If I holds, since all active players at n are active at n+1 and x = ∑ x
we have a contradiction. Therefore part a) is proved. Part b) is proved
noticing that (1) implies that if x(n) = x(n+l), then x (n) = x (n+1) V i ∈ N
1 i
∩ N+1. But since all active players at n will be active at n+1 and y > O we
reach a contradiction. Therefore x(n) < x(n+l), A.2. plus (1) show that x (n)
i
> X (n+1) V i ∈ N ∩ N+1. Finally if i & N+1 but i ∈ N x (n) > x (n+1) = O .—
i i i ■
If A. 2 does not hold, Proposition 1 fails as the following examples
-which refer to the Cournot model- show.
Example 1.- (Seade (1980)). Let p = x °8, C = x . Using the first order
----------------------------------- i i
conditions of profit maximization, it is easily seen that x (1) < x (2).
i i
2
Example 2.- p = a - bx, C = ex + d/2 x with a > c, d < O, d + 2b > O and
--------------------------------- i i i
d + b < O. (Total costs will be negative for x large enough, but this can be
i
(a - c)n
fixed). Then x = ------------ so x is decreasing with n if b + d < O. On the
b+d+nb
other hand second order conditions are fulfilled if d+2b > O. A graphical
argument similar to Example 2 can be found in Mc Manus (1964).
We now turn to study how payoffs change with entry.
Proposition 2: Under A.l, A.2 and A.3
a) U (n) ≥ U (n+1).
i i
b) If y > O the above inequalities are strict.
17