<p>(i) Is not complete. For instance, (10,5) is not <span class="math-container">$\succeq$</span> (9,6), because <span class="math-container">$10&gt;9$</span>, but <span class="math-container">$5&lt;6$</span>. However, (9,6) is also not <span class="math-container">$\succeq$</span> (10,5) for the same reason. Hence, there exists a pair of bundles <span class="math-container">$A,B$</span> such that neither <span class="math-container">$A\succeq B$</span> nor <span class="math-container">$B \succeq A$</span>. Thus, it is not rational.</p>
<p>(ii) and (iii) are both rational. You can either see this by showing transitivity and completeness directly or you exploit a central result in microeconomics: A preference <span class="math-container">$\succeq$</span> can be represented by a utility function <span class="math-container">$u$</span> only if it is rational.
For (ii), you can define <span class="math-container">$u(x_1,x_2)=x_1$</span> such that <span class="math-container">$x_2$</span> doesn't matter for comparisons as the preference states. For (iii), you can define Leontief preferences, <span class="math-container">$u(x_1,x_2)=\min(x_1,x_2)$</span>.</p>
<p>(iv) is a funny case, because it shows why the result we exploited above is just &quot;only if&quot; and not &quot;if and only if.&quot; <a href="https://en.wikipedia.org/wiki/Lexicographic_preferences" rel="noreferrer">Lexicographic preferences</a> are rational, but they are not representable by a standard utility function.</p>
<p>For any two bundles either the first dimension differs, or the first one is equal and the second one differs or both are equal. So for any two bundles we have either <span class="math-container">$\succeq$</span> or <span class="math-container">$\preceq$</span> or both, i.e., the preferences are complete. They are also transitive because <span class="math-container">$X\succeq Y$</span> implies that either <span class="math-container">$x_1&gt;y_1$</span> or <span class="math-container">$x_1=y_1$</span> and <span class="math-container">$x_2\geq y_2$</span>
and <span class="math-container">$Y\succeq Z$</span> implies that either <span class="math-container">$y_1&gt;z_1$</span> or <span class="math-container">$y_1=z_1$</span> and <span class="math-container">$y_2\geq z_2$</span>. Therefore, either <span class="math-container">$x_1&gt;z_1$</span> or <span class="math-container">$x_1=z_1$</span> and <span class="math-container">$x_2\geq z_2$</span>, implying <span class="math-container">$X\succeq Z$</span>.</p>

CC BY-SA 4.0
2020-11-13T18:33:35.000Z
(+1) But my understanding is that lexicographic preferences are not representable by *any* utility function, whether "standard" or not.
Herr K.
CC BY-SA 4.0
2020-11-13T20:02:49.033Z
Yes, by "standard" I meant any function defined within some metric space I am aware of. My imagination is mostly limited to the reals, and mathematicians are usually much more creative. Maybe there is something in some crazy space where the usual rules do not apply.
Bayesian
CC BY-SA 4.0
2020-11-13T12:56:01.593Z
2020-11-13T12:56:01.593Z
Bayesian
<p>I am taking advanced micro course this semester. In one of the problems we need to determine whether the preference relation is rational (i.e. complete and transitive). Since we have not really discussed the set of two-dimensional real vectors <span class="math-container">$X=R^2$</span> (non-negative), I am wondering whether these preference relations are indeed rational, and if yes, how can it be shown</p>
<p>(i) The relation ≽ defined by <span class="math-container">$(x_1,x_2)$</span> ≽ <span class="math-container">$(y_1,y_2)$</span> if and only if <span class="math-container">$x_1≥y_1$</span> and <span class="math-container">$x_2≥y_2$</span><br />
(ii) The relation ≽ defined by <span class="math-container">$(x_1,x_2)$</span> ≽ <span class="math-container">$(y_1,y_2)$</span> if and only if <span class="math-container">$x_1≥y_1$</span><br />
(iii) The relation ≽ defined by <span class="math-container">$(x_1,x_2)$</span> ≽ <span class="math-container">$(y_1,y_2)$</span> if and only if <span class="math-container">$\min\{x_1,x_2\} ≥ \min\{y_1,y_2\}$</span><br />
(iv) The relation ≽ defined by <span class="math-container">$(x_1,x_2)$</span> ≽ <span class="math-container">$(y_1,y_2)$</span> if and only if <span class="math-container">$x_1&gt;y_1$</span> or <span class="math-container">$x_1=y_1$</span> and <span class="math-container">$x_2≥y_2$</span></p>

