<p>The utility function $U'$ is a monotonic transformation  of $U$ if there is a strictly monotonically increasing function $f$ for which
$$
\forall x,y: \ f(U(x,y)) = U'(x,y) 
$$
Larger values of $U(x,y)$ should always get larger values assigned. But with the function you outlined this is not the case if $U(x_1,y_1) \leq U(x_2,y_2)$ and
$$
\frac{U(x_2,y_2)}{U(x_1,y_1)} \leq \frac{x_1}{x_2}.
$$
Basically the order between bundles $(x_1,y_1)$ and $(x_2,y_2)$ may change depending on the ratio of their original utilities and the ratio of $x_1$ and $x_2$. An example has been provided by @Oliv. </p>

CC BY-SA 3.0
2016-02-06T06:45:15.980Z
2016-02-06T06:45:15.980Z
Giskard
<p>Take any utility function $f(x)$. Just because $x\geq0$ doesn't mean $f(x)&gt;0$. There are plenty (infinitely many) of utility functions that are sometime negative and sometimes positive. </p>

<p>Consider one $U = x-5$ This is person has exactly the same preference ordering as someone with preferences $U = x-10$, $U = x$, and $U = x+5$. Now imagine multiplying these utility functions by x. This makes the negative utility values positive at different values of x. We can tell that because the utility functions have the same ordering of the utility values of x.
<a href="https://i.sstatic.net/kckNK.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kckNK.png" alt="enter image description here"></a></p>

<p>Now consider the those same functions transformed by multiplying by x. 
<a href="https://i.sstatic.net/hDszm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hDszm.png" alt="enter image description here"></a>
We can see that these transformed functions now actually prefer very low values of x to very high ones. </p>

CC BY-SA 3.0
2016-02-06T12:25:24.477Z
2016-02-07T12:35:53.637Z
2016-02-07T12:35:53.637Z
BKay
<p>Because $x$ is not a constant, and therefore multiplying $U(x,y)$ by $x$ does not necessarily preserve the ordering between bundles.</p>

<p>For instance, consider the function $U(x,y)=x+y$, and the bundles $(x_0,y_0)=(2,0)$ and $(x_1,y_1)=(1,2)$. We have
\begin{equation*}
U(x_0,y_0)=2 &lt; U(x_1,y_1)=3
\end{equation*}
but
\begin{equation*}
x_0 U(x_0,y_0) = 4 &gt; x_1 U(x_1,y_1) = 3
\end{equation*}
Hence multiplying $U$ by $x$ changes the preferences, therefore is not a monotonic transformation.</p>

<p>As a side note, multiplying the utility function by a constant provides a monotonic transformation if the constant is positive, and not only nonnegative as you assumed (if the constant is zero the orderings are obviously not preserved).</p>

CC BY-SA 4.0
2021-04-30T13:56:45.747Z
Perfect explanation yet simple to understand. That's what we need.
j3141592653589793238
CC BY-SA 3.0
2016-02-06T06:44:26.000Z
2016-02-06T06:44:26.000Z
Oliv
<p>If $x$, is a commodity and it is s.t. $x \geq 0$ (it is always nonnegative) how will multiplying it times a utility function $U(x,y)$ NOT yield a simple monotonic transformation?</p>

<microeconomics><utility><preferences>
Given a Utility function, U(x,y), why is multiplying U(x,y) * x not a monotonic transformation?
bloopton
# Given a Utility function, U(x,y), why is multiplying U(x,y) * x not a monotonic transformation?

(Asked by: username_0 on 2016-02-06)

If $x$, is a commodity and it is s.t. $x \geq 0$ (it is always nonnegative) how will multiplying it times a utility function $U(x,y)$ NOT yield a simple monotonic transformation?

**Tags:** microeconomics, utility, preferences

---

## 3 answers

### Answer by username_3 on 2016-02-06. Score: 6 | Accepted answer

Because $x$ is not a constant, and therefore multiplying $U(x,y)$ by $x$ does not necessarily preserve the ordering between bundles.

For instance, consider the function $U(x,y)=x+y$, and the bundles $(x\_0,y\_0)=(2,0)$ and $(x\_1,y\_1)=(1,2)$. We have
\begin{equation\*}
U(x\_0,y\_0)=2 < U(x\_1,y\_1)=3
\end{equation\*}
but
\begin{equation\*}
x\_0 U(x\_0,y\_0) = 4 > x\_1 U(x\_1,y\_1) = 3
\end{equation\*}
Hence multiplying $U$ by $x$ changes the preferences, therefore is not a monotonic transformation.

As a side note, multiplying the utility function by a constant provides a monotonic transformation if the constant is positive, and not only nonnegative as you assumed (if the constant is zero the orderings are obviously not preserved).

> **Comments:**
> **username_4**: Perfect explanation yet simple to understand. That's what we need.  

---

### Answer by username_1 on 2016-02-06. Score: 4

The utility function $U'$ is a monotonic transformation of $U$ if there is a strictly monotonically increasing function $f$ for which
$$
\forall x,y: \ f(U(x,y)) = U'(x,y)
$$
Larger values of $U(x,y)$ should always get larger values assigned. But with the function you outlined this is not the case if $U(x\_1,y\_1) \leq U(x\_2,y\_2)$ and
$$
\frac{U(x\_2,y\_2)}{U(x\_1,y\_1)} \leq \frac{x\_1}{x\_2}.
$$
Basically the order between bundles $(x\_1,y\_1)$ and $(x\_2,y\_2)$ may change depending on the ratio of their original utilities and the ratio of $x\_1$ and $x\_2$. An example has been provided by @Oliv.


---

### Answer by username_2 on 2016-02-06. Score: 3

Take any utility function $f(x)$. Just because $x\geq0$ doesn't mean $f(x)>0$. There are plenty (infinitely many) of utility functions that are sometime negative and sometimes positive.

Consider one $U = x-5$ This is person has exactly the same preference ordering as someone with preferences $U = x-10$, $U = x$, and $U = x+5$. Now imagine multiplying these utility functions by x. This makes the negative utility values positive at different values of x. We can tell that because the utility functions have the same ordering of the utility values of x.
[![enter image description here](https://i.sstatic.net/kckNK.png)](https://i.sstatic.net/kckNK.png)

Now consider the those same functions transformed by multiplying by x.
[![enter image description here](https://i.sstatic.net/hDszm.png)](https://i.sstatic.net/hDszm.png)
We can see that these transformed functions now actually prefer very low values of x to very high ones.


---

