NetNews Usenet Archive 1992 #19/NN_1992_19.iso/spool/sci/math/stat/1801

Text File | 1992-09-03 | 3.3 KB | 79 lines

Newsgroups: sci.math.stat
Path: sparky!uunet!zaphod.mps.ohio-state.edu!sol.ctr.columbia.edu!The-Star.honeywell.com!umn.edu!thompson
From: [e-mail] (T. Scott Thompson)
Subject: Re: Simple Proof that c=median minimizes E[ |X-c| ] needed.
Message-ID: <[e-mail]>
Sender: [e-mail] (Usenet News Administration)
Nntp-Posting-Host: daphne.socsci.umn.edu
Reply-To: [e-mail]
Organization: Economics Department, University of Minnesota
References: <[e-mail]> <[e-mail]>
Date: Fri, 4 Sep 1992 02:01:11 GMT
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[e-mail] (USENET News System) writes:

>In article <[e-mail]>, [e-mail] (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.
>>

[explanation for the discrete case deleted]

>In the continuous case the derivative exists nowhere but a limiting
>argument could be used. (This might be over my head :-).

I think that the continuous version is more straightforward since then
the objective function is _everywhere_ differentiable.  To get a feel
for the continuous case consider that any solution to the problem

             min  E(|x-c|)
              c

must satisfy a first-order condition provided E(|x-c|) is
differentiable in c.  When x is continuous this function is _always_
differentiable since the smooth distribution for x smooths out the
kink point in the absolute value function.  Differentiability only
fails if Pr( x = c ) is strictly positive.

The trick is to compute the derivative of E(|x-c|).  We can commute
the derivative operator and the expectation integral (showing that
this is OK is the technically difficult part of the proof that your
professor may have had in mind) in order to derive

       d/dc E(|x-c|) = E( d/dc |x-c| )

                     = E [ 1{ x < c } - 1{ x > c } ]

                     = E 1{ x < c } - E 1{ x > c}

                     = Pr( x < c ) - Pr( x > c )

where 1{ ... } is the indicator function taking the value +1 if "..."
is true and zero otherwise.  To see why this is true, simply note that
the derivative of |x-c| with respect to c equals +1 if x < c and -1 if
x > c.  We can ignore the case x = c, since this has zero probability,
and since any reasonable extension of the definition of
differentiability to handle this case would yield a value for the
derivative somewhere between -1 and +1 at the point x = c.
Technically, to verify the first equality you must check that the
conditions of the "Lebesgue dominated convergence theorem" are
satisfied.  You can look this up in any advanced probability theory
text.

Now notice that the derivative is zero if and only if

                Pr( x < c ) = Pr( x > c )

and this equation is solved by choosing c = median(x).  The solution
is unique if the probability density of x is positive on some
neighborhood of the median.

--
T. Scott Thompson              email:  [e-mail]
Department of Economics        phone:  [phone]
University of Minnesota        fax:    [phone]