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The following is a problem I am dealing with related to Weak Axiom of Revealed Preference. I have given my solution below to the situation. What I am not getting is how is WARP not violated?

A law firm looking to hire to fill three positions gets applications from Andrew, Barbara and Celia.

The law firm's set of alternatives is the set of possible hiring decisions:

$ X = \{ \phi, \{a\}, \{b\}, \{c\}, \{a,b\}, \{b,c\}, \{a,c\}, \{a,b,c\} \} $

For any $Y \subset \{a,b,c\}$, define the power set of Y as

$ 2^{Y} \equiv \{Z | Z \subset Y \}$.

$ 2^{Y}$ is the set of hiring decisions that the firm can make when it receives applications from the lawyers in Y.

The law firm's budget sets $ B \in \mathcal{B} $ are the sets of hiring decisions it can make after receiving applications from some combination of Andrew, Barbara and Celia :

$\mathcal{B} = \{2^{Y} |Y \subset \{a, b, c \} \}$

1) When it receives applications from Andrew and Barbara, it will choose to hire Barbara (and not Andrew):-

$ C(2^{\{a,b\}}) = \{b\} $

2) When it receives applications from Barbara and Celia, it will choose to hire Celia (and not Barbara):

$ C(2^{\{b,c\}}) = \{c\} $

The following is the problem I am confused in : Q) What restrictions does the weak axiom place on the firm's hiring decision $C(2^{\{a,b,c\}})$ when it receives applications from Andrew, Barbara and Celia?

My Solution:

According to Mas-Colell et al (Definition 1.C.1) , the Weak Axiom of Revealed Preference says that if x is ever chosen when y is available then there can be no budget set containing both alternatives for which y is chosen and x is not.

So based on my understanding of WARP, in my situation above, when Andrew and Barbara apply the firm chooses Barbara , i.e. $ Barbara \succsim_R Andrew$ and when Barbara and Celia apply, the firm chooses Celia i.e. : $ Celia \succsim_R Barbara $

Here we see that since Barbara is not chosen over Celia , WARP is violated. Because WARP would imply that Barbara is chosen everywhere when Barbara is a choice in the set. So when Andrew , Barbara and Celia apply, and WARP violates the above relation given , the firm would hire only Andrew.

What I am not getting is how is WARP not violated?

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Because WARP would imply that Barbara is chosen everywhere when Barbara is a choice in the set.

This is not true. Instead WARP would say that it is not possible for Andrew to be chosen whenever both Andrew and Barbara are in the choice set, which is different than your claim above.

Just because $b$ is chosen from the set $\{a,b\}$ doesn't mean that $b$ should always be chosen, e.g. from the set $\{b,c\}$. It just means that $b$ is revealed-preferred to $a$, but it doesn't mean that $b$ is preferred to all other possible alternatives (in particular, $c$). For example, the preference $c\succ b\succ a$ is perfectly consistent with the revealed choice pattern of the firm.

Write out the two conditions explicitly: $$ \begin{align} C(\color{red}{\varnothing},\color{red}{\{a\}},\{b\},\color{red}{\{a,b\}})&=\{b\}\tag{1}\\ C(\color{red}{\varnothing},\color{red}{\{b\}},\{c\},\color{red}{\{b,c\}})&=\{c\}\tag{2} \end{align} $$ $(1)$ just says that whenever the four choices --- hire no one, hire $a$ only, hire $b$ only, hire both --- are present, the firm hires $b$ only. Likewise, $(2)$ says whenever the four choices --- hire no one, hire $b$ only, hire $c$ only, hire both --- are present, the firm hires $c$ only. These two are consistent with WARP (for $(1)$, the x in your quoted definition is $a$ and y is $b$; for $(2)$, x is $b$ and y is $c$).

For restrictions on $C(2^{\{a,b,c\}})$, writing it out explicitly, we get $$ C(\color{red}{\varnothing},\color{red}{\{a\}},\color{red}{\{b\}},\color{black}{\{c\}},\color{red}{\{a,b\}},\color{black}{\{a,c\}},\color{red}{\{b,c\}},\color{black}{\{a,b,c\}})=\{\{a\},\{a,c\},\{a,b,c\}\} $$ where the elements colored in red are revealed to be inferior according to $(1)$ and $(2)$.

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