We also know that there must exist the smallest-valueAg, denoted by di,g, such that
∆ui,g(ei,g, di,g) ≥ 0 and ∆ui,g(ei,g, di,g + 1) < 0. In the following, we derive the last inequality of (2.11),∆ui,g(ei,g, Ag) < 0 for Ag > di,g, from∆ui,g(ei,g, di,g+ 1) < 0. We denote(di,g + 1) by d′i,g for short. There are two possible relationships among e∗g(d′i,g), ei,g, andec,g. Ifei,g < ec,g < e∗g(d′i,g), as e∗g(Ag) is increasing with Ag, we knowei,g <
ec,g < e∗g(Ag) for Ag > di,g. According to (A.12), this implies ∆ui,g(ei,g, Ag) < 0 for
With the substitution ofx and x′ and the use of (A.14), the last inequality in (A.13)
becomes
The last inequality can be proved easily from the first-order differentiation. Therefore, ifei,g < e∗g(di,g + 1) ≤ ec,g, ∆ui,g(ei,g, di,g+ 1) < 0 implies ∆ui,g(ei,g, di,g+ 2) < 0.
We may continue to consider the possible relationships amonge∗g(di,g+ 2), ei,g, andec,g, and apply the same proof procedure. Finally, we get∆ui,g(ei,g, Ag) < 0 for Ag > di,g. The first inequality of (2.11), ∆ui,g(ei,g, Ag) > 0 for Ag < di,g, can be derived from
∆ui,g(ei,g, di,g) ≥ 0 in a similar way. So we do not repeat the proof here.
With all of the above derivations, there exists unique di,g such that ui,g(ei,g, Ag) is maximized atAg = di,g, and (2.11) is satisfied.
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