The interesting paper Calvo and Obstfeld (1988) uses two-stage optimization on an OLG model which then reduces to a standard representative agent framework.

First stage optimization consists on a static optimization which makes the optimal allocation between different cohorts vertically(c.f equation (9) in the paper.) Authors solve this first stage problem as:

$$\mathcal{L}=u\left[c\left(t-n,t\right)\right]\Delta\left(n\right)P\left(n\right)e^{\rho n}+\lambda\left[C\left(t\right)-\int_{0}^{\infty}c\left(t-n,t\right)P\left(n\right)d\left(n\right)\right]$$

where $C\left(t\right)=\int_{0}^{\infty}c\left(t-n,t\right)P\left(n\right)dn$.

$n, P(n), \Delta\left(n\right)$ are given in the paper and not relevant for my question at this moment.

Normally, in this paper, the dynamics of capital accumulation are given as follows:


In fact, my question is trivial but I could not be sure.

Is the part with bracket in Lagrangian comes from $\dot{K}=0$ which gives us the equality $Y\left[K\left(t\right)\right]=C\left(t\right)=\int_{0}^{\infty}c\left(t-n,t\right)P\left(n\right)dn$ ?

As the Lagrangian is for a "static" problem, I think it makes sense but I can not be sure if this is the case.

Any suggestion or hints are welcome.


That's a coincidence, because they assume nondepreciating capital. If $\delta>0$ was positive we'd have \begin{align} \dot K = Y - C - \delta K \end{align} which gives \begin{align} \dot K = 0 \quad \Rightarrow\quad C = Y - \delta K. \end{align}

But the static optimization would still read \begin{align} &U(C(t)):=\max_{c(t-n,n)}\int^\infty_0 u(c(t-n,n))dn \\ &s.t.\quad \int^\infty_0 u(c(t-n,n))dn \leq C(t) \end{align} Actually they define an indirect utility function with subject to a resource constraint. In every period (thus static) the sum of individual consumption cannot exceed aggregate consumption.

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    $\begingroup$ thanks for your answer. But in this case, I don't understand the fact that authors define the aggregate consumption in page 416. So in this case, why they put a constraint on this first stage on aggregate consumption ? We know that $c\left(t\right)=\int_{0}^{\infty}z\left(t-n,t\right)P\left(n\right)dn$. So, why the constraint does not vanish ? $\endgroup$ Aug 3 '15 at 1:38
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    $\begingroup$ (Haven't read the paper. This is how I get the math). They define aggregate consumption to make the dynamic optimization handy. You can easily see the point if you switch stages. You solve over the periods $t\in[0,\infty)$ for an optimal level of aggregate consumption $\{C(t)\}_{t\geq 0}$ (dynamic optimization). Then you allocate in every period $C(t)$ over the residents, making $C(t)$ a resource constraint in every period. Cause this holds for every period it boils down to a static problem in every period $t$. Dynamic and static optimization are seperated. There is no cennection between... $\endgroup$
    – clueless
    Aug 3 '15 at 7:17
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    $\begingroup$ ...the law of motion $\dot K$ and the static problem. Disclaimer: this is my interpretation. But it makes sense to me. $\endgroup$
    – clueless
    Aug 3 '15 at 7:19
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    $\begingroup$ @cluless Sorry to bother you, I have a last question in order to take your advice. Is it legitmate to put also "another" constraint as $C(t)\geq Z(t)$ where $Z(t)=\int_{0}^{\infty}z\left(t-\tau,t\right)P\left(\tau\right)d\tau=\rho\int_{T}^{t}e^{-\rho\left(t-s\right)}\underbrace{\left\{ \int_{0}^{\infty}c\left(s-\tau,t\right)P\left(\tau\right)d\tau\right\} }ds$. which is a stock variable. (it is a habit stock, as in Ryder and Heal (1972)) $Z$ is a variable that I define here. Thanks so much in advance for your help ! :) $\endgroup$ Aug 3 '15 at 12:46
  • $\begingroup$ No idea. You may specify the question in a seperate thread. But, why not impose a secod constraint? $\endgroup$
    – clueless
    Aug 4 '15 at 13:46

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