Post on 30-Dec-2015
description
1
What happens to the location estimator if we minimize with a power
other that 2?
Robert J. Blodgett
Statistic Seminar - March 13, 2008
Outline
1. Affect of different exponents2. Definition3. End points of the path4. Repeated exponents5. Bounds6. Directions of the path7. Outliers
When data points come from a symmetric unimodal distribution, the clear location estimator is the maximum. Different exponents are compared for three such distributions. For each exponent, 10,000 simulations had 20 points each.
The curves show number of simulations within ¼, ½, and 1 of the maximum, 0.
1. Affect of different exponents
Uniform on (-5, 5)
t ot 1
0
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7000
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9000
10000
power
1. 0 2. 0 3. 0 5. 0 7. 5 10. 0
Normal – N(0, 1)t ot 1
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power
1. 0 2. 0 3. 0 5. 0 7. 5 10. 0
Double Exponentialt ot 1
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power
1. 0 2. 0 3. 0 5. 0 7. 5 10. 0
The lp-norm estimator minimizes the following sum as m(p) changes with x
k … x
1 .
2. Definition
K
j
p
j pmx1
)(
for 1< p < ∞
0)()(1
1
pmxsignpmx j
K
j
p
j
Setting the derivative with respect to m(p) equal to zero gives
.
Expression (1)
Why can we take a derivative?
Since p > 1, the limit equals 0.
10
11
0
lim
0
lim
10
11
0
lim
0
lim
1
1
pif
pifx
xx
x
x
pif
pifx
xx
x
x
pp
pp
As m(p) increases, each term in expression (1) decreases. Hence, for each exponent, there is a unique minimizing m(p). The minimizing point varies continuously with the exponent.
The next step is to locate the end points.
The limit median is the limit of the minimizing points as the exponent approaches 1.
Properties: (1) For an odd number of data point or when the middle two are equal, the middle one is the limit median. (2) Otherwise, the limit median is between the two middle data points.
3. End points of the path
D. Jackson (1921), “Notes on the median of a set of numbers,” Bul. Amer. Mathemat. Soc. 27, pp. 160-164.
Pair the jth and (K-j+1)st points. Let xL
denote the lower point and xU
the upper
point. The limit median is at the solution to
. pairsall
Upairsall
L mxxm
oThe following bound shows how to assign the minimum when the exponent approaches infinity. The minimum and the midrange are within
11
11
21
1
1
1
p
p
K
KR
where R denotes the range and K > 1.
When the exponent approaches 1, the term in parenthesis approaches 1. Consequently, this bound would include all the data points then. When exponent approaches infinity, this term approaches zero. Consequently, the minimizing point approaches the midrange as the exponent approaches infinity.
As an example, a path is constructed for the data set with the five points 0, 0, 2, 2.1, and 4.
expon
1
2
3
4
5
6
7
8
9
10
val
0 1 2 3 4
For how many exponents can have the same minimum?
4. Repeated Exponents
Let aj = sign (x
j – m) and b
j = |x
j-m|
where the sum is over all points the same distance from m.
The following theorem bounds the number of repartitions.
If (1) each aj is a non-zero, real number and (2) bN >…> b1 > 0, then the number of changes in sign of the aj-values is greater than or equal to the number of x-values where ajbj
x equals zero.
The following proof is similar to one in Laguerre (1883).
Proof. (By induction on the number of changes in sign.) If there are no changes in sign, then no x-value makes the sum zero.
For the induction step assume the aj-values have C changes in sign and the result holds for any exponential sum with fewer changes in sign. Let a1 to aw all have the same sign and let aw+1 have the opposite sign.
x
w
jN
jj
x
w
x
j
N
jj b
babba
11
w
j
x
w
jN
jj
x
w
jN
jj b
b
b
ba
b
ba
dx
dlog
11
Since bN >…> b1 > 0, when j > w, log(bj/bw) > 0. Thus, for j > w, the jth coefficient of this derivative has the same sign as aj. When j < w, log(bj/bw) < 0. Thus, for j < w, the jth coefficient of this derivative has the opposite sign from aj. Hence, the sign no longer changes at j = w. Therefore, the coefficients of this derivative have at most C – 1 changes in sign. By the induction hypothesis, this derivative has at most C – 1 x-values where it equals 0. By Rolles theorem, the original sum can have at most C x-values where it equals 0. ■
A similar proof of Descartes’ rule of signs works for polynomials. Exponential sums and polynomial like sums where the exponents maybe any real numbers are closely related.
