8 Conclusion
Sander Scholtus
Previous
In this article, we proposed a new formulation of the
error localisation problem which can take the distinction between hard and soft
edits into account. In addition, we showed that a modified version of the
branch-and-bound algorithm of De Waal and Quere (2003) can be used to solve
this new error localisation problem. It was suggested that this new algorithm
can be used to increase the quality of automatic editing. This suggestion was
confirmed by the empirical results reported in Section 7, although it should be
stressed that these results were obtained with data containing synthetic
errors. An application is currently being investigated of the new error
localisation algorithm to realistic data.
It remains an open problem how the costs of soft edit
failures may best be modelled, i.e.,
how the term in (3.1) should be defined. The different
results with approaches C and D in Section 7 demonstrate that the quality
of automatic error localisation may be improved by a suitable choice of failure
weights. It will be interesting to see to what extent the quality of automatic
editing may be improved further by experimenting with different combinations of
failure weights confidence weights and the balancing parameter in (3.1).
Other forms of than (3.2) could also be considered, including
forms that depend on the sizes of the soft edit failures. As mentioned in
Section 3, it is intuitively appealing to take the amounts by which soft edits
are failed into account in the error localisation problem, so that larger soft
edit failures yield higher values of One interesting choice for could be the Mahalanobis distance of soft edit
failures, as suggested by Hedlin (2003) in a different context. It should be
noted that the algorithm from Section 6 may be used to solve the error
localisation problem for all choices of that can be expressed as (reasonably
well-behaved) functions of One simply uses the appropriate expression for
as the target function in problem (5.1). On
the other hand, if depends explicitly on the sizes of the soft
edit failures, then we have to resort to a more complex approach. In
particular, the information provided by Theorems 1 and 2 is no longer
sufficient, because we now need to know not only which soft edits will be
failed after imputation but also the amounts by which they will be failed. An
approach for solving the error localisation problem in this more complicated
situation can be found in Scholtus (2011).
In summary, it remains to be seen how the theoretical results
outlined in this article should be applied to obtain the best results in
practice. Nevertheless, given that subject-matter experts use the conceptual
difference between hard and soft edits during manual editing, it seems evident
that the new error localisation algorithm has the potential to increase the
quality of automatic editing.
Acknowledgements
The views expressed in this article are those of the
author and do not necessarily reflect the policies of Statistics Netherlands.
The author would like to thank: Jeroen Pannekoek and Ton de Waal for
their many helpful suggestions; Edwin de Jonge and Mark van der Loo for
programming the branch-and-bound algorithm in R; Sevinç Göksen for collecting
the evaluation results presented in Section 7; and finally the Associate Editor
and two anonymous referees for their constructive comments.
Appendix A Proofs
A.1 Proof of Theorem 1
In order to prove Theorem 1, it is convenient to prove
first an auxiliary lemma. Suppose that is obtained from by eliminating We define, for each edit the index set of the edit(s) in from which it has been derived. That is to
say, we define if is obtained by copying the edit and we define if is obtained by eliminating a variable from the
pair of edits
Lemma 1.
Consider the situation of Theorem 1 for and suppose that has been eliminated to obtain from Let be a representing set of the index sets belonging to all Then there exists a value for that, together with the original values of the
variables that are involved in satisfies all edits in except those in
Proof (of
Lemma 1). By construction, contains all indices of failed edits from which do not involve Hence, the only way for the lemma to be false
would be if there existed two edits that involve say and with and so that it is not possible to find a value for
that satisfies both edits simultaneously. In
this case, an implied edit in is generated by eliminating from and Moreover, by the fundamental property given at
the end of Section 2.3, this implied edit must be failed by the original values
of the other variables, i.e., the
implied edit must be an element of But this would contradict the assumption that is a representing set of for all Hence, it is impossible to find such a pair of
edits, and the lemma follows.
The proof of Theorem 1 now proceeds by induction on the
number of treated variables For the statement is trivial. For the theorem follows as a special case of Lemma
1; note that We suppose therefore that the statement has
been proved for all and we consider the case with
If is obtained from by fixing a variable to its original value,
and is a representing set of the sets for the failed edits from then by construction is also a representing set of the sets for the failed edits from Thus, in this case, the statement for follows immediately from the induction
hypothesis.
Hence, we are left with the case that is obtained from by eliminating a variable, say We define, for each the index set of the edit(s) from from which is derived, as above. Next, we use to construct a set by applying the following procedure to each
•
If is obtained by copying (so and then we add to
•
If is obtained by eliminating from and (so and then we add to if contains an element of and we add to otherwise.
It is easy to see that this procedure produces a
representing set of the index sets for all
According to Lemma 1, there exists a value for which, together with the original values of
the variables that have not been treated, satisfies
the edits in except those in That is to say, can be partitioned similarly to as where contains the edits with indices in Moreover, it is not difficult to see that the
above procedure implies that is a representing set of the index sets for all Hence, the induction hypothesis establishes
that, given the original values of the variables that have not been eliminated and given the chosen value for there exist values for the other eliminated
variables that satisfy all the original edits except those in This shows that the statement holds for and completes the proof of Theorem 1.
A.2 Proof of Theorem 2
To prove Theorem 2, we start again with an auxiliary
lemma. Analogous to the numerical case, when is obtained from by eliminating we define the index set of edits in from which the edit is derived. To be precise, we define if is obtained by copying the edit and we define if is obtained by eliminating a variable from the
set of edits
Lemma 2.
Consider the situation of Theorem 2 for and suppose that has been eliminated to obtain from Let be a representing set of the index sets belonging to all Then there exists a value for that, together with the original values of the
variables that are involved in satisfies all edits in except those in
Proof (of
Lemma 2). By construction, contains all indices of failed edits from which do not involve Hence, the only way for the lemma to be false
would be if there existed edits that involve say with so that it is not possible to find a value for
that satisfies these edits simultaneously,
given the values of the other variables. Clearly, this could only happen if since otherwise any value for outside would work. We may assume without loss of
generality that is a minimal set having this property.
Furthermore, it must hold in this case that for all variables involved in the original value of is contained in all sets In other words, must satisfy properties (2.8) and (2.9). This
means that would generate an implied edit in which, by the fundamental property given at
the end of Section 2.3, must be failed by the original values of the remaining
variables. However, this would contradict the assumption that is a representing set of for all This completes the proof of Lemma 2.
The proof of Theorem 2 is now completely analogous to
that of Theorem 1, with Lemma 2 taking the role of Lemma 1.
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