which yield an unambiguously negative sign for @p: Summing up results, we
have Lemma 3. □
A.4 Appendix 4. Proof of Proposition 4.
By definition, the pure strategy equilibrium no MAI requires p = 0 and z = 0,
so that in this case β* = 0:5: This collection of strategies is a Nash equilibrium
if and only if i) EZ¼ < E-Z¼ and ii) EZyh < EZyh for all h. By Lemma 2,
requirement i) is satisfied if and only if ° > °l, while ii) is necessarily true for
p = 0 (this is checked by inspection of (5) and (6)).
The pure strategy equilibrium full MAI requires p =1and z =1; so that
in this case β* = βι∙ This collection of strategies is a Nash equilibrium if and
only if iii) EZ¼ > E-Z¼ and iv) EZyh > E-Zyh for all h. By Lemma 2, iii) is
satisfied if and only if °<°u ; while iv) is necessarily true for p =1 (check (5)
and (6)).
The mixed strategy equilibrium partial MAI exists if, at p 2 (0; 1) and
z 2 (0; 1), v) EZ¼ = E-Z¼ and vi) EZyh = E-Z yh for one h. By Lemma 2, v)
is satisfied if and only if °l <°<°u , while, vi) is true by Lemma 3 . Results
are summarized in Proposition 4. □
A.5 Appendix 5. Proof of Proposition 5.
When βl < ° it must be checked whether β*, as given by (20), is still higher
than °: If this is not the case, then in the expression for ¢E in (19) β must
be replaced by °. It is straightforward to show that β* < ° if and only if
° 2 (°*,°l), where
15 — 12βι + 4β2
(29)
4 5 — βι
At ° = °L we have that β* = °. Moreover, one checks that °u > °L and that
°ι R °L if and only if β R 8-4P6.
Consider first β* > °; i.e., ° = (°%°l). Note that in this case Lemma 3
holds unchanged. An equilibrium with partial MAI requires ° < β* < 1/2 which
corresponds to °L <°<°u by the above argument and Lemma 2. By the same
argument as in the proof of Proposition 4, ° < °u is also necessary and sn∏'icieιιl
for the emergence of a no MAI equilibrium. To check for the existence of an
equilibrium with residual MAI, we have to check whether ¢E > 0 when β = °.
It is easy to show that a necessary and su∏cient condition for an equilibrium
with residual MAI is °>°L .
Consider then β* < °; i.e. ° 2 (°β°Lβ In such a case, there cannot be
a partial MAI because there is no country with βh >°that splits the set of
potential participants into a subset of actual participants and one of outsiders.
No MAI is surely an equilibrium, because °<°L <°u. An equilibrium with
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