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Suppose that there are two types of outcomes, i.e. $X=X_1 \cup X_2$ with $X_1 \cap X_2=∅$. All outcomes in $X_2$ are the same to the decision maker (he doesn't understand these kind of products). He asks a friend for advice and then chooses between the recommended option in $X_2$ and the available options in $X_1$. Formally, $x\sim y$ for all $x,y \in X_2$ and

$$ C(A)=max_≿ \{(A∩ X_1)∪ R(A∩ X_2)\} $$

Show that this choice procedure satisfies WARP if $R$ satisfies WARP.


In these types of questions, I use as a definition of WARP:

A choice function $C$ satisfies the weak axiom of revealed preference if for all $Y,Z \in \mathcal{M}(X)$, $$Z\subset Y \quad\text{and}\quad C(Y)\cap Z\neq ∅ \quad \text{implies}\quad C(Z)=C(Y)∩ Z.$$


Can anyone help me to handle this proof? I easily handled the case with not partitioned set, but I couldn't get it done with this one.

Thanks in advance.

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  • $\begingroup$ Please define what $R$ is. Is it the original preference relation over $X$? $\endgroup$ Dec 26, 2023 at 17:42

1 Answer 1

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Assume $Z \subset Y$ and $C(Y) \cap Z \ne \emptyset$

We need to show that $C(Z) = C(Y) \cap Z$.

($\subset$) Let $x \in C(Z) = \arg\max_{\succeq}\{(Z \cap X_1) \cup R(Z \cap X_2)\}$

We need to show that $x \in C(Y)$. Let $y \in C(Y) \cap Z$ (which exists as the latter is non-empty by assumption).

  1. If $y \in X_1$, then $y \in Z \cap X_1$, so $x \succeq y$. By definition of $y$, we have that $y \succeq w$ for all $w \in Y \cap X_1$. By transitivity, $x \succeq w$ for all $w \in Y \cap X_1$. Also $y \succeq w$ for all $w \in R(Y \cap X_2)$. Again by transitivity $x \succeq w$ for all $w \in R(Y \cap X_2)$. Conclude that $x \in \arg\max_{\succeq}\{(Y \cap X_1) \cup R(Y \cap X_2)\}$, so $x \in C(Y)$.

  2. If $y \in R(Y \cap X_2)$. Then $Z \cap X_2 \subset Y \cap X_2$ and $R(Y \cap X_2) \cap (Z \cap X_2) \ne \emptyset$ as the latter contains $y$. As $R$ satisfies WARP, we have that $R(Z \cap X_2) = R(Y \cap X_2) \cap (Z \cap X_2)$. As $y$ is in the latter, we have $y \in R(Z \cap X_2)$. As $x \in C(Z)$, we can conclude that $x \succeq y$ once more. As in point 1 above, we can use this to show that $x \in C(Y)$.

($\supset$) Let $y \in C(Y) \cap Z$. We want to show that $y \in C(Z)$.

We have that $y \in \arg\max_{\succeq}((Y \cap X_1) \cup R(Y \cap X_2))\}$.

  1. As $y \succeq x$ for all $x \in Y \cap X_1$, we ahve that $y \succeq x$ for all $x \in Z \cap X_1$.

  2. Let $w \in R(Y \cap X_2)$ and $z \in R(Z \cap X_2)$. By definition, we have that $y \succeq w$ and by assumption $w \sim z$. As such, $y \succeq z$. This holds for all $z \in R(Z \cap X_2)$.

Conclude that $y \succeq x$ for all $x \in (Z \cap X_1) \cup R(Z \cap X_2)$, so $y \in C(Z)$.

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