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A Model of Adversarial Committees
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Now we can take N and
> 0 such that
N !
(N − αN +1)!( αN −1)!
γ
αN −1
N
(1−γ
N
)
N − αN +1
(N +1)·γN −1
i= αN
N !
(N −i)!i!
γ
i
N
(1−γ
N
)
N −i
< / αN ,
that is,
N !· αN
(N − αN +1)!( αN −1)!
γ
αN −1
N
(1−γ
N
)
N − αN +1
(N +1)·γN −1
i= αN
N !
(N −i)!i!
γ
i
N
(1−γ
N
)
N −i
< .
Since α ∈ [1/2, 1),
N − αN +1
αN
< 1. Hence the numerator in the above expression can now
be rewritten as
N !· αN
(N − αN +1)!( αN −1)!
γ
αN −1
N
(1 − γ
N
)
N − αN +1
≥
N !
(N − αN )!( αN −1)!
γ
αN −1
N
(1 − γ
N
)
N − αN +1
.
Using the right-hand side of the above inequality we get
>
N !· αN
(N − αN +1)!( αN −1)!
γ
αN −1
N
(1−γ
N
)
N − αN +1
(N +1)·γN −1
i= αN
N !
(N −i)!i!
γ
i
N
(1−γ
N
)
N −i
≥
1
(N +1)·γ
N
−1
i= αN
(N − αN )!( αN −1)!
(N −i)!i!
γ
N
1−γ
N
i− αN +1
.
A representative term of the denominator in the right-hand side of the above inequalitycan be written as
(N − αN )!( αN −1)!
(N − (N +1)·γ
N
+1)!( (N +1)·γ
N
−1)!
γ
N
1−γ
N
(N +1)·γ
N
− αN
=
(N − αN )···(N − (N +1)·γ
N
+2)
( (N +1)·γ
N
−1)··· αN
γ
N
1−γ
N
(N +1)·γ
N
− αN
=
N − αN
αN
· · · · ·
N − (N +1)·γ
N
+2
(N +1)·γ
N
−2
·
1
(N +1)·γ
N
−1
γ
N
1−γ
N
(N +1)·γ
N
− αN
.
Note that since lim
N →∞
γ
N
> α ≥
12
,
γ
N
1−γ
N
> 1. Now we claim that lim
N →∞
x
N
N
= ∞ if
x > 1. To see this, let x = 1 + h where h > 0. Then we have
x
N
N
=
(1+h)
N
N
=
1
N
1
N
+
N !
(N −1)!
h +
N !
(N −2)!2!
h
2
+ · · · +
N !
(N −1)!
h
N −1
+ h
N
=
1
N
+
(N −1)!(N −1)!
h +
(N −1)!
(N −2)!2!
h
2
+ · · · +
(N −1)!(N −1)!
h
N −1
+
h
N
N
=
1
N
+ h +
N −1
2
h
2
+ · · · + h
N −1
+
h
N
N
>
N −1
2
h
2
.
Thus, lim
N →∞
x
N
N
= ∞ as claimed. This completes the proof.
2
References
D. Austen-Smith and J. Banks, Information Aggregation, Rationality and the Con-
dorcet Jury Theorem, Amer. Polit. Sci. Rev. 90 (1995) 34–45.
29
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Now we can take N and
> 0 such that
N !
(N − αN +1)!( αN −1)!
γ
αN −1
N
(1−γ
N
)
N − αN +1
(N +1)·γN −1
i= αN
N !
(N −i)!i!
γ
i
N
(1−γ
N
)
N −i
< / αN ,
that is,
N !· αN
(N − αN +1)!( αN −1)!
γ
αN −1
N
(1−γ
N
)
N − αN +1
(N +1)·γN −1
i= αN
N !
(N −i)!i!
γ
i
N
(1−γ
N
)
N −i
< .
Since α ∈ [1/2, 1),
N − αN +1
αN
< 1. Hence the numerator in the above expression can now
be rewritten as
N !· αN
(N − αN +1)!( αN −1)!
γ
αN −1
N
(1 − γ
N
)
N − αN +1
≥
N !
(N − αN )!( αN −1)!
γ
αN −1
N
(1 − γ
N
)
N − αN +1
.
Using the right-hand side of the above inequality we get
>
N !· αN
(N − αN +1)!( αN −1)!
γ
αN −1
N
(1−γ
N
)
N − αN +1
(N +1)·γN −1
i= αN
N !
(N −i)!i!
γ
i
N
(1−γ
N
)
N −i
≥
1
(N +1)·γ
N
−1
i= αN
(N − αN )!( αN −1)!
(N −i)!i!
γ
N
1−γ
N
i− αN +1
.
A representative term of the denominator in the right-hand side of the above inequality
can be written as
(N − αN )!( αN −1)!
(N − (N +1)·γ
N
+1)!( (N +1)·γ
N
−1)!
γ
N
1−γ
N
(N +1)·γ
N
− αN
=
(N − αN )···(N − (N +1)·γ
N
+2)
( (N +1)·γ
N
−1)··· αN
γ
N
1−γ
N
(N +1)·γ
N
− αN
=
N − αN
αN
· · · · ·
N − (N +1)·γ
N
+2
(N +1)·γ
N
−2
·
1
(N +1)·γ
N
−1
γ
N
1−γ
N
(N +1)·γ
N
− αN
.
Note that since lim
N →∞
γ
N
> α ≥
1
2
,
γ
N
1−γ
N
> 1. Now we claim that lim
N →∞
x
N
N
= ∞ if
x > 1. To see this, let x = 1 + h where h > 0. Then we have
x
N
N
=
(1+h)
N
N
=
1
N
1
N
+
N !
(N −1)!
h +
N !
(N −2)!2!
h
2
+ · · · +
N !
(N −1)!
h
N −1
+ h
N
=
1
N
+
(N −1)!
(N −1)!
h +
(N −1)!
(N −2)!2!
h
2
+ · · · +
(N −1)!
(N −1)!
h
N −1
+
h
N
N
=
1
N
+ h +
N −1
2
h
2
+ · · · + h
N −1
+
h
N
N
>
N −1
2
h
2
.
Thus, lim
N →∞
x
N
N
= ∞ as claimed. This completes the proof.
2
References
D. Austen-Smith and J. Banks, Information Aggregation, Rationality and the Con-
dorcet Jury Theorem, Amer. Polit. Sci. Rev. 90 (1995) 34–45.
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