condition for the inequality to hold is that 1 - γ/α ≤ 0 or α ≤ γ.
If these conditions are meet,

∞

t

1

(1 + αj)

γ/α

dj = −

(1 + αt)

1−γ/α

α − γ

(20)

To see that these expressions are positive, notice α - γ < 0, cancelling out the negative sign at the

beginning of the right hand side of (20). Furthermore, notice that (20) is decreasing in t. Since

these are the only two properties of the discount function needed to prove the monotonicity results

in Lemmas 1, 2, 3, and 4(a), they also apply to hyperbolic discounting, with e

−ρt()

replaced by

(20). It is also straightforward to derive bounds on p ( ) with similar interpretations for those in

Lemmas 2, 3, and 4 for the exponential case.

Since conditions (b), (c), (d), and (e) of Lemma 4 depend on the upper bound of p (k), we

derive them here. To simplify notation, set H(k) ≡ −

(1+αt(k))

1−γ/α

α−γ

.

dp(k)t(k)H(k)

dk

= H(k) p (k)t(k) + t (k)p(k) + p(k)t(k)H (k),

(21)

where

H (k) = −αt (k)

1 − γ/α

α − γ

1 + αt(k)

−γ/α

(22)

Setting (21) greater than or equal to zero and isolating p (k) yields

p (k) ≥ −

H(k)t (k)p(k) + H (k)t(k)p(k)

H(k)t(k)

.

(23)

Consider necessary conditions for p (k) ≤ 0. Equation (23) implies then that

− H(k)t (k)p(k) − H (k)t(k)p(k) ≤ 0

(24)

19

By the integral test (cf. Rudin (1976)), since (18) is decreasing in t, these conditions also imply that the inﬁnite

sum of (18) converges in the discrete time case as well.

23