Furthermore, one easily checks that when ° = 1/2, then β* = 11 7+12+¾fg4β is
always larger than 7/4. Hence, there must exist °l and °u, 0 < °* < °l < °u <
1/2, such that βl < β* < 0.5 if and only if °l < ° < °u. The values of °l and
°u are found to be, respectively,
|
°l = |
2 βl - 4 + 12 V(132β2 - 276βl + 141), |
(21) |
|
°u = |
-2 + 6^(21 + 12β2 - 30βl). |
(22) |
|
Next, we see that |
<9ΔE 3 1 _ |
(23) |
|
~=4+°-2β>0 |
for β < 0.5. Consider then β = 0.5. In this case, ΔE = 0 iff. ° = °u. Thus, by
(23), it must be that ΔE > 0 if and only if ° < °u. Likewise, consider β = βl.
Then, ΔE = 0 iff. ° = °l. Thus, by (23), it must be that ΔE < 0 if and only
if ° > °l. The above findings are summarized in Lemma 2. □
A.3 Appendix 3. Proof of Lemma 3.
Since EZyh = E_ Zyh for the country with bargaining power β*, we have that
(p´ (β* - °)(1 - β* + °) - ( — ɔ (β* (1 - β*)) = 0. (24)
∖Z∕ ∖ 1 — z J
Solving (24) for p we obtain
zβ* (1 - β*)
(25)
(β* - °)(1 - β* + °)(1 - z) + zβ* (1 - β*).
It is directly seen from (25) that p is in the range (0,1) for any z in (0,1) and
that limp = 0 and limp = 1. It can also be checked that since β* < 1/2, then
z→0 z→1
p > z. Finally, we can show that p is monotonically rising with z. To do that,
we need to recall that β* = 0.5 - z (0.5 - βl) (this comes from (7)). >From
total differentiation of p with respect to z we have @p = pz + pj- β). Partial
derivatives are easily evaluated as follows:
pz
p^*
βZ
β* (1 - β*)(β* - °)(1 - β* + °)
[(β* - °)(1 - β* + °) (1 - z) + zβ* (1 - β*)]2
-z° (1 - z) [(1 - 2β*) (1 + °) + 2β*2]
[(β* - °)(1 - β* + °) (1 - z) + zβ* (1 - β*)]2
> 0,
< 0,
- (0.5 - βl ) < 0,
(26)
(27)
(28)
23