EDITOR’S PROOF
244 G. Serra
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too long to develop here (the detailed calculations are reported in previous versions
of this paper). With the appropriate algebra we find that
P(vR=V|primary)=
⎧
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪⎩
1
2 ifπRI∈(^0 ,π]
πRIq^2 +q−^12 q^2 −πRIq+^12 πRI ifπRI∈(π,^12 )
1
2 q+
1
4 ifπRI=
1
2
πRIq−πRIq^2 +^12 q^2 +^12 πRI ifπRI∈(^12 ,π)
πRI ifπRI∈[π, 1 )
I can now calculate the value of interest,S. The values above are used to
calculateS≡P(vR=V|primary)−P(vR=V|leadership), remembering that
P(vR=V|leadership)=πRI. With some algebra and noting the continuity ofS
atπRI=π,πRI=^12 andπRI=π, we find that
S=
⎧
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎩
1
2 −πRI forπRI∈(^0 ,π]
πRIq^2 −πRIq−^12 q^2 −^12 πRI+q forπRI∈[π,^12 ]
−πRIq^2 +πRIq+^12 q^2 −^12 πRI forπRI∈[^12 ,π]
0forπRI∈[π, 1 )
which are the values we were looking for.
Now we need to analyze the sign ofS.IfπRI∈( 0 ,π]we have thatS=^12 −
πRI> 0 ⇔πRI<^12 , but that is satisfied becauseπRI≤πand I have already noted
thatπ<^12 .IfπRI∈[π,^12 ]we have thatS=πRIq^2 −πRIq−^12 q^2 −^12 πRI+q>
0 ⇔πRI<^2 q−q
2
1 + 2 q− 2 q^2 (noting that 1+^2 q−^2 q
(^2) >0) which is satisfied because
1
2 <
2 q−q^2
1 + 2 q− 2 q^2 .IfπRI∈[
1
2 ,π)we have thatS=−πRIq
(^2) +πRIq+ 1
2 q
(^2) − 1
2 πRI>
0 ⇔πRI< q
2
1 − 2 q+ 2 q^2 which is satisfied becauseπ=
q^2
1 − 2 q+ 2 q^2. And finally ifπRI∈
[π, 1 )we haveS=0. So we have indeedS>0forπRI∈( 0 ,π]∪[π,^12 ]∪[^12 ,π)
andS=0forπRI∈[π, 1 ), as the lemma claims.
A.5 Proof of Lemma 3
I calculate the differential ofSwith respect toπRIand check its sign. IfπRI∈( 0 ,π),
∂S
∂πRI=−1 which is strictly negative. IfπRI∈(π,
1
2 ),
∂S
∂πRI=q
(^2) −q− 1
2 which
is strictly negative forq∈(^12 , 1 ).IfπRI∈(^12 ,π),∂π∂SRI=−q^2 + 2 q−1 which is
strictly negative forq∈(^12 , 1 ).SoSis decreasing withπRIin all those intervals.
Sis non-differentiable atπRI=πandπRI=^12 , but is continuous at both points,
and is therefore decreasing just like their neighboring points. HenceSdecreases
withπRIwhenπRI∈( 0 ,π)∪{π}∪(π,^12 )∪{^12 }∪(^12 ,π).