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- Newsgroups: sci.math.stat
- Subject: Re: Simple Proof that c=median minimizes E[ |X-c| ] needed.
- Message-ID: <1992Sep3.221035.17952@massey.ac.nz>
- From: news@massey.ac.nz (USENET News System)
- Date: Thu, 3 Sep 92 22:10:35 GMT
- References: <3SEP199213440863@utkvx2.utk.edu>
- Organization: Massey University
- Lines: 28
-
- In article <3SEP199213440863@utkvx2.utk.edu>, menees@utkvx2.utk.edu (Menees, Bill) writes:
- >
- > I'm a senior math major taking my first p&s course, and this problem
- > has come up and it intrigues me. My prof. has a proof for it, but he said it
- > was way over my head. Does anyone know of a proof suitable for a senior
- > undergrad? Thanks in advance.
- >
- If you are a senior math major then you should have no difficulty
- in following the proof. I suspect the real problem is that it's way
- over your prof.'s head. :-)
-
- Well, I must admit that I'm going to cheat. :-) Let's prove the result
- for a discrete distribution only.
-
- When c != any possible x
- d/dc (sum|x-c|) = number of x's greater than c - number of x's less than c
- otherwise this does not exist.
- This derivative is decreasing if c < median and increasing if c > median
- and as sum|x-c| is continuous and piecewise linear, the minimum either
- occurs where there is a discontinuity in the derivative, or where it is
- zero over an interval. The latter case corresponds to a median. In the
- former case we choose the discontinuity such that on either side
- |number of x's greater than c - number of x's less than c| is smallest.
-
- In the continuous case the derivative exists nowhere but a limiting
- argument could be used. (This might be over my head :-).
-
-
-