I am currently studying econ models, and trying to calculate the FOC for different kind of models. I use J.Gali "Monetary Policy, Inflation and the Business Cycle - An Introduction to the New Keynesian Framework".

In the second chapter, the classical monetary model households have the following objective function :

$ E_0\sum^\infty_{t=0} \beta^t U(C_t, N_t)$

under the budget constraints

$P_t C_t + Q_t B_t \le B_{t-1}+W_tN_t-T_t$

I completely understand how to find the FOC under this model.

However, in the next chapter, households have the same utility function $ E_0\sum^\infty_{t=0} \beta^t U(C_t, N_t)$. However, here $ C_t \equiv(\int^1_0 C_t(i)^{1-\frac{1}{\varepsilon}}di)^{\frac{\varepsilon}{\varepsilon-1}} $. Households try to maximize their utility under $P_tC_t+Q_tB_t\le B_{t-1}+W_tN_t+T_t$

Where $P_tC_t=\int^1_0P_t(i)C_t(i)di$

If I understand well, in this kind of model we have a continuum of goods $i$ and households face an optimal allocation of consumption expenditures problem. Before maximizing their utility depending on both $C_t$ and $N_t$, they have to maximize $C_t$ for any given expenditure level ($\int^1_0P_t(i)C_t(i)\equiv Z_t$ where $Z_t$ is the expenditure level), so the maximization problem yields

L= [$ \int^1_0 C_t(i)^{1-\frac{1}{\varepsilon}}di]^{\frac{\varepsilon}{\varepsilon-1}} - \lambda(\int^1_0P_t(i)C_t(i)di- Z_t) $

Intuitively, I'd replace $\int^1_0 C_t(i)$ by $C_t$ and do $\frac{dL}{dC_t}=0$ which would give me something like $\lambda_tP_t = \frac{\varepsilon}{\varepsilon -1}C_t^{\frac{\varepsilon}{\varepsilon -1}-1}$

But here you see I don't really take in account the integrals, and the answer given in the book is : $C_t(i)^{-{1/\varepsilon}} C_t^{1/\varepsilon}=\lambda_t P_t(i)$

I don't know how to find this and I'm guessing it's linked to the integrals derivations. What is the process behind this ?

  • $\begingroup$ Did you look at the chapter appendix? $\endgroup$ Dec 24, 2023 at 13:57
  • $\begingroup$ Yep, and actually what I've written here is in appendix for some part. It just gives the results, not the demonstrations. $\endgroup$
    – krauuuus
    Dec 24, 2023 at 13:58

1 Answer 1


The Lagrangian is the following: $$ L = \left(\int_0^1 C_t(i)^{1 - \frac{1}{\varepsilon}}di \right)^{\frac{\varepsilon}{\varepsilon-1}} - \lambda\left(\int_0^1 P_t(i) C_t(i) di - Z_t\right) $$

Let $\left(\int_0^1 C_t(i)^{1 - \frac{1}{\varepsilon}}\right)^{\frac{\varepsilon}{\varepsilon-1}} = C_t$.

The first order conditions are given by: $$ \frac{\varepsilon}{\varepsilon-1}\left(\int_0^1 C_t(j)^{1 - \frac{1}{\varepsilon}} dj \right)^{\frac{1}{\varepsilon-1}}\left(1 - \frac{1}{\varepsilon}\right)C_t(i)^{-\frac{1}{\varepsilon}} - \lambda P_t(i) = 0 $$

This can be rewritten as: $$ \begin{align*} &C_t^{\frac{1}{\varepsilon}} C_t(i)^{-\frac{1}{\varepsilon}} = \lambda P_t(i),\\ \leftrightarrow &\left(\frac{C_t(i)}{C_t}\right)^{-\frac{1}{\varepsilon}} = \lambda P_t(i). \end{align*} $$

  • $\begingroup$ Thank's @tdm ! So if I understand it well, the integrals doesn't have any impact on the derivation process ? $\endgroup$
    – krauuuus
    Feb 2 at 12:29
  • 1
    $\begingroup$ Yes, I always struggled myself with this 'leap of faith' derivation... but that's how they do it :-) $\endgroup$
    – tdm
    Feb 2 at 14:22

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