<ol>
<li><p>Concerning your first question, a simple symmetry will do (I think).
<span class="math-container">$$
\begin{align*}
&amp;x/2  + z/2  \sim y/2  + z/2 ,\\
\to &amp;x/2  \sim y/2, &amp;&amp;\text{(subtract $z/2$)}\\
\to &amp;x/2  + y/2  \sim y \text{ and } x \sim y/2  + x/2, &amp;&amp;\text{(add $y/2$ and $x/2$ respectively)}\\
\to &amp; x \sim y &amp;&amp;\text{(by transitivity of the above)}
\end{align*}
$$</span>
Using a similar reasoning, we see that <span class="math-container">$x \succeq y$</span> is equivalent to <span class="math-container">$x/2 \succeq y/2$</span></p>
</li>
<li><p>For your second question, note that by repeated application of <span class="math-container">$x \sim y \to x/2 \sim y/2$</span> we get:
<span class="math-container">$$
x \sim y \to x/2 \sim y/2 \to \ldots \to x/2^n \sim y/2^n.
$$</span>
Now, let us show that for all <span class="math-container">$k \in \mathbb{N}$</span>, <span class="math-container">$x \sim y \to kx/2^n \sim ky/2^n$</span>. We proceed by induction. For <span class="math-container">$k = 1$</span> we have the result as <span class="math-container">$x \sim y \to x/2^n \sim y/2^n$</span>. Assume it holds up to <span class="math-container">$k$</span> and consider the case <span class="math-container">$k+1$</span>.
Then,
<span class="math-container">$$
\begin{align*}
&amp;x \sim y \to kx/2^n \sim ky/2^n \text{ and } x/2^n \sim y/2^n,\\
\to &amp;kx/2^n + x/2^n \sim ky/2^n + x/2^n \text{ and } ky/2^n + x/2^n \sim ky/2^n + y/2^n,&amp;&amp;\\
\to &amp;(k+1)x/2^n \sim ky /2^n + x/2^n \text{ and } ky/2^n + x/2^n \sim (k+1)y/2^n\\
\to &amp;(k+1)x/2^n \sim (k+1)y/2^n &amp;&amp; \text{(by transitivity)}
\end{align*}
$$</span>
Note that similarly, we can show that <span class="math-container">$x \succeq y$</span> implies <span class="math-container">$kx/2^n \succeq ky/2^n$</span>. For all <span class="math-container">$k, n \in \mathbb{N}$</span>.</p>
</li>
<li><p>Let us first show that there is a <span class="math-container">$n \in \mathbb{N}$</span> such that <span class="math-container">$(1,0) \preceq (0,2^n)$</span>. If not, then for all <span class="math-container">$n \in \mathbb{N}$</span>, <span class="math-container">$(1,0) \succ (0, 2^n)$</span>.
The latter implies that (multiplying boths sides by <span class="math-container">$1/2^n$</span> and using steps 1 and 2 above),
<span class="math-container">$$
(1/2^n, 0) \succeq (0,1)
$$</span>
Taking the limit for <span class="math-container">$n \to \infty$</span>, continuity gives that,
<span class="math-container">$$
(0,0) \succeq (0,1),
$$</span>
which contradicts monotonicity.
This shows that there is some <span class="math-container">$K \in \mathbb{R}_+$</span> such that <span class="math-container">$(1,0) \preceq (0,K)$</span>. Now let,
<span class="math-container">$$
k = \inf \{k \in \mathbb{R}| (1,0) \preceq (0,k)\}
$$</span>
Note that the set <span class="math-container">$\{k \in \mathbb{R}| (1,0) \preceq (0,k)\}$</span> is bounded from below by <span class="math-container">$0$</span> as <span class="math-container">$(1,0) \succ (0,0)$</span> by monotoncity, so this infimum is well defined. Also, by continuity, the upper contour sets are closed, so,
<span class="math-container">$(1,0) \preceq (0,k)$</span>. Given that <span class="math-container">$k$</span> is the infimum, we in fact have <span class="math-container">$(1,0) \sim (0,k)$</span>.</p>
</li>
</ol>

