<p>I wish I could leave this as a comment, but the first formulation is a solution to a maximisation problem of the form
<span class="math-container">$$
\max_{C_t} \sum_{t} \beta^{t} u(C_t) \\
\text{s.t. } \quad C_t +K_{t+1} \leq (1-\delta)K_{t} + w_{t} + r_tK_t \quad \forall t 
$$</span></p>

<p>The second formulation is a solution to a maximisation problem of the form
<span class="math-container">$$
\max_{C_t} \sum_{t} \beta^{t} u(C_t) \\
\text{s.t. } \quad \sum_{t} C_t \leq \sum_{t} (A_0 +Y_t) 
$$</span></p>

<p>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.</p>

CC BY-SA 4.0
2020-01-30T00:11:12.503Z
May ask you what do you mean with a sequential and time zero problem?
Alex Ruiz
CC BY-SA 4.0
2020-01-30T17:40:01.263Z
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.
Walrasian Auctioneer
CC BY-SA 4.0
2020-01-28T00:44:09.120Z
2020-01-28T00:44:09.120Z
Walrasian Auctioneer
<p>I have a question in regards to the use of Lagrange multipliers in macro models.</p>

<p>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:</p>

<p><span class="math-container">$$
L=\sum_{t=0}^{\infty}\bigg\{\beta^t u(C_t)-\lambda_t[C_t+K_{t+1}-(1-\delta)K_t-w_t-r_tK_t]\bigg\}
$$</span></p>

<p>and here the Lagrange multiplier is indexed in time and written as <span class="math-container">$\lambda_t$</span>.</p>

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

<p><span class="math-container">$$
L=\sum_{t=0}^{\infty}[u(C_t)+\lambda(A_0+Y_t-C_t)]
$$</span></p>

<p>and here the Lagrange multiplier is not indexed in time, so we write it just as <span class="math-container">$\lambda$</span>. </p>

<p>Hence, my question is when is it appropriate to use each type of multiplier? </p>

<p>Thank you.</p>

CC BY-SA 4.0
2020-01-26T05:46:48.560Z
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 $t$.
mark leeds
CC BY-SA 4.0
2020-01-28T18:24:43.400Z
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.
mark leeds
<macroeconomics><microeconomics><mathematical-economics><theory><open-economy-macro>
When to use a lagrange multiplier?
Alex Ruiz
# When to use a lagrange multiplier?

(Asked by: username_0 on 2020-01-25)

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.

**Tags:** macroeconomics, microeconomics, mathematical-economics, theory, open-economy-macro
> **Comments:**
> **username_1**: 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 $t$.  
> **username_1**: 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.  

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## 1 answers

### Answer by username_2 on 2020-01-28. Score: 1

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.

> **Comments:**
> **username_0**: May ask you what do you mean with a sequential and time zero problem?  
> **username_2**: 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.  

---

