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.

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

1 Answer 1


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 $y \in C(Z)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.