s} If X is a set #X is the cardinality of X. DEFINITION: A stick s is a MEDIAN of the set S if #(s+)>=n/2 and #(s-)>=n/2. EXAMPLE. If S consists of two sticks s<s' then both s and s' are medians. EXERCISE 1. Prove the "Fundamental Theorem of Median Theory".Every finite preordered set has at least one median. (not completely trivial). A slightly fancier way of doing this is to take the ordered set consisting of the equivalence classes {s} with respect to =. Then #{s} is a given distribution etc. Another example, going back to the thing that started all this, is an ordered set of men M where m'>=m if m' has had at least as many sexual partners as m. Now here is a "real world" example. The pre-ordered set is a set of Gifts ordered "chronologically" as follows; G_1=a partridge in a pear tree < G_2= a turtle dove <G_3= _<. . . .<G_12= a drummer drumming Depending on how one interprets the song there are two possibilities. I. S consists of 66 gifts; 1 partridge, 2 turtle doves,. . ,12 drummers. II. S consists of 364 gifts; 12 partrigdes, 22 turtle doves, . . ,12 drummers. EXERCISE 2; Find the median gift for both interpretations I and II. The link below is provided for people (like me) who don't remember which gifts go with which days. http://www.night.net/christmas/12-Days.html David
participants (4)
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David Gale -
Henry Baker -
James Propp -
Mike Speciner