Efficient revealed preference check of whether two preferences over lotteries are the same?

Question

I want to find conditions under which two utility functions can be known to be linear transformations of each other.

Consider a (possibly finite) arbitrary set of outcomes $X$ (Not necessarily interpreted as commodities), and a specific utility function $u^*:X\to \mathbb R$. We want to find out whether an agent's utility function is a linear transformation of $u^*$ or not. Consider a set of lotteries $\mathcal L$ on $X$. Now suppose we give an agent choices between subsets $C\subseteq \mathcal L$. What I want is: What choices do we need to give to the agent to know that his utility function is a linear transformation of some utility function $u^*$? Note: we don't need to check whether the agent has rational preferences. We assume from the outset that he has a vnM utility funciton.

This is simply revealed preference, but I don't recall reading about anything about this specifically.

Answer 1

Perhaps I'm way off base on this, but given the relatively strong requirements imposed in the question, I'm tempted to say that we can never truly be able to state with certainty that some individual's utility function $u_i$ is absolutely a linear transformation of some other function $u^{*}$ considered ex ante. While depending on the specific $u_i$, $u^{*}$ and initial questions asked, it could be possible to dismiss the two as not satisfying the transformation requirement, but I don't think one could ever actually confirm it.

The underlying reason for this thinking is the inclusion of vNM utilities and lotteries over outcomes. Since the former assumption imposes cardinality on our utility framework, there is always the possibility of discrepancies between $u_i$ and $u^{*}$. Even considering just one good (money, for example) any sequence of observed choices made by some individual following a utility function $u_i$ might only differ from how an individual following $u^{*}$ behaves in the "gaps." No matter how fine the probabilities and dollar amounts get, we can always come up with some alternative $u'$ which replicates the decisions up till that point. (We can always consider, for example, some $u^{'}$ that's a piecewise on some interval where $u^{*}$ is smooth.

Of course, as our ability to observe smaller and smaller intervals increases from repeated choice observational data, the possible deviations between $u^{*}$ and some alternative $u'$ trends towards zero. But again, finding out that all observations thusfar suggest the individual's utility $u_i$ is a linear transformation of $u^{*}$ doesn't guarantee that there isn't some interval of the function where they're different.

It should also be pointed out that this really only considers extreme cases- for all experimental purposes we only need a function that approximates an individual's utility function. But I do think it's tougher to truly prove one is a linear transformation of the particular $u^{*}$ in mind.

Answer 2

First, we can rephrase your question in the following way:

Given a linear $u^*: \mathcal L \to \mathbb R$ and an expected utility preference $\succsim$, what is the minimal subset of $\succsim$ needed to ensure that it is represented by $u^*$.$^1$

You can do it in $|X| - 1$ observations, so long as every dimension of $\mathcal L$ is "represented" in the observations. Let's explore how this can be done, and hopefully in the process, make my claim coherent.

For the moment assume that $X$ is finite. Let $b,w$ be the best and worst elements of $X$ according to $u^*$; wlog assume that $u^*(b) = 1$ and $u^*(w) = 0$. Of course it must be that $b \succsim w$, or we have a counter-example to the hypothesis that $\succsim$ is represented by $u^*$. Now for each other element $x \in X$ we need only to know if $$u^*(x)b + (1-u^*(x))w \sim x$$

If this is true then by the properties of EU, any representation (also normalized on $b,w$) must set $u(x)$ to be $u^*(x)u(b) + (1-u^*(x))u(w) = u^*(x)1 + (1-u^*(x))0 = u^*(x)$.

Now, if $X$ is infinite, there may not be a best and worst element, you will have to deal with cases where the element $x$ is above or below the bounds of the normalization. This issue arrises also in the proof of the mixture space theorem and is dealt with in the same way---take mixtures with the extreme elements, and figure out the weights accordingly.

It is easy to show that any less data than the above leaves a degree of freedom which can be exploited. In other words, you must check each dimension but it suffices to do this with a single observation.

For what its worth this is a specific case of a much more general result: a linear function is completely determined by its behavior on basis vectors! $X$ is your basis for $\mathcal L$.

[1] I could be getting your question wrong. Perhaps you mean that $u^*$ is over $X$ not $\mathcal L$ and you do not have access to cardinal properties of $u^*$. If this is the case, the answer still stands, but the question seems pretty meaningless.