Turn Taking: Intertemporal Cooperation and Symmetry through Intratemporal Asymmetry by Sau-Him Paul Lau and Vai-Lam Mui (2010)
8
Other Proofs (Not for Publication)
The following result is useful in subsequent analysis. From (3) and δ ∈ (0, 1), we have à ! à ! h+l h−l h + δl = (1 + δ) + (1 − δ) 2 2 Ã
!
h+l > (1 + δ) . 2
(B1)
Proof of Lemma 1. We obtain three preliminary results useful for the proof of Lemma 1. First, define U (p) = p2 (t) + (1 − p)2 (s) + p (1 − p) (h + l)
(B2)
as a player’s current-period payoff when both players choose Tough with probability p. Using (B2), it is straightforward to obtain !
Ã
h i h+l h+l − U (p) = p2 + (1 − p)2 + 2p (1 − p) − U (p) 2 2 2
=p
Ã
!
Ã
!
h+l h+l − t + (1 − p)2 −s 2 2
(B3)
> 0. Second, from Table 2 and (B2), we have Vi (p, p) ³
´
³
= p2 [t + δVi (p, p)]+(1 − p)2 [s + δVi (p, p)]+p (1 − p) h + δV L +(1 − p) p l + δV H = U (p) + δ
nh
i
³
p2 + (1 − p)2 Vi (p, p) + p (1 − p) V H + V L
Therefore, we obtain
Vi (p, p) =
³
U (p) + δp (1 − p) V H + V L h
1 − δ p2 + (1 − p)2 23
i
´
.
´o
.
(B4)
(B5)
´
From (6), (7), (B3) and (B5), we have V
H
L
+V −Vi (p, p) = 2
=
=
n
h
1 − δ p2 + (1 − p)2
{1 − δ [1 − 2p (1 − p)]}
(1 − δ)
³
V H +V L 2
´
³
V H +V L 2
h
´
io ³
V H +V L 2
³
h
´
h
=
1−δ
− U (p)
p2
³
i
³
´i
− U (p) + δp (1 − p) V H + V L
1 − δ p2 + (1 − p)2 h+l h2
³
i
− U (p) + δp (1 − p) V H + V L
V H +V L 2
h
h
− U (p) + δp (1 − p) V H + V L
1 − δ p2 + (1 − p)2
1 − δ p2 + (1 − p)2
+ 2δp (1 − p)
h
´
+ (1 − p)2
> 0.
i i
´i
(B6)
Third, h > l implies V H > V L . This inequality, together with (B6), leads to VH >
VH +VL > Vi (p, p) . 2
(B7)
We now prove Lemma 1. Using (6) and (7), we have V H = h + δl + δ 2 V H .
(B8)
From (B4), we have nh
i
³
´o
p2 + (1 − p)2 Vi (p, p) + p (1 − p) V H + V L . (B9) We prove Lemma 1 by showing that (a) the coefficient of δ 2 in (B8) is larger than that in (B9), and (b) h + δl in (B8) is larger than s + δU (p) in (B9). First, (B7) implies that s+δVi (p, p) = s+δU (p)+δ 2
h
i
V H = p2 + (1 − p)2 + 2p (1 − p) V H h
i
³
´
> p2 + (1 − p)2 Vi (p, p) + p (1 − p) V H + V L .
Second, (4), (B1) and (B3) imply that
Ã
!
h+l h + δl − [s + δU (p)] > (1 + δ) − [s + δU (p)] 2 =
Ã
!
"
#
h+l h+l −s +δ − U (p) 2 2 > 0. 24
´i
This proves Lemma 1. Proof of Lemma 2. For the accommodating case, l > t. According to (B7), V H > Vi (p, p). Combining these results, we obtain (9a). Proof of Lemma 3. The following analysis applies to the symmetric equilibrium supported by the TTIR strategy with p∗1 = p∗2 = p∗ and V1∗ = V2∗ = V ∗ . From (8), (9), and the first equality of (12), we have 1 − p∗ =
(V H
V L − t − δV ∗ > 0. − s − δV ∗ ) + (V L − t − δV ∗ )
(B10)
Also, (6), (7) and the first equality of (12) imply p∗ − 0.5 =
h − l − (1 + δ) (s − t) . 2 (1 + δ) [(V H − s − δV ∗ ) + (V L − t − δV ∗ )]
(B11)
It is easy to see that the denominator of (B11) is positive. For the numerator, either t − s ≥ 0 or t − s < 0 can be consistent with t > l. If t − s ≥ 0, it is easy to conclude from (3) that the numerator of (B11) is positive. If t − s < 0, then (4), t > l and 0 < δ < 1 imply h − l − (1 + δ) (s − t) > h − l − 2 (s − t) "Ã
!
