# Arrow-Debreu Pricing, Planner's Problem

Consider a two period, single good, $2$ agent model. Time beings in perios $0$ in a known state (state $0$) but in period $1$ the world may find itself in any one of two states $s = 1,2$ with probabilities $\pi_s = 1/2$ for each state. Both of the consumers agree on the probabilities.

Each consumer as a constant relative risk aversion utility function with utility index $u^{i}(c_{s}^{i}) = \frac{(c^{i}_{s})^{1-\gamma^i}-1}{1-\gamma^i}$, $i = A,B$. Specifically, we assume that $\gamma^{A} > 0$ and $\gamma^{B} = 0$.

In addition to the consumption good there are $J=2$ financial securities with period $1$ payoffs $D$ (a $J\times S$ matrix) given by $$D = \begin{pmatrix} 1 & 0\\ -1 & 4\\ \end{pmatrix}$$ Thus, the first security pays off $1$ in both states and the second security pays off $-1$ in state $1$ and $4$ in state $2$. There is a spot market for these securities in period $0$ at prices $p_j > 0$. Each security is in zero net supply so that if one consumer is a buyer then the other consumer must be the seller. There are no short-sell constraints on the securities.

Consumers have no period zero endowment and do not consume in the first period so that $e_0^{i} = c_{0}^{i} = 0$, $i = A,B$.

The consumers are not allowed to short-sell the consumption good. I.e., we impost the restriction that $c_{s}^{i}\geq 0$,$\forall i = A,B$ and $s = 1,2$. There is no production in this economy so each agent is exogenously endowed with period $1$ endowments of: $e^{A} = (6,2)$ and $e^{B} = (6,6)$ in states $s = 1$, and $2$, for each agent respectively.

a.) Carefully write out the second consumer's maximization problem including the relevant budget constraints. Let $\lambda_{s}^{B}$ be the Lagrange multipliers on the budget constraints. Carefully write out the first-order conditions for this consumer including the inequality and complementary slackness conditions.

b.) Solve for the contract curve for this economy.

c.) Verify that the Arrow security prices for this economy are $q_1 = q_2$ and that the optimal allocation of goods is $c^{A} = (4,4)$ and $c^{B} = (8,4)$.

d.) Find the consumers' security portfolios $\theta^{A}$ and $\theta^{B}$.

e.) Normalize the Arrow security prices so that $q_1 = q_2$ and find the security prices $p_1$ and $p_2$.

Solution a.) The utility maximization problem for $B$ is \begin{align*} \max_{c^{B},\theta}\mathbb{E}\left[u(c^{B})\right] = \frac{1}{2}(c_1^{B} - 1) + \frac{1}{2}(c_2^{B} - 1) \ \ \text{s.t.} \ \ &p_1\theta_{1}^{B} + p_2\theta_{2}^{B} = 0, s = 0\\ &c_1^{B}\leq 6 + \theta_{1}^{B} - \theta_{2}^{B}, s = 1\\ &c_{2}^{B}\leq 6 + \theta_{1}^{B} + 4\theta_{2}^{B}, s =2; c_1^{B}\geq 0, c_2^{B}\geq 0 \end{align*} We have the Lagrangian, $$\mathcal{L}(c^{B},\theta^{B},\lambda^{B}) = \frac{1}{2}(c_1^{B} - 1) + \frac{1}{2}(c_2^{B} - 1) + \lambda_{0}^{B}(0- p_1 \theta_1^{B} - p_2 \theta_2^{B}) + \lambda_{1}^{B}(6 + \theta_1^{B} - \theta_2^{B} - c_1^{B}) + \lambda_2^{B}(6 + \theta_{1}^{B} + 4\theta_2^{B} - c_2^{B})$$ The First-order conditions are: \begin{align*} &\lambda_{0}^{B}: p_1\theta_1^{B} + p_2 \theta_{2}^{B} = 0; \lambda_{0}^{B}\geq 0\\ &\lambda_1^{B}: c_1^{B}\leq 6 + \theta_1^{B} - \theta_2^{B};\lambda_1^{B}\geq 0, \lambda_1^{B}(6+\theta_1^{B} - \theta_2^{B} - c_1^{B}) = 0\\ &\lambda_2^{B}: c_2{B} \leq 6+\theta_{1}^{B} + 4\theta_2^{B}, \lambda_2^{B}\geq 0, \lambda_2^{B}(6+\theta_{1}^{B} + 4\theta_2^{B} - c_2^{B}) = 0\\ &c_1^{B}: \frac{1}{2}-\lambda_1^{B}\leq 0; c_1^{B}\geq 0, c_1^{B}\left(\frac{1}{2} - \lambda_{1}^{B}\right) = 0\\ &c_2^{B}: \frac{1}{2}-\lambda_2^{B}\leq 0; c_2^{B}\geq 0, c_2^{B}\left(\frac{1}{2} - \lambda_{2}^{B}\right) = 0\\ &\theta_1^{B}: - \lambda_0^{B}p_1 + \lambda_1^{B} + \lambda_2^{B} = 0\\ &\theta_2^{B}: - \lambda_0^{B}p_2 - \lambda_1^{B} + 4\lambda_2^{B} = 0 \end{align*}

