<p>I think I have an answer. It had previously been shown that there exists an individual <span class="math-container">$i^*$</span> such that <span class="math-container">$F((\succsim^i)_{i \in N}) = \succsim^{i^*}$</span> for <strong>any</strong> preference profile <span class="math-container">$(\succsim^i)_{i \in N}$</span>. Now, if we let <span class="math-container">$f((\succsim^i)_{i \in N}) = x$</span>, that means the only restriction we are imposing on <span class="math-container">$(\succsim^i)_{i \in N}$</span> is that not all individuals rank <span class="math-container">$y$</span> as the best alternative, because then by unanimity the social choice rule would select <span class="math-container">$y$</span>. In particular, we are allowing <span class="math-container">$(\succsim^i)_{i \in N}$</span> to be a preference profile in which all individuals rank <span class="math-container">$x$</span> as the best alternative. Now consider the profile <span class="math-container">$(\trianglerighteq)_{i \in N}$</span>, which by definition leaves the relative ordering of <span class="math-container">$x$</span> and <span class="math-container">$y$</span> the same as it is in <span class="math-container">$(\succsim^i)_{i \in N}$</span>. By Step 1 <span class="math-container">$f((\trianglerighteq)_{i \in N}) \in \{x, y \}$</span>. If <span class="math-container">$f((\trianglerighteq)_{i \in N}) = y$</span>, then that would need to hold even in the case where all individuals rank <span class="math-container">$x$</span> as the best alternative, thus violating unanimity. Therefore <span class="math-container">$f((\trianglerighteq)_{i \in N})$</span>.</p>
<p>Does this make sense to you?</p>

CC BY-SA 4.0
2024-07-14T20:12:47.833Z
2024-07-14T20:14:50.743Z
2024-07-14T20:14:50.743Z
Viktor Glushkov
<p>I'm self-studying Osborne and Rubinstein's <em>Models in Microeconomic Theory</em> and I'm struggling to understand a step in their proof of the Gibbard-Satterthwaite theorem. I won't provide the full proof, but the passages of interest are the two following steps (<span class="math-container">$f$</span> denotes a social choice rule):</p>
<blockquote>
<p><strong>Step 1:</strong> If <span class="math-container">$f((\succsim^i)_{i \in N}) = x$</span> and <span class="math-container">$(\trianglerighteq^i)_{i \in N}$</span> differs from <span class="math-container">$(\succsim^i)_{i \in N}$</span> only in that for some individual <span class="math-container">$j$</span> the rank of some alternative <span class="math-container">$y$</span> is higher in <span class="math-container">$\trianglerighteq^j$</span> than it is in <span class="math-container">$\succsim^j$</span>, then <span class="math-container">$f((\trianglerighteq^i)_{i \in N}) \in \{x, y \}$</span>.</p>
</blockquote>
<blockquote>
<p><strong>Step 2:</strong> Let <span class="math-container">$f((\succsim^i)_{i \in N}) = x$</span> and let <span class="math-container">$y$</span> be another alternative. By Step 1 we have <span class="math-container">$f((\trianglerighteq^i)_{i \in N}) = x$</span>, where <span class="math-container">$\trianglerighteq^i$</span> is obtained from <span class="math-container">$\succsim^i$</span> by moving all alternatives except <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to below these two alternatives (retaining the order of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>).</p>
</blockquote>
<p>In Step 2, what allows us to conclude that <span class="math-container">$f((\trianglerighteq^i)_{i \in N}) = x$</span> instead of <span class="math-container">$f((\trianglerighteq^i)_{i \in N}) = y$</span>?</p>

<mathematical-economics><social-choice>
Proof of the Gibbard-Satterthwaite theorem in Osborne and Rubinstein
Viktor Glushkov
# Proof of the Gibbard-Satterthwaite theorem in Osborne and Rubinstein

(Asked by: username_0 on 2024-07-14)

I'm self-studying Osborne and Rubinstein's *Models in Microeconomic Theory* and I'm struggling to understand a step in their proof of the Gibbard-Satterthwaite theorem. I won't provide the full proof, but the passages of interest are the two following steps ($f$ denotes a social choice rule):

> **Step 1:** If $f((\succsim^i)\_{i \in N}) = x$ and $(\trianglerighteq^i)\_{i \in N}$ differs from $(\succsim^i)\_{i \in N}$ only in that for some individual $j$ the rank of some alternative $y$ is higher in $\trianglerighteq^j$ than it is in $\succsim^j$, then $f((\trianglerighteq^i)\_{i \in N}) \in \{x, y \}$.

> **Step 2:** Let $f((\succsim^i)\_{i \in N}) = x$ and let $y$ be another alternative. By Step 1 we have $f((\trianglerighteq^i)\_{i \in N}) = x$, where $\trianglerighteq^i$ is obtained from $\succsim^i$ by moving all alternatives except $x$ and $y$ to below these two alternatives (retaining the order of $x$ and $y$).

In Step 2, what allows us to conclude that $f((\trianglerighteq^i)\_{i \in N}) = x$ instead of $f((\trianglerighteq^i)\_{i \in N}) = y$?

**Tags:** mathematical-economics, social-choice


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## 1 answers

### Answer by username_0 on 2024-07-14. Score: 0

I think I have an answer. It had previously been shown that there exists an individual $i^\*$ such that $F((\succsim^i)\_{i \in N}) = \succsim^{i^\*}$ for **any** preference profile $(\succsim^i)\_{i \in N}$. Now, if we let $f((\succsim^i)\_{i \in N}) = x$, that means the only restriction we are imposing on $(\succsim^i)\_{i \in N}$ is that not all individuals rank $y$ as the best alternative, because then by unanimity the social choice rule would select $y$. In particular, we are allowing $(\succsim^i)\_{i \in N}$ to be a preference profile in which all individuals rank $x$ as the best alternative. Now consider the profile $(\trianglerighteq)\_{i \in N}$, which by definition leaves the relative ordering of $x$ and $y$ the same as it is in $(\succsim^i)\_{i \in N}$. By Step 1 $f((\trianglerighteq)\_{i \in N}) \in \{x, y \}$. If $f((\trianglerighteq)\_{i \in N}) = y$, then that would need to hold even in the case where all individuals rank $x$ as the best alternative, thus violating unanimity. Therefore $f((\trianglerighteq)\_{i \in N})$.

Does this make sense to you?


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