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**Sample text**

M := E(~t) > 1. Finally, the one-stage return r(x, a) is supposed to be a "utility of consumption," so that r(x, a) := u(x - a) 38 3. Finite-Horizon Problems for some utility function u, and we wish to maximize the expected total discounted utility from consumption: J(Jr, x) := E: [~l atu(xt - at)] , x E X:= [0,00), with discount factor a > O. 14)], for all x E X and t = N -1, ... ,0, o max [u(x - a) A(x) + aEJt+1(a~t)]. 2) The solution depends of course on the particular form of u. Here we consider the case of functions u with so-called isoelastic marginal utility, that is, u(x) = (b/,y)x 1 , where band r are positive constants, and r < 1.

2) Suppose that these functions are measurable and that, for each t = 0, ... 2) for all x E X; that is, VX E X and t = 0, ... 3) Then the (deterministic Markov) policy 7l'* = {fa, ... , J*(x) = Jo(x) = J(7l'*, x) VX EX. 4) Proof. , if t = 0, 1, ... 2 Dynamic programming 25 C t (-7r, x) is called the "cost-to-go" or cost from time t onwards when using the policy Jr and Xt = x. 2) J(Jr, x) = CO(Jr,x). 7) To prove the theorem, we shall show that, for all x E X and t = 0, ... e, Jr*, Ct (Jr*, x) Jt(x).

Let Jo, J 1 , ... 1 ) and for t = N - 1, N - 2, ... ,0, Jt(X) := min A(x) [C(X, a) + ixr Jt+1(y)Q(dylx, a)] . 2) Suppose that these functions are measurable and that, for each t = 0, ... 2) for all x E X; that is, VX E X and t = 0, ... 3) Then the (deterministic Markov) policy 7l'* = {fa, ... , J*(x) = Jo(x) = J(7l'*, x) VX EX. 4) Proof. , if t = 0, 1, ... 2 Dynamic programming 25 C t (-7r, x) is called the "cost-to-go" or cost from time t onwards when using the policy Jr and Xt = x. 2) J(Jr, x) = CO(Jr,x).