CC BY-SA 4.0
2024-12-24T07:31:47.917Z
Thank you for your answer! You've made everything clear to me now.
Joseph Basford
CC BY-SA 4.0
2024-12-23T10:48:36.263Z
2024-12-23T10:48:36.263Z
tdm
<p>I am going over an old problem set exercise and cannot figure out how certain conclusions follow in the proof my professor has given. There are several other questions on this topic on here (e.g., <a href="https://economics.stackexchange.com/questions/7089/existence-of-linear-representation">this question</a> and <a href="https://economics.stackexchange.com/questions/32748/linear-utility">this other question</a>), but my question is about parsing the particular proof my professor has presented.</p>
<p>Suppose I have a rational (i.e. complete and transitive) preference relation <span class="math-container">$(\succeq)$</span> over <span class="math-container">$\mathbb{R}^2$</span>.</p>
<p>Consider the following axioms.</p>
<ol>
<li><span class="math-container">$\textbf{MON}$</span>: <span class="math-container">$a\geq b$</span> (pointwise) implies <span class="math-container">$a\succeq b$</span>. Also, if <span class="math-container">$a\neq b$</span> and <span class="math-container">$a\geq b$</span>, then <span class="math-container">$a\succ b$</span>.</li>
<li><span class="math-container">$\textbf{ADD}$</span>: <span class="math-container">$(a_1,a_2)\succeq (b_1,b_2)$</span> implies that <span class="math-container">$(a_1+t,a_2+s)\succeq (b_1+t,b_2+s)$</span> for any <span class="math-container">$(t,s)\in \mathbb{R}^2$</span>.</li>
<li><span class="math-container">$\textbf{CON}$</span>: The upper and lower contour sets are closed.</li>
</ol>
<p>I want to prove these axioms hold if and only if <span class="math-container">$(\succeq)$</span> has a linear representation, i.e., if there is <span class="math-container">$u:\mathbb{R}^2\to \mathbb{R}$</span> that represents preferences which has form <span class="math-container">$$u((x_1,x_2))=\alpha_1x_1+\alpha_2x_2$$</span> where <span class="math-container">$\alpha_1,\alpha_2\in [0,\infty)$</span>.</p>
<p>Proving such a representation implies the axioms is trivial.</p>
<p>To prove the reverse implication, we want to take a similar approach to the proof of the vNM representation theorem and first prove that independence holds. Thence, we can conclude that a linear representation exists by Milnor and Herstein's mixture space theorem. In any case, rationality and continuity already get us a representation, so I just want to show it can be made linear.</p>
<p>First, I claim that <span class="math-container">$x\sim y \iff \frac{1}{2}x+\frac{1}{2}z\sim \frac{1}{2}y+\frac{1}{2}z$</span> for any <span class="math-container">$x,y,z\in \mathbb{R}^2$</span>. My professor argues in the following way. By <span class="math-container">$\textbf{ADD}$</span>, adding <span class="math-container">$\frac{1}{2}z$</span> to both sides, <span class="math-container">$$\frac{1}{2}x+\frac{1}{2}z\sim \frac{1}{2}y+\frac{1}{2}z \iff \frac{1}{2}x\sim \frac{1}{2}y$$</span> Further, again by <span class="math-container">$\textbf{ADD}$</span>, adding <span class="math-container">$\frac{1}{2}y$</span> to both sides, <span class="math-container">$$\frac{1}{2}x\sim \frac{1}{2}y\iff \frac{1}{2}x+\frac{1}{2}y\sim y$$</span> From here, my professor claims that this shows <span class="math-container">$$\frac{1}{2}x\sim \frac{1}{2}y \iff x\sim y$$</span> Why does this follow?</p>
<p>Additionally, after showing <span class="math-container">$x\sim y \iff \frac{1}{2}x+\frac{1}{2}z\sim \frac{1}{2}y+\frac{1}{2}z$</span>, he claims we can repeatedly apply this relation <span class="math-container">$2^n$</span> times, for <span class="math-container">$n\in \mathbb{N}$</span>, to obtain <span class="math-container">$$x\sim y \iff \alpha x+(1-\alpha)z\sim\alpha y+(1-\alpha)z$$</span> for any <span class="math-container">$\alpha=\frac{k}{2^n}$</span> with <span class="math-container">$0\leq k \leq 2^n$</span>. Why is this true? How can we access <span class="math-container">$\alpha&gt;\frac{1}{2}$</span> by applying this relation?</p>
<p>After extending by <span class="math-container">$\textbf{CON}$</span> to any <span class="math-container">$\alpha$</span>, he concludes that setting <span class="math-container">$z=0$</span> gives that <span class="math-container">$$x\sim y \iff \lambda x\sim \lambda y$$</span> for any <span class="math-container">$\lambda&gt;0$</span>. Then, by <span class="math-container">$\textbf{CON}$</span> and <span class="math-container">$\textbf{MON}$</span>, <span class="math-container">$(1,0)\sim(0,k)$</span> for some <span class="math-container">$k\in \mathbb{R}$</span>. Why does such a <span class="math-container">$k$</span> exist? For <span class="math-container">$k\leq 0$</span>, <span class="math-container">$(1,0)\geq (0,k)$</span>, but for <span class="math-container">$k&gt;0$</span> they cannot be ordered by pointwise inequality, so how do we conclude such a <span class="math-container">$k$</span> exists?</p>
<p>In summary, my questions are,</p>
<ol>
<li>Why is it that <span class="math-container">$x\sim y\iff \frac{1}{2}x\sim \frac{1}{2}y$</span>?</li>
<li>How do we then show that <span class="math-container">$$x\sim y \iff \alpha x+(1-\alpha)z\sim\alpha y+(1-\alpha)z$$</span> for any <span class="math-container">$\alpha=\frac{k}{2^n}$</span> with <span class="math-container">$0\leq k \leq 2^n$</span>?</li>
<li>Why is it true that <span class="math-container">$(1,0)\sim(0,k)$</span> for some <span class="math-container">$k\in \mathbb{R}$</span>?</li>
</ol>

