In "Section IV Shopping" of Kőszegi / Rabin (A model of reference-dependent preferences, QJE 2006), the example of consumer buying a pair of shoes is given.

They claim that "her disutility from spending on the shoes is between $(1+\eta)p$ and $(1+ \eta \lambda)p$" (p. 1146).

How do they get this result?

I can see where the second term comes from, namely buying at $p$ when expectations were to buy at 0, which generates money-related utility of

$$ -p - \eta \lambda p = - (1+\eta \lambda) p$$

But the first term is beyond me: If she buys at $p$, but expected to buy at some $p' > p$, I would obtain

$$ - p + \eta(p' - p) = -(1+\eta)p + \eta p' $$

which is not the same as the $(1+\eta)p$ stated in the paper.

What am I missing here? Thank you.


1 Answer 1


I think it is just a matter of definition. The disutility from spending on the shoes is not the utility obtained by purchasing the shoes: it is the difference between the utility obtained when she does not buy the shoes and the utility she gets when she buys them.

Suppose for instance that she expects to buy the shoes at price $p$. Her "money" utility of buying the shoes equals \begin{equation*} U(\text{Buying}) = -p \end{equation*} since there is no gain-loss utility in that case, and $-p$ is her final wealth (normalizing initial wealth to zero).

However, if she finally decides not to buy the shoes, her money utility equals \begin{equation*} U(\text{Not Buying}) = \eta p \end{equation*} since her final wealth equals $0$, and in addition she experiences a gain-utility of $p$ with respect to the reference point (spending $p$).

Therefore the disutility from buying is \begin{equation*} U(\text{Not Buying}) - U(\text{Buying}) = (1+\eta) p \end{equation*}

I hope it helps.

  • $\begingroup$ Makes sense! When you say after buying the final wealth is $1-p$, you seem to assume the initial wealth is 1, which I think isn't stated anywhere. I doesn't matter since it cancels out anyway, but I would say it's just $-p$ instead of $1-p$. $\endgroup$
    – bonifaz
    Commented May 15, 2015 at 9:23
  • $\begingroup$ @bonifaz You are right, sorry, they even normalize initial endowment to (0,0). But as you wrote it does not matter since initial wealth cancels out. I'll edit my answer right now. $\endgroup$
    – Oliv
    Commented May 15, 2015 at 10:14

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