<p>Start by writing the Lagrangian (as it sounds like you have already done):</p>

<p><span class="math-container">$$L=\sum_{t=0}^{\infty} \{\beta^t(\ln C_t^y + \ln C_t^o) + \lambda_t (C_t^y +\frac{C_t^o}{1+n} +k_{t+1}(n+1)-k_t-f(k_t))\}.$$</span></p>

<p>First order conditions are given by 
<span class="math-container">$$\frac{\partial L}{\partial C_t^y} = \frac{\beta^t}{C_t^y} +\lambda_t = 0$$</span>
<span class="math-container">$$\frac{\partial L}{\partial C_t^o} = \frac{\beta^t}{C_t^o} +\frac{\lambda_t}{1+n} = 0$$</span>
<span class="math-container">$$\frac{\partial L}{\partial k_{t+1}} = \lambda_{t-1}(1+n) -  \lambda_t(1 + f'(k_t)) = 0$$</span>
<span class="math-container">$$\frac{\partial L}{\partial \lambda_{t}} = C_t^y +\frac{C_t^o}{1+n} +k_{t+1}(n+1)-k_t-f(k_t)=0.$$</span></p>

<p>First notice the first two equations give us 
<span class="math-container">$$\frac{C_t^y}{C_t^o} = 1+n.$$</span>
Also notice that the first equation implies <span class="math-container">$\lambda_t = -\frac{\beta^t}{C_t^y}$</span>. Substituting this into our third equation gives 
<span class="math-container">$$\frac{\beta^{t+1}}{C_{t+1}^y}(1+n) = \frac{\beta^{t}}{C_{t}^y}(1+n)(1 + f'(k_t)$$</span>
or equivalently
<span class="math-container">$$\frac{\beta C_t^n}{C_{t+1}^y} = (1 + f'(k_t)).$$</span></p>

<p>Finally, we can write our system of four equations with three equations, having removed <span class="math-container">$\lambda$</span> from the system:
<span class="math-container">$$\frac{C_t^y}{C_t^o} = 1+n$$</span>
<span class="math-container">$$\frac{\beta C_t^y}{C_{t+1}^y} = (1 + f'(k_t)).$$</span>
<span class="math-container">$$C_t^y +\frac{C_t^o}{1+n} +k_{t+1}(n+1)-k_t-f(k_t)=0.$$</span></p>

<p>Dropping time subscripts, i.e. applying the steady-state condition, gives
<span class="math-container">$$\frac{C^y}{C^o} = 1+n$$</span>
<span class="math-container">$$\beta = (1 + f'(k)).$$</span>
<span class="math-container">$$C^y +\frac{C^o}{1+n} +nk-f(k)=0.$$</span></p>

<p>Hence, you are left with a system of three equations with three unknowns <span class="math-container">$C^y$</span>, <span class="math-container">$C^o$</span>, and <span class="math-container">$k$</span>. Solve the system to arrived at a steady-state solution.</p>

CC BY-SA 4.0
2019-03-15T21:14:06.640Z
Very clear! Thanks for the answer, it was very helpful
Chaos
CC BY-SA 4.0
2019-03-15T21:15:29.043Z
Good luck with your PhD studies (or very advanced undergraduate studies)!
dlnB
CC BY-SA 4.0
2019-03-15T04:19:20.090Z
2019-03-15T19:11:38.053Z
2019-03-15T19:11:38.053Z
dlnB
<p>Assume we have a model of OVG where there are 2 overlapping generations, <em>youngs</em> and <em>olds</em>, the agents are two period living. The utility function is logaritmic, and the production function is Cobb-Douglas.</p>

<p>I have to show that the solution for the social planner does not coincide with the equilibrium from the market.
The social planner solves:
<span class="math-container">$$\max  \ln(c_t^y)+\beta \ln(c_{t}^o)$$</span>
<span class="math-container">$$s.t.$$</span>
<span class="math-container">$$c_t^y+\frac{c_t^o}{1+n}+k_{t+1}(1+n)-k_t=f(k_t)$$</span>
Where the units are expressed in per capita terms and <span class="math-container">$n$</span> represents the population growth rate.
 The FOC for the social planner is 
<span class="math-container">$$\frac{c_{t}^o}{c_t^y}=\beta(1+n)$$</span>
We know that in steady state the level of consumption is actually constant i.e.
<span class="math-container">$c_t^o=c^o$</span> and <span class="math-container">$c^y_t=c^y$</span>.
What I am not getting is if I should replace this on the constraint in order to obtain the values of S.S. or if I need to take the FOC with respect to the <span class="math-container">$k$</span>.
I am a little bit lost, since I've already solved the problem for the market economy but I am having a hard time figuring out how to proceed.
Any hint will be greatly appreciated.</p>

