Adaptive procedures for non-parameter tests

x<-c(51.9,56.9,45.2,52.3,59.5,41.4,46.4,45.1,53.9,42.9,41.5,55.2,32.9,54.0,45.0)
y<-c(59.2,49.1,54.4,47.0,55.9,34.9,62.2,41.6,59.3,32.7,72.1,43.8,56.8,76.7,60.3)

drive4=function(x,y){
   n1=length(x)
   n2=length(y)
   n=n1+n2
   cb=(1:n)/(n+1)

   const=(n1*n2)/(n*(n-1))#p550

   p1=phi1(cb)
   var1=const*sum(p1^2)

   p2=phi2(cb)
   var2=const*sum(p2^2)

   p3=phi3(cb)
   var3=const*sum(p3^2)

   p4=phi3(cb)
   var4=const*sum(p4^2)

   vars=c(var1,var2,var3,var4)
   allxy=c(x,y)

   rall=rank(allxy)/(n+1)
   ind=c(rep(0,n1),rep(1,n2))

   s1=sum(ind*phi1(rall))
   s2=sum(ind*phi2(rall))
   s3=sum(ind*phi3(rall))
   s4=sum(ind*phi4(rall))

   tests=c(s1,s2,s3,s4)
   ztests=tests/sqrt(vars)

list(vars=vars,tests=tests,ztests=ztests)
}

phi1=function(u){
phi1=2*u-1
phi1
}

phi2=function(u){
phi2=sign(2*u-1)
phi2
}

phi3=function(u){
n=length(u)
phi3=rep(0,n)
for(i in 1:n){
   if(u[i]<=0.25){phi3[i]=4*u[i]-1}
   if(u[i]>0.75){phi3[i]=4*u[i]-3}
}
phi3
}

phi4=function(u){
n=length(u)
phi4=rep(0.5,n)
for(i in 1:n){
if(u[i]<=0.5){phi4[i]=4*u[i]-3/2}
}
phi4
}
drive4(x,y)

v<-c(x,y)
sv<-sort(v)
q<-quantile(sv, c(0.05, 0.25,0.5,0.75,0.95))

U005<-c()
M05<-c()
L005<-c()
U05<-c()
L05<-c()

for(i in 1:30){
if (sv[i]<q[1]){L005[i]=sv[i]}
if (sv[i]>q[2]&sv[i]<q[4]){M05[i]=sv[i]}
if (sv[i]>q[4]){U005[i]=sv[i]}
if (sv[i]>q[3]){U05[i]=sv[i]}
if (sv[i]<q[3]){L05[i]=sv[i]}
}

L005
M05
U005
U05
L05


Um005<-mean(U005,na.rm=TRUE)
Um005

Mm05<-mean(M05,na.rm=TRUE)
Mm05

Um05<-mean(U05,na.rm=TRUE)
Um05

Lm05<-mean(L05,na.rm=TRUE)
Lm05

Lm005<-mean(L005,na.rm=TRUE)
Lm005

Q1<-(Um005-Mm05)/(Mm05-Lm005)
Q1

Q2=(Um005-Lm005)/(Um05-Lm05)
Q2

derivative of even and odd function

derivative of an even function is an odd function. 
derivative of an odd function is an even function.

Variance of Signum Function

Functional and composite function

Variance for the statistics of the general rank score test

median of a series of numbers minus its median is always zero

1. suppose we only have 3 (odd) number of values x1,x2, x3 ordered already

so the median is x2.

The  new sequence will be x1-x2, 0, x3-x2.

We see the new median is 0.

2. suppose we have 4 (even) number of values y1, y2, y3, y4, ordered already.

the median is (y2+y3)/2.

The new sequence is : y1-(y2+y3)/2, y2-(y2+y3)/2, y3-(y2+y3)/2, y4-(y2+y3)/2.

The new median is: (y2-(y2+y3)/2+ y3-(y2+y3)/2) which is also 0.

This observation is  useful when we estimate the shift estimator of a Sign Scores test.