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.

  • $\begingroup$ 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. $\endgroup$
    – dlnB
    Mar 15, 2019 at 18:27

1 Answer 1


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.

  • $\begingroup$ Very clear! Thanks for the answer, it was very helpful $\endgroup$
    – Chaos
    Mar 15, 2019 at 21:14
  • $\begingroup$ Good luck with your PhD studies (or very advanced undergraduate studies)! $\endgroup$
    – dlnB
    Mar 15, 2019 at 21:15

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