Flexible versus dedicated technology adoption in the presence of a public firm.

PROOF OF LEMMA 1. Note that [partial derivative][sigma].sub.2]/[partial derivative][gamma] < 0 and [partial derivative][[sigma].sub.3]/[partial derivative][gamma] < 0. Further, from Equations 2 and 3, we obtain [[sigma].sub.2] [absolute value of [sub.[gamma]=0] = [0.06a.sup.2], [[sigma].sup.2]][sub.[gamma][right arrow] 1] = [0.042a[.sup.2], [[sigma].sub.3][absolute value of [sub.[gamma]=0] = [0.0977a.sup.2], [[sigma]][sub.3]][sub.[gamma][right arrow]1 = 0, and [[sigma].sub.3][absolute value of [sub.[gamma]=0] > [[sigma].sub.]][sub.[gamma]=0, > [[sigma].sub.2]][sub.[gamma]=0], while [[sigma].sub.2] [absolute value of [sub.[gamma][right arrow] 1 > [[sigma].sub.3]][sub.[gamma][right arrow]1] = 0. Therefore, [[sigma].sub.2] and [[sigma].sub.3] must cross. Setting Equations 2 and 3 equal we obtain [[gamma].sup.*] = 0.2432, where [[sigma].sub.2] and [[sigma].sub.3] cross. The result then follows immediately. QED.

PROOF OF PROPOSITION 1. Lemma 1 establishes that the relevant critical value for s in the mixed duopoly is min {[[sigma].sub.2], [[sigma].sub.3]}; in particular, for [gamma] < [[gamma].sup.*] the relevant critical value is given by [[sigma].sub.2], and for [gamma] [greater than or equal to] [[gamma].sup.*] it is given by [[sigma].sub.3], [[gamma].sup.*] = 0.2432. Thus, we need to show that [[sigma].sub.2] < [[sigma].sub.1] for [gamma] < [[gamma].sup.*] and [[sigma].sub.3] < [[sigma].sub.1] for [gamma] [greater than or equal to] [[gamma].sup.*]. Note that [partial derivative][[sigma].sub.1]/[partial derivative][gamma] < 0, [partial derivative][[sigma].sub.2]/partial derivative][gamma] < 0, and [partial derivative][[sigma].sub.3]/[partial derivative][gamma] < 0. Further, from Equations 1 and 2, we obtain [[sigma].sub.1][[absolute value of [sub.[gamma]=0] = 0.0937[a.sup.2] and [[sigma].sub.2]], respectively. [[sigma].sub.1] = [[sigma].sub.2] at [gamma] = 0.4593 > [[gamma].sup.*] and [[sigma].sub.2][absolute value of [sub.[gamma]=0] < [[sigma].sub.1]][sub.[gamma]=0]. Therefore, [[sigma].sub.2] < [[sigma].sub.1] when [gamma] < [[gamma].sup.*]. Similarly, from Equations 1 and 3 we obtain [[sigma].sub.1][[absolute value of [sub.[gamma][right arrow]1] = 0.0221[a.sup.2] and [[sigma].sub.3]][sub.[gamma][right arrow]1] = 0, respectively. [[sigma].sub.1] = [[sigma].sub.3] at [gamma] = 0.0393 < [[gamma].sup.*] and [[sigma].sub.3][absolute value of [sub.[gamma][right arrow]l < [[sigma].sub.1]1][sub.[gamma][right arrow]1. Therefore, [[sigma].sub.3] < [[sigma].sub.1] when [gamma] > [[gamma].sup.*], and we have shown that min{[[sigma].sub.2], [[sigma].sub.3]} < [[sigma].sub.1]. The rest of the proposition follows from the relevant equilibrium conditions. QED.

PROOF OF LEMMA 2. From Equations 5 and 6,

[[sigma].sub.6] - [[sigma].sub.5] [a.sup.2][f.sub.5,6][([gamma])/200([[gamma].sup.2]-3).sup.2]([[gamma].sup.2] 1)[(8[[gamma].sup.2] - 15).sup.2].

