# When to use a lagrange multiplier?

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.

• 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 Jan 26 '20 at 5:46
• 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 Jan 28 '20 at 18:24

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.

• May ask you what do you mean with a sequential and time zero problem? – Alejandro Ruiz Jan 30 '20 at 0:11
• 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 Jan 30 '20 at 17:40