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A Model of Campaign Spending in Repeated Elections
Unformatted Document Text:  Thus, for any incumbent I, µ(I) = a 1 . After observing the incumbent’s first election spending and his vote share, the challenger forms a belief about the type of incumbent that may have generated a 1 and v 1 . Let ρ(I|a 1 , v 1 ) be that probability. The challenger’s entrance strategy is a map: σ : R 2+ → {enter, not enter} Thus, for the challenger (whose quality, Q, is common knowledge), σ(a 1 , v 1 ) is the challenger’s (non-)entrance, having observed the incumbent’s spending in the firstelection and the vote share, given a low quality challenger entered in the first election. After the challenger has decided whether to enter, the incumbent decides how much to spend in the last election. The incumbent’s second-election spending strategyis a map: γ : I × {H,L} → R + where γ(I, Q) = a 2 is the amount spent in the second (and last) election given the action of the challenger in the second election. Thus, a set of strategies is (σ ∗ , µ ∗ , γ ∗ ) . Since the challenger has incomplete information about the preferences of the in- cumbent, an appropriate equilibrium concept for this game is sequential equilibrium(Kreps and Wilson 1982). Loosely, incumbents and challengers maximize their ex-pected payoffs at every point in time, where the expectation is with respect to thechallenger’s beliefs, updated by Bayes’ rule when possible. 3 Equilibrium In this model, all incumbents separate in their second period (optimal) spending, asthere no further actions to influence. However, incumbents may pool or separate intheir first period spending. 3.1 Pooling Equilibrium I characterize the pooling equilibrium, where the weak incumbent imitates the strongincumbent. Proposition 1 (Pooling) There exists an equilibrium (σ ∗ , µ ∗ , γ ∗ ) in which: σ ∗ (a 1 , v 1 ) = {not enter} ∀v ≥ ˆv σ ∗ (a 1 , v 1 ) = {enter} ∀v < ˆv µ ∗ (I) = a S,L1 ∀I γ ∗ (I, Q) = a I,Q2 where a I,Q2 is the amount raised and spent by incumbent I against challenger Q. I construct strategies by solving the game backwards. 9

Authors: Goodliffe, Jay.
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background image
Thus, for any incumbent I, µ(I) = a
1
.
After observing the incumbent’s first election spending and his vote share, the
challenger forms a belief about the type of incumbent that may have generated a
1
and v
1
. Let ρ(I|a
1
, v
1
)
be that probability. The challenger’s entrance strategy is a
map:
σ : R
2
+
→ {enter, not enter}
Thus, for the challenger (whose quality, Q, is common knowledge), σ(a
1
, v
1
)
is the
challenger’s (non-)entrance, having observed the incumbent’s spending in the first
election and the vote share, given a low quality challenger entered in the first election.
After the challenger has decided whether to enter, the incumbent decides how
much to spend in the last election. The incumbent’s second-election spending strategy
is a map:
γ : I
× {H,L} → R
+
where γ(I, Q) = a
2
is the amount spent in the second (and last) election given the
action of the challenger in the second election. Thus, a set of strategies is (σ
, µ
, γ
)
.
Since the challenger has incomplete information about the preferences of the in-
cumbent, an appropriate equilibrium concept for this game is sequential equilibrium
(Kreps and Wilson 1982). Loosely, incumbents and challengers maximize their ex-
pected payoffs at every point in time, where the expectation is with respect to the
challenger’s beliefs, updated by Bayes’ rule when possible.
3
Equilibrium
In this model, all incumbents separate in their second period (optimal) spending, as
there no further actions to influence. However, incumbents may pool or separate in
their first period spending.
3.1
Pooling Equilibrium
I characterize the pooling equilibrium, where the weak incumbent imitates the strong
incumbent.
Proposition 1
(Pooling) There exists an equilibrium (σ
, µ
, γ
) in which:
σ
(a
1
, v
1
) =
{not enter} ∀v ≥ ˆv
σ
(a
1
, v
1
) =
{enter}
∀v < ˆv
µ
(I) = a
S,L
1
∀I
γ
(I, Q) = a
I,Q
2
where a
I,Q
2
is the amount raised and spent by incumbent I against challenger Q.
I construct strategies by solving the game backwards.
9


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