Solution b.) We have $$c_1^{A} + c_2^{B} = 12 \ \ \text{and} \ \ c_2^{A} + c_2^{B} = 8$$ We want to find the set of pareto optimal points (contact curve). We are looking for $$MRS^{A} = MRS^{B}$$ Note that $u^{i} = \pi_1 u(c_1^{i}) + \pi_2 u(c_2^{i})$ Thus, $$MRS^{i} = \frac{\pi_1 u^{i \prime}(c_1^{i})}{\pi_2 u^{i\prime}(c_2^{i})}$$ $$MRS^{A} = \left(\frac{c_1^{A}}{c_2^{A}}\right)^{-\gamma_A} = 1 = MRS^{B}$$ $c_1^{A} = c_2^{A}$ is the equation of the contract curve.

Attempted solution c.) This is where I am stuck, I know that we have to solve for the Planner's problem: $$\mathcal{L} = \sum_{i=A,B} \eta^{i}u^{i}(c_1^{i}) + \lambda_1\left(\sum_{i}e_1^{i} - \sum_{i}c_1^{i}\right) + \lambda_2\left(\sum_{i}e_2^{i} - \sum_{i}c_2^{i}\right)$$ but I am not sure how to proceed any further. Any suggestions would be greatly appreciated. This is my first time solving a planners problem and I am a Mathematics grad student so do not have much exposure to economics.

• Why do you think you need to solve the planner's problem to solve (c)? I think you need to show that the proposed prices and allocations are consistent with equilibrium. That is, the allocations are feasible and, given the prices, solve each consumer's utility maximisation problem. Mar 3, 2017 at 3:10
• Well my professor said that we should solve it, since similar problems ask to solve the Planner's problem Mar 3, 2017 at 3:11
• I don't see how the planner's problem should give you the state (Arrow security) prices, but that might just be my having forgotten how to solve GE models with securities markets. In any case -- why are you not able to solve the constrained optimisation as you did with the second consumer's problem? Mar 3, 2017 at 3:17
• You should post an answer to your own question. You should also be able to accept your own answer as well. Mar 5, 2017 at 15:41
• @TheoreticalEconomist Ok will do momentarily Mar 5, 2017 at 15:42

