Proof of Proposition. 18 In this section we give an algorithm to prove the following proposition.
Proof of Proposition. 4. The first part follows directly from the subgame Γ1,G (see payoffs given in figure 1). After loosing to G1, brand would launch a generic if ΠT1 + ΠT1 − θ + δ(ΠT1 + ΠT1) ≥ 0 0 ΠD0 +δΠT0. Rearranging the terms gives the required result θ ≤ (ΠT1 +ΠT0 −ΠD0)+δ·(ΠT1 +ΠT1 − 0 ΠT0) = θ∗∗(κ). Similarly, the second part follows from the subgame Γ1,B. After winning against 0 G1, the brand launches a generic in post-patent period if ΠM + δ(ΠT1 + ΠT1 − θ) ≥ ΠM + δ(ΠT0). Rearranging gives the required result θ ≤ (ΠT1 + ΠT1 − ΠT0) = θ∗(κ). Q Appendix C. Extension to the Game Tree j 0 2
Proof of Proposition. 2. Substituting the instrument values into the welfare function, we can evaluate the critical discount factor δt2: δt2 = WDt WCt f − = f f WDt − WNt Ξ1(n, γ)2Ξ2(n, γ) BNt (n)Ξ3(n, γ)Ξ4(n, γ) where Ξ1(n, γ) = αΞ1 (γ) + αΞ1 (γ)n + αΞ1 (γ)n2 with αΞ1 (γ) = 2(36 − 72γ + 59γ2 − 26γ3 + 6γ4) αΞ1 (γ) = γ(60 − 98γ + 65γ2 − 18γ3) αΞ1 (γ) = γ2(3 − 2γ)(4 − 3γ) Ξ2(n, γ) = αΞ2 (γ) + αΞ2 (γ)n + αΞ2 (γ)n2 + αΞ2 (γ)n3 with 1 αΞ2 (γ) = Γ(0)(64 − 96γ + 48γ2 − 8γ3 + 2γ4 − 3γ5) αΞ2 (γ) = γ(128 − 192γ + 96γ2 − 24γ3 + 20γ4 − 9γ5) αΞ2 (γ) = γ2(32 − 32γ + 12γ2 − 16γ3 + 9γ4) αΞ2 (γ) = γ5(4 − 3γ) Ξ3(n, γ) = αΞ3 (γ) + αΞ3 (γ)n + αΞ3 (γ)n2 + αΞ3 (γ)n3 with αΞ3 (γ) = 2Γ(0)(48 − 88γ + 54γ2 − 9γ3 − 2γ4) αΞ3 (γ) = γ(192 − 328γ + 172γ2 − 4γ3 − 13γ4) αΞ3 (γ) = 2γ2(2 + γ)(3 − 2γ)(4 − 3γ) αΞ3 (γ) = γ4(4 − 3γ) and Ξ4(n, γ) = αΞ4 (γ) + αΞ4 (γ)n + αΞ4 (γ)n2 + αΞ4 (γ)n3 with αΞ4 (γ) = 2Γ(0)(96 − 272γ + 326γ2 − 213γ3 + 78γ4 − 12γ5) αΞ4 (γ) = γ(576 − 1656γ + 2012γ2 − 1300γ3 + 439γ4 − 60γ5) αΞ4 (γ) = 2γ2(144 − 350γ + 337γ2 − 148γ3 + 24γ4) αΞ4 (γ) = γ3(4 − 3γ)(12 − 15γ + 4γ2) Similarly, we can express WDs − WCs B f (n)2Ψ1(n, γ) δsf = f = f f WDs − WNf N Ψ2(n, γ) where Ψ1(n, γ) = αΨ1 (γ) + αΨ1 (γ)n + αΨ1 (γ)n2 + αΨ1 (γ)n3 with 0 920γ4 − 1500γ5 + 486γ6 − 75γ7) 1 488γ4 − 4772γ5 + 1440γ6 − 195γ7 2 12γ4 − 21γ5) αΨ1 (γ) = Γ(0)(640 − 2240γ + 3712γ2 − 3936γ3 + 2 αΨ1 (γ) = γ(1536 − 6080γ + 11136γ2 − 12544γ3 + 9 ) αΨ1 (γ) = 8γ2(1 − γ)(72 − 224γ + 314γ2 − 252γ3 + 1 3 αΨ1 (γ) = 16γ3(1 − γ)4(4 − 3γ) and Ψ2(n, γ) = αΨ2 (γ)+αΨ2 (γ)n+αΨ2 (γ)n2+αΨ2 (γ)n3+αΨ2 (γ)n4+αΨ2 (γ)n5+αΨ2 (γ)n6+ αΨ2 (γ)n7 with αΨ2 (γ) = 4Γ(0)3(35328 − 216320γ + 619648γ2 − 1099712γ3 + 1351200γ4 − 1215472γ5 + 823136γ6 − 422760γ7 + 162306γ8 − 44499γ9 + 7848γ10 − 675γ11) αΨ2 (γ) = 4γΓ(0)2(214016 − 1338624γ + 3897600γ2 − 7007296γ3 + 8694080γ4 − 7867696γ5 + 5333968γ6 − 2724272γ7 + 1030752γ8 − 275313γ9 + 46629γ10 − 3780γ11) αΨ2 (γ) = γ2Γ(0)(2166784 − 13846528γ + 41058816γ2 − 74952448γ3 + 94113152γ4 − 85835584γ5 + 58323648γ6 − 29631712γ7 + 11045456γ8 − 2873016γ9 + 467586γ10 − 35883γ11) αΨ2 (γ) = γ3(2949120 − 19290112γ + 58413056γ2 − 108592896γ3 + 138375168γ4 − 127473216γ5 + 86926912γ6 − 43953856γ7 + 16141904γ8 − 4089648γ9 + 640548γ10 − 46737γ11) αΨ2 (γ) = γ4(1152000 − 7170048γ + 20475648γ2 − 35512576γ3 + 41630528γ4 − 34601856γ5 + 20694304γ6 − 8792064γ7 + 2533320γ8 − 445176γ9 + 36045γ10) αΨ2 (γ) = γ5(4 − 3γ)(63232 − 330240γ + 775168γ2 − 1075904γ3 + 970672γ4 − 583872γ5 + 228220γ6 − 52740γ7 + 5481γ8) αΨ2 (γ) = 8γ6(1 − γ)(4 − 3γ)2(216 − 728γ + 1050γ2 − 812...
Proof of Proposition. 7 The first and second order derivatives of Jr with respect to Cr are ∂Jr ∂Cr ∂2Jr = λ − λpFR (Cr − Gr) − πe (1 − FR(Cr)) (E.1) ∂C2 = −λpfR (Cr − Gr) + πefR(Cr) (E.2) r
