I am stuck on the following exercise, related to preference relations and von-Neumann-Morgenstern utility function.

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?


2 Answers 2


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.

  • $\begingroup$ I rejected the edit because I assume WLOG x > y. So I meant to say what I wrote in the proof. :-) $\endgroup$
    – ml0105
    Aug 13, 2015 at 17:26
  • $\begingroup$ So how does $u(x,0)>u(y,0)$ and $u(x,1)>u(y,1)$ imply that $I(x)\cap I(y)=\emptyset$? $\endgroup$ Aug 13, 2015 at 18:40
  • $\begingroup$ Actually, whoops. The edit you proposed was correct. We would have $u(x, 0) > u(y, 1)$ WLOG. Sorry for that minor error! $\endgroup$
    – ml0105
    Aug 13, 2015 at 18:47
  • 1
    $\begingroup$ ‎Thanks. So, this impossibility result has nothing to do with probabilities or with von-Neumann-Morgenstern. $\endgroup$ Aug 14, 2015 at 5:15
  • $\begingroup$ 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. $\endgroup$
    – ml0105
    Aug 14, 2015 at 5:18

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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