so that —πz1 (ε)/ — π2 (ε) must decline in ε everywhere. Suppose that
—πz1 (ε)/ — π'2 (ε) > 1 at the first intersection of πi (ε) and π2 (ε). Hence it
must be less than 1 at the second intersection and greater than 1 at the
third intersection. But the latter condition contradicts the condition that
—πz1 (ε)/ — πz2(ε) declines. Hence a third intersection cannot exist. Also
π1 (ε) > π2 (ε) before the first intersection and after the second intersection
and vice versa in between follows. ■
Appendix G: Proof of Proposition 9
Differentiate the first order condition (15) with respect to ε. This yields
. de .
f(⅛)-dε = —%π (ɛ); V i,ε.
(30)
Multiply this equation by -1, take logs and differentiate with respect to
ε. This yields
fz z(⅛)
fl (⅛)
de,∙
-j→
dε
d2ei∕dε2
dei / dε
= —c&;
V i, ε.
(31)
First, we prove the first part of equation (21). Multiply equation (31) by
dei∕dε and aggregate across investors. This yields
^ fZ(e^) fdeΛ2 ( ∖. y ∕o9∖
Af(⅛) Ы = v ε (32)
since ∑i d2ei∕dε2 = 0.
In the HARA-case with 7 > —œ,
42