This is positive as [f.sub.5,6]([gamma]) < 0, where

[f.sub.5.6]([gamma]) = -15300 + 66600[gamma], - 78135[[gamma].sup.2] - 39900[[gamma].sup.3] + 111331[[gamma].sup.4] 14380[[gamma].sup.5] - 49792[[gamma].sup.6] + 13120[[gamma].sup.7] - 1920[[gamma].sup.9] - 512[[gamma].sup.10], and the denominator is negative as [lim.sub.[gamma][right arrow]1] < 0. QED.

PROOF OF LEMMA 3. Note that [[sigma].sub.4][absolute value of [sub.[gamma]=0] = 0.0937[a.sup.2], [[sigma].sub.6]][sub.[gamma]=0] = 0.0978[a.sup.2], [[sigma].sub.4][absolute value of [sub.[gamma]=1] = 0.0246[a.sup.2], and [lim.sub.[gamma][right arrow]1] [[sigma].sub.6] = [infinity]. Therefore, [[sigma].sub.4][absolute value of [sub.[gamma]=0] < [[sigma].sub.6]][sub.[gamma]=0] and [[sigma].sub.4] | [sub.[gamma]=1] < [lim.sub.[gamma][right arrow] 1 [[sigma].sub.6] = [infinity]. [[sigma].sub.6] reaches its minimum at [gamma] = 0.6689, whereas [[sigma].sub.4] [absolute value of [sub.[gamma]=0.6689] = 0.0393[a.sup.2] and [[sigma].sub.6][sub.[gamma]=0.6689] = 0.0388[a.sup.2], meaning that [[sigma].sub.4][absolute value of [sub.[gamma]=0.6689] > [[sigma].sub.6]][sub.[gamma]=0.6689. Hence, [[sigma].sub.4] and [[sigma].sub.6] must cross twice: Setting [[sigma].sub.4] and [[sigma].sub.6] equal, we find that they cross at [gamma].sub.1] = 0.0056 and at [[gamma].sub.2] = 0.6755. The rest of the lemma follows. QED.

PROOF OF PROPOSITION 2. Follows from Lemma 3 and the necessary conditions for equilibrium. QED.

PROOF OF LEMMA 4. Using Equations 1 and 4 we obtain

[[sigma].sub.1] - [[sigma].sub.4] = [a.sup.2][gamma][f.sub.1,4]([gamma])/2[(3 + [gamma]).sup.2][(4 + 3[gamma]).sup.2][(24 - 11[[gamma].sup.2]).sup.2],

where

[f.sub.l,4]([gamma]) = 576 + 168[gamma] - 1608[[gamma].sup.2] - 488[[gamma].sup.3] + 646[[gamma].sup.4] 20[[gamma].sup.6] + 81[[gamma].sup.7] [??] 0 for [gamma] > [[gamma].sup.**] = 0.6442.

The rest of the lemma follows immediately. QED.

PROOF OF LEMMA 5. Note that [[sigma].sub.6] [absolute value of [sub.[gamma]=0] = 0.1[a.sup.2], [[sigma].sub.2][sub.[gamma]=0] = 0.06[a.sup.2], [[sigma].sub.6] [absolute value of [gamma][right arrow]1] = [infinity], and [[sigma].sub.2][sub.[gamma][right arrow]1] = 0.042[a.sup.2]. Further, [partial derivative][[sigma].sub.2]/ [partial derivative][gamma] < 0 and [partial derivative][[sigma].sub.6]/[partial derivative][gamma] [??] 0 for [gamma] [??] 0.6669. Setting [[sigma].sub.2] and [[sigma].sub.6] equal, we find that they cross at [gamma] = 0.3133 and at [gamma] = 0.8172. It is then obvious that [[sigma].sub.2] < [[sigma].sub.6] when [gamma] [less than or equal to] 0.3133 and when [gamma] [greater than or equal to] > 0.8172, and [[sigma].sub.2] > [[sigma].sub.6] when [gamma] [member of] (0.3133, 0.8172). The rest of the lemma follows from the equilibrium conditions. QED.

