# Unequal probability inverse sampling Section 7. Discussion

The selection problem can therefore be resolved for all cases, with or without replacement and with equal or unequal probabilities. The proposed solution based on the elimination method respects the inclusion probabilities exactly, which is not true for Ohlsson’s sequential sampling. The implementation is especially simple, since the program provides an ordered sequence of occupations to propose until the objective has been met.

The estimation issue is slightly more difficult. For the unequal probability sampling without replacement, we must make do with a heuristic solution. As well, it can be seen that, in the second stage, there tends to be lower inclusion probabilities in enterprises that have many occupations. This should lead us to select with greater probabilities the enterprises that may have a larger number of occupations, to avoid selecting occupations with probabilities that are too unequal.

## Acknowledgements

The author wishes
to thank Pierre Lavallée for submitting this interesting problem and providing
thoughtful comments on an earlier version of this article. The author also
thanks Audrey-Anne Vallée for her meticulous proofreading, and a referee and
writer of *Survey Methodology* for their pertinent remarks, which made it
possible to improve this article.

## Appendix

#

# Load sampling package, which
contains the function inclusionprobabilities().

#

library(sampling)

#

# The function returns a vector with the
sequence numbers of the eliminations.

# The last (resp. first) unit
eliminated is the first (resp. last)

# component of the vector.

# The function therefore provides the
numbers of the units to be presented

# successively for the inverse
selection.

# The argument x is the vector of
values of the auxiliary variable used to calculate

# the inclusion probabilities.

#

elimination<-function(x)

pikb=x/sum(x)

M = length(pikb)

n = sum(pikb)

sb = rep(1, M)

b = rep(1, M)

res=rep(0, M)

for (i in 1:(M)) {

a = inclusionprobabilities(pikb, M - i)

v = 1 - a/b

b = a

p = v * sb

p = cumsum(p)

u = runif(1)

for (j in 1:length(p)) if (u < p[j])

break

sb[j] = 0

res[i]=j

}

res[M:1]

}

#

# 500,000 simulations with a size in
a list of size M=20.

# By taking the first m components of
vector v, we obtain a sample

# of size m.

#

M=20

x=runif(M)

Pik=array(0,c(M,M))

#

# Calculate the inclusion
probabilities for all sample sizes from 1 to 20.

#

for(i in 1:M)
Pik[i,]=inclusionprobabilities(x, i)

rowSums(Pik)

SIM=50000

SS=array(0,c(M,M))

for(i in 1:SIM)

{

S=array(0,c(M,M))

v=elimination(x)

for(i in 1:M) S[i,v[1:i]]=1

SS=SS+S

}

SS=SS/SIM

#

# Compare actual and empirical
inclusion probabilities.

#

Pik

SS

SS-Pik

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