STOCHASTIC

MINIMAX POLICIES TOR SELLING AN ASSET AND DOLLAR AVERAGING 389
Theorem 2(a) that the critical value becomes zero when n becomes so large that exceeding
the previous maximum price before time is up is precluded.
THEOREM 5. The critical value, zt (n), is a decreasing function ofn.
PROOF. The proof is by induction. Suppose that zt+i{n) is decreasing in n. Furthermore,
suppose that L,+1[n — a, z] — az S Lt+i[n + b, z] + bz for all n and all z such
that OSzl zf+i(»). For n < 0, it is defined that z*(n) = z*(0). For z such that
and all n:
O ^ g z?+i(n + b)
L,+i[n — 2a, z] — az — L,+i[n + b — a, z] + bz,
L,+i[n — a, z] fe L,+i[n + b, z] + (o + b)z,
Lt+i[n — a, z] — az g Li+i[n + b,z] + bz.
The first inequality follows because by assumption zt+i(n + b) g zt+i(n — a);
the second follows by applying (14) to both sides and using (12b) if n — a < 0 or
n + b < 0. Suppose that the last inequality holds with equality for some
z fe z*+i{n + b).
This z must then be z*{n) since Max [Lt+i(n — a, z) — az, Li+i{n + 6, z) + 62]
must strictly increase with z for larger z; Ll+i[n + b, z] being invariant with z for
z S zt+i(n + b). Also, Max [L,+l(n — a, z) — az, L,+1{n + b, z) + bz] must strictly
decrease with z for smaller z because Ll+i(n — a, z) — az is strictly decreasing in z.
If equality does not hold for some z such that z,+i{n + b) I2S 1 then z*(n) = 1.
In any case, L,+i[n — a,z] — az £ L,+l(n + b,z) + bz for all n and z: 0 g z g z*{n)
as required. It will next be shown that z*(n) cannot increase with n. This is true immediately,
if z*{n) = 1. Suppose next that z such that zt+i(n + b) g z < 1 satisfies
the equation L,+i(n — a, z) — az = Lt+i(n + b, z) + bz. Take n > n and let z
satisfy Lt+i(n — a, z') — az' £ L,+i(n + b, z) + 6z' with equality unless z' = 1.
It will next be shown that having z > z is impossible. First, note that if z' > z, then
L,+i(n — a, z) & L,+i(n — a, z) because from Theorem 2(c) L,+i is a decreasing
function of z. Also, Li+i(n + b, z ) = Li+i(n + 6, z) because
z > z ^ zf+i(n + 6) £ z?+i(n' + 6).
Using these relations and the definitions of z and z above gives:
L,+i(n' — a,z) — Ll+i(n + b, z) a ii+i(re' — a, z) — L,+i(n' + b, z)
fe (a + b)z
> Ll+i(n — a,z) — Ll+i(n + b,z).
Having the left-hand expression greater than the last term on the right-hand side, however,
contradicts the fact established in Theorem 1 that L,+i is convex. Therefore, it
must be true that z ^ z. Thus, it has been shown that z*(n) is a decreasing function
of n. To complete the proof by induction it is necessary to show that zj-i(n) is a decreasing
function of n and that LT(n — a, e) — az fe LT(n + b, z) + bz for all z
such that 0 g z S z?-i(n) and all n.
LT-i(n, z) = Minos,s, Max [LT(n — a, x) — ax, LT(n + b, x) + bx]
= MinoS,s, Max [a — n — ax, bx] 0 S n S o
[—ax, bx] n > a.
3. MODELS OF OPTION STRATEGY 587