<p>Consider the locations (1) $(0.000001,1)$ and (2) $(0.0000005,10)$. $U\left(x_1,y_1\right) = 1.1$.$U\left(x_2,y_2\right) = 10.05$. However, $x_2 &lt; x_1$, so this is not a lexicographic ordering. It is only with the additional constraint that the values of $x$ and $y$ be integers $\in[0,1000]$ that the function you proposes has this attribute. Because real values of $x$ and $y$ implies that $x_i / y_i$ can be made arbitrarily large or small, no linear function can guarantee a lexicographic ordering over all possible values of $x$ and $y$.  That is, specify a utility function of the form $$U = \beta \cdot X + Y$$
 If $X$ has to be an integer then as long as $\beta&gt;1000$ this will be lexicographic on the domain of the problem because a one unit increase in $X$ will always pay off more utility than 1,000 units of $Y$. But if $X$ is a real number, the problem changes. Consider any arbitrarily value of $\beta &gt; 1000$ that would have satisfied the integer problem. For changes in $X &gt; 1/\beta$ this continues to work (preserve the lex. prefs)  in the real number problem. But for changes in $\Delta  X &lt; 1000 / \beta$:
$$ U(X + \Delta  X,0) = \beta (X+\Delta  X) &lt; \beta X + 1000= U(X,1000)$$
Because by construction $\beta \cdot \Delta X &lt; 1000$. So these preferences are not lexicographic over arbitrary values of $X$ and $Y$. </p>

CC BY-SA 3.0
2015-08-13T14:32:32.277Z
2015-08-13T14:32:32.277Z
BKay
<p>We can say more generally that lexical preferences are not representable using a continuous utility function. Lexical preferences are not continuous. Note the definition of a continuous preference relation.</p>

<p>The preference relation $\succeq$ is continuous if for any sequences of consumption bundles $(x_{i})_{i \in \mathbb{N}}$ and $(y_{i})_{i \in \mathbb{N}}$ with $x_{i} \to x$, $y_{i} \to y$, and $x_{i} \succeq y_{i}$ for each $i \in \mathbb{N}$, then $x \succeq y$. That is, continuity preserves the relation at the limit point.</p>

<p>Consider $(x_{i})_{i \in \mathbb{N}}$ defined by $x_{i} = (\frac{1}{2^{i}}, 0)$ and $(y_{i})_{i \in \mathbb{N}}$ defined by $y_{i} = (0, 1)$. Clearly, $x_{i} \succeq y_{i}$ for each $i \in \mathbb{N}$. However, $x_{i} \to (0, 0)$ while $y_{i} \to (0, 1)$. So this preference relation is not preserved at the limit point.</p>

<p>Even more generally, no utility function represents the lexical preference relation. I prove for $\mathbb{R}_{+}^{2}$, but this argument extends to $\mathbb{R}_{+}^{n}$ by projecting into $\mathbb{R}_{+}^{2}$. </p>

<p><strong>Proof</strong>: Suppose to the contrary that some utility function $u : \mathbb{R}_{+}^{2} \to \mathbb{R}$ represents $\succ_{lex}$. We thus have $u(x, 1) &gt; u(x, 0)$, as $(x, 1) \succ (x, 0)$. We construct the interval $I(x) = [u(x, 0), u(x, 1)]$. Now for any two distinct $x, y \in \mathbb{R}_{+}$, $I(x) \cap I(y) = \emptyset$ as we have either $x &gt; y$ or $y &gt; x$ (so WLOG, we have $(x, 0) \succ (y, 1)$). </p>

<p>Define $\mathbb{I} = \{ I(x) : x \in \mathbb{R}_{+} \}$, and let $\phi : \mathbb{R}_{+} \to \mathbb{I}$ be given by $\phi(x) = I(x)$. Observe that $\phi$ is an injection, as each $I(x), I(y)$ are disjoint for distinct $x, y$.</p>

<p>Note that $\mathbb{Q}$ is dense in $\mathbb{R}$. So there exists a rational number in each interval. Define $\tau : \mathbb{I} \to \mathbb{Q}$ such that $\tau(I(x))$ returns a rational number contained in $I(x)$. So $\tau$ is an injection. We have $\tau$ composed with $\phi$ an injection, which implies $|\mathbb{R}_{+}| \leq |\mathbb{Q}_{+}|$, a contradiction. QED.</p>

