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Consider the following two utility functions:

$EU(p)=\sum_i u_ip_i$

$EU^2(p)=(\sum_i u_ip_i)^2$.

In preference theory, $EU$ and $EU^2$ are equivalent because they represent the same preference. A preference is a complete and transitive binary relation over the set of alternatives.

However, obviously, in likelihood-estimation with logit/luce choice rule, $EU$ and $EU^2$ give different probabilities. For example, the probability of choosing $p$ over $q$ can be:

$L(p)=\frac{EU(p)}{EU(p)+EU(q)}$

OR:

$L(p)=\frac{EU^2(p)}{EU^2(p)+EU^2(q)}$

That is, for exactly the same preference, two different likelihood functions are viable.

How do econometricians justify the usage of $EU$ instead of $EU^2$?

Related questions:

Does concavity of the utility function has any bite?

When can one safely talk about decreasing marginal utility?

The related posts are about cardinal utility or the strength of preference. However, those arguments cannot be simply applied here because:

The "cardinal utility" applies to the lower case $u$, which is unique up to an affine transformation, but it does not apply to EU as a whole, which is unique up to a positive monotonic transformation.

dodo
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    Where do you see anyone use EU^2? There are risk-loving utility functions but I think your question is fundamentally why we are maximizing utility instead of squared utility? – RegressForward Apr 09 '22 at 02:23
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    @RegressForward (EU)^2 is equivalent to EU. The risk-loving utility you are mentioning is probably E(U^2). – dodo Apr 09 '22 at 03:51
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    When one wants to rely on the ratios of utilities, one clearly cannot say that monotonic transformations do not matter. One can argue for cardinal utility values, one could say that the utility the choice model is using is merely a tool with nice theoritecial properties, or one could say that utility models are full of gas. So what exactly are you asking here? I feel like you should specify the exact model you are looking at, and perhaps also the situation you think the model would fare poorly in. – Giskard Apr 09 '22 at 08:28
  • Why do you claim that "$EU$ and $EU^2$ give different probabilities"? The first and second order conditions for both problems are the same, and so are their solutions. You can consider any even power, or any increasing transformation of $EU$. – Bertrand Apr 09 '22 at 12:11
  • @Giskard For EU, the "cardinal utility" applies to the lower case u, which is unique up to an affine transformation, but it does not apply to EU as a whole, which is unique up to a positive monotonic transformation. Let me update the question. I will deeply appreciate you if there is a cardinal utility argument for EU as a whole. – dodo Apr 09 '22 at 17:22
  • @Bertrand They give different probabilities while doing likelihood estimation. See the update. – dodo Apr 09 '22 at 17:23
  • Can you define precisely what you mean by $\mathbb{E}(U)$ and $\mathbb{E}(U^2)$ are equivalent? – Richard Hardy Apr 10 '22 at 07:06
  • @RichardHardy It is not about $\mathbb E(U^2)$. It is about $[\mathbb E(U)]^2$. They are equivalent in preference theory because they represent the same binary relation. I updated the precise definition in the question. – dodo Apr 10 '22 at 23:18
  • I suspect they are not equivalent when the probabilities are fixed. It seems you are changing probabilities together with utilities to argue for equivalence. That does not seem reasonable to me, but perhaps I am missing something about preference theory (which I hardly know anything about). – Richard Hardy Apr 11 '22 at 05:33

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