Conditions to use the Lagrangian method

Question

I have seen that the prices and $\text{MU}_{i}$ are assumed to be positive (or, the preferences monotonic). This is always mentioned when a utility maximization problem is being solved with the Lagrangian method. Is there a reason for that? If the $\lambda$ exists, don't we just get a solution regardless of all these conditions?

Answer 1

Prices

Prices can be assumed to be positive simply because they typically are positive. There can be some rare examples, such as oil price futures on April 20, 2020. Nonetheless, such examples are quite rare.

However, explicitly assuming prices are positive is not that common. For example, Varian Microeconomic Analysis 3rd edition does not assumes that for many consumer problems (eg see chapter 9).

Some texts like to also explicitly make prices non-negative, but it is not ubiquitous.

Marginal Utility

This is because typical microeconomic problems assume that ‘more is always better’ for consumer (which implies nonsatiation).

If you assume that more is always better you need to assume $U’>0$.

If you choose some utility function for which $U’<0$ will be violated because this implies that at some point consumer consumes so much that she or he won’t want to consume more.

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If the λ exists, don't we just get a solution regardless of all these conditions?

There are some more conditions that have to be satisfied for optimization problem to have a solution(s) such as that the functions are differentiable etc. However, the assumption above do not have much with existence of solution per se (although depending on exact problem they might be necessary). Rather you would want to impose some restrictions on the problem if they are reasonable to reduce the number of possible solutions.

For example, if prices are very rarely negative it might not be interesting to consider potential solutions where prices are negative.

Likewise, while there are certainly cases where humans more is not always better (eg food consumption), if we look at some composite goods or large variety of goods it’s not unreasonable to assume more is always better. There are probably very few people in the world who would not want to increase consumption of something. Some people might prefer traveling more, spending more time in some nice resort, traveling more comfortably, having better education, more healthcare and so on. Most people will prefer more to less, save some exceptions (e.g. Buddhist monks?).

Consequently, assuming $U’>0$ is reasonable assumption outside some specialized problems where you might want not to assume more is always better. You might not seen problems like that because undergraduate or intermediate books and even many graduate books won’t always cover these more specialized cases, but they certainly exist. So it is possible to find problems where it’s not assumed that marginal utility is always positive, they are just very rare.

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Conditions for the existence of Lagrangian

Here are the conditions for existence of Lagrangian multiplier (following Sydsaeter et al Further Mathematics for Economic Analysis pp 153-154):

Given a problem:

$$ \max f(x) \text{s.t.} \begin{cases} \displaystyle g_j(\mathbf{x}) =0, j=1,...,r \\ \displaystyle h_k(\mathbf{x}) \leq 0, j=1,...,s \end{cases}$$

The Lagrangian multipliers will exist (and thus you could apply Lagrangian) if $f,g,h$ are all $C^1$ in some open set $A$ in $\mathbb{R}$, and suppose that $\mathbf{x^*}$ is a local extreme point in the problem above over A. Then there exist numbers $\alpha, \lambda_1,...,\lambda_r, \mu_1,...,\mu_s$ that are not all 0, such that:

• $\alpha \geq 0$
• $\alpha \nabla f(\mathbf{x^*}) = \sum \lambda_j \nabla g_j (\mathbf{x^*}) + \sum \mu_k \nabla h_k (\mathbf{x^*})$
• for each $k=1,...,s$ one has $\mu_k \geq 0 $, and $\mu_k=0$ if $h_k(\mathbf{x^*})<0$

So existence of the Lagrangian multiplier does not require objective function to be strictly increasing nor does it require all parameters of the constraint $g$ and $h$ to be non-negative.

When it comes to existence of solution Lagrangian multiplier will have solution (see Sydsaeter et al pp 146) $\mathbf{x^*} = (x_1^*,...,x_n^*)$ if $f, g$ and $h$ are $C^1$ functions, $r<n$, standard Khun-Tucker conditions hold and constant qualification is satisfied (gradient vectors $\nabla g_j (\mathbf{x^*}, 1\leq j \leq m$ corresponding to those constraints that are active at $\mathbf{x}^*$ are linearly independent, and subject to the above mentioned requirements for existence of multipliers, then there will be some unique numbers $\mathbf{\lambda}, \mathbf{\mu}$ such that:

• $\alpha \nabla f(\mathbf{x^*}) = \sum \lambda_j \nabla g_j (\mathbf{x^*}) + \sum \mu_k \nabla h_k (\mathbf{x^*})$
• $\mu_k \geq 0 $ and $\mu_k = 0$ if $h_k(\mathbf{x^*})<c_k, k =1,...,R$

and if the Lagrangain is concave in $\mathbf{x}$ and there is admissible $\mathbf{x^*}$ that satisfies conditions above, then $\mathbf{x^*}$ solves the maximization problem.

Note none of the conditions above require all parameters of $h$ or $g$ to be positive or $f'_i$ to be everywhere strictly non-negative (of course the function needs to be concave in the neighbourhood of some local maximum unless there is corner solution, but there is no reason why it should be non-negative everywhere.

Answer 2

Consider the following utility maximization problem

$$\max_{x_1,...,x_K} U(x) \\ s.t. \sum_{k=1}^K p_kx_k \leq I \\ x_k \geq 0,$$

you can set up a Lagrangian function

$$\mathcal L(x_1,...,x_K,\lambda,\delta_1,...,\delta_K) = U(x) + \lambda(I - \sum_k p_kx_k) + \sum_k\delta_k x_k$$

with the first-order conditions

$$\frac{\partial \mathcal L}{\partial x_k} = \frac{\partial U}{\partial x_k}-\lambda p_k +\delta_k = 0$$

Consider then a candidate for a solution $x' = (x'_1,...,x'_K)$. Let us assume that there is some $j$ such that $\partial U(x')/\partial x_j >0$. This states that for some good at the candidate solution the agent is non-satiated. Assuming further that $p_j>0$ this implies that

$$\frac{\frac{\partial U(x')}{\partial x_j} + \delta_j}{p_j}=\lambda > 0$$

because $\delta_j \geq 0$ from KKT conditions of solution candidate. This implies that the income constraint must be binding $\sum_k p_k x'_k = I$.

Let us now assume further that there is some $x'_s$ for which $p_s'<0$ this would then imply that

$$\frac{\partial U(x')}{\partial x_s}=\lambda p_s -\delta_s,$$ where the LHS is strictly negative because $\lambda>0$ and $\delta_s\geq 0$ implying that the marginal utility $\frac{\partial U(x')}{\partial x_s}<0$. The intuition here is that negative price introduces a money making machine and unless the agent gets negative marginal utility from $x_s$ at $x'$ the agent could relax the income constraint while increasing utility. An example would be a person getting positive marginal utility from working with the price of work being $-w$ the negative of the wage rate.

If $p_s=0$ then

$$\frac{\partial U(x')}{\partial x_s} = -\delta_s$$

where $\delta_s\geq 0$ so either the person would be getting negative marginal utility from $x_s$ or simply 0. However, if $x'_s>0$ it is further known that $\delta_s=0$ implying that $\frac{\partial U(x')}{\partial x_s}=0$. The intuition here is simply that if the price is 0 the satisfaction of the income constraint is unaffected by $x'_s$. If $x'_s>0$ the agent can increase or decrease $x'_s$ still satisfying the constraint $x'_s\geq 0$. Hence if the agent has negative marginal utility the agent can decrease $x'_s$ to get higher utility while still satisfying all constraints. If the agent has positive marginal utility the agent can increase $x'_s$ to get higher utility again while still satisfying all constraints.