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Sep 3, 1992, 2:44:00 PM9/3/92

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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.

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.

Sep 3, 1992, 6:10:35 PM9/3/92

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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 :-).

Sep 3, 1992, 10:01:11 PM9/3/92

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ne...@massey.ac.nz (USENET News System) writes:

>In article <3SEP1992...@utkvx2.utk.edu>, men...@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.

>>

[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: thom...@atlas.socsci.umn.edu

Department of Economics phone: (612) 625-0119

University of Minnesota fax: (612) 624-0209

Sep 3, 1992, 9:55:05 PM9/3/92

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Here are some ideas off the top of my head, that should get you going in

the right direction. You will need some decent calculus skills, but

that's about it.

the right direction. You will need some decent calculus skills, but

that's about it.

Case 1. X is a real-valued continuous random variable with density f(x)

such that xf(x) -> 0 as x -> +infinity and as x -> -infinity. Write

E[|X-c|] as the sum of two integrals, one going from -infinity to c and

the other going from c to +infinity, where neither integral involves

abs. val's anymore. Differentiate this sum with respect to c and

explore...

Case 2. x_1, x_2, ... x_n are fixed real numbers (i.e. you have a

sample of size n). Then c=the sample median minimizes the sum of

absolute deviations |x_1-c| + |x_2-c| + ... + |x_n-c|. You can do the

cases n=2, 3, and 4 "by hand" by graphing the sum of abs. deviations as

a function of c---it is piecewise linear with bends at the data points

x_1, x_2, ... What do you have to do to give a rigorous proof for n

data points now?

Case 3. X is a discrete random variable taking on the values x_1, x_2,

.... with probabilities p_1, p_2, .... The ideas you explored above

should be useful in proving that the median minimizes E[|X-c|] for this

case.

-BJ

Sep 4, 1992, 12:02:05 PM9/4/92

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The original poster might also be interested in the fact that

c = median also solves the problem

c = median also solves the problem

min E( |x-c| - |x| )

c

The advantage of this characterization over the original one is that

since |x-c| - |x| is a bounded function of x, the expectation

E( |x-c| - |x| ) always exists, while E( |x-c| ) might be infinite.

Sep 5, 1992, 6:45:29 PM9/5/92

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In article <3SEP1992...@utkvx2.utk.edu> men...@utkvx2.utk.edu

(Menees, Bill) writes:

(Menees, Bill) writes:

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

cha...@umnstat.stat.umn.edu

Sep 6, 1992, 5:32:35 PM9/6/92

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common location statistics. I'll state it for samples

to avoid technicalities.

Suppose we have a sample x_1, x_2, . . .,x_n.

Given a positive real number p define m to minimise

Sum |x_i - m|**p

then we get

p --> 0 mode

p = 1 median

p = 2 mean

p --> infinity midrange

Neat, eh?

--

Murray A. Jorgensen [ m...@waikato.ac.nz ] University of Waikato

Department of Mathematics and Statistics Hamilton, New Zealand

__________________________________________________________________

'Tis the song of the Jubjub! the proof is complete,

if only I've stated it thrice.'

Sep 12, 1992, 8:30:11 PM9/12/92

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> 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 <1992Sep5.2...@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.

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