After many emails, my professor conceded to posting a solution to this problem, the funny thing is that he does not use the confusing $\mathcal{L}$ equation he wrote previously in class.
Solution c.) We have the Planner's problem $$\max_{c^{A},c^{B}}\{\mathbb{E}[u^{A}(c^{A})] + \eta \mathbb{E}[u^{B}(c^{B})]$$ such that the resource constraints \begin{align*} c_1^{A} + c_1^{B} &= e_1^{A}+e_1^{B} = 12\\ c_2^{A} + c_2^{B} &= e_2^{A} + e_2^{B} = 8 \end{align*} $\eta$ is the Negishi weights - which we will deal with later. Thus we have, $$\max_{c_1^{A},c_2^{A}}\{\left[ \frac{1}{2}\frac{(c_1^{A})^{1- \gamma^{A} -1}}{1 - \gamma^{A}} + \frac{1}{2}\frac{(c_2^{A})^{1 - \gamma^A}-1}{1-\gamma^{A}}\right] + \eta\left[\frac{1}{2}((12-c_1^{A})-1) + \frac{1}{2}((8- c_2^{A}) - 1)\right]\}$$ We have the FOC: \begin{align*} &c_1^{A}: \frac{1}{2}(1 - \gamma^{A})(c_1^{A})^{-\gamma^{A}} - \frac{1}{2}\eta = 0\\ &c_2^{A}: \frac{1}{2}(1 - \gamma^{A})(c_2^{A})^{-\gamma^{A}} - \frac{1}{2}\eta = 0\\ \end{align*} We see that $c_1^{A} = c_2^{A}$. In order to solve the problem for $\eta$, the solution must also satisfy the "lifetime budget constraint" of each agent. Since there are only $2$ agents, it is sufficient to solve the problem for just $1$ agent: $$\Rightarrow \sum_{s=1}^{2}u^{i\prime}(c_s^{i})c_s^{i} = \sum_{s}u^{i\prime}(c_{s}^{i})e_{s}^{i} , \ i = A , B$$ The first sum is the marginal value of consumption and the second sum is the marginal value of endowments. The lifetime budget constraints implies for $A: u^{A\prime}(c_1^{A})c_1^{A} + u^{A\prime}(c_2^{A}) = u^{\prime}(c_1^{A})e_1^{A} + u^{\prime}(c_2^{A})e_2^{A}$ but since $c_1^{A} = c_2^{A}$, $u^{A\prime}(c_1^{A}) = u^{A\prime}(c_2^{A})$. So we get $$c_1^{A} + c_2^{A} = e_1^{A} + e_2^{A} = 8 \Rightarrow c_1^{A} = c_2^{A} = 4 \Rightarrow c_1^{B} = 12 - c_1^{A} = 8 \ \text{and} \ c_2^{B} = 8 - c_1^{A} = 4$$ Note from the FOC's, we have $$\eta = (1-\gamma^{A})(c_1^{A})^{-\gamma^{A}})$$ So if we specify $\gamma^{A}$, we could solve for the specific value of $\eta$. In our example we don't need to do this since $\gamma^{B} = 0$ is a degenerate case.
Solution d.) From the FOC's in part (a) we have \begin{align*} 6 + \theta_1^{B} - \theta_2^{B} &= c_1^{B} = 8\\ 6 + \theta_1^{B} + 4\theta_2^{B} = c_2^{B} = 4 \end{align*} Thus we have $\theta_1^{B} = 6/5$ and $\theta_2^{B} = -4/5$. Since $\theta_{s}^{A} + \theta_{s}^{B} = 0$, $s= 1,2$ (zero net supply) We get $$\theta_1^{A} = -6/5 \ \text{and} \ \theta_2^{A} = -4/5$$
Solution e.) We know that $$p = Dq \Rightarrow \begin{pmatrix} p_1\\ p_2 \end{pmatrix} = \begin{pmatrix} 1 & 1\\ -1 & 4\\ \end{pmatrix}\begin{pmatrix} q_1\\ q_2 \end{pmatrix}$$ Then \begin{align*} p_1 &= q_1 + q_2\\ p_2 &= -q_1 + 4q_2 \end{align*} We know that $q_1 = q_2$ but we do not know the sum because there was no period zero consumption in this model from which to price the bond. Let's choose $q_1 + q_2 = 1$ then $q_1 = q_2 = 1/2$ so $p_1 = 1$ and $p_2 = 3/2$.