<p>Consider lotteries over <span class="math-container">$\{x,y,z\}$</span>. Let <span class="math-container">$u(x)=0, u(y)=\frac{1}{2}, u(z)=1$</span>. Consider the nonlinear transformation f(t)=t^2. Let <span class="math-container">$v:=f\circ u$</span>, so <span class="math-container">$v(x)=0,v(y)=\frac{1}{4}, v(z)=1$</span>. </p>

<p>Consider two lotteries,  <span class="math-container">$P=(0,1,0)$</span> and <span class="math-container">$Q=(\frac{1}{2},0,\frac{1}{2})$</span>.</p>

<p><span class="math-container">$$E_P[u]=\frac{1}{2}=E_Q[u]$$</span></p>

<p><span class="math-container">$$E_P[v]=\frac{1}{4}&lt;\frac{1}{2}=E_Q[v]$$</span></p>

<p>In general, let <span class="math-container">$\succeq$</span> be a preference over lotteries <span class="math-container">$\Delta(X)$</span>.  Let <span class="math-container">$U$</span> be a utility function of <span class="math-container">$\succeq$</span> that has the EU form, so <span class="math-container">$U(P)=E_P[u]$</span> for some <span class="math-container">$u$</span>. Take any increasing transformation <span class="math-container">$f$</span> and define <span class="math-container">$V(P):=f(U(P))$</span> then <span class="math-container">$V$</span> will also be a utility function for <span class="math-container">$\succeq$</span>, that is <span class="math-container">$V(P)\geq V(Q)\iff P\succeq Q$</span>. However, unless <span class="math-container">$f$</span> is a positive affine transformation, that is <span class="math-container">$f(x)=Ax+B$</span> where <span class="math-container">$A&gt;0$</span> then <span class="math-container">$V$</span> will not be an expected utility function. That is, there will not exists a <span class="math-container">$v$</span> such that <span class="math-container">$V(p)=E_p[v]$</span>. </p>

CC BY-SA 4.0
2019-10-22T09:30:03.633Z
Thank you for your answer. Is $\Delta(X)$ just a notation or are there some meaning behind  $\Delta$?
Aqqqq
CC BY-SA 4.0
2019-10-22T10:09:45.733Z
Should $V(p)=E_p[v]$ be $V(P)=E_P[v]$ instead (in the last paragraph)? Why "there will not exists a $v$ such that $V(p)=E_p[v]$"?
Aqqqq
CC BY-SA 4.0
2019-10-21T15:07:01.807Z
2019-10-21T15:13:45.603Z
2019-10-21T15:13:45.603Z
user24629
<p>According to a textbook, in the context of uncertainty (e.g. in lottery), <strong>if the preference is complete</strong>, a monotone increasing but nonlinear transformation of a utility function would not represent the same preferences. Why is it so? </p>

<p>An example of such preference would be appreciated.</p>

CC BY-SA 4.0
2019-10-20T18:40:14.240Z
This claim is false. Please identify the textbook and the exact page.
Giskard
CC BY-SA 4.0
2019-10-21T07:36:10.853Z
It is from slide of my professor, which he claim to be from a textbook.
Aqqqq
CC BY-SA 4.0
2019-10-21T09:41:33.540Z
I'm voting to close this question as off-topic because it is based upon a false claim whose source cannot be identified.
Giskard
CC BY-SA 4.0
2019-10-21T12:44:34.237Z
@Giskard Would it be the case in case of uncertainty (e.g. in lottery)?
Aqqqq
CC BY-SA 4.0
2019-10-21T14:32:24.403Z
Maybe if the OP posts the exact argument used to make this claim, we can at least have a discussion about why it is false.
heh
CC BY-SA 4.0
2019-10-21T15:09:17.967Z
Possible duplicate of [Why might a monotone increasing but nonlinear transformation of a utility function not represent the same preferences?](https://economics.stackexchange.com/questions/32359/why-might-a-monotone-increasing-but-nonlinear-transformation-of-a-utility-functi)
Art
CC BY-SA 4.0
2019-10-21T20:09:45.420Z
@Art no it is not a duplicate. See the bold faced part.
Aqqqq
CC BY-SA 4.0
2019-10-22T02:03:19.137Z
The same answer applies though...
Art
<microeconomics><utility><preferences>
Why is a monotone increasing but nonlinear transformation of a utility function not represent the same preferences if the preference is complete?
Aqqqq
# Why is a monotone increasing but nonlinear transformation of a utility function not represent the same preferences if the preference is complete?

(Asked by: username_0 on 2019-10-20)

According to a textbook, in the context of uncertainty (e.g. in lottery), **if the preference is complete**, a monotone increasing but nonlinear transformation of a utility function would not represent the same preferences. Why is it so?

An example of such preference would be appreciated.

**Tags:** microeconomics, utility, preferences
> **Comments:**
> **username_1**: This claim is false. Please identify the textbook and the exact page.  
> **username_0**: It is from slide of my professor, which he claim to be from a textbook.  
> **username_1**: I'm voting to close this question as off-topic because it is based upon a false claim whose source cannot be identified.  
> **username_0**: @Giskard Would it be the case in case of uncertainty (e.g. in lottery)?  
> **username_2**: Maybe if the OP posts the exact argument used to make this claim, we can at least have a discussion about why it is false.  
> **username_3**: Possible duplicate of [Why might a monotone increasing but nonlinear transformation of a utility function not represent the same preferences?](https://economics.stackexchange.com/questions/32359/why-might-a-monotone-increasing-but-nonlinear-transformation-of-a-utility-functi)  
> **username_0**: @Art no it is not a duplicate. See the bold faced part.  
> **username_3**: The same answer applies though...  

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## 1 answers

### Answer by username_4 on 2019-10-21. Score: 1

Consider lotteries over $\{x,y,z\}$. Let $u(x)=0, u(y)=\frac{1}{2}, u(z)=1$. Consider the nonlinear transformation f(t)=t^2. Let $v:=f\circ u$, so $v(x)=0,v(y)=\frac{1}{4}, v(z)=1$.

Consider two lotteries, $P=(0,1,0)$ and $Q=(\frac{1}{2},0,\frac{1}{2})$.

$$E\_P[u]=\frac{1}{2}=E\_Q[u]$$

$$E\_P[v]=\frac{1}{4}<\frac{1}{2}=E\_Q[v]$$

In general, let $\succeq$ be a preference over lotteries $\Delta(X)$. Let $U$ be a utility function of $\succeq$ that has the EU form, so $U(P)=E\_P[u]$ for some $u$. Take any increasing transformation $f$ and define $V(P):=f(U(P))$ then $V$ will also be a utility function for $\succeq$, that is $V(P)\geq V(Q)\iff P\succeq Q$. However, unless $f$ is a positive affine transformation, that is $f(x)=Ax+B$ where $A>0$ then $V$ will not be an expected utility function. That is, there will not exists a $v$ such that $V(p)=E\_p[v]$.

> **Comments:**
> **username_0**: Thank you for your answer. Is $\Delta(X)$ just a notation or are there some meaning behind  $\Delta$?  
> **username_0**: Should $V(p)=E_p[v]$ be $V(P)=E_P[v]$ instead (in the last paragraph)? Why "there will not exists a $v$ such that $V(p)=E_p[v]$"?  

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