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2015-08-13T17:26:30.543Z
I rejected the edit because I assume WLOG x > y. So I meant to say what I wrote in the proof. :-)
ml0105
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2015-08-13T18:40:58.190Z
So how does $u(x,0)>u(y,0)$ and $u(x,1)>u(y,1)$ imply that $I(x)\cap I(y)=\emptyset$?
Erel Segal-Halevi
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2015-08-13T18:47:16.113Z
Actually, whoops. The edit you proposed was correct. We would have $u(x, 0) > u(y, 1)$ WLOG. Sorry for that minor error!
ml0105
CC BY-SA 3.0
2015-08-14T05:15:12.007Z
‎Thanks. So, this impossibility result has nothing to do with probabilities or with von-Neumann-Morgenstern.
Erel Segal-Halevi
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2015-08-14T05:18:13.240Z
It's a more general result in topology that happens to come up in consumer theory. I view consumer theory as applied topology with an economics intuition.
ml0105
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2015-08-14T08:03:36.290Z
Do you know of a general reference to this question of representing ordinal as cardinal? http://math.stackexchange.com/q/1396145/29780
Erel Segal-Halevi
CC BY-SA 3.0
2015-08-14T19:31:08.713Z
The works of Arrow and Debreu are probably the best places to look. I'm neither a topologist nor a consumer theorist. This set of slides might be a good starting piont though: http://www.econ.ucla.edu/iobara/lecturepreferenceandutility201a.pdf
ml0105
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2015-08-13T16:52:20.767Z
2015-08-13T18:46:43.403Z
2015-08-13T18:46:43.403Z
ml0105
<p>I am stuck on the following exercise, related to preference relations and <a href="https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem" rel="noreferrer">von-Neumann-Morgenstern utility function</a>.</p>
<blockquote>
<p>A farmer wants to dig a well in a square field <span class="math-container">$[0,1000]\times[0,1000]$</span>. The preferences of the farmer on the possible locations are lexicographic, i.e:</p>
<ul>
<li>If <span class="math-container">$x_1&lt;x_2$</span> then <span class="math-container">$(x_1,y_1)\prec(x_2,y_2)$</span> for all <span class="math-container">$y_1,y_2$</span>.</li>
<li>If <span class="math-container">$x_1=x_2=x$</span>, then <span class="math-container">$(x,y_1)\prec(x,y_2)$</span> iff <span class="math-container">$y_1 &lt; y_2$</span>.</li>
</ul>
<p>Initially, assume that the well location must have integer coordinates.
Is there a preference relation on lotteries, that satisfies the von-Neumann-Morgenstern axioms, and extends the lexicographic preference relation? If so, what is a linear utility function that represents this relation?</p>
</blockquote>
<p>I think the answer is yes, and a possible linear utility function is: <span class="math-container">$u(x,y)=100000x + y$</span>.</p>
<blockquote>
<p>Now, assume that the well location can have real coordinates. Prove that there is no linear utility function that represents the preference relation on lotteries.
Which one of von-Neumann-Morgenstern axioms is violated by the preference relation on lotteries?</p>
</blockquote>
<p>Here I am stuck. I don't understand why the utility function I suggested above doesn't work? And what axiom is violated here?</p>

<utility><preferences>
Lexicographic preference relation cannot be represented by a utility function
Erel Segal-Halevi
# Lexicographic preference relation cannot be represented by a utility function

(Asked by: username_0 on 2015-08-13)

I am stuck on the following exercise, related to preference relations and [von-Neumann-Morgenstern utility function](https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem).

> A farmer wants to dig a well in a square field $[0,1000]\times[0,1000]$. The preferences of the farmer on the possible locations are lexicographic, i.e:
>
> * If $x\_1<x\_2$ then $(x\_1,y\_1)\prec(x\_2,y\_2)$ for all $y\_1,y\_2$.
> * If $x\_1=x\_2=x$, then $(x,y\_1)\prec(x,y\_2)$ iff $y\_1 < y\_2$.
>
> Initially, assume that the well location must have integer coordinates.
> Is there a preference relation on lotteries, that satisfies the von-Neumann-Morgenstern axioms, and extends the lexicographic preference relation? If so, what is a linear utility function that represents this relation?

I think the answer is yes, and a possible linear utility function is: $u(x,y)=100000x + y$.

> Now, assume that the well location can have real coordinates. Prove that there is no linear utility function that represents the preference relation on lotteries.
> Which one of von-Neumann-Morgenstern axioms is violated by the preference relation on lotteries?

Here I am stuck. I don't understand why the utility function I suggested above doesn't work? And what axiom is violated here?

**Tags:** utility, preferences


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## 2 answers

### Answer by username_2 on 2015-08-13. Score: 17 | Accepted answer

We can say more generally that lexical preferences are not representable using a continuous utility function. Lexical preferences are not continuous. Note the definition of a continuous preference relation.

The preference relation $\succeq$ is continuous if for any sequences of consumption bundles $(x\_{i})\_{i \in \mathbb{N}}$ and $(y\_{i})\_{i \in \mathbb{N}}$ with $x\_{i} \to x$, $y\_{i} \to y$, and $x\_{i} \succeq y\_{i}$ for each $i \in \mathbb{N}$, then $x \succeq y$. That is, continuity preserves the relation at the limit point.

