26
-⅛E yields (1 - e~rT )∫ sF(s, τ)e-rsds = TE - E, so that
0
—E = - Te ~rT (1 - e ~rT ) -1E - TE + E = Ed, where d = (1 + r - rT - rT ( erT -1)-1 ) =
dr
(1 + r)(erτ -1) -rTerτ
rT 1 '
er -1
numerator and the
To apply the L’Hopital’s rule for d, we differentiate its
denominator with respect to rT and get
(1 + r)erT -erT
d = -------—
erT
- rTe rT (1 + r) - 1
------- which gives that lim d =--------= r > 0 . Hence, we
rT→0 1
have - (E +rE) = = E[1 +r -rT - rT(e rT -1)-1]< 0. Now the overall term
rT rT
B1 =-(1 - )( pf (T ) + V ) - (1 + r - rT - -r- ) E < 0, so that T1H
er -1 er -1
<0. Q.E.D.
Appendix 3. Proof of Lemma 2:
Note first that we can write
A.3.1. F(T,τ) -rE =
T
∫F(s,τ)e-rsds
0
F (T ,τ )
T
∫F(s,τ)e-rsds
r
(1-e-rT)
TT
If FT > (≤) 0, then ∫ F(T,τ)e -rs ds > (≤)∫F(s,τ)e-rsds ⇔
00
FT"(1 -e-rT)>(≤)∫F(s,τ)e-rsds ⇔ T f(T∙τ) >(≤) r-rT)
0 ∫ F(s, τ)e-rsds ( e )
0
Hence, F(T,τ)-rE > (≤) 0 as FT(T,τ) > (≤) 0. Q.E.D.
Appendix 4. Proof of the Theorem: the sign of TτH = the sign of FτT
Recall from the text that the cross-derivative of equation [9] can be written as
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