PROOF OF LEMMA 6. [partial derivative][[sigma].sub.3]/[partial derivative][gamma] < 0 and [partial derivative][[sigma].sub.5]/[partial derivative][gamma] < 0. Furthermore, [[sigma].sub.3] [absolute value of [sub.[gamma]=0]] = [0.0977a.sup.2], [[sigma].sub.3][absolute value of [sub.[gamma][right arrow]1] = 0, [[sigma].sub.5] [absolute value of [sub.[gamma]=0] = [0.06a.sup.2], and [[sigma] [absolute value of [sub.[gamma][right arrow]1] = [0.008a.sup.2], so that [[sigma].sub.3] [absolute value of [sub.[gamma]=0] = [0.06a.sup.2] while [[sigma].sub.3] [absolute value of [sub.[gamma][right arrow]1] = 0 [[sigma].sub.5] [absolute va lue of [sub.[gamma][right arrow]1. Therefore, [[sigma].sub.5] and [[sigma].sub.3] cross at a critical value of [gamma], [[gamma].sup.***] = 0.3133. Thus, if [gamma] [less than or equal to] [[gamma].sup.***], [[sigma].sub.5] > [[sigma].sub.3]. The rest of the lemma follows from the equilibrium conditions. QED.

PROOF OF PROPOSITION 3. As shown in Lemma 4, for (DE, FMS) or (FMS, DE) to be equilibria in the private duopoly, [[sigma].sub.1] < s < [[sigma].sub.4] must hold; this can only happen for [gamma] > [[gamma].sup.**] = 0.644205. Recall that (DE, FMS) is an equilibrium in the mixed duopoly if [[sigma].sub.2] < s < [[sigma].sub.6]. We know that [partial derivative][[sigma].sub.2]/[partial derivative][gamma] < 0 and [partial derivative][[sigma].sub.4]/[partial derivative][gamma] < 0 and that [[sigma].sub.2] [absolute value of [sub.[gamma]=0] = [0.06a.sup.2], [[sigma].sub.2] [absolute value of [sub.[gamma][right arrow]1] = [0.042a.sup.2], [[sigma].sub.4] [absolute value of [sub.[gamma]=0] = [0.9375a.sup.2], and [[sigma].sub.4] [absolute value of [sub.[gamma][right arrow]1] = [0.02459a.sup.2]. Therefore, [[sigma].sub.2] [absolute value of [sub.[gamma]=0] < [[sigma].sub.4] [ absolute value of [sub.[gamma]=0] while [[sigma].sub.2] [absolute value of [sub.[gamma][right arrow]1] > [[sigma].sub.4] [absolute value of [sub.[gamma][right arrow]1]. Thus, they must cross. Setting [[sigma].sub.2] and [[sigma].sub.4] equal, we know that [[sigma].sub.2] [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] [[sigma].sub.4] for [gamma] [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] 0.450595. Therefore, for [gamma] >[[gamma].sub.**], [[sigma].sub.2] > [[sigma].sub.4], implying that [[sigma].sub.1] < s < [[sigma].sub.4] and [[sigma].sub.3] < s < [[sigma].sub.6] can not hold simultaneously. Furthermore, recall that (FMS, DE) is an equilibrium in the mixed duopoly if [[sigma].sub.3] < s < [[sigma].sub.5]. We know that [partial derivative][[sigma].sub.1]/[partial derivative][gamma] < 0 and [partial derivative][[sigma].sub.5]/[gamma] < 0 and that [[sigma].sub.1] [absolute value of [sub.[gamma]=0] = [0.09375a.sup.2], [[sigma].sub.5] [absolute value of [sub.[gamma]=0] = [0.06a.sup.2], [[sigma].sub.1] [absolute value of [sub.[gamma][right arrow]1] = [0.06a.sup.2], and [[sigma].sub.5] [absolute value of [sub.[gamma][right arrow]1] = [0.009328a.sup.2]. Thus, [[sigma].sub.1] > [[sigma].sub.5] for any [gamma] and therefore [[sigma].sub.1] < s < [[sigma].sub.4] and [[sigma].sub.3] < s < [[sigma].sub.5]. The rest of the proposition follows. QED.

Appendix 3: Welfare Analysis