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Bayesian Equilibria and Shifting Power Bargaining Games
Unformatted Document Text:  condition for the inequality to hold is that 1 - γ/α ≤ 0 or α ≤ γ. 19 If these conditions are meet, ∞ t 1 (1 + αj) γ/α dj = − (1 + αt) 1−γ/α α − γ (20) To see that these expressions are positive, notice α - γ < 0, cancelling out the negative sign at the beginning of the right hand side of (20). Furthermore, notice that (20) is decreasing in t. Since these are the only two properties of the discount function needed to prove the monotonicity results in Lemmas 1, 2, 3, and 4(a), they also apply to hyperbolic discounting, with e −ρt() replaced by (20). It is also straightforward to derive bounds on p ( ) with similar interpretations for those in Lemmas 2, 3, and 4 for the exponential case. Since conditions (b), (c), (d), and (e) of Lemma 4 depend on the upper bound of p (k), we derive them here. To simplify notation, set H(k) ≡ − (1+αt(k)) 1−γ/α α−γ . dp(k)t(k)H(k) dk = H(k) p (k)t(k) + t (k)p(k) + p(k)t(k)H (k), (21) where H (k) = −αt (k) 1 − γ/α α − γ 1 + αt(k) −γ/α (22) Setting (21) greater than or equal to zero and isolating p (k) yields p (k) ≥ − H(k)t (k)p(k) + H (k)t(k)p(k) H(k)t(k) . (23) Consider necessary conditions for p (k) ≤ 0. Equation (23) implies then that − H(k)t (k)p(k) − H (k)t(k)p(k) ≤ 0 (24) 19 By the integral test (cf. Rudin (1976)), since (18) is decreasing in t, these conditions also imply that the infinite sum of (18) converges in the discrete time case as well. 23

Authors: Gochal, Joseph. and Levy, Jack.
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condition for the inequality to hold is that 1 - γ/α ≤ 0 or α ≤ γ.
If these conditions are meet,
t
1
(1 + αj)
γ/α
dj = −
(1 + αt)
1−γ/α
α − γ
(20)
To see that these expressions are positive, notice α - γ < 0, cancelling out the negative sign at the
beginning of the right hand side of (20). Furthermore, notice that (20) is decreasing in t. Since
these are the only two properties of the discount function needed to prove the monotonicity results
in Lemmas 1, 2, 3, and 4(a), they also apply to hyperbolic discounting, with e
−ρt()
replaced by
(20). It is also straightforward to derive bounds on p ( ) with similar interpretations for those in
Lemmas 2, 3, and 4 for the exponential case.
Since conditions (b), (c), (d), and (e) of Lemma 4 depend on the upper bound of p (k), we
derive them here. To simplify notation, set H(k) ≡ −
(1+αt(k))
1−γ/α
α−γ
.
dp(k)t(k)H(k)
dk
= H(k) p (k)t(k) + t (k)p(k) + p(k)t(k)H (k),
(21)
where
H (k) = −αt (k)
1 − γ/α
α − γ
1 + αt(k)
−γ/α
(22)
Setting (21) greater than or equal to zero and isolating p (k) yields
p (k) ≥ −
H(k)t (k)p(k) + H (k)t(k)p(k)
H(k)t(k)
.
(23)
Consider necessary conditions for p (k) ≤ 0. Equation (23) implies then that
− H(k)t (k)p(k) − H (k)t(k)p(k) ≤ 0
(24)
19
By the integral test (cf. Rudin (1976)), since (18) is decreasing in t, these conditions also imply that the infinite
sum of (18) converges in the discrete time case as well.
23


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