<microeconomics><mathematical-economics><preferences><decision-theory>
Additive Utility Representation
Joseph Basford
# Additive Utility Representation

(Asked by: username_0 on 2024-12-23)

I am going over an old problem set exercise and cannot figure out how certain conclusions follow in the proof my professor has given. There are several other questions on this topic on here (e.g., [this question](https://economics.stackexchange.com/questions/7089/existence-of-linear-representation) and [this other question](https://economics.stackexchange.com/questions/32748/linear-utility)), but my question is about parsing the particular proof my professor has presented.

Suppose I have a rational (i.e. complete and transitive) preference relation $(\succeq)$ over $\mathbb{R}^2$.

Consider the following axioms.

1. $\textbf{MON}$: $a\geq b$ (pointwise) implies $a\succeq b$. Also, if $a\neq b$ and $a\geq b$, then $a\succ b$.
2. $\textbf{ADD}$: $(a\_1,a\_2)\succeq (b\_1,b\_2)$ implies that $(a\_1+t,a\_2+s)\succeq (b\_1+t,b\_2+s)$ for any $(t,s)\in \mathbb{R}^2$.
3. $\textbf{CON}$: The upper and lower contour sets are closed.

I want to prove these axioms hold if and only if $(\succeq)$ has a linear representation, i.e., if there is $u:\mathbb{R}^2\to \mathbb{R}$ that represents preferences which has form $$u((x\_1,x\_2))=\alpha\_1x\_1+\alpha\_2x\_2$$ where $\alpha\_1,\alpha\_2\in [0,\infty)$.

Proving such a representation implies the axioms is trivial.

To prove the reverse implication, we want to take a similar approach to the proof of the vNM representation theorem and first prove that independence holds. Thence, we can conclude that a linear representation exists by Milnor and Herstein's mixture space theorem. In any case, rationality and continuity already get us a representation, so I just want to show it can be made linear.

First, I claim that $x\sim y \iff \frac{1}{2}x+\frac{1}{2}z\sim \frac{1}{2}y+\frac{1}{2}z$ for any $x,y,z\in \mathbb{R}^2$. My professor argues in the following way. By $\textbf{ADD}$, adding $\frac{1}{2}z$ to both sides, $$\frac{1}{2}x+\frac{1}{2}z\sim \frac{1}{2}y+\frac{1}{2}z \iff \frac{1}{2}x\sim \frac{1}{2}y$$ Further, again by $\textbf{ADD}$, adding $\frac{1}{2}y$ to both sides, $$\frac{1}{2}x\sim \frac{1}{2}y\iff \frac{1}{2}x+\frac{1}{2}y\sim y$$ From here, my professor claims that this shows $$\frac{1}{2}x\sim \frac{1}{2}y \iff x\sim y$$ Why does this follow?

Additionally, after showing $x\sim y \iff \frac{1}{2}x+\frac{1}{2}z\sim \frac{1}{2}y+\frac{1}{2}z$, he claims we can repeatedly apply this relation $2^n$ times, for $n\in \mathbb{N}$, to obtain $$x\sim y \iff \alpha x+(1-\alpha)z\sim\alpha y+(1-\alpha)z$$ for any $\alpha=\frac{k}{2^n}$ with $0\leq k \leq 2^n$. Why is this true? How can we access $\alpha>\frac{1}{2}$ by applying this relation?

