<p>Assume <span class="math-container">$Z \subset Y$</span> and <span class="math-container">$C(Y) \cap Z \ne \emptyset$</span></p>
<p>We need to show that <span class="math-container">$C(Z) = C(Y) \cap Z$</span>.</p>
<p>(<span class="math-container">$\subset$</span>)
Let <span class="math-container">$x \in C(Z) = \arg\max_{\succeq}\{(Z \cap X_1) \cup R(Z \cap X_2)\}$</span></p>
<p>We need to show that <span class="math-container">$x \in C(Y)$</span>. Let <span class="math-container">$y \in C(Y) \cap Z$</span> (which exists as the latter is non-empty by assumption).</p>
<ol>
<li><p>If <span class="math-container">$y \in X_1$</span>, then <span class="math-container">$y \in Z \cap X_1$</span>, so <span class="math-container">$x \succeq y$</span>. By definition of <span class="math-container">$y$</span>, we have that <span class="math-container">$y \succeq w$</span> for all <span class="math-container">$w \in Y \cap X_1$</span>. By transitivity, <span class="math-container">$x \succeq w$</span> for all <span class="math-container">$w \in Y \cap X_1$</span>. Also <span class="math-container">$y \succeq w$</span> for all <span class="math-container">$w \in R(Y \cap X_2)$</span>. Again by transitivity <span class="math-container">$x \succeq w$</span> for all <span class="math-container">$w \in R(Y \cap X_2)$</span>. Conclude that <span class="math-container">$x \in \arg\max_{\succeq}\{(Y \cap X_1) \cup R(Y \cap X_2)\}$</span>, so <span class="math-container">$x \in C(Y)$</span>.</p>
</li>
<li><p>If <span class="math-container">$y \in R(Y \cap X_2)$</span>. Then <span class="math-container">$Z \cap X_2 \subset Y \cap X_2$</span> and <span class="math-container">$R(Y \cap X_2) \cap (Z \cap X_2) \ne \emptyset$</span> as the latter contains <span class="math-container">$y$</span>. As <span class="math-container">$R$</span> satisfies WARP, we have that <span class="math-container">$R(Z \cap X_2) = R(Y \cap X_2) \cap (Z \cap X_2)$</span>. As <span class="math-container">$y$</span> is in the latter, we have <span class="math-container">$y \in R(Z \cap X_2)$</span>. As <span class="math-container">$x \in C(Z)$</span>, we can conclude that <span class="math-container">$x \succeq y$</span> once more. As in point 1 above, we can use this to show that <span class="math-container">$x \in C(Y)$</span>.</p>
</li>
</ol>
<p>(<span class="math-container">$\supset$</span>) Let <span class="math-container">$y \in C(Y) \cap Z$</span>. We want to show that <span class="math-container">$y \in C(Z)$</span>.</p>
<p>We have that <span class="math-container">$y \in \arg\max_{\succeq}((Y \cap X_1) \cup R(Y \cap X_2))\}$</span>.</p>
<ol>
<li><p>As <span class="math-container">$y \succeq x$</span> for all <span class="math-container">$x \in Y \cap X_1$</span>, we ahve that <span class="math-container">$y \succeq x$</span> for all <span class="math-container">$x \in Z \cap X_1$</span>.</p>
</li>
<li><p>Let <span class="math-container">$w \in R(Y \cap X_2)$</span> and <span class="math-container">$z \in R(Z \cap X_2)$</span>. By definition, we have that <span class="math-container">$y \succeq w$</span> and by assumption <span class="math-container">$w \sim z$</span>. As such, <span class="math-container">$y \succeq z$</span>. This holds for all <span class="math-container">$z \in R(Z \cap X_2)$</span>.</p>
</li>
</ol>
<p>Conclude that <span class="math-container">$y \succeq x$</span> for all <span class="math-container">$x \in (Z \cap X_1) \cup R(Z \cap X_2)$</span>, so <span class="math-container">$y \in C(Z)$</span>.</p>

CC BY-SA 4.0
2023-12-28T09:38:15.467Z
2023-12-28T09:38:15.467Z
tdm
<p>Suppose that there are two types of outcomes, i.e. <span class="math-container">$X=X_1 \cup X_2$</span> with <span class="math-container">$X_1 \cap X_2=∅$</span>. All outcomes in <span class="math-container">$X_2$</span> 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 <span class="math-container">$X_2$</span> and the available options in <span class="math-container">$X_1$</span>. Formally, <span class="math-container">$x\sim y$</span> for all <span class="math-container">$x,y \in X_2$</span> and</p>
<p><span class="math-container">$$ C(A)=max_≿ \{(A∩ X_1)∪ R(A∩ X_2)\} $$</span></p>
<p>Show that this choice procedure satisfies WARP if <span class="math-container">$R$</span> satisfies WARP.</p>
<hr />
<p>In these types of questions, I use as a definition of WARP:</p>
<p>A choice function <span class="math-container">$C$</span> satisfies the weak axiom of revealed preference if for all <span class="math-container">$Y,Z \in \mathcal{M}(X)$</span>,
<span class="math-container">$$Z\subset Y \quad\text{and}\quad  C(Y)\cap Z\neq ∅  \quad \text{implies}\quad  C(Z)=C(Y)∩ Z.$$</span></p>
<hr />
<p>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.</p>
<p>Thanks in advance.</p>

CC BY-SA 4.0
2023-12-26T17:42:51.397Z
Please define what $R$ is. Is it the original preference relation over $X$?
Walrasian Auctioneer
<microeconomics><decision-theory><choice-theory>
Proving the Choice with Recommendations
homo-economitux
# Proving the Choice with Recommendations

(Asked by: username_0 on 2023-12-24)

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.

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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 $Y,Z \in \mathcal{M}(X)$,
$$Z\subset Y \quad\text{and}\quad C(Y)\cap Z\neq ∅ \quad \text{implies}\quad C(Z)=C(Y)∩ Z.$$

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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.

**Tags:** microeconomics, decision-theory, choice-theory
> **Comments:**
> **username_1**: Please define what $R$ is. Is it the original preference relation over $X$?  

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## 1 answers

### Answer by username_2 on 2023-12-28. Score: 3 | Accepted answer

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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