I understand the distant definition of LNS but I don't get how to actually apply it to given utility functions like u=x1/x2 or u=x1-x2 or of any form? Is there structured math-y way to check if they represent LNS preferences? I digged the internet whole day but couldn't find how to solve these kind of questions. Any help would be highly appreciated. thank you!


1 Answer 1


You apply the definition.

There might be multiple ways for a utility function to represent LNS preferences. A surefire case is when there is a strictly monotonic term in a single good. For example, $u(\mathbf{x}) = f(\mathbf{x}_{-i}) + x_i$ (does this form ring any bell?) where $f$ is a horrendously complicated function of all goods except good $i$ is always LNS because of the linear term in $x_i$: whatever might happen with the other goods, the agent is always happier if you add just a bit of good $i$, or using the definition:

$$\forall \mathbf{x} \in X, \varepsilon > 0, \exists \mathbf{x}' : \|\mathbf{x} - \mathbf{x}'\| \leq \varepsilon \land \mathbf{x}' \succ \mathbf{x}$$

In this case, $\mathbf{x}' = \mathbf{x} + \varepsilon e_i$, where $e_i = (0, ..., 1, ..., 0)$ i.e. it is only 1 in position $i$.

I am sure some user will be delighted in pointing out an exception, but let's say this is true for all reasonably horrendous $f$. (It might be true for any $f$, I did not take the time to check.)

Another way of ensuring LNS is to check that the function has no local maxima or plateaus. This is easy to define and check in the univariate case, but becomes less trivial when dealing with multiple goods. There may be satiation for a single good, while not violating LNS: $u(x, y) = (x-2)^2 + \log(y)$.

So is there a "mathy" way? You apply the definition. Math is not always calculation-y. Unfortunately or fortunately, depends on how you see it.

  • 1
    $\begingroup$ $x'_{-i}=x_{-i}$, so exceptions to this should not be expected. $\endgroup$
    – VARulle
    Dec 14, 2022 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.