After extending by $\textbf{CON}$ to any $\alpha$, he concludes that setting $z=0$ gives that $$x\sim y \iff \lambda x\sim \lambda y$$ for any $\lambda>0$. Then, by $\textbf{CON}$ and $\textbf{MON}$, $(1,0)\sim(0,k)$ for some $k\in \mathbb{R}$. Why does such a $k$ exist? For $k\leq 0$, $(1,0)\geq (0,k)$, but for $k>0$ they cannot be ordered by pointwise inequality, so how do we conclude such a $k$ exists?

In summary, my questions are,

1. Why is it that $x\sim y\iff \frac{1}{2}x\sim \frac{1}{2}y$?
2. How do we then show that $$x\sim y \iff \alpha x+(1-\alpha)z\sim\alpha y+(1-\alpha)z$$ for any $\alpha=\frac{k}{2^n}$ with $0\leq k \leq 2^n$?
3. Why is it true that $(1,0)\sim(0,k)$ for some $k\in \mathbb{R}$?

**Tags:** microeconomics, mathematical-economics, preferences, decision-theory

---

## 1 answers

### Answer by username_1 on 2024-12-23. Score: 2 | Accepted answer

1. Concerning your first question, a simple symmetry will do (I think).
   $$
   \begin{align\*}
   &x/2 + z/2 \sim y/2 + z/2 ,\\
   \to &x/2 \sim y/2, &&\text{(subtract $z/2$)}\\
   \to &x/2 + y/2 \sim y \text{ and } x \sim y/2 + x/2, &&\text{(add $y/2$ and $x/2$ respectively)}\\
   \to & x \sim y &&\text{(by transitivity of the above)}
   \end{align\*}
   $$
   Using a similar reasoning, we see that $x \succeq y$ is equivalent to $x/2 \succeq y/2$
2. For your second question, note that by repeated application of $x \sim y \to x/2 \sim y/2$ we get:
   $$
   x \sim y \to x/2 \sim y/2 \to \ldots \to x/2^n \sim y/2^n.
   $$
   Now, let us show that for all $k \in \mathbb{N}$, $x \sim y \to kx/2^n \sim ky/2^n$. We proceed by induction. For $k = 1$ we have the result as $x \sim y \to x/2^n \sim y/2^n$. Assume it holds up to $k$ and consider the case $k+1$.
   Then,
   $$
   \begin{align\*}
   &x \sim y \to kx/2^n \sim ky/2^n \text{ and } x/2^n \sim y/2^n,\\
   \to &kx/2^n + x/2^n \sim ky/2^n + x/2^n \text{ and } ky/2^n + x/2^n \sim ky/2^n + y/2^n,&&\\
   \to &(k+1)x/2^n \sim ky /2^n + x/2^n \text{ and } ky/2^n + x/2^n \sim (k+1)y/2^n\\
   \to &(k+1)x/2^n \sim (k+1)y/2^n && \text{(by transitivity)}
   \end{align\*}
   $$
   Note that similarly, we can show that $x \succeq y$ implies $kx/2^n \succeq ky/2^n$. For all $k, n \in \mathbb{N}$.
3. Let us first show that there is a $n \in \mathbb{N}$ such that $(1,0) \preceq (0,2^n)$. If not, then for all $n \in \mathbb{N}$, $(1,0) \succ (0, 2^n)$.
   The latter implies that (multiplying boths sides by $1/2^n$ and using steps 1 and 2 above),
   $$
   (1/2^n, 0) \succeq (0,1)
   $$
   Taking the limit for $n \to \infty$, continuity gives that,
   $$
   (0,0) \succeq (0,1),
   $$
   which contradicts monotonicity.
   This shows that there is some $K \in \mathbb{R}\_+$ such that $(1,0) \preceq (0,K)$. Now let,
   $$
   k = \inf \{k \in \mathbb{R}| (1,0) \preceq (0,k)\}
   $$
   Note that the set $\{k \in \mathbb{R}| (1,0) \preceq (0,k)\}$ is bounded from below by $0$ as $(1,0) \succ (0,0)$ by monotoncity, so this infimum is well defined. Also, by continuity, the upper contour sets are closed, so,
   $(1,0) \preceq (0,k)$. Given that $k$ is the infimum, we in fact have $(1,0) \sim (0,k)$.

> **Comments:**
> **username_0**: Thank you for your answer! You've made everything clear to me now.  

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