I'm reading over some models and I found a paper in progress that uses this model $$U_i(c_i, n, \theta_i) = \log(c_i) + \log(n) + \theta_i$$

where $c_i \text{ and } n \geq 0$ , but when you consider that when $0 < c_i < 1$, $log(c_i)$ becomes negative, wouldn't this be wrong to describe the model then. Can't you just replace it with a function such that $f(c_i)$ is concave and twice differentiable?


1 Answer 1

  1. Is the utility function wrong?

Utility functions usually represent ordinal preferences. The exact value of the function at any given input does not matter. What matters is whether that value is greater than, less than or equal to the value at some other input.

For example, if all I know is that $U_i(e^{-1}, e^{-1}, 0) = -2$ (as in your example), this in and of itself does not convey any information at all. Now if I am also told that $U_i(1, 1, 0) = 0$, then all I know is that the point $(1, 1, 0)$ is preferred over the point $(e^{-1}, e^{-1}, 0)$, no more no less. Similarly, if I know that $U_i(e^{-2}, e^{-2}, 0) = -4$, then I know that $(1, 1, 0)$ is preferred over $(e^{-1}, e^{-1}, 0)$, which in turn is preferred over $(e^{-2}, e^{-2}, 0)$.

When I am given the complete specification of the utility function as $U_i(c_i, n, \theta) = \log c_i + \log n + \theta$, now I can compare any two points in the domain of the function by comparing the relative values at the two points and seeing which one is greater.

To conclude, the exact value of the utility function does not matter, including whether that value is positive or negative. Only the relative positioning of different points in the domain matters. So no, the utility function (based on the information given by you and usual practice) is not wrong.

  1. Can we replace it with another function?

Yes, any function of the form $f_i = g(U_i)$ such that $g$ is strictly increasing in the range of $U_i$ would work. This is because, we would have

$$f_i(x) \geq f_i(y) \Leftrightarrow U_i(x) \geq U_i(y)$$

So the relative positioning of the points in the domain remains the same. Therefore both $f_i$ and $U_i$ represent the same preferences.

Hope this answers your question.

  • 1
    $\begingroup$ Very much so, thank you very much for the clear explanation. $\endgroup$ Aug 9, 2023 at 9:55
  • 6
    $\begingroup$ I disagree. A utility function like $u(c) = \log c$ will rarely denote a purely ordinal utility function. This function is most likely used in a context with risk, meaning that it is a (cardinal) Bernoulli utility function with its concavity reflecting risk aversion. This means that you can't rescale it with an arbitrary increasing function $g$, but only with a linear (affine) one. $\endgroup$
    – VARulle
    Aug 9, 2023 at 12:46
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    $\begingroup$ @VARulle A univariate function like that is most likely cardinal, but this utility fn isn't univariate. So without further info we can't really say. However, I still believe there is chance that it is cardinal (therefore I used "usually"). For the sake of more completeness, I will edit the answer to add a section on cardinal utility. $\endgroup$ Aug 10, 2023 at 12:18
  • $\begingroup$ The context was as consumption, I believe it must have been ordinal. $\endgroup$ Aug 18, 2023 at 10:23
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    $\begingroup$ @HansBrecker, well, you could say that "the increase is higher for A than for B" in a pure technical sense. But this wouldn't have any meaningful interpretation. The only valid conclusion would be that both 1 and 2 prefer B to A. $\endgroup$
    – VARulle
    Aug 24, 2023 at 15:08

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