CC BY-SA 4.0
2019-03-15T18:27:21.233Z
The problem is that you are choosing three variables, not two, (plus the multiplier) i.e. you must choose $C_t^y,C_t^o,$ and $k_{t+1}$ (and of course for $\lambda_t$). This will give you four first-order conditions rather than 3. Eliminate $\lambda$ from the system and then impose steady state. See my solution below.
dlnB
<macroeconomics><economic-growth><steady-state>
Overlapping Generations model: Social Planner solution
Chaos
# Overlapping Generations model: Social Planner solution

(Asked by: username_0 on 2019-03-12)

Assume we have a model of OVG where there are 2 overlapping generations, *youngs* and *olds*, the agents are two period living. The utility function is logaritmic, and the production function is Cobb-Douglas.

I have to show that the solution for the social planner does not coincide with the equilibrium from the market.
The social planner solves:
$$\max \ln(c\_t^y)+\beta \ln(c\_{t}^o)$$
$$s.t.$$
$$c\_t^y+\frac{c\_t^o}{1+n}+k\_{t+1}(1+n)-k\_t=f(k\_t)$$
Where the units are expressed in per capita terms and $n$ represents the population growth rate.
The FOC for the social planner is
$$\frac{c\_{t}^o}{c\_t^y}=\beta(1+n)$$
We know that in steady state the level of consumption is actually constant i.e.
$c\_t^o=c^o$ and $c^y\_t=c^y$.
What I am not getting is if I should replace this on the constraint in order to obtain the values of S.S. or if I need to take the FOC with respect to the $k$.
I am a little bit lost, since I've already solved the problem for the market economy but I am having a hard time figuring out how to proceed.
Any hint will be greatly appreciated.

**Tags:** macroeconomics, economic-growth, steady-state
> **Comments:**
> **username_1**: The problem is that you are choosing three variables, not two, (plus the multiplier) i.e. you must choose $C_t^y,C_t^o,$ and $k_{t+1}$ (and of course for $\lambda_t$). This will give you four first-order conditions rather than 3. Eliminate $\lambda$ from the system and then impose steady state. See my solution below.  

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## 1 answers

### Answer by username_1 on 2019-03-15. Score: 1 | Accepted answer

Start by writing the Lagrangian (as it sounds like you have already done):

$$L=\sum\_{t=0}^{\infty} \{\beta^t(\ln C\_t^y + \ln C\_t^o) + \lambda\_t (C\_t^y +\frac{C\_t^o}{1+n} +k\_{t+1}(n+1)-k\_t-f(k\_t))\}.$$

First order conditions are given by
$$\frac{\partial L}{\partial C\_t^y} = \frac{\beta^t}{C\_t^y} +\lambda\_t = 0$$
$$\frac{\partial L}{\partial C\_t^o} = \frac{\beta^t}{C\_t^o} +\frac{\lambda\_t}{1+n} = 0$$
$$\frac{\partial L}{\partial k\_{t+1}} = \lambda\_{t-1}(1+n) - \lambda\_t(1 + f'(k\_t)) = 0$$
$$\frac{\partial L}{\partial \lambda\_{t}} = C\_t^y +\frac{C\_t^o}{1+n} +k\_{t+1}(n+1)-k\_t-f(k\_t)=0.$$

First notice the first two equations give us
$$\frac{C\_t^y}{C\_t^o} = 1+n.$$
Also notice that the first equation implies $\lambda\_t = -\frac{\beta^t}{C\_t^y}$. Substituting this into our third equation gives
$$\frac{\beta^{t+1}}{C\_{t+1}^y}(1+n) = \frac{\beta^{t}}{C\_{t}^y}(1+n)(1 + f'(k\_t)$$
or equivalently
$$\frac{\beta C\_t^n}{C\_{t+1}^y} = (1 + f'(k\_t)).$$

Finally, we can write our system of four equations with three equations, having removed $\lambda$ from the system:
$$\frac{C\_t^y}{C\_t^o} = 1+n$$
$$\frac{\beta C\_t^y}{C\_{t+1}^y} = (1 + f'(k\_t)).$$
$$C\_t^y +\frac{C\_t^o}{1+n} +k\_{t+1}(n+1)-k\_t-f(k\_t)=0.$$

Dropping time subscripts, i.e. applying the steady-state condition, gives
$$\frac{C^y}{C^o} = 1+n$$
$$\beta = (1 + f'(k)).$$
$$C^y +\frac{C^o}{1+n} +nk-f(k)=0.$$

Hence, you are left with a system of three equations with three unknowns $C^y$, $C^o$, and $k$. Solve the system to arrived at a steady-state solution.

> **Comments:**
> **username_0**: Very clear! Thanks for the answer, it was very helpful  
> **username_1**: Good luck with your PhD studies (or very advanced undergraduate studies)!  

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