<p>One reason I could think of is regarding convexity of the function (and hence risk aversion). For example, <span class="math-container">$u(x) = x$</span> is risk neutral, but <span class="math-container">$v(x) = \ln(u(x)) = \ln(x)$</span> is risk averse.</p>

<p>See also <a href="https://en.wikipedia.org/wiki/Risk_aversion#Utility_of_money" rel="nofollow noreferrer">this</a>.</p>

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2019-10-21T10:01:23.347Z
2019-10-21T10:24:40.220Z
2019-10-21T10:24:40.220Z
Art
<p>According to a textbook, a monotone increasing but nonlinear transformation of a utility function might not represent the same preferences. Why is it so?</p>

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

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2019-10-21T09:09:58.617Z
Maybe I am underthinking this, but the transformation has to be _strictly_ monotone increasing. I think it is simple to see why, as only the strictly monotone function would keep the ordering of any preference relation intact.

But you probably meant something else?
IMA
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2019-10-21T09:27:43.587Z
Thank you. I somehow thought that monotone increasing the same as strictly monotone increasing.
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<microeconomics><utility><preferences>
Why might a monotone increasing but nonlinear transformation of a utility function not represent the same preferences?
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# Why might a monotone increasing but nonlinear transformation of a utility function not represent the same preferences?

(Asked by: username_0 on 2019-10-21)

According to a textbook, a monotone increasing but nonlinear transformation of a utility function might not represent the same preferences. Why is it so?

An example of such preference would be appreciated.

**Tags:** microeconomics, utility, preferences

> **Comments:**
> **username_1**: Maybe I am underthinking this, but the transformation has to be _strictly_ monotone increasing. I think it is simple to see why, as only the strictly monotone function would keep the ordering of any preference relation intact.  But you probably meant something else?  
> **username_0**: Thank you. I somehow thought that monotone increasing the same as strictly monotone increasing.  

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## 1 answers

### Answer by username_2 on 2019-10-21. Score: 1 | Accepted answer

One reason I could think of is regarding convexity of the function (and hence risk aversion). For example, $u(x) = x$ is risk neutral, but $v(x) = \ln(u(x)) = \ln(x)$ is risk averse.

See also [this](https://en.wikipedia.org/wiki/Risk_aversion#Utility_of_money).


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