# Proving the Choice with Recommendations

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.

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

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 $$y \in C(Z)$$.