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.

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    $\begingroup$ This claim is false. Please identify the textbook and the exact page. $\endgroup$ – Giskard Oct 20 '19 at 18:40
  • $\begingroup$ It is from slide of my professor, which he claim to be from a textbook. $\endgroup$ – Aqqqq Oct 21 '19 at 7:36
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    $\begingroup$ I'm voting to close this question as off-topic because it is based upon a false claim whose source cannot be identified. $\endgroup$ – Giskard Oct 21 '19 at 9:41
  • $\begingroup$ @Giskard Would it be the case in case of uncertainty (e.g. in lottery)? $\endgroup$ – Aqqqq Oct 21 '19 at 12:44
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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 $Q=(\frac{1}{2},0,\frac{1}{2})$.



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 $V(p)=E_p[v]$.

  • $\begingroup$ Thank you for your answer. Is $\Delta(X)$ just a notation or are there some meaning behind $\Delta$? $\endgroup$ – Aqqqq Oct 22 '19 at 9:30
  • $\begingroup$ 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 $V(p)=E_p[v]$"? $\endgroup$ – Aqqqq Oct 22 '19 at 10:09

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