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The social planner problem is \begin{align} \max_{\lbrace x_i,y_i,x_j,y_j \rbrace } &\theta_i U_i(x_i,y_i) +\theta_j U_i(x_j,y_j) \\ & \text{s.t.}\\ &x_i+x_j = \omega_i^x + \omega_j^x\\ &y_i+y_j = \omega_i^y + \omega_j^y\\ \end{align} where $\theta_i,\theta_j$ are weights.

As I understand it, with a proper choice of weights, one can turn the Social Planner's problem into

  1. A Lagrangian optimization problem that solves for a Pareto optimal allocation.
  2. A Lagrangian optimization problem that solves for an allocation that is a Walrasian equilibrium.

Is this accurate? If so, what constraints are there on the weights for each case?

Stan Shunpike
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  • Is there any particular reason why there are different wealths set aside for good x and good y for agents i and j? – Kitsune Cavalry Oct 23 '15 at 23:52
  • @KitsuneCavalry I think these are the initial endowments for the Walrasian equilibrium. – Giskard Oct 24 '15 at 06:42
  • This statement seems accurate (given the usual conditions on $U$) and you can arrive at the 1. conclusion by straightforward calculation, and at the 2. by thinking a little bit about the calculation used in the 1. case. – Giskard Oct 24 '15 at 06:44
  • I'm trying to think if property one holds if the utility function for either agent is locally satiated. I think it does, but I can't work out the Lagrangian in my head. – Kitsune Cavalry Oct 24 '15 at 07:07
  • If you work out the first order conditions, you can combine the conditions and obtain two sets of conditions, the interpersonal equity conditions and the Pareto optimality conditions. The Pareto optimality conditions can be derived no matter what weights you choose. For the latter question, I then recommend the second welfare theorem. – HRSE Oct 26 '15 at 10:36

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