# Why is a monotone increasing but nonlinear transformation of a utility function not represent the same preferences if the preference is complete?

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.

• This claim is false. Please identify the textbook and the exact page. – Giskard Oct 20 '19 at 18:40
• It is from slide of my professor, which he claim to be from a textbook. – Aqqqq Oct 21 '19 at 7:36
• I'm voting to close this question as off-topic because it is based upon a false claim whose source cannot be identified. – Giskard Oct 21 '19 at 9:41
• @Giskard Would it be the case in case of uncertainty (e.g. in lottery)? – Aqqqq Oct 21 '19 at 12:44
• – Art Oct 21 '19 at 15:09

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