
Peter J. Hammond and Horst Zank
68
We now prove by backward induction that (2.31) holds at every node n of T. At any
terminal node n with a consequence y ∈ Y one has
so (2.31) holds trivially.
As the induction hypothesis, suppose that
�(T , n
′
) = C
(F(T , n
′
for every
∈ N
+1
(T , n
. Now, rule (2.30) states that
T , n) =∪
n
′
∈β(T ,n)
�(T , n
. Together
with (2.33) and the induction hypothesis, this implies that
This proves the relevant backward induction step, so (2.31) holds for all nodes n in tree T.
□
2.4.7 Time Inconsistency and Hyperbolic Discounting
2.4.7.1 Timed Consequences and Discounting
Following Samuelson (1937), Koopmans (1960)