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Unformatted text preview: The Necessity of the Transversality Condition at Inﬁnity: A (Very) Special Case Peter Ireland∗ EC720.01 - Math for Economists Boston College, Department of Economics Fall 2010 Consider a discrete-time, inﬁnite horizon model that characterizes the optimal consumption of an exhaustible resource (like oil or coal). Let time periods be indexed by t = 0, 1, 2, ..., and let xt , t = 0, 1, 2, ..., denote the stock of the exhaustible resource that remains at the beginning of period t. Let ct , t = 0, 1, 2, ..., denote the amount of this resource that is consumed during period t. Since no new units of the resource are ever created, the amount consumed simply subtracts from the available stock according to xt − ct ≥ xt+1 for all t = 0, 1, 2, ..., where the inequality constraint (which will always bind at the optimum) simply recognizes that the resource can be freely disposed of. The optimization problem then involves choosing sequences {ct }∞ and {xt }∞ to maximize utility t=0 t=1 from consuming the resource over the inﬁnite horizon, given by ∞ ￿ β t ln(ct ), t=0 where the discount factor lies between zero and one, 0 < β < 1, subject to the constraint xt − ct ≥ xt+1 for all t = 0, 1, 2, ..., taking as given the level of the initial resource stock x0 > 0. Strictly speaking, we could also add nonnegativity constraints ct ≥ 0 for all t = 0, 1, 2, ... and xt ≥ 0 for all t = 1, 2, 3, ... to the statement of the problem, but the assumption of log utility, which implies that the marginal utility of consumption becomes inﬁnite as the level of consumption approaches zero, also implies that these constraints will never bind at the optimum. ∗ c Copyright ￿2010 by Peter Ireland. Redistribution is permitted for educational and research purposes, so long as no changes are made. All copies must be provided free of charge and must include this copyright notice. 1 We now know that there are at least three ways of deriving the necessary conditions describing a solution to this problem: using the Kuhn-Tucker theorem and the Lagrangian, using the maximum principle and the Hamiltonian, or using dynamic programming and the Bellman equation. If we choose to use the Kuhn-Tucker theorem, then we would start by deﬁning the Lagrangian for the problem as L= ∞ ￿ t β ln(ct ) + t=0 ∞ ￿ t=0 ˜ λt+1 (xt − ct − xt+1 ). This deﬁnition of the Lagrangian casts the problem in “present value” form, in the sense ˜ that λt measures the present value at t = 0 of having an additional unit of the resource available at the end of period t or the beginning of period t + 1. Alternatively, we can use the new variable ˜ λt+1 = β −t λt+1 , ˜ to replace λt with β t λt and write the Lagrangian in “current value” form as L= ∞ ￿ β t ln(ct ) + t=0 ∞ ￿ t=0 β t λt+1 (xt − ct − xt+1 ). According to the Kuhn-Tucker theorem, if the sequences {c∗ }∞ and {x∗ }∞ solve the t t=0 t t=1 dynamic optimization problem, then there exists a sequence {λ∗ }∞ of value for the t t=1 Lagrange multipliers such that together, these three sequences satisfy: a) The ﬁrst-order condition for c∗ , t βt − β t λ∗+1 = 0 t c∗ t or more simply 1 = λ∗+1 t c∗ t (1) for all t = 0, 1, 2, .... b) The ﬁrst-order condition for x∗ , t β t λ∗+1 − β t−1 λ∗ = 0 t t or more simply βλ∗+1 = λ∗ t t (2) x∗ − c∗ ≥ x∗+1 t t t (3) for all t = 1, 2, 3, .... c) The constraint 2 d) The nonnegativity condition λ∗+1 ≥ 0 t (4) for all t = 0, 1, 2, .... e) The complementary slackness condition λ∗+1 (x∗ − c∗ − x∗+1 ) = 0 t t t t (5) for all t = 0, 1, 2, .... Note that the ﬁrst-order condition (1) for consumption implies that λ∗+1 = t 1 >0 c∗ t and so, by extension, the complementary slackness condition (5) will always bind. Note also that we can use (1) to “solve out” for the Lagrange multipliers in (2), using λ∗+1 = t and λ∗ = t 1 c∗ t 1 c∗−1 t . Hence, the solution to the original dynamic optimization problem can be characterized by ﬁnding solutions to a system of two diﬀerence equations in the two unknown sequences {c∗ }∞ and {x∗ }∞ . The ﬁrst diﬀerence equation comes from the ﬁrst-order conditions t t=0 t t=1 and can be written as βλ∗+1 = λ∗ t t or or β 1 =∗ ∗ ct c t −1 c∗ = β c∗−1 t t or c∗+1 = β c∗ t t (6) This optimality condition can be interpreted as one that indicates that it is optimal to equate the marginal rate of intertemporal substitution β /c∗ t 1/c∗−1 t to the intertemporal price, which is ﬁxed at unity by the technological assumption that the exhaustible resource can be stored across periods without depreciation. 