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A two agent planner problem where agents receive stochastic endowments, and the planner aims to find constrained efficient allocations while respecting incentive compatibility constraints. The problem involves calculating transition probabilities, agent preferences, and deriving expressions for the planner's problem and the agents' value functions. The document also discusses the use of a grid and iterative methods to find optimal decision rules and value functions.
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Consider the two agent planner’s problem studied by Kocherlakota. Suppose the two agents receive stochastic endowment streams, such that the aggregate endowment is 1 in every period. Endowments cannot be stored or disposed of. Each period, agent 1’s endowment is drawn from a set S = { 0. 5 − ε, 0. 5 , 0 .5 + ε}. Denote the elements of this set s 1 , s 2 and s 3. Let st denote the draw from S at date t. Let st^ = (s 0 , s 1 , s 2 , ...st) ∈ St^ denote the history of the economy up to and including date t. Denote agent 1’s endowment y(st). Thus agent 2’s endowment is 1 − y(st). Let the date zero probability of history st^ be denoted π(st). Assume that π(st|st−^1 ) = π(st|st− 1 ) (i.e. productivity shocks are first order Markov). Assume that transition probabilities are given by the matrix:
Γ =
q 1 − q 0 1 −p 2 p^
1 −p 2 0 1 − q q
where the element Γi,j is equal to π(sj |si), and p ∈ (0, 1) , q ∈ (0, 1). Assume that preferences for agent 1 are given by
t β
t P st^ π(s
t) log(c(st)).
Preferences for agent 2 are similar. Note that in equilibrium, the consumption of the type two agent is given by 1 − c(st). Assume that the planner cares about both types equally.
(a) Suppose β = 0. 96 , p = 0. 9 , and ε = 0. 25. For these parameter values, compute the set of possible values for Vi(st).
(b) Formulate (as a Lagrangian) the planner’s problem now that the planner has to worry about the commitment problem. Use the Marcet and Marimon trick to formulate the problem in terms of the sum of the past values of the multipliers on the participation constraints. Take first order conditions. Now you will solve the model recursively, as we outlined in class. The state variable will be xt =
z(st−^1 ), st
where z(st−^1 ) is the ratio of the sum of the values of multipliers for the history st−^1. Since neither t nor st^ is a state variable in the recursive formulation, we can simply write x = (z− 1 , s). (c) Create a grid on z− 1. Let Z = (z 1 ,z 2 , ...zn) denote the n points on this grid. To start, set n = 11. To space the grid points “evenly” so that zi < 1 , i ≤ 5 , z 6 = 1, zj > 1 , j ≥ 6 , use an exponential spacing formula:
zi = zi− 1
μ zn z 1
¶ (^) n− (^11) 1 < i < n
How should one choose the end points for this grid? (hint: think about the possible range for the ratio of marginal utilities in autarky) (d) Now you need initial guesses for decision variables, for the law of motion of the state variables, and for the value functions Wi(x). Use the problem without enforcement constraints to produce these initial guesses given the parameter values above: W (^) i^0 (x), z^0 (x), c^0 (x), v i^0 (x) ∀x ∈ S × Z. (v i^0 (x) is the ratio of the multiplier on agent i’s incentive compatibility constraint to the sum of multipliers on this constraint).
(a) Take x 1 the first point on the grid over X = Z × S. Check whether agent 1 has an incentive to default at this grid point. If he does set solve numerically for c^1 (x 1 ) and z^1 (x 1 ) to satisfy (i) the first order condition for c, and (ii) the incentive compatibility constraint for agent 1 with equality, where current period utility is given by c^1 (x 1 ) and continuation utility for each possible s^0 is given by W 10 (z^1 (x 1 ), s^0 ). Note that z^1 (x 1 ) does not necessarily lie on the grid on Z. Thus you will want to interpolate (linearly) to evaluate the function W 10 (., s^0 ) in between grid points. (b) If agent 1 does not want to default, check incentive compatibility for agent 2. If agent 2 wants to default, solve for c^1 (x 1 ) and z^1 (x 1 ) following an analogous procedure to the one above. (c) If neither agent wants to default, z^1 (x 1 ) ≡ z^1 (z− 1 = z 1 , s = s 1 ) = z 1 , v 11 (x 1 ) = v 21 (x 1 ) = 0, and c^1 (x 1 ) satisfies the FOC for c. (d) Compute and store the new guess for the value functions
W 11 (x 1 ) = u(c^1 (x 1 )) + β
s^0
π(s^0 |s 1 )W 10 (z^1 (x 1 ), s^0 )