# How to handle multiple lagrange multipliers in a maximization problem?

Let's assume a standard household maximization problem of the form: \begin{align} \underset{C_t}{max} \sum_{t=0}^{\infty} \beta^t U(C_t) \end{align} subject to a standard Budget constraint: \begin{align} C_t + b_t \leq b_{t-1}\frac{R_{t-1}}{\Pi_t} + Y_t \end{align} and a borrowing limit: \begin{align} b_t \leq \bar{b} \end{align}

I would write the lagrangian like this: \begin{align} L = \sum_{t=0}^{\infty} \beta^tU(C_t) - \lambda_t(C_t + b_t - b_{t-1}\frac{R_{t-1}}{\Pi_t} - Y_t) - \mu_t(b_t-\bar{b}) \end{align}

If I now derive the Lagrangian by bondholdings $$b_t$$, I get the expression: \begin{align} \lambda_t = \lambda_{t+1}\frac{R_t}{\Pi_{t+1}} - \mu_t \end{align}

Now, as per usual, I can easily get rid of the $$\lambda$$ multipliers by taking the derivative of the Lagrangian with respect to consumption and substituting them out, so I end up with: \begin{align} U'(C_t) = \beta U'(C_{t+1})\frac{R_t}{\Pi_{t+1}} - \mu_t \end{align} But how do I get rid of the other multiplier and if I can't substitute with anything, how do I handle an Euler Equation like this? If I were to build a model with this Euler Equation, would the multiplier just stay and enter my model as a parameter I have to define?

• Could you please clarify a couple of points? 1) Are $R,\:\Pi$ and $Y$ exogenous? 2) Do you also require $C_t\geq 0$? Jan 22, 2023 at 18:33
• I think usually they would be endogenous in a larger model, but I only care about the household problem here, so I guess we can assume them exogenous. I would say $C_t \geq 0$ as well, but I'm really just interested in how one would handle the multiplier. Jan 23, 2023 at 11:50
• The discount factor multiplies the whole per-period Lagrangian, not just the utility function. Jan 23, 2023 at 20:23
• (I should have thought of this when I posted my first comment.) Do you allow borrowing $b_t$ to be negative, ie to become saving / investment on the part of the household? Jan 24, 2023 at 18:33
• @AlecosPapadopoulos I don't actually understand why that is. At least here it doesn't seem to make a difference, but in general how is that justified? If I look at how the Lagrangian is generally constructed $L(x,\lambda) = f(x) + \lambda g(x)$, the discount factor is only part of $f(x)$, why does it multiply the whole Lagrangian? Feb 7, 2023 at 14:20

This is less than a full answer but hopefully will be of some help.

A general observation is that complex optimization problems do not always have an analytical solution. So although you show a substitution that eliminates $$\lambda_t$$, it cannot be assumed that there exists a substitution that will eliminate $$\mu_t$$. On the other hand it would be inappropriate to treat $$\mu_t$$ as a parameter to be defined. It has a definite meaning already as the shadow price associated with the borrowing limit, so we must have $$\mu_t>0$$ if $$b_t=\bar{b}$$ and $$\mu_t=0$$ if the borrowing constraint is slack at time $$t$$.

A further observation is that, assuming as we would normally expect that $$U$$ is increasing in $$C$$, for an optimum the budget constraint will always be binding, since if it were slack in any period, the infinite sum could be increased by increasing $$C_t$$ in that period. So I will refer to it below as the budget equality.

There is one circumstance in which the problem can be solved. That is if the borrowing constraint is initially slack, and calculation of the time paths of $$C_t$$ and $$b_t$$ using:

$$U'(C_t) = \beta U'(C_{t+1})\frac{R_t}{\Pi_{t+1}}$$

and the budget equality shows that the borrowing constraint happens to be always slack. (The details of the calculation will depend on the functional form of $$U$$, the values of the exogenous quantities, and the terminal condition.) In that special case $$\mu_t$$ is always zero so can be ignored. That scenario would be more likely if the value of $$\dfrac{R_t}{\Pi_{t+1}}$$, the return on borrowing, is always very low.

If that circumstance does not hold, the problem appears very difficult. The combination of an infinite horizon and a set of exogenous variables which may not be constant over time is unusual. One reason why the household problem is sometimes modelled in an infinite framework is:

"the household problem becomes stationary in the sense that the problem at date t is exactly the same as that at t+1" (Hendricks (2005))

Stationarity does not mean that the values of variables will be constant over time, but it often means that they take a regular progression converging to a limit value, and this can facilitate the solution of infinite horizon problems. In this case, however, unless additional assumptions are made the problem never becomes stationary because we have no reason to assume, even for the far distant future, that $$R_{t+1}=R_t$$, or similarly for $$\Pi_{t+1}$$ or $$Y_{t+1}$$.

When we allow for debt, somebody else does the lending. So our debt is an asset for somebody else. From the lender's point of view, a transversality condition arises, related to the holding of assets at infinity without consuming them.

So I would suggest to study the model from the point of view of the lender, and see what this may imply for the debt-multiplier on the part of the borrower.

Just remember that while the transversality condition in a finite horizon model is clear-cut (consume everything before you die), the case is more subtle in an infinite horizon problem.