3 The second diﬀerence equation comes from the binding constraint and can be written as x∗+1 = x∗ − c∗ . t t t (7) We know that in general, two boundary conditions are needed to pin down a unique solution to this system of two diﬀerence equations. One boundary condition is the initial condition x0 given. (8) In a ﬁnite-horizon version of the problem, second boundary condition would be given by the complementary slackness condition on the nonnegativity constraint x∗ +1 ≥ 0 for T the terminal value of the stock, which we know from our more general analysis can be written as β T λ∗ +1 x∗ +1 = 0. (9) T T Moreover, in the ﬁnite-horizon version of the problem, we could show that this transversality condition will hold because the nonnegativity constraint on the terminal value of the stock binds at the optimum: x∗ +1 = 0. T Intuitively, with a ﬁnite horizon, if a strictly positive amount of the exhaustible resource remains at the end of period T , then a higher level of utility could be achieved by consuming that positive amount of the resource at one or more periods t = 0, 1, ..., T . For the inﬁnite-horizon version of the problem, our more general analysis suggests that the relevant terminal, or transversality, condition is given by lim β T λ∗ +1 x∗ +1 = 0. T T T →∞ (10) Notice that the ﬁrst-order condition (2) implies that β T λ∗ +1 is going to be constant at the T optimum. So in this special case, (10) will hold because lim x∗ +1 = 0. T T →∞ Intuitively, with an inﬁnite horizon, it will never be optimal for the stock of the exhaustible resource to be run all the way down to zero over any ﬁnite period of time, since that would entail zero consumption from that point onward. On the other hand, if the stock is not exhausted in the limit, then there is a sense in which the resource is not being consumed “fast enough,” in a way that parallels our argument for the ﬁnite-horizon case when some amount of the resource remains at the end of the horizon. This model is simple enough, in fact, that we can prove formally that (10) is a necessary condition for the inﬁnite-horizon case. The proof involves two steps. 4 Step one is to argue that {β t λ∗+1 x∗+1 }∞ is a nonincreasing sequence. To show this, note t t t=0 that for all t = 1, 2, 3, ..., (2), (3), (4), and the nonnegativity of c∗ imply that t β t λ∗+1 x∗+1 − β t−1 λ∗ x∗ = β t λ∗+1 (x∗+1 − x∗ ) ≤ β t λ∗+1 (x∗ − c∗ − x∗ ) = −β t λ∗+1 c∗ ≤ 0. t t tt t t t t t t t t t Step two is to argue that inf β t λ∗+1 x∗+1 = 0. t t t To show this, suppose to the contrary that there exists an ε > 0 such that β t λ∗+1 x∗+1 ≥ ε t t for all t = 0, 1, 2, .... Since (1) implies that λ∗+1 > 0 for all t = 0, 1, 2, ..., (2) implies t that this last condition requires that x∗+1 ≥ γ t for all t = 0, 1, 2, ..., where γ equals ε divided by the constant, positive value of β t λ∗+1 t along the optimal path. But, in this case, we can deﬁne new sequences {c∗ }∞ and t t=0 {x∗∗ }∞ with t t=1 c∗∗ = c∗ + γ , 0 0 c∗∗ = c∗ for all t = 1, 2, 3, ..., t t and x∗∗ = x∗ − γ for all t = 1, 2, 3, ... t t that satisfy all of the constraints from the original problem, but yield a higher level of utility, contradicting the assumption that {c∗ }∞ and {x∗ }∞ solve the problem. t t=0 t t=1 Taken together, {β t λ∗+1 x∗+1 }∞ and inf t β t λ∗+1 x∗+1 = 0 require that (10) hold at the optit t t=0 t t mum, completing the proof. The proof turns out to be relative straightforward for this simple problem, but becomes much more diﬃcult to apply in other cases that are only slightly more complicated. Even for the Ramsey model with log utility and Cobb-Douglas production, for instance, this proof does not generalize. For a much more elaborate proof that does apply to that version of the Ramsey model, see Ivar Ekeland and Jose Alexandre Scheinkman, “Transversality Conditions for Some Inﬁnite Horizon Discrete Time Optimization Problems,” Mathematics of Operations Research, Vol. 11 (May 1986), pp.216-229. For the sake of completeness, let’s wrap up by completely characterizing the solution to the original dynamic optimization problem. Once again, that solution must satisfy the diﬀerence equations c∗+1 = β c∗ t t 5 (6) and x∗+1 = x∗ − c∗ t t t (7) x0 given (8) lim β T λ∗ +1 x∗ +1 = 0 T T (10) together with the initial condition and the transversality condition T →∞ where the latter requires that lim x∗ +1 = 0. T T →∞ To start, consider (7), which implies x∗ = x0 − c∗ , 1 0 x∗ = x∗ − c∗ = x0 − c∗ − c∗ , 2 1 1 0 1 x∗ = x∗ − c∗ = x0 − c∗ − c∗ − c∗ 3 2 2 0 1 2 or, after repeating T times, x∗ T = x0 − T −1 ￿ c∗ t t=0 or, after repeating inﬁnitely many times and using the transversality condition 0 = x0 − Rewritten as ∞ ￿ ∞ ￿ c∗ . t t=0 c∗ = x0 , t t=0 this result conﬁrms our intuition about the implications of the transversality condition: it shows that over the inﬁnite horizon, it is optimal to consume the entire resource stock, otherwise, a higher level of utility could be achieved while still satisfying all of the constraints. Now use (6), which implies that c∗ = β t c∗ t 0 to pin down the level of the consumption path from x0 = ∞ ￿ c∗ t = c∗ 0 t=0 ∞ ￿ βt = t=0 c∗ 0 . 1−β Evidently, it is optimal to start by consuming c∗ = (1 − β )x0 0 at t = 0 and then to allow consumption to decrease proportionally according to (6) for all t = 1, 2, 3, .... 6 ...
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