Corrected Proof of Lemma 2 in
“Implementation with Contingent Contracts”
Rahul Deb and Debasis Mishra1
October 7, 2015
There is a mistake in the proof of Lemma 2 (the statement of which is correct) in Deb and
Mishra (2014) - the partition created in Lemma 2 is not f -ordered partitioned as claimed.
We are very grateful to Kiho Yoon who brought this error to our attention.
Below is the corrected proof of Lemma 2.
Lemma 2: Suppose the type space is finite and f is an acyclic scf. Then, the type space
can be f -ordered partitioned.
Corrected Proof: We define a new relation Bf as follows: for every vi , vi0 , we say vi Bf vi0
if there is a finite sequence {vi1 , . . . , vik } such that vi ≡ vi1 f vi2 f . . . f vik ≡ vi0 , with at
least one of the above relations strict (f ). Now, we do the proof in several steps.
Step 1. First, we show that Bf is acyclic. Consider a sequence vi1 , . . . , vik such that vi1 Bf
. . . Bf vik . Assume for contradiction vik Bf vi1 . So, we get vi1 Bf . . . Bf vik Bf vik+1 ≡ vi1 .
Between any consecutive types vij and vij+1 in this sequence, we have vij f . . . f vij+1 , with
strict relation holding for at least one. Hence, we get a finite sequence vi1 f . . . f vik+1 ≡ vi1
with strict relation holding for at least one. Acyclicity of f implies that vi1 f vi1 , contradicting the reflexivity of f .
Step 2. Next, choose any nonempty subset Vi0 ⊆ Vi , and let M (Vi0 ) ⊆ Vi0 be the set of
maximal elements of Vi0 corresponding to the relation Bf . Formally
M (Vi0 ) := {vi | vi ∈ Vi0 and there is no vi0 ∈ Vi0 such that vi0 Bf vi }.
Since Bf is acyclic, M (Vi0 ) is non-empty.
Step 2(a). vi0 f vi for every vi , vi0 ∈ M (Vi0 ). To see this, if vi0 f vi , then vi0 Bf vi ,
contradicting vi ∈ M (Vi0 ).
Step 2(b). vi0 f vi for every vi ∈ M (Vi0 ) and every vi0 ∈ (Vi0 \ M (Vi0 )). Assume for contradiction vi0 f vi . Since vi0 ∈
/ M (Vi0 ), there exists a vi00 such that vi00 Bf vi0 . Hence, there is a
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Rahul Deb: University of Toronto, [email protected]; Debasis Mishra: Indian Statistical Institute, Delhi, [email protected]
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finite sequence, vi00 f . . . f vi0 f vi , with strict relation holding at least once. As a result,
we have vi00 Bf vi , contradicting the fact that vi ∈ M (Vi0 ).
Step 3. Now, we recursively define a partition. We set Vi1 = M (Vi ). Having defined,
(Vi1 , . . . , Vik−1 ), we define Rk := Vi \ (Vi1 ∪ . . . ∪ Vik−1 ). If Rk = ∅, we stop, else, we set
Vik := M (Rk ). Suppose (Vi1 , . . . , ViK ) is the partition created. We show that (Vi1 , . . . , ViK )
satisfies Property P1 and P2. To do that, pick vi , vi0 ∈ Vij for some j. Since Vij = M (Rj ),
by Step 2(A), we have vi0 f vi . So, Property P1 is satisfied.
Similarly, pick vi ∈ Vij and vi0 ∈ (Vij+1 ∪ . . . ∪ ViK ). Since Vij ≡ M (Rj ) and (Vij+1 ∪ . . . ∪
ViK ) ≡ Rj \ Vij , by Step 2(B), we get vi0 f vi . So, Property P2 is satisfied.
References
Deb, R. and D. Mishra (2014): “Implementation with Contingent Contracts,” Econometrica, 82, 2371–2393.
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