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Phil Carmody ha scritto:
>
> > 2. what is the median of {0, 1} and why.
> >
> Useless. It depends on the domain you're working in.
the median of the set of values { x1, x2} does not depend "on the "domain" you're working in." Anyway what do mean by "domain"? We are working with finite sets.
It can depend on the comparer you use, in general. But not for n= 2. For n=2, it is always coincident with the mean.
If we want to discuss about something and you take the freedom to insult someone without reason, you should at least show that youk know what we are talking about, otherwise you may end up not making a good show of yourself.
Anyway, let me recall some basics:
For a discrete numerical set {X1, ... , Xn} ,
MEDIAN {X1, ... , Xn}
where by x(i), I mean the i-th order statistic, ie the i-th value in the ordered value (ordered according to any custom comparer).
So the median does have an optimality property similar to the mean. But in order to make this property a "characterizing" one, the assumption is made that, for even cardinalities, the average of the 2 (n>1) central terms is to be taken. In fact, as Phil noted - in case n is even - all points between the 2 central terms (end point included) are point of minimum for the sum of absolute deviations, as clearly they are contained in all the intervals of type x(i) - x(n-i+1), i=0,...,n\2.
[ For categorical (say, "nonnumerical") data, the definition needs slight adjustments. ]
MEDIAN {0, 1} = .5 MEDIAN {0, 2, 1} = 1 MEDIAN {0, 1, 3, 2} = 1.5 MEDIAN {0, 4, 3, 1, 2} = 2
in the above cases MEDIAN() and AVG() are coincident due to the "symmetry" of the finite support of the discrete uniform distribution.
For n=1,2 or any case of symmetric support, AVG and MEDIAN are always coincident,
i.e., MEDIAN {X1, X2} = AVG {X1, X2} for all X1, X2.
(For symmetric supports AVG() and MEDIAN() carry the same information.)
The above concept of MEDIAN gets easily generalized to the concept of QUANTILE [eg, quartile, percentile,...] Received on Sun Sep 24 2006 - 05:03:19 CDT