Secure Implementation Experiments: Do Strategy-proof Mechanisms ...

for the profile v= ( v1, v2 )
f
f
f
is denoted by f( v) = ( y ( v), t ( v))
, where y ( v ) is the level of the
f f f
public good and t ( v) = ( t 1
( v), t 2
( v))
is the transfer vector.
Definition 3. The SCF f is strategy-proof if
f f f f
1 1 2 1 1 2 1 1 2 1 1 2
v ( y ( v , ~ v )) + t ( v , ~ v ) ≥ v ( y ( ~ v , ~ v )) + t ( ~ v , ~ v ) for all ~ v ∈V
and ~ v ∈ V ; and
f f f f
2 1 2 2 1 2 2 1 2 2 1 2
1 1
2 2
v ( y ( v , v )) + t ( v , v ) ≥ v ( y ( v , v )) + t ( v , v ) for all ~ v ∈V
and ~ v ∈ V .
1 1
2 2
By the Revelation Principle (Gibbard, 1973), strategy-proofness is necessary for dominant
strategy implementation, and therefore also for secure implementation. However, the following
additional condition, called the rectangular property, is necessary for secure implementation.
Definition 4. The SCF f satisfies the rectangular property if for all vv , ~ ∈ V, if
f f f f
1 1 2 1 1 2 1 1 2 1 1 2
v ( y ( v , ~ v )) + t ( v , ~ v ) = v ( y ( ~ v , ~ v )) + t ( ~ v , ~ v ) and
f f f f
2 1 2 1 1 2 2 1 2 2 1 2
v ( y ( ~ v , v )) + t ( ~ v , v ) = v ( y ( ~ v , ~ v )) + t ( ~ v , ~ v ) ,
then f ( v1, v2 ) = f(~ v , ~ v )
1 2 .
Saijo et al. (2003) show that the rectangular property is necessary and sufficient for sure
implementation: 7
7 To see why the rectangular property is necessary for secure implementation intuitively, suppose that the direct
revelation mechanism g = f securely implements the SCF f. Let n = 2 and ( v1, v2 ) be the true preference profile.
Suppose u1( f( v1, ~ v2)) = u1( f( ~ v1, ~ v2)) , i.e.,
f f f f
(*) v1( y ( v1, ~ v2)) + t1 ( v1, ~ v2) = v1( y ( ~ v1, ~ v2)) + t1 ( ~ v1, ~ v2)
.
In other words, agent 1 is indifferent between reporting the true preference v 1 and reporting another preference ~ v 1
when agent 2’s report is ~ v 2 . Since reporting v 1 is a dominant strategy by strategy-proofness, it follows from (*) that
f f f f
v1( y ( ~ v1, ~ v2)) t1 ( ~ v1, ~ v2) v1( y ( v1, ~ v2)) t1 ( v1, ~ f
f
+ = + v2)
≥ v1( y ( v1′ , ~ v2)) + t1 ( v1′
, ~ v2) for all v1′ ∈V1,
that is, reporting ~ v 1 is one of agent 1’s best responses when agent 2 reports ~ v 2 .
Next suppose that u2( f( ~ v1, v2)) = u2( f( ~ v1, ~ v2))
, i.e.,
f f f f
(**) v2( y ( ~ v1, v2)) + t1 ( ~ v1, v2) = v2( y ( ~ v1, ~ v2)) + t2 ( ~ v1, ~ v2)
.
f f f f
By using an argument similar to the above, it is easy to see that v2( y ( ~ v1, ~ v2)) + t2 ( ~ v1, ~ v2) = v2( y ( ~ v1, v2)) + t2 ( ~ v1, v2)
f
f
≥ v2( y ( ~ v1, v2′ )) + t1 ( ~ v1, v2′
) for all v2′ ∈V2 ,
14