STOCHASTIC

388 GORDON PTE
PROOF. Suppose/(X) S g{x) forx = x. Then,
gfa') = MaxoS.Sr \gi(x')] £ MaxoS g(x) for all x as stated.
THEOREM 3. Lt(n, z) is a strictly decreasing function of t, given n and z, if
n < (T - t)a.
PROOF. The proof is by induction. Assume that the r.esult holds for t + 1. Let
h,+i(n, x) = Max [Lt+\(n — a, x) — ox, £i+i(re + b, x) + bx].
Let Lt = Minos,s, h,+l(n, x) = ht+l(n, x). If n + b /i1+2(n, x ).
Since L,+i(n, z) cannot exceed h,+1(n, x') it follows that L,(n, z) > Lt+i(n, z). The
case must now be considered where n + b fe [T — (t + l)]a. In this case Lemma 1
cannot be strictly applied because Ll+i(n + b, x) = Lt+i{n + b, x) = 0. However,
it does follow by similar reasoning that h,+i(n, x ) £ h,l+t{n, x ). Suppose that equality
holds. Then hl+i = ht+i = bx, and furthermore, since bx is a minimum, inspection
of ht+i and hl+i shows that L,+i(n — a, x) — ax' = bx' and L,+i(n — a, x) — ax =
bx'. The latter equalities, however, imply that Ll+i(n — a, x) = L,+2(n — a, x)
which contradicts the assumption that L,+i > L(+2 for n — a < [T — (t + l)]o.
This is impossible because the latter condition must be satisfied if n < (T — t)a.
Therefore, it must be true for this case as well that hi+i{n, x) > h,+i(n, x) since
equality is impossible. Thus, Lt > L,+i, if n < (T — t)a. From part (a) of Theorem
2 and (12) it follows that LT-i > Zr for n < (T — t)a = a because Lr-i > 0, Lr = 0
if 0 S n < a and Lr_i = £i-i(0, z) — n > LT = —?i if re < 0. Therefore, the result
holds for T — 1 and hence by induction for all t. This completes the proof.
The characterization of the dependence of the minimax variable regret on t, n and
2 is now complete. It remains to characterize the optimal policy with respect to sale of
the asset. First it will be shown that the optimal policy is of the following form. At
any time there exists a critical value which depends on n. All of the asset which is held
above this critical value will be sold. If less than this critical value is held, none will be
sold.
THEOREM 4. At any time the minimax policy is of the form: x* = z, if z S z, (re),
x* = z*(n), if z > z*{n).
PROOF. Using (14) and the notation introduced in the proof of Theorem 3, the
minimax policy at t minimizes ht+i(n, x) with respect to x on the range 0 S x S z.
Moreover, from Theorem 1 and the fact that the maximum of a set of convex functions
is convex it follows that ht+i(n, x) is convex in x given n. Let Zi*(w) be the x which
minimizes hl+i(n, x) on the range, 0 S x S 1. It follows immediately that z*(n)
minimizes ht+i(n, x) on 0 £ x S z for all2 £ z*(n) as required. It remains to show
that 2 minimizes ht+i(n, x) on 0 S x S z for all z < z*(n). The proof of this is by
contradiction. Suppose for some 2 such that x < 2 < z*(n), that ht+i(n, x) <
h,+i{n, 2). Set X = (z - z*)/(x' - z*) so that 0 < X < 1 and \x + (1 - \)z* = z.
It follows that XA,+i(n, x ) -f- (1 — \)hl+i(n, z ) < ht+i(n, 2), but this is impossible
because hi+\(n, x) is convex. This completes the proof. Next, the critical values defined
in the last theorem are characterized with respect to their dependence on re. In
particular, it is shown that they are decreasing functions of n. It is already known from
586 PART V. DYNAMIC MODELS