<p>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.</p>
<p>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.</p>
<p>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 &quot;Introduction to Modern Economic Growth.&quot;</p>

CC BY-SA 4.0
2024-05-12T18:40:31.000Z
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?
bruno
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2024-05-12T18:56:53.017Z
Essentially, yes. One usually makes assumptions that guarantee that both theorems apply.
Michael Greinecker
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2024-05-12T19:01:00.623Z
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.
bruno
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2024-05-12T20:10:01.543Z
AK models are some kind of endogenous growth models. But I don't remember much more about them.
Michael Greinecker
CC BY-SA 4.0
2024-05-12T18:37:03.430Z
2024-05-12T18:37:03.430Z
Michael Greinecker
<p>Is there any difference between solving the social planner's problem or the representative household's in the Ramsey-Cass-Koopmans model?</p>
<p>The social planner chooses a plan <span class="math-container">$\left\{c_t, l_t, k_{t+1}\right\}_{t=0}^{\infty}$</span> so as to maximize utility subject to the resource constraint of the economy, taking initial <span class="math-container">$k_0$</span> as given:
<span class="math-container">$$
\begin{gathered}
\max \mathcal{U}_0=\sum_{t=0}^{\infty} \beta^t U\left(c_t, 1-l_t\right) \\
c_t+k_{t+1} \leq(1-\delta) k_t+F\left(k_t, l_t\right), \quad \forall t \geq 0 \\
c_t \geq 0, \quad l_t \in[0,1], \quad k_{t+1} \geq 0 ., \quad \forall t \geq 0 \\
k_0&gt;0 \text { given }
\end{gathered}
$$</span>
This is called the social planner's problem.</p>
<p>Given a price sequence <span class="math-container">$\left\{R_t, w_t\right\}_{t=0}^{\infty}$</span>, household <span class="math-container">$j$</span> chooses a plan <span class="math-container">$\left\{c_t^j, l_t^j, k_{t+1}^j\right\}_{t=0}^{\infty}$</span> so as to maximize lifetime utility subject to its budget constraints
<span class="math-container">$$
\begin{gathered}
\max \mathcal{U}_0^j=\sum_{t=0}^{\infty} \beta^t U\left(c_t^j, 1-l_t^j\right) \\
\text { s.t. } c_t^j+a_{t+1}^j \leq\left(1+R_t\right) a_t^j+w_t l_t^j \\
c_t^j \geq 0, \quad l_t^j \in[0,1], \quad a_{t+1}^j \geq \underline{a}_{t+1}
\end{gathered}
$$</span>
This is called the household's problem.</p>

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2024-05-12T17:51:18.387Z
Can you write down what these problems are meant to be?
Michael Greinecker
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2024-05-12T18:15:13.267Z
@MichaelGreinecker I edited.
bruno
CC BY-SA 4.0
2024-05-12T21:43:50.970Z
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(?)
Brennan
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2024-05-12T21:53:43.900Z
@Brennan yes but under some conditions I think
bruno
CC BY-SA 4.0
2024-05-13T04:50:08.000Z
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.
Brennan
CC BY-SA 4.0
2024-05-13T07:40:35.410Z
I think "having riskless bonds isn't necessarily the same as having access to capital" makes sure that the first theorem of welfare economics holds, because of the following reasons: - Markets are complete.
- Perfect information. But I am not sure whether I am right or not.
bruno
<macroeconomics><economic-growth><endogenous-growth>
Social Planner vs Representative Household
bruno
# Social Planner vs Representative Household

(Asked by: username_0 on 2024-05-12)

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.

**Tags:** macroeconomics, economic-growth, endogenous-growth

> **Comments:**
> **username_1**: Can you write down what these problems are meant to be?  
> **username_0**: @username_1 I edited.  
> **username_2**: 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(?)  
> **username_0**: @username_2 yes but under some conditions I think  
> **username_2**: 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.  
> **username_0**: I think "having riskless bonds isn't necessarily the same as having access to capital" makes sure that the first theorem of welfare economics holds, because of the following reasons: - Markets are complete. - Perfect information. But I am not sure whether I am right or not.  

---

## 1 answers

### Answer by username_1 on 2024-05-12. Score: 2 | Accepted answer

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."

> **Comments:**
> **username_0**: 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?  
> **username_1**: Essentially, yes. One usually makes assumptions that guarantee that both theorems apply.  
> **username_0**: 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.  
> **username_1**: AK models are some kind of endogenous growth models. But I don't remember much more about them.  

---