.11
1111
jjj uK
jj
K
j
ux
j
xuK
jj
x
j
K
jj yabababa
Let uj = (log bj)/(log b1) and y = (b1)x.
Repeated exponents also has been disguised as a number theory problem. One example from number theory is the following equalities for j = 0, 1, 2, 3, 4, or 5.
1j + 12j + 21j + 43j + 52j + 63j
= 3j + 7j + 28j + 36j + 57j + 61j
Notice that there are 6 changes in sign.
5. Bounds
Order the data points and let
cj = (xj + xK-j+1)/2.
Let c- = minimum cj and
c+ = maximum cj.
For all p in (1, ∞), m(p) is in [c-, c+].
Proof. When m > cj, xj – m is negative and |xj – m| |xK-j+1 – m|. Thus, the contribution of this pair to expression (1) is negative. When m is greater than all cj-values, the contribution from each pair is negative (including the median if K is odd.) Thus, expression (1) is not zero. Similarly if m less than all cj. ■
It also follows that1) If c- < c+, then m(p) is in (c-, c+) for all p in (1, ∞).2) The limit median is in [c-, c+].3) The Winsorized mean and the usual median are both in [c-, c+].
The following graphs show the ratio of the actual length of an interval covered divided by c+ - c-. The curves show the number out of 10,000 simulations with points from N(0, 1) below .5, .7, and .9.
Even size groupspr t 5
0
1000
2000
3000
4000
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6000
7000
8000
9000
10000
gr p_ s i ze
4 6 8 10 12 14 16 18 20
Odd size groupspr t 5
0
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7000
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gr p_ s i ze
5 7 9 11 13 15 17 19
The extra curve is for the number below 1.
In the example of data set 0, 0, 2, 2.1, and 4, the pairs of data points are as follows. First, 0 and 4 with average 2; next, 0 and 2.1 with average 1.05; and finally 2. Consequently, (c-, c+) = (1.05, 2). Any minimum in this interval has at most 2 exponents.
The next two results contain the idea that once the high or low terms dominate the expression, they will continue to dominate it.
1. If (1) each aj is a non-zero, real number, (2) bK >…> b1 > 0, and
0
K
sj
q
jjba
K
j
x
jjba1
(3) for s = 1, …, K,
then > 0 for x q.
2. If (1) each aj is a non-zero, real number, (2) bK >…> b1 > 0, and
01
s
j
q
jjba(3) for s = 1, …, K,
then
K
j
x
jjba1
> 0 for 0 < x q.
6. Directions of the path
For p in (1, ∞), direct calculations give
Taking the sign gives the direction the estimate moves. From the directions, iteration gives the points where the path turns.
.)1(
)(log
1
2
1
1
K
j
p
j
jj
K
j
p
j
mxp
mxsignmxmx
dp
dm
The following result helps find the direction at p = 1 or infinity.
Let S denote the entire data set and M its minimizing point at p. Let S* denote the data set with xj omitted and M* its minimizing point at p.
If S has more than one data point and p is in (1, ∞), then (1) M > xj if and only if M* > xj, (2) M = xj if and only if M* = xj, and (3) M < xj if and only if M* < xj.
For the exponent at infinity the following result gives the direction of approach.
If (1) the cj-values are not all identical and (2) cF denotes the average of the pair with the largest range whose average does not equal c1, then for sufficiently large p both m(p) and cF are on the same side of c1.
A similar result holds when near p = 1 when K is odd or the middle two data points are equal.
K
jj
jj
K
j
mx
mxsignmx
dp
dm
p
1
1
1
2
2
log
1
lim
When K is even and the middle two data points are different, the following expression can indicate the direction at the limit median.
expon
1
2
3
4
5
6
7
8
9
10
val l
1. 050 1. 617 2. 000
The example data set has a turning point at an exponent of 2.12. There is at most two exponents for each value of the estimator.
7. Outliers
This section contains two graphs of simulation results. Six data points were taken from a N(0, 1). After calculating their location estimator, an outlier was added. The location estimator was recalculated. At each exponent, the number of the 10,000 simulations with differences below 1/8, 1/4 and 1/2 are shown.
Outlier at 2.
t ot 1
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power
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Outlier at 3
The extra curve is for differences below 1.