🚀 go-pugleaf

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.

In article <1992Sep5.224529.12961@news2.cis.umn.edu> I replied:

> 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, ... [and]
> 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.

The promised proof, if anyone is still interested.

Suppose c is a median and c < d (the case c > d follows by symmetry).

  E{|X - d| - |X - c|} >= (d - c)[P(X <= c) - P(X > c)} >= 0

the first inequality following because |x - d| - |x - c| is equal to d - c
for x <= c and greater than - (d - c) elsewhere, and the second inequality
following by the definition of "median".

There is a similar proof that the mean minimizes expected squared deviation
-- that shouldn't use any calculus either.

-- 
Charles Geyer
School of Statistics
University of Minnesota
charlie@umnstat.stat.umn.edu