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point, that is (T t + 1) = 0. Thus, (T t + 1) = C 1 (1 ) T P t
j wm(j) =
0 ! C 1 = (1 ) T P t
i=0
j wm(j) , and hence:
(s) = (1 ) T P t
j=s
j wm(j) ; s = 1; : : : ; T t + 1:
Now we can substitute this result in the second foc, and we would get:
"
(s + 1) = (s)
#
1
s (1 ) @x (; s)
+ f x (1 ) T P t
j wm(j)
@w m (s)
m
(s + 1) = (s)
b s ; s = 0; : : : ; T
t
j=s+1
This …rst order condition can be solved for the optimal value of the costate variable
(s). This equation of di¤erences has the following solution:
s 1 Xs 1 s j 1 1
(s) = 0 b j ; s = 1; : : : ; T t + 1:
j=0
where 0 is a certain constant. As the value function ends at T
j=0
t, an increase
in w m (T t) cannot a¤ect the payo¤, so that the transversality condition has a
T t+1
1
free end point, that is (T t + 1) = 0. Thus, (T t + 1) =
0
P T t j T t 1
j=0 bj = 0 !
0 = () T t+1 P T t j
T t 1
j=0
bj , and hence:
(s) = T P t
j=s
= T P t
=
j (s 1) b j
j=s
j s 1 1
(1 )
TP
t
j=s
"
s (1 ) @x (; s)
+ f x (1 ) T P t
@w m (s)
j s 1 j + f x
@x (; s)
@w m (s)
(1 ) T P t
j=s
j=s+1
j s 1 T P t
j=s+1
j wm(j) #
j wm(j)
Substituting (s) into the …rst foc to eliminate the costate variable:
(1
s m
e m (s)) m+1 = s P
(1 ) n P
k i (w m (s)) x i + n k i (w m (s)) x i (s + 1)
i=1
i=1
n
P
+ (s + 1) f x k i (w m (s)) x i
(1
i=1
n
m
P
e m (s)) m+1 = (1 ) k i (w m (s)) x i (1 + s + s) + s
i=1
(1 ) n
1
P
e m (s) = 1
k i (w m (s)) x i [1 + s + s] + s
m i=1
25
(m+1)