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Article #287Re: Simple Proof that c=median minimizes E[ |X-c| ] needed.
From: Charles Geyer
Date: Sat, 05 Sep 1992 22:45
Date: Sat, 05 Sep 1992 22:45
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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. Because this might be a homework problem, I won't post a proof right now, but there is a one-line proof that requires no calculus and no probability beyond the fact that expection is a positive linear operator, i. e. E(aX + Y) = a E(X) + E(Y) and 0 <= X implies 0 <= E(X) It works for a general probability distribution using the definition that c is a median if 1/2 <= P(c <= X) and 1/2 <= P(X <= c) I'll post the proof in a week or so. -- Charles Geyer School of Statistics University of Minnesota charlie@umnstat.stat.umn.edu
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Newsgroups: sci.math.stat Path: arc1-ca.pugleaf.net!sparky!uunet!munnari.oz.au!spool.mu.edu!umn.edu!charlie From: charlie@umnstat.stat.umn.edu (Charles Geyer) Subject: Re: Simple Proof that c=median minimizes E[ |X-c| ] needed. Message-ID: <1992Sep5.224529.12961@news2.cis.umn.edu> Sender: news@news2.cis.umn.edu (Usenet News Administration) Nntp-Posting-Host: isles.stat.umn.edu Organization: School of Statistics, University of Minnesota References: <3SEP199213440863@utkvx2.utk.edu> Date: Sat, 5 Sep 1992 22:45:29 GMT Lines: 30