dPiw ∂2∏i(wi,w-i,p)
dwi ∂wi2
1-
FOC
( 2 (⅛)'
+ Piw
d2Piw
dw2
dgi(wi)A + dPjw d d2gi(wi)Pw
dwi J dwi I dw2 i
- 2Λh

We wish to show that this expression Ω is negative given that dPW / dwi > 0.
To to that, when there is no ambiguity, in order to simplify expressions, we use
the following notations:
P≡ Piw = Pid
P о ≡ dΡW
dwi
P00 ≡ d2PW
dw2
i
0 _ dgi(wi)
g dwi
00 _ d2gi(wi)
g ≡ dw2
Using these notations, we have equivalently:
Ω = (-2 (P0)2 + PP"ɔ (1 - g0) + P0 (g00P - 2Λh (P0)2} . (59)
We now need to compute P0, P00, g0 and g00 at any point (i.e. not only at the
symmetric candidate equilibrium).
First let compute P0 and P00. Recall that
P = Piw
exp
∕wi -Λw pw
∖ μw
/Wj∙ -ΛwPjwλ
exp eχp( —μwj)
1...n
We have, using again the implicit function theorem
μW p (1 — p )
1 + λw P (1 - P ).
Note that:
∂P
Pо ∂wi
P = i _ ∂P
1 ∂Pw
[P (1 - P)]0 = (1 - 2P) P0
Therefore, after simplifications, we get:
32
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