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A Model of Split-ticket Voting with Uncertainty
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| | Unformatted Document Text:
The voter whose ideal point is x
i
=
12
is indifferent between θ
DD
and θ
RR
and between
θ
DR
and θ
RD
. And since she is closer to θ
DR
than θ
DD
,
ψ
1
(
1
2
) = 2π
dd
π
dr
(u
i
(θ
DD
) − u
i
(θ
DR
) + u
i
(θ
DR
) − u
i
(θ
DD
))
+2(π
2
dd
+ π
2
dr
)(u
i
(θ
DD
) − u
i
(θ
DR
))
= 2(π
2
dd
+ π
2
dr
)(u
i
(θ
DD
) − u
i
(θ
DR
)) < 0.
Then, since ψ
1
(x
i
) is continuous and strictly decreasing and the signs change, by the
Intermediate Value Theorem we can conclude that there is z in (
θ
DD
+θ
DR
2
,
12
) such
that ψ
1
(z ) = 0.
Let ψ
2
(x
i
) and ψ
3
(x
i
) be
ψ
2
(x
i
) = Eu
i
(dr) − Eu
i
(rd) = 2π
dd
π
rd
(u
i
(θ
DD
) − u
i
(θ
RD
)) + 2π
dd
π
dr
(u
i
(θ
DR
) − u
i
(θ
DD
))
+2(π
dd
π
rr
+ π
dr
π
rd
)(u
i
(θ
DR
) − u
i
(θ
RR
)) + 2π
dr
π
rr
(u
i
(θ
DD
) − u
i
(θ
RR
))
+2π
rd
π
rr
(u
i
(θ
RR
) − u
i
(θ
RD
))
(9)
ψ
3
(x
i
) = Eu
i
(rd) − Eu
i
(rr) = 2π
dd
π
dr
(u
i
(θ
DD
) − u
i
(θ
DR
))
+2(π
rd
π
rr
+ π
dr
π
rd
+ π
dd
π
rr
)(u
i
(θ
RD
) − u
i
(θ
RR
)).
(10)
With the condition π
dr
= π
rd
and π
dd
= π
rr
, we can rewrite (9) as
ψ
2
(x
i
) = 2(π
dd
π
rd
+ π
dd
π
rr
+ π
dr
π
rd
+ π
rd
π
rr
)(u
i
(θ
DR
) − u
i
(θ
RD
)).
By taking a derivative, we can easily see ψ
2
(x
i
) is a strictly decreasing function.
By construction, θ
DR
and θ
RD
are the same distance from
12
. So, when x
i
=
12
,
u
i
(θ
DR
) = u
i
(θ
RD
), which means that z such that ψ
2
(x
i
) = 0 is
12
.
Rewrite equation (10) as
ψ
3
(x
i
) = 2π
dd
π
dr
(u(|θ
DD
− x
i
|) − u(|θ
DR
− x
i
|))
+2(π
rd
π
rr
+ π
dd
π
rr
+ π
dr
π
rd
)(u(|θ
RD
− x
i
|) − u(|θ
RR
− x
i
|)),
and the same logic used for ψ
1
(x
i
) can be applied to show that ψ
3
(x
i
) is continuous
and strictly decreasing in x
i
∈ [0, 1]. Also, since it has a positive value at
12
, according
to Lemma 1, by the Intermediate Value Theorem again, we know that there is z
in
the interval (
12
,
θ
RD
+θ
RR
2
) such that ψ
3
(x
i
) = 0.
Next, we claim that z
= 1 − z . By definition of z ,
ψ
1
(z ) = 2(π
dd
π
dr
+ π
dd
π
rr
+ π
dr
π
rd
)(u(|θ
DD
− z |)
−u(|θ
DR
− z |)) + 2π
rd
π
rr
(u(|θ
RD
− z |) − u(|θ
RR
− z |)) = 0. (11)
19
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The voter whose ideal point is x
i
=
1
2
is indifferent between θ
DD
and θ
RR
and between
θ
DR
and θ
RD
. And since she is closer to θ
DR
than θ
DD
,
ψ
1
(
1
2
) = 2π
dd
π
dr
(u
i
(θ
DD
) − u
i
(θ
DR
) + u
i
(θ
DR
) − u
i
(θ
DD
))
+2(π
2
dd
+ π
2
dr
)(u
i
(θ
DD
) − u
i
(θ
DR
))
= 2(π
2
dd
+ π
2
dr
)(u
i
(θ
DD
) − u
i
(θ
DR
)) < 0.
Then, since ψ
1
(x
i
) is continuous and strictly decreasing and the signs change, by the
Intermediate Value Theorem we can conclude that there is z in (
θ
DD
+θ
DR
2
,
1
2
) such
that ψ
1
(z ) = 0.
Let ψ
2
(x
i
) and ψ
3
(x
i
) be
ψ
2
(x
i
) = Eu
i
(dr) − Eu
i
(rd) = 2π
dd
π
rd
(u
i
(θ
DD
) − u
i
(θ
RD
)) + 2π
dd
π
dr
(u
i
(θ
DR
) − u
i
(θ
DD
))
+2(π
dd
π
rr
+ π
dr
π
rd
)(u
i
(θ
DR
) − u
i
(θ
RR
)) + 2π
dr
π
rr
(u
i
(θ
DD
) − u
i
(θ
RR
))
+2π
rd
π
rr
(u
i
(θ
RR
) − u
i
(θ
RD
))
(9)
ψ
3
(x
i
) = Eu
i
(rd) − Eu
i
(rr) = 2π
dd
π
dr
(u
i
(θ
DD
) − u
i
(θ
DR
))
+2(π
rd
π
rr
+ π
dr
π
rd
+ π
dd
π
rr
)(u
i
(θ
RD
) − u
i
(θ
RR
)).
(10)
With the condition π
dr
= π
rd
and π
dd
= π
rr
, we can rewrite (9) as
ψ
2
(x
i
) = 2(π
dd
π
rd
+ π
dd
π
rr
+ π
dr
π
rd
+ π
rd
π
rr
)(u
i
(θ
DR
) − u
i
(θ
RD
)).
By taking a derivative, we can easily see ψ
2
(x
i
) is a strictly decreasing function.
By construction, θ
DR
and θ
RD
are the same distance from
1
2
. So, when x
i
=
1
2
,
u
i
(θ
DR
) = u
i
(θ
RD
), which means that z such that ψ
2
(x
i
) = 0 is
1
2
.
Rewrite equation (10) as
ψ
3
(x
i
) = 2π
dd
π
dr
(u(|θ
DD
− x
i
|) − u(|θ
DR
− x
i
|))
+2(π
rd
π
rr
+ π
dd
π
rr
+ π
dr
π
rd
)(u(|θ
RD
− x
i
|) − u(|θ
RR
− x
i
|)),
and the same logic used for ψ
1
(x
i
) can be applied to show that ψ
3
(x
i
) is continuous
and strictly decreasing in x
i
∈ [0, 1]. Also, since it has a positive value at
1
2
, according
to Lemma 1, by the Intermediate Value Theorem again, we know that there is z
in
the interval (
1
2
,
θ
RD
+θ
RR
2
) such that ψ
3
(x
i
) = 0.
Next, we claim that z
= 1 − z . By definition of z ,
ψ
1
(z ) = 2(π
dd
π
dr
+ π
dd
π
rr
+ π
dr
π
rd
)(u(|θ
DD
− z |)
−u(|θ
DR
− z |)) + 2π
rd
π
rr
(u(|θ
RD
− z |) − u(|θ
RR
− z |)) = 0. (11)
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