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Is there any difference between solving the social planner's problem or the representative household's in the Ramsey-Cass-Koopmans model?

The social planner chooses a plan $\left\{c_t, l_t, k_{t+1}\right\}_{t=0}^{\infty}$ so as to maximize utility subject to the resource constraint of the economy, taking initial $k_0$ as given: $$ \begin{gathered} \max \mathcal{U}_0=\sum_{t=0}^{\infty} \beta^t U\left(c_t, 1-l_t\right) \\ c_t+k_{t+1} \leq(1-\delta) k_t+F\left(k_t, l_t\right), \quad \forall t \geq 0 \\ c_t \geq 0, \quad l_t \in[0,1], \quad k_{t+1} \geq 0 ., \quad \forall t \geq 0 \\ k_0>0 \text { given } \end{gathered} $$ This is called the social planner's problem.

Given a price sequence $\left\{R_t, w_t\right\}_{t=0}^{\infty}$, household $j$ chooses a plan $\left\{c_t^j, l_t^j, k_{t+1}^j\right\}_{t=0}^{\infty}$ so as to maximize lifetime utility subject to its budget constraints $$ \begin{gathered} \max \mathcal{U}_0^j=\sum_{t=0}^{\infty} \beta^t U\left(c_t^j, 1-l_t^j\right) \\ \text { s.t. } c_t^j+a_{t+1}^j \leq\left(1+R_t\right) a_t^j+w_t l_t^j \\ c_t^j \geq 0, \quad l_t^j \in[0,1], \quad a_{t+1}^j \geq \underline{a}_{t+1} \end{gathered} $$ This is called the household's problem.

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  • $\begingroup$ Can you write down what these problems are meant to be? $\endgroup$ Commented May 12 at 17:51
  • $\begingroup$ @MichaelGreinecker I edited. $\endgroup$
    – bruno
    Commented May 12 at 18:15
  • $\begingroup$ Typically the social planner's problem includes welfare weights such that your utility function would be: $$\sum_{i=1}^n \lambda_i \sum_{t=0}^\infty \beta^t U(c_t^i, 1-l_t^i)$$ Though I think this would be isomorphic to maximizing each of their individual lifetime utilities(?) $\endgroup$
    – Brennan
    Commented May 12 at 21:43
  • $\begingroup$ @Brennan yes but under some conditions I think $\endgroup$
    – bruno
    Commented May 12 at 21:53
  • $\begingroup$ Symmetric weights $\lambda_i = \lambda_j$ for $i\neq j$, surely. Not sure what else is required. Also, having riskless bonds isn't necessarily the same as having access to capital. For example you ensure capital is non-negative but are agnostic about the lower bound of assets, which could be negative implying the individual can borrow. I imagine this is just a mistake though, $R_t$ can just be the interest less the depreciation. $\endgroup$
    – Brennan
    Commented May 13 at 4:50

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If we assume that there is a single household (there are some subtleties in how one defines a representative household), then the Pareto optimality is the same as solving the planner's problem.

The first welfare theorem gives conditions under which every competitive equilibrium is Pareto optimal, and the second welfare theorem gives conditions under which every Pareto optimum is a competitive equilibrium after a suitable redistribution of endowments. With one agent, there is nothing to redistribute.

There are two caveats in how the welfare theorems differ from the usual models with finitely many commodities known from microeconomics courses. To make the proof of the first welfare theorem go through, you have to be able to rewrite the recursive problem as a sequence problem. In particular, prices have to be such that total values are finite. For the second welfare theorem, one needs some more technical assumptions to ensure that the infinite-dimensional version of the separating hyperplane theorem, the geometric Hahn-Banach theorem, applies. For suitable assumptions and a proof, you can take a look at Chapter 5 of Acemoglu's "Introduction to Modern Economic Growth."

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  • $\begingroup$ So, the two approaches are identical, because there are no market imperfections, so the first welfare theorem holds: the competitive, decentralized equilibrium is a solution to the planner problem? $\endgroup$
    – bruno
    Commented May 12 at 18:40
  • $\begingroup$ Essentially, yes. One usually makes assumptions that guarantee that both theorems apply. $\endgroup$ Commented May 12 at 18:56
  • $\begingroup$ I have another question actually, this is out of context but what is the difference between the AK and Ramsey Models, is AK a version of Ramsey with a specific production function? Acemoglu has AK in his book but not Ramsey. $\endgroup$
    – bruno
    Commented May 12 at 19:01
  • $\begingroup$ AK models are some kind of endogenous growth models. But I don't remember much more about them. $\endgroup$ Commented May 12 at 20:10

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