Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

570 Nonparametric and Robust Statistics

Remark 10.1.1(Natural Nonparametric Estimators). Functionals induce non-
parametric estimators naturally. LetX 1 ,X 2 ,...,Xndenote a random sample from
some distribution with cdfF(x)andletT(F) be a functional. Letx 1 ,x 2 ,...,xnbe
a realization of this sample. Recall that the empirical distribution function of the
sample is given by

F̂n(x)=n−^1 [#{xi≤x}], −∞<x<∞. (10.1.1)

Hence,Fnis a discrete cdf that puts mass (probability) 1/nat eachxi. Because
F̂n(x)isacdf,T(F̂n) is well defined. Furthermore,T(F̂n) depends only on the
sample; hence, it is a statistic. We callT(F̂n)theinduced estimatorofT(F).
For example, ifT(F) is the mean of the distribution, then it is easy to see that
T(F̂n)=x; see Exercise 10.1.3.
For another example, consider the median. Note thatFˆnis a discrete cdf; hence,
we use the general definition of a median of a distribution that is given in Definition
1.7.2 of Chapter 1. Let ˆθdenote the usual sample median which is defined in
expression (4.4.4); that is,θˆ=x((n+1)/2)ifnis odd whileˆθ=[x(n/2)+x((n/2)+1)]/ 2
ifnis even. To show thatθˆsatisfies Definition 1.7.2, note that:

Ifnis even, thenFˆn(θˆ)=1/2.

Ifnis odd then

n−^1 #{xi<θˆ}=^12 −n^1 ≤ 1 /2andFn(θˆ)≥ 1 / 2.

Thus in either case, by Definition 1.7.2,θˆis a median ofFˆn. Note that whenn
is even any point in the interval (X(n/2),X((n/2)+1)) satisfies the definition of a
median.

We begin with the definition of a location functional.

Definition 10.1.1.LetXbe a continuous random variable with cdfFX(x)and pdf
fX(x). We say thatT(FX)is alocation functionalif it satisfies

IfY=X+a,thenT(FY)=T(FX)+a, for alla∈R, (10.1.2)

IfY=aX;thenT(FY)=aT(FX), for alla =0. (10.1.3)

For example, suppose T is the mean functional; i.e.,T(FX)=E(X). Let
Y =X+a;thenE(Y)=E(X+a)=E(X)+a. Secondly, ifY =aX,then
E(Y)=aE(X). Hence the mean is a location functional. The next example shows
that the median is a location functional.

Example 10.1.1. LetF(x)bethecdfofXand letT(FX)=FX−^1 (1/2) be the
median functional ofX. Note that another way to state this isFX(T(FX)) = 1/2.
LetY =X+a. It then follows that the cdf ofY isFY(y)=FX(y−a). The
following identity shows thatT(FY)=T(FX)+a:

FY(T(FX)+a)=FX(T(FX)+a−a)=FX(T(FX)) = 1/ 2.