11
th that8)
∣ ,. [(1 β)2 + χ]3u2 г l./ ∖ (1 β)2uu2 and
F(0) = jʌ___—__—__, lim F(ε) = 2___an__ and
σ⅛1 -β )4 i σj
(1 -β)2u2 < F(ε) < [(1 -β)2÷χ]3u2
σ2 σ2μ(1 -β )4
We are now ready to prove:
PROPOSITION 3.2: (1 ~β)2u2 < ɛ* < [(1 ~β)2 + χ]3u2
σ2 σ2μ(1 -β )4
Proof: The left-hand side of (3.8) is a 45-degree straight line through the origin. Since
F(0) = [(1 β) + χ] u and dF < 0, these two functions must intersect at one and only
σ2μ(1 -β )4 əɛ
one point. Moreover, since
u (1 ~β) < F(ε) < [(1 ~β) +χ] u , the intersection occurs at a
σ2μ σ22(1 -β )4
value of ɛ that is bounded between (1 β ) u and [(1 β) +χ] u
σ2 σμ(1 -β )4
Figure 3.1 illustrates the argument graphically. Clearly, a solution for ɛ exists and is
unique.
8) These statements are demonstrated in Appendix B to this paper.