Stage 1. Firm 1 has the option to make a strategic investment k>0 Stage 2. Firms 1 and 2 play a Nash game, choosing strategies 1, 32 E R, resulting profts丌 Let the reaction functions be G1=y1(32, k), 92=2(v1) Suppose there is a Nash equilibrium yi(k), y* (k) in stage 2 satisfying stability condition We have dr2[i(k),2(k)O2(k),(k)]y(k) ak upper oni(m,A)<0.m,边2<0 By equilibrium condition yi=y12(3i),kl, we have 孤-盖 Hence, to reduce the incentive of entry, we need 2k>0, which requires k>0 We also have d1lyi(k), y2(k),k ayi(k), y2(k), k) ay(k), ayi(k), y(k),k d dy ak The first term is the strategic effect; the second term is the direct effect. By equilibrium condition y2=j2[i(32, k)], we find that the strategic effect is positive if 22<0 Example 2.9. Suppose firm 1 is the incumbent and firm 2 is a new entrant. Let C1(y)=c1(k)y, C2(g)=C2y p(y)=a-y where a>0 and c2>0 are constants, and cI(k)>0 and d(k )<0. We have f 1=a-c1(k)-y2Stage 1. Firm 1 has the option to make a strategic investment k > 0. Stage 2. Firms 1 and 2 play a Nash game, choosing strategies y1, y2 ∈ R, resulting profits π1(y1, y2, k) and π2(y1, y2). Let the reaction functions be yˆ1 = ˆy1(y2, k), yˆ2 = ˆy2(y1). Suppose there is a Nash equilibrium [y∗ 1(k), y∗ 2(k)] in stage 2 satisfying stability condition:     ∂yˆ1 ∂y2 ∂yˆ2 ∂y1     < 1. We have dπ2[y∗ 1(k), y∗ 2(k)] dk = ∂π2[y∗ 1(k), y∗ 2(k)] ∂y1 ∂y∗ 1(k) ∂k . Suppose ∂π1(y1, y2, k) ∂y2 < 0, ∂π2(y1, y2) ∂y1 < 0. By equilibrium condition y∗ 1 = ˆy1[ˆy2(y∗ 1), k], we have ∂y∗ 1 ∂k = ∂yˆ1 ∂k 1 − ∂yˆ1 ∂y2 ∂yˆ2 ∂y1 . Hence, to reduce the incentive of entry, we need ∂y∗ 1 ∂k > 0, which requires ∂yˆ1 ∂k > 0. We also have dπ1[y∗ 1(k), y∗ 2(k), k] dk = ∂π1[y∗ 1(k), y∗ 2(k), k] ∂y2 ∂y∗ 2(k) ∂k + ∂π1[y∗ 1(k), y∗ 2(k), k] ∂k . The first term is the strategic effect; the second term is the direct effect. By equilibrium condition y∗ 2 = ˆy2[ˆy1(y∗ 2, k)], we find that the strategic effect is positive if ∂yˆ2 ∂y1 < 0. Example 2.9. Suppose firm 1 is the incumbent and firm 2 is a new entrant. Let C1(y) = c1(k)y, C2(y) = c2y pd (y) = a − y, where a > 0 and c2 ≥ 0 are constants, and c1(k) ≥ 0 and c0 1(k) < 0. We have † yˆ1 = 1 2 [a − c1(k) − y2], yˆ2 = 1 2 (a − c2 − y1). 2 — 13