#
(B12)
h
i
h+l =2 − s + t − l > 0. 2 Combining (B10), (B11), and (B12), we have (18). From the second equality of (11), we have h
i
p∗ (δ) V L (δ) − t − δV ∗ (δ, p∗ (δ)) = [1 − p∗ (δ)] V H (δ) − s − δV ∗ (δ, p∗ (δ)) , where the dependence of p∗ on δ, and V ∗ on δ and p∗ (δ) are written explicitly. Differentiating the above expression with respect to δ, and rearranging, gives (19). Consider the denominator of (19). Combining (18) and (B21) below, the third term in the denominator of (19) is positive. The first two terms in the denominator of (19) are positive when (8) and (9) hold. Therefore, we have ³
´
³
´
V H − s − δV ∗ + V L − t − δV ∗ + δ (1 − 2p∗ )
∂V ∗ > 0. ∂p∗
(B13)
Consider the numerator of (19). First, (18) implies p∗ > 1 − p∗ > 0. 25
(B14)
Second, using (6) and (7), we have Ã
!
Ã
∂V ∗ ∂V ∗ ∂V L ∂V H −δ −V∗ − −δ −V∗ ∂δ ∂δ ∂δ ∂δ
!
∂V L ∂V H − ∂δ³ ∂δ ´ H −∂ V − V L
= =
∂δ h−l > 0. = (1 + δ)2
(B15)
Third, using (B6), we have ( "
#)
∂ VH +VL − V ∗ (δ, p∗ (δ)) δ ∂δ 2 h
h+l 2
h
⎧
i
⎫
⎬ δ h+l − U (p∗ ) ∂ ⎨ 2 h i = ∂δ ⎩ 1 − δ (p∗ )2 + (1 − p∗ )2 ⎭ i
− U (p∗ )
(B16)
=n h io2 > 0. 1 − δ (p∗ )2 + (1 − p∗ )2
∂ (δV H ) L H H Fourth, (7) implies ∂V∂δ = = δ ∂V∂δ + V H , and (6) implies ∂V∂δ = ∂δ ∂ (δV L ) L = δ ∂V∂δ + V L . Using these relationships, (B15) and (B16), we have ∂δ
´i ∂V ∗ ∂ h ³ H ∂V L −δ −V∗ = δ V −V∗ ∂δ ∂δ ∂δ " Ã
!#
" Ã
∂ ∂ VH +VL VH −VL −V∗ + = δ δ ∂δ 2 ∂δ 2 " Ã !# VH −VL ∂ > δ ∂δ 2 ⎡ ³
´
³
!#
´⎤
H ∂ δV L 1 ⎣ ∂ δV ⎦ = − 2 ∂δ ∂δ
1 = 2
Ã
∂V L ∂V H − ∂δ ∂δ
!
(B17)
> 0.
Therefore, (B14), (B15), and (B17) imply that ∗
p
Ã
!
Ã
∂V L ∂V H ∂V ∗ ∂V ∗ −δ − V ∗ − (1 − p∗ ) −δ −V∗ ∂δ ∂δ ∂δ ∂δ Ã
!
Ã
! !
