Following the notation of Mas-Collel, Whinston, and Green, consider a family of budget sets $\mathcal{B}=\{\{x,y\},\{x,y,z\}\}$. To make the example concrete, let's let

  • $x$ be a book
  • $y$ be a left shoe
  • $z$ be a right shoe

For motivation, let's suppose the individual prefers going for a walk over reading a book. However, to go for a walk, he needs two shoes. If he can only have one shoe, then he would rather choose the book instead.

Thus when faced with Budget set $\{x,y\}$, the consumer chooses the the book: $C(\{x,y\})=\{x\}$. However when faced with all three alternatives, he chooses the shoes: $C(\{x,y,z\})=\{y,z\}$.

As we constructed it, this choice rule $C$ violates the Weak Axiom of Revealed Preference:

Suppose there exists some budget set $B_1 \in \mathcal{B}$ with $x,y \in B_1$ and $x \in C(B_1)$. Then for any $B_2 \in \mathcal{B}$ with $x,y \in B_2$ and $y\in C(B_2)$, we must also have $x \in C(B_2)$.

In our case, since $C(\{x,y\})=x$, then WARP would require $x \in C(\{x,y,z\})$ since $y \in C(\{x,y,z\})$. However, it seems we were able to motivate a sensible reason why $C(\{x,y,z\})=\{y,z\}$.

Side note: I have left prices of the goods out of this story. (Hopefully this is acceptable; to be fair, MWG's chapter on choice rules doesn't mention prices either. (I assume we should imagine the price of these goods are all 0, or all the same price -- would that be the right way to look at it?)

This example brings me to the question posed in the title: Can we have sensible choice behavior that violates WARP?

  • $\begingroup$ So if you let x be in C(x,y,z) everything would be fine, but the way you set up his preferences are that he literally does not want the book if he has the shoes. While you can construct specific pathological examples where that works, in general, why wouldn't he want both books and shoes? His inability to consume them at the same time isn't a problem (I think). $\endgroup$
    – VCG
    Sep 15, 2016 at 14:05

2 Answers 2


I think that your example relies on a misunderstanding of the meaning of the choice function. $C(\{x,y,z\})$ does not contain the elements that the decision-maker would choose simultaneously if he could. He is only allowed to select one element at a time (otherwise, why not consuming all three items ?). Therefore $C(\{x,y,z\}=\{y,z\}$ does not mean that he selects both shoes. It means that, if you ask him to choose one element inside $\{x,y,z\}$, he might either pick $y$ or pick $z$, being indifferent between the two. In that case, your story would not make much sense, since the value of a single shoe would be the same in both choice sets.

Regarding your question, there are some well-known violations of WARP that rely on behavioral biases, in particular the attraction effect and the compromise effect. These expressions describe situations where the comparison between two goods is affected by the nature of a third (unselected) good in the choice set. You can easily find some references (experimental evidence, models) on these phenomena.


Even if the choice function set up you describe is so that

$C(\{\{x\},\{y\}\}) = \{x\}$ and $C(\{\{x\},\{y, z\}\}) = \{y, z\}$

then $\{y, z\}$ you will note was unaffordable in the first scenario ($p_z = \infty$). So WARP is not violated.


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