Consider $(x\_{i})\_{i \in \mathbb{N}}$ defined by $x\_{i} = (\frac{1}{2^{i}}, 0)$ and $(y\_{i})\_{i \in \mathbb{N}}$ defined by $y\_{i} = (0, 1)$. Clearly, $x\_{i} \succeq y\_{i}$ for each $i \in \mathbb{N}$. However, $x\_{i} \to (0, 0)$ while $y\_{i} \to (0, 1)$. So this preference relation is not preserved at the limit point.

Even more generally, no utility function represents the lexical preference relation. I prove for $\mathbb{R}\_{+}^{2}$, but this argument extends to $\mathbb{R}\_{+}^{n}$ by projecting into $\mathbb{R}\_{+}^{2}$.

**Proof**: Suppose to the contrary that some utility function $u : \mathbb{R}\_{+}^{2} \to \mathbb{R}$ represents $\succ\_{lex}$. We thus have $u(x, 1) > u(x, 0)$, as $(x, 1) \succ (x, 0)$. We construct the interval $I(x) = [u(x, 0), u(x, 1)]$. Now for any two distinct $x, y \in \mathbb{R}\_{+}$, $I(x) \cap I(y) = \emptyset$ as we have either $x > y$ or $y > x$ (so WLOG, we have $(x, 0) \succ (y, 1)$).

Define $\mathbb{I} = \{ I(x) : x \in \mathbb{R}\_{+} \}$, and let $\phi : \mathbb{R}\_{+} \to \mathbb{I}$ be given by $\phi(x) = I(x)$. Observe that $\phi$ is an injection, as each $I(x), I(y)$ are disjoint for distinct $x, y$.

Note that $\mathbb{Q}$ is dense in $\mathbb{R}$. So there exists a rational number in each interval. Define $\tau : \mathbb{I} \to \mathbb{Q}$ such that $\tau(I(x))$ returns a rational number contained in $I(x)$. So $\tau$ is an injection. We have $\tau$ composed with $\phi$ an injection, which implies $|\mathbb{R}\_{+}| \leq |\mathbb{Q}\_{+}|$, a contradiction. QED.

> **Comments:**
> **username_2**: I rejected the edit because I assume WLOG x > y. So I meant to say what I wrote in the proof. :-)  
> **username_0**: So how does $u(x,0)>u(y,0)$ and $u(x,1)>u(y,1)$ imply that $I(x)\cap I(y)=\emptyset$?  
> **username_2**: Actually, whoops. The edit you proposed was correct. We would have $u(x, 0) > u(y, 1)$ WLOG. Sorry for that minor error!  
> **username_0**: ‎Thanks. So, this impossibility result has nothing to do with probabilities or with von-Neumann-Morgenstern.  
> **username_2**: It's a more general result in topology that happens to come up in consumer theory. I view consumer theory as applied topology with an economics intuition.  
> **username_0**: Do you know of a general reference to this question of representing ordinal as cardinal? http://math.stackexchange.com/q/1396145/29780  
> **username_2**: The works of Arrow and Debreu are probably the best places to look. I'm neither a topologist nor a consumer theorist. This set of slides might be a good starting piont though: http://www.econ.ucla.edu/iobara/lecturepreferenceandutility201a.pdf  

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### Answer by username_1 on 2015-08-13. Score: 6

Consider the locations (1) $(0.000001,1)$ and (2) $(0.0000005,10)$. $U\left(x\_1,y\_1\right) = 1.1$.$U\left(x\_2,y\_2\right) = 10.05$. However, $x\_2 < x\_1$, so this is not a lexicographic ordering. It is only with the additional constraint that the values of $x$ and $y$ be integers $\in[0,1000]$ that the function you proposes has this attribute. Because real values of $x$ and $y$ implies that $x\_i / y\_i$ can be made arbitrarily large or small, no linear function can guarantee a lexicographic ordering over all possible values of $x$ and $y$. That is, specify a utility function of the form $$U = \beta \cdot X + Y$$
If $X$ has to be an integer then as long as $\beta>1000$ this will be lexicographic on the domain of the problem because a one unit increase in $X$ will always pay off more utility than 1,000 units of $Y$. But if $X$ is a real number, the problem changes. Consider any arbitrarily value of $\beta > 1000$ that would have satisfied the integer problem. For changes in $X > 1/\beta$ this continues to work (preserve the lex. prefs) in the real number problem. But for changes in $\Delta X < 1000 / \beta$:
$$ U(X + \Delta X,0) = \beta (X+\Delta X) < \beta X + 1000= U(X,1000)$$
Because by construction $\beta \cdot \Delta X < 1000$. So these preferences are not lexicographic over arbitrary values of $X$ and $Y$.


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