∂V ∗ ∂V ∗ ∂V L ∂V H −δ − V ∗ − (1 − p∗ ) −δ − V ∗ > 0. > (1 − p ) ∂δ ∂δ ∂δ ∂δ (B18) ∗
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That is, the numerator of (19) is negative. Combining (B13) and (B18) gives Lemma 3. Proof of Proposition 2. 14 To show the existence of p∗ ∈ (0, 1) satisfying (12) for the mutual-tough case, we define the following continuous function g (p) =
[V H
V H − s − δV (p) −p − s − δV (p)] + [V L − t − δV (p)]
(B19)
over p ∈ [0, 1], where V (p) = Vi (p, p) according to (10). It is easy to observe that g (.) is a well-defined function for every p ∈ [0, 1], and the solution to (12) is defined by g (p) = 0. Using (B5) and (4) to (7), we can show that for all δ ∈ (δ T T , 1), g (0) =
[V H
=
V H − s − δV (0) −0 − s − δV (0)] + [V L − t − δV (0)] VH −s−δ
³
s 1−δ
´
V H + V L − s − t − 2δ
and g (1) =
[V H =
³
s 1−δ
´
> 0,
(B19a)
V H − s − δV (1) −1 − s − δV (1)] + [V L − t − δV (1)] h
− VL−t−δ
³
t 1−δ
V H + V L − s − t − 2δ
´i ³
t 1−δ
´
< 0.
(B19b)
Applying the Intermediate Value Theorem, we know that there exists a p ∈ [0, 1] such that g (p) = 0. 15 Moreover, g (p) = 0 does not hold at p = 0 or p = 1, as observed in (B19a) and (B19b). Therefore, we conclude that the solution to (12) exists in the interval (0, 1). The proof of the uniqueness of the solution to (12) for the mutual-tough case is also similar to that of Proposition 1, except that the analysis holds only for δ ∈ (δ T T , 1) when t > l, whereas it holds for all δ ∈ (0, 1) when t < l. Proof of Proposition 3. At the symmetric equilibrium supported by the 14 An
important feature that distinguishes the proof of Proposition 2 from that of Proposition 1 is that condition (9a) does not hold for all p ∈ (0, 1) for the mutualtough case, but it does hold for p = p∗ when δ > δ T T . 15 We use the Intermediate Value Theorem instead of the Fixed Point Theorem because f (p), as defined in (A1), is not necessarily a mapping from [0, 1] to [0, 1] for the mutual-tough case.
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TTIR strategy, the second equality of (11) can be rewritten as h
i
h
i
p∗ (θ) V L (θ) − t − δV ∗ (p∗ (θ)) = [1 − p∗ (θ)] V H (θ) − s − δV ∗ (p∗ (θ)) , (B20) H L ∗ ∗ ∗ where the dependence of V on θ, V on θ, p on θ, and V on p (θ) are explicitly stated. It is clear from (6), (7), (10) and (24) that V ∗ depends on θ via p∗ (θ) only. Differentiating (B20) with respect to θ, and simplifying, we obtain (26) in the main text. The sign of various components of (26) are obtained as follows. We know from (3), (5), (6), (7), (24), (25), and the first equality of (A2) when p∗1 = p∗2 = p∗ and V1∗ = V2∗ = V ∗ that ∗
∂V (1 + δ) t − (h + δl) = = ∗ ∂p (1 − δp∗ )2 (1 + δ)
h
(1 + δ) t − (1 + δ)
³
h+l 2 ∗ 2
´
+ (1 − δ)
(1 − δp ) (1 + δ)
³
h−l 2
´i
< 0,
(B21)
H
∂V λ = > 0, ∂θ (1 + δ) (1 + θ)2 and
³
∂ VH +VL ∂θ
´
= 0.
(B22)
(B23)
Moreover, (8) and (9) hold for t < l and δ ∈ (0, 1) or for t > l and δ ∈ (δ T T , 1). When 0.5 < p∗ < 1, both the denominator and numerator of (26) are positive. Thus, 0.5 < p∗ < 1 is a sufficient condition for ∂p∗ > 0. ∂θ
(B24)
Moreover, since 2p∗ (1 − p∗ ) is decreasing in p∗ when 0.5 < p∗ < 1, we conclude from (23) and (B24) that 0.5 < p∗ < 1 is a sufficient condition for ∂E (D) ∂p∗ ∂E (D) × = ∂θ ∂p∗ ∂θ to be positive. Since 0.5 < p∗ < 1 is satisfied for game G∞ with t > l and δ ∈ (δ T T , 1) according to Lemma 3, this proves part (a). Since V H > V L , it is easy to see from (B20) that V H − s −δV ∗ > V L − t − δV ∗ when t ≥ s. Consequently, p∗ > 1 − p∗ and thus, 0.5 < p∗ < 1. This proves part (b).
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