🚀 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.
> 
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 :-).