<microeconomics><preferences>
Rational preferences/individual decision-making theory
user30845
# Rational preferences/individual decision-making theory

(Asked by: username_0 on 2020-11-13)

I am taking advanced micro course this semester. In one of the problems we need to determine whether the preference relation is rational (i.e. complete and transitive). Since we have not really discussed the set of two-dimensional real vectors $X=R^2$ (non-negative), I am wondering whether these preference relations are indeed rational, and if yes, how can it be shown

(i) The relation ≽ defined by $(x\_1,x\_2)$ ≽ $(y\_1,y\_2)$ if and only if $x\_1≥y\_1$ and $x\_2≥y\_2$  
(ii) The relation ≽ defined by $(x\_1,x\_2)$ ≽ $(y\_1,y\_2)$ if and only if $x\_1≥y\_1$  
(iii) The relation ≽ defined by $(x\_1,x\_2)$ ≽ $(y\_1,y\_2)$ if and only if $\min\{x\_1,x\_2\} ≥ \min\{y\_1,y\_2\}$  
(iv) The relation ≽ defined by $(x\_1,x\_2)$ ≽ $(y\_1,y\_2)$ if and only if $x\_1>y\_1$ or $x\_1=y\_1$ and $x\_2≥y\_2$

**Tags:** microeconomics, preferences

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## 1 answers

### Answer by username_1 on 2020-11-13. Score: 8

(i) Is not complete. For instance, (10,5) is not $\succeq$ (9,6), because $10>9$, but $5<6$. However, (9,6) is also not $\succeq$ (10,5) for the same reason. Hence, there exists a pair of bundles $A,B$ such that neither $A\succeq B$ nor $B \succeq A$. Thus, it is not rational.

(ii) and (iii) are both rational. You can either see this by showing transitivity and completeness directly or you exploit a central result in microeconomics: A preference $\succeq$ can be represented by a utility function $u$ only if it is rational.
For (ii), you can define $u(x\_1,x\_2)=x\_1$ such that $x\_2$ doesn't matter for comparisons as the preference states. For (iii), you can define Leontief preferences, $u(x\_1,x\_2)=\min(x\_1,x\_2)$.

(iv) is a funny case, because it shows why the result we exploited above is just "only if" and not "if and only if." [Lexicographic preferences](https://en.wikipedia.org/wiki/Lexicographic_preferences) are rational, but they are not representable by a standard utility function.

For any two bundles either the first dimension differs, or the first one is equal and the second one differs or both are equal. So for any two bundles we have either $\succeq$ or $\preceq$ or both, i.e., the preferences are complete. They are also transitive because $X\succeq Y$ implies that either $x\_1>y\_1$ or $x\_1=y\_1$ and $x\_2\geq y\_2$
and $Y\succeq Z$ implies that either $y\_1>z\_1$ or $y\_1=z\_1$ and $y\_2\geq z\_2$. Therefore, either $x\_1>z\_1$ or $x\_1=z\_1$ and $x\_2\geq z\_2$, implying $X\succeq Z$.

> **Comments:**
> **username_2**: (+1) But my understanding is that lexicographic preferences are not representable by *any* utility function, whether "standard" or not.  
> **username_1**: Yes, by "standard" I meant any function defined within some metric space I am aware of. My imagination is mostly limited to the reals, and mathematicians are usually much more creative. Maybe there is something in some crazy space where the usual rules do not apply.  

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