From (S), we have x'.™ = x%--ι > x ⇒ kn-2 > 3x. If (D) holds with equality, i.e. x = xm, then
the LHS of equality [14] equals to zero and the equality is satished if and only if kn-2 — x =
2χm ⇒ kn-1 = x + 2xm. Work in the same way inductively to period 1. In period 1, we have
x™ = x^-1 = ■■ = '∙l'-2 > x ⇒ k > nx and as (D) holds with equality kn-ι = x + (n — 1) xm. The
entry deterring capacity k1 is implicitly dehned by
— (x, β1 (T)) = — ( kL-Ξ. , oʌ for t = 2,..,n (16)
∂x^ V ,f ( )J ∂xψ \— — 1 , 1 , , ( )
and we conclude that k1 ∈ (nx, x + (n — 1) Tm].
Step 5. In a subgame without entry
∂v / m ∂V k k — xψ ∖ rλ
dx™ '1 ,o) — ∂W ( n — 1 , 0J=0 (17)
and x™ = k∕n < xm for all t = 1,..., n. Hence, the entire capacity will be used in an equilibrium
without entry. No capacity is left idle.
Step 6. Working backward to period 0, the incumbent would install k1 to deter entry in all markets.
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