<ol>
<li>Is the utility function <em>wrong</em>?</li>
</ol>
<p>Utility functions <em>usually</em> 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.</p>
<p>For example, if all I know is that <span class="math-container">$U_i(e^{-1}, e^{-1}, 0) = -2$</span> (as in your example), this in and of itself does not convey <em>any information at all</em>. Now if I am also told that <span class="math-container">$U_i(1, 1, 0) = 0$</span>, then all I know is that the point <span class="math-container">$(1, 1, 0)$</span> is preferred over the point <span class="math-container">$(e^{-1}, e^{-1}, 0)$</span>, no more no less. Similarly, if I know that <span class="math-container">$U_i(e^{-2}, e^{-2}, 0) = -4$</span>, then I know that <span class="math-container">$(1, 1, 0)$</span> is preferred over <span class="math-container">$(e^{-1}, e^{-1}, 0)$</span>, which in turn is preferred over <span class="math-container">$(e^{-2}, e^{-2}, 0)$</span>.</p>
<p>When I am given the complete specification of the utility function as <span class="math-container">$U_i(c_i, n, \theta) = \log c_i + \log n + \theta$</span>, 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.</p>
<p>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.</p>
<ol start="2">
<li>Can we replace it with another function?</li>
</ol>
<p>Yes, any function of the form <span class="math-container">$f_i = g(U_i)$</span> such that <span class="math-container">$g$</span> is strictly increasing in the range of <span class="math-container">$U_i$</span> would work. This is because, we would have</p>
<p><span class="math-container">$$f_i(x) \geq f_i(y) \Leftrightarrow U_i(x) \geq U_i(y)$$</span></p>
<p>So the relative positioning of the points in the domain remains the same. Therefore both <span class="math-container">$f_i$</span> and <span class="math-container">$U_i$</span> represent the same preferences.</p>
<p>Hope this answers your question.</p>

CC BY-SA 4.0
2023-08-09T09:55:36.463Z
Very much so, thank you very much for the clear explanation.
Hans Brecker
CC BY-SA 4.0
2023-08-09T12:46:07.660Z
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.
VARulle
CC BY-SA 4.0
2023-08-10T12:18:00.460Z
@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.
Ishan Kashyap Hazarika
CC BY-SA 4.0
2023-08-18T10:23:16.580Z
The context was as consumption, I believe it must have been ordinal.
Hans Brecker
CC BY-SA 4.0
2023-08-19T10:37:20.207Z
@VARulle , given ordinal preferences. Even though the numbers themselves don't say anything. Can you say something about the percent change? For example by comparing the same utility function with different values against each other

Suppose 2 consumers, first we compare their utility functions consuming bundle A, they both have different values. $U_1(A) = 0.5$, $U_2(A) = 1$
Then we look at the percentage change when consuming bundle B. 
$U_1(B) = 0.8$, $U_2(B) = 1.1$. Therefore the % change for A = 60% and for B = 10%.

Can we say that the increase is higher for A than for B?
Hans Brecker
CC BY-SA 4.0
2023-08-24T15:08:55.047Z
@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.
VARulle
CC BY-SA 4.0
2023-08-09T09:50:47.913Z
2023-08-09T09:50:47.913Z
Ishan Kashyap Hazarika
<p>I'm reading over some models and I found a paper in progress that uses this model
<span class="math-container">$$U_i(c_i, n, \theta_i) = \log(c_i) + \log(n) + \theta_i$$</span></p>
<p>where <span class="math-container">$c_i \text{ and }  n \geq 0$</span> , but when you consider that when <span class="math-container">$0 &lt; c_i &lt; 1$</span>, <span class="math-container">$log(c_i)$</span> becomes negative, wouldn't this be wrong to describe the model then. Can't you just replace it with a function such that <span class="math-container">$f(c_i)$</span> is concave and twice differentiable?</p>

<mathematical-economics><utility>
Does this utility function work?
Hans Brecker
# Does this utility function work?

(Asked by: username_0 on 2023-08-09)

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?

**Tags:** mathematical-economics, utility

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## 1 answers

### Answer by username_1 on 2023-08-09. Score: 2 | Accepted answer

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.

2. 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.

> **Comments:**
> **username_0**: Very much so, thank you very much for the clear explanation.  
> **username_2**: 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.  
> **username_1**: @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.  
> **username_0**: The context was as consumption, I believe it must have been ordinal.  
> **username_0**: @VARulle , given ordinal preferences. Even though the numbers themselves don't say anything. Can you say something about the percent change? For example by comparing the same utility function with different values against each other  Suppose 2 consumers, first we compare their utility functions consuming bundle A, they both have different values. $U_1(A) = 0.5$, $U_2(A) = 1$ Then we look at the percentage change when consuming bundle B.  $U_1(B) = 0.8$, $U_2(B) = 1.1$. Therefore the % change for A = 60% and for B = 10%.  Can we say that the increase is higher for A than for B?  
> **username_2**: @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.  

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