Kőszegi - Rabin (2006) model problem

Question

First of all, that's a homework question, and I will try to make it as useful to future readers as possible.

Problem

So, I was given a problem about the model described by Botond Kőszegi and Matthew Rabin in "A Model of Reference-Dependent Preference" in 2006:

A person deals with two goods: tea ($c_1 \in \{0, 1\}$), and money ($c_2 \in R$). $$ c_1 = \begin{cases}1, \mbox{if he buys tea}\\ 0, \mbox{if he does not buy tea} \end{cases} $$ His consumption utility is given by: $$m(c) = v \times c_1 + c_2$$ His gain-loss utility is given by: $$ n(x | r) = \begin{cases} (x - r), \mbox{if } x \ge r\\ \lambda \times (x - r), \mbox{if } x \lt r \end{cases} $$ Tea costs $P_H = 30$; also, let $v = 13, \lambda = 4$.

There are multiple tasks, but I'm struggling even with the first one:

Prove that the reference point $r = (c_1, c_2) = (1, -30)$ with probability 1 is his personal equilibrium.

Do not forget that in the gain-loss utility in case of deviation from expectations, the lost utility should be considered - as it looks in the consumption utility.

Confusion 1: What does the second sentence even mean and how do I consider the lost utility??

My attempt

First of all, let's define what we're looking for. According to page 1143 of the paper:

A selection $\{F_l \in D_l\}_{l \in R}$ is a personal equilibrium (PE) if for all $l \in R$ and $F'_l \in D_l$ $U(F_l | \int F_l dQ(l)) \ge U(F'_l | \int F_l dQ(l))$.

So, provided that the probability is 1, I can get rid of the integrals and prove that $U(r | r) \ge U(x | r) \forall x \ne r$, or, given that there are only two possible choices (right?), that $U((1, -30) | (1, -30)) \ge U((0, 0) | (1, -30))$.

Confusion 2: does he really have only two choices and is the second choice $(0, 0)$?

Next, compute $U(r | r)$. As per p. 1138 of the same work:

We assume that overall utility has two components: $u(c|r) = m(c) + n(c|r)$, where $m(c)$ is "consumption utility" typically stressed in economics, and $n(c|r)$ is "gain-loss utility."

$$ U(r | r) = U((1, -30) | (1, -30)) = m(1, -30) + n((1, -30) | (1, -30)) $$

As per the same page:

We also assume that gain-loss utility is separable: $n(c|r) = \sum_{k=1}^K n_k(c_k | r_k)$.

So I can do:

$$ m(1, -30) = v \times 1 - 30 = 13 - 30 = -17\\ n((1, -30) | (1, -30)) = n(1 | 1) + n(-30 | -30) = (1 - 1) + (-30 + 30) = 0\\ \mbox{Thus, } U(r | r) = -17 + 0 = -17 $$

Now for the second option:

$$ U((0, 0) | (1, -30)) = m(0, 0) + n((0, 0) | (1, -30)) = 0 + n(0 | 1) + n(0 | -30) = \lambda \times (0 - 1) + (0 + 30) = 30 - \lambda = 30 - 4 = 26\\ \mbox{Thus, } U(x | r) = 0 + 26 = 26 $$

Confusion 3: So, I got $U(r | r) \lt U(x | r)$, which is quite the opposite of what I need to prove...

UPDATE

I'm being told by a fellow student that one should multiply the second part of the gain-loss utility by $v$ (!) to get:

$$ n(x | r) = \begin{cases} (x - r), \mbox{if } x \ge r\\ v \times \lambda \times (x - r), \mbox{if } x \lt r \end{cases} $$

So that the computation of $U(x | r)$ I get:

$$ U((0, 0) | (1, -30)) = \lambda \times v \times (0 - 1) + (0 + 30) = -22 $$

...and the same for $U((1, -30) | (1, -30))$.

They also say that this is what the "Do not forget that in the gain-loss utility in case of deviation from expectations, the lost utility should be considered - as it looks in the consumption utility." part of the task tells me to do. Unfortunately, I still have no idea what this part means or why I should multiply only the second part of the function by exactly $v$. I mean, why would you just change the function out of nowhere?

Question

Where am I wrong? And why do I need to multiply by $v$, and do I actually need to do this?

Answer 1

You can interprete $c_1 \in \{0,1\}$ as "I have the tea" and "I don't have the tea", and this tea is valued $v=13$. $c_2$ is just a wealth transfer, i.e., it is $-p$ if the tea is bought and $0$ if the purchase is rejected. Gain-loss utility is assesed separately in each dimension. Such that $$n(c|r) = n_1(c_1|r_1) + n_2(c_2|r_2).$$

In your case the reference point is $\widehat r =(1,-30)$. In words, the buyer expects to purchase the good at price 30. In a PE, it must be that given this reference point $$U (\widehat r| \widehat r) \geq U ( c | \widehat r) \quad \forall c.$$ That is, the buyer prefers to execute his plan.

This leads to utility $$U (\widehat r| \widehat r) = v * 1 - 30 + 0 = 13-30=-17.$$ Suppose instead that the buyer does not buy the good. In pure strategies and with a fixed price, there are only two options "buy at price 30", $(1,30)$ and not buying, $(0,0)$. That is, she does not spend $p$ of her wealth and does not get the good with value $v$, $$U ((0,0)| \widehat r) = v * 0 - 0 \:+ \quad \lambda(v*0-v*1) \:+ \quad (0-(-30)), $$ where the first part is the consumption utility from not buying, zero. The second part is the loss-utility from not having the good weighted by the loss-aversion parameter $\lambda$. He does not have the good, so he feels a loss of $v$, the value of the good. The third part is the gain-utility from saving 30 bucks. Hence, $$U ((0,0)| \widehat r) =30 - \lambda v = -22$$ with the given parameters. We can verify the given PE.

$v$ is the value of the good. Think of this setting as separable consumption utility $$m (c) = m_1 (c_1) + m_2(c_2)$$ with $m_1 (c_1) = v c_1$, i.e., if you have the good you garner $v$ and otherwise zero. The money just enters linearly $m_2 (c_2) = c_2$. Gain-loss utility is always evaluated with respect to consumption utility, see the paper. That is, actually $$n_i (c_i|r_i) = \mu [m_i (c_i) - m_i(r_i)]$$ with the given "universal gain-loss function".

EDIT: To make this useful for all, let me mention the paper corresponding to this model: Heidhues and Kőszegi (TE, 2014). See how they define the product and the money dimension, $k^v =v b$, where $b$ is the $c_1$ of your model (buying or not buying), and $k^p=-bp$. This paper has interesting insights on the impact of consumer loss aversion on pricing and it supersedes their earlier paper.