<p>Since <span class="math-container">$u(x)$</span> represents <span class="math-container">$\succsim$</span>, <span class="math-container">$\succsim$</span> must be rational. By completeness, <span class="math-container">$\neg (x\succsim y) \iff x \prec y$</span>. Therefore, the claim that <span class="math-container">$$v(x) \geq v(y) \iff x\succsim y, \forall x, y$$</span> is equivalent to its contrapositive (for both sides):
<span class="math-container">$$v(x) &lt; v(y) \iff x \prec y, \forall x, y$$</span>
Pick any <span class="math-container">$x\prec y$</span>, we should have <span class="math-container">$u(x) &lt; u(y)$</span>. However, suppose your transformation <span class="math-container">$f$</span> is constant (non-decreasing, but not strictly increasing), <span class="math-container">$f(u(x)) = f(u(y))$</span>. That is, not <span class="math-container">$v(x) &lt; v(y)$</span>, which means <span class="math-container">$v(\cdot)$</span> does not represent <span class="math-container">$\succsim$</span>.</p>

CC BY-SA 4.0
2024-08-24T05:21:09.427Z
Ok just to be clear if we write down the preference relation set $\succsim$ where $x \succsim y $, iff $(x,y) \in$  $\succsim$, so for $x \succ y$ to be true it should be that $(x,y) \in$  $\succsim$ and $(y,x) \not\in$ $\succsim$
hr08
CC BY-SA 4.0
2024-08-24T10:15:56.360Z
also what i have learned is $x \succ y \iff (x \succsim y) \wedge \neg(y \succsim x)$, so I didnt understand the negation of weak preference
hr08
CC BY-SA 4.0
2024-08-24T20:32:50.680Z
If I understand correctly, your two comments are saying the same thing. You are correct, what you mentioned was the definition of a weak preference **without any premise**. However, I was using a property here: if $\succsim$ can be represented by $u(\cdot)$, $\succsim$ must be **complete**, which means for any $x, y$: $\neg(x\succsim y) \iff x \prec y$.
DiZ
CC BY-SA 4.0
2024-08-24T20:37:38.237Z
It seems like Simon's lecture notes did not mention this property, but I believe it is easy to verify yourself. The idea is that $>$ possesses transitivity and completeness, so does $\succsim$.
DiZ
CC BY-SA 4.0
2024-08-25T03:44:20.740Z
Thanks alot got it.
hr08
CC BY-SA 4.0
2024-08-23T20:25:12.290Z
2024-08-23T20:25:12.290Z
DiZ
<p>I was going through the <strong><a href="http://www.econ.ucla.edu/sboard/teaching/econ11_09/econ11_09_lecture2.pdf" rel="nofollow noreferrer">Theorem 2</a></strong> :</p>
<p><em>Suppose <span class="math-container">$u(x)$</span> represents agents preferences, <span class="math-container">$\succsim$</span> and <span class="math-container">$f : \mathbb{R} \rightarrow \mathbb{R} $</span> is a strictly increasing function. Then the new new utility function <span class="math-container">$v(x) = f(u(x))$</span> also represents the agent's preferences <span class="math-container">$\succsim$</span>. So mathematically:
<span class="math-container">$v(x) \geq v(y) \iff x \succsim y$</span></em></p>
<p>I am having trouble understanding this above double implication statement.</p>
<p>Suppose <span class="math-container">$f$</span> is not strictly increasing (i.e just a non decreasing function) and if we do monotonic (not strictly) transformation of the utility (like <a href="https://economics.stackexchange.com/a/18512/24444">this</a>) <span class="math-container">$v : z \mapsto 0$</span> then for two alternatives <span class="math-container">$x,y$</span> let <span class="math-container">$x \succ y$</span> but their utility values <span class="math-container">$v(u(x)) = v(u(y))$</span> are same. So shouldn't we consider it satisfying the equivalence relation above
Since for <span class="math-container">$x \succsim y$</span> we have <span class="math-container">$x\succ y$</span> true and for <span class="math-container">$u(x) \geq u(y)$</span> we have <span class="math-container">$u(x) = u(y)$</span> true making the equivalence hold true for non decreasing function also.</p>
<p>I know this is wrong , since the ordering is not maintained after the transformation but Im not able to understand it in the form of mathematical statement of double implication, what am I doing wrong here.</p>

<utility><preferences><choice-theory>
Non decreasing function for utility transformation
hr08
# Non decreasing function for utility transformation

(Asked by: username_0 on 2024-08-23)

I was going through the **[Theorem 2](http://www.econ.ucla.edu/sboard/teaching/econ11_09/econ11_09_lecture2.pdf)** :

*Suppose $u(x)$ represents agents preferences, $\succsim$ and $f : \mathbb{R} \rightarrow \mathbb{R} $ is a strictly increasing function. Then the new new utility function $v(x) = f(u(x))$ also represents the agent's preferences $\succsim$. So mathematically:
$v(x) \geq v(y) \iff x \succsim y$*

I am having trouble understanding this above double implication statement.

Suppose $f$ is not strictly increasing (i.e just a non decreasing function) and if we do monotonic (not strictly) transformation of the utility (like [this](https://economics.stackexchange.com/a/18512/24444)) $v : z \mapsto 0$ then for two alternatives $x,y$ let $x \succ y$ but their utility values $v(u(x)) = v(u(y))$ are same. So shouldn't we consider it satisfying the equivalence relation above
Since for $x \succsim y$ we have $x\succ y$ true and for $u(x) \geq u(y)$ we have $u(x) = u(y)$ true making the equivalence hold true for non decreasing function also.

I know this is wrong , since the ordering is not maintained after the transformation but Im not able to understand it in the form of mathematical statement of double implication, what am I doing wrong here.

**Tags:** utility, preferences, choice-theory


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## 1 answers

### Answer by username_1 on 2024-08-23. Score: 3 | Accepted answer

Since $u(x)$ represents $\succsim$, $\succsim$ must be rational. By completeness, $\neg (x\succsim y) \iff x \prec y$. Therefore, the claim that $$v(x) \geq v(y) \iff x\succsim y, \forall x, y$$ is equivalent to its contrapositive (for both sides):
$$v(x) < v(y) \iff x \prec y, \forall x, y$$
Pick any $x\prec y$, we should have $u(x) < u(y)$. However, suppose your transformation $f$ is constant (non-decreasing, but not strictly increasing), $f(u(x)) = f(u(y))$. That is, not $v(x) < v(y)$, which means $v(\cdot)$ does not represent $\succsim$.

> **Comments:**
> **username_0**: Ok just to be clear if we write down the preference relation set $\succsim$ where $x \succsim y $, iff $(x,y) \in$  $\succsim$, so for $x \succ y$ to be true it should be that $(x,y) \in$  $\succsim$ and $(y,x) \not\in$ $\succsim$  
> **username_0**: also what i have learned is $x \succ y \iff (x \succsim y) \wedge \neg(y \succsim x)$, so I didnt understand the negation of weak preference  
> **username_1**: If I understand correctly, your two comments are saying the same thing. You are correct, what you mentioned was the definition of a weak preference **without any premise**. However, I was using a property here: if $\succsim$ can be represented by $u(\cdot)$, $\succsim$ must be **complete**, which means for any $x, y$: $\neg(x\succsim y) \iff x \prec y$.  
> **username_1**: It seems like Simon's lecture notes did not mention this property, but I believe it is easy to verify yourself. The idea is that $>$ possesses transitivity and completeness, so does $\succsim$.  
> **username_0**: Thanks alot got it.  

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