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I have a question in regards to the use of Lagrange multipliers in macro models.

I have seen that when writing a Lagrange there are two ways to write the Lagrange multiplier: 1) Indexed or 2) Unindexed. For example, in order to solve the Ramsey model in discrete time, in the simplest possible version, one would write the Lagrange as follows:

$$ L=\sum_{t=0}^{\infty}\bigg\{\beta^t u(C_t)-\lambda_t[C_t+K_{t+1}-(1-\delta)K_t-w_t-r_tK_t]\bigg\} $$

and here the Lagrange multiplier is indexed in time and written as $\lambda_t$.

But when solving a simple lifetime consumption problem for an individual, one would write the Lagrange as follows:

$$ L=\sum_{t=0}^{\infty}[u(C_t)+\lambda(A_0+Y_t-C_t)] $$

and here the Lagrange multiplier is not indexed in time, so we write it just as $\lambda$.

Hence, my question is when is it appropriate to use each type of multiplier?

Thank you.

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    $\begingroup$ Hi: My guess is that the first model assumes that $\lambda$ is constant and the second model assumes that it's changing over time so it has a different value at each time $t$. $\endgroup$ – mark leeds Jan 26 at 5:46
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    $\begingroup$ My comment above is wrong. The answer below tells me that the $\lambda$ without the $t$ arises because it's a one constraint problem. The case where $\lambda$ is subscribted with $t$ arises because there is a constraint at each time $t$. Thanks to Walsrian Auctioneer for insight. $\endgroup$ – mark leeds Jan 28 at 18:24
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I wish I could leave this as a comment, but the first formulation is a solution to a maximisation problem of the form $$ \max_{C_t} \sum_{t} \beta^{t} u(C_t) \\ \text{s.t. } \quad C_t +K_{t+1} \leq (1-\delta)K_{t} + w_{t} + r_tK_t \quad \forall t $$

The second formulation is a solution to a maximisation problem of the form $$ \max_{C_t} \sum_{t} \beta^{t} u(C_t) \\ \text{s.t. } \quad \sum_{t} C_t \leq \sum_{t} (A_0 +Y_t) $$

In a way, the first is a sequential problem, while the second is a time zero problem. In some instances the approaches are equivalent, but I'm not an expert on this.

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  • $\begingroup$ May ask you what do you mean with a sequential and time zero problem? $\endgroup$ – Alejandro Ruiz Jan 30 at 0:11
  • $\begingroup$ I mean the distinction between all trade happening today by trading time indexed contracts, versus only trading today's contracts, leaving the market, then returning the next day. $\endgroup$ – Walrasian Auctioneer Jan 30 at 17:40

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