<p>Recall the definition:</p>

<blockquote>
  <p>The function $u: X \rightarrow \mathbb{R}$ represents $\succeq$ on $X$ if for any $x,y \in X$, then $x \succeq y \iff u(x) \geq u(y)$</p>
</blockquote>

<p>We can show that if $u: X \rightarrow \mathbb{R}$ represents $\succeq$ on $X$, then for any strictly increasing function, $f: \mathbb{R} \rightarrow \mathbb{R}$,  the function $v(x) = f(u(x))$ also represents $\succeq$</p>

<p>(Short) <strong>Proof:</strong> For any $x,y \in X$, if $u$ represents $\succeq$, then $x \succeq y \iff u(x) \geq u(y)$ by definition.</p>

<p>Because $v$ is strictly increasing, $f(u(x)) \geq f(u(y)), \implies v(x) \geq v(y)$.</p>

<p>Thus $x \succeq y \iff v(x) \geq v(y), \implies v$ represents $\succeq$</p>

<hr>

<p>In your case, you have $u(x,y) = x^\frac{1}{3}y^\frac{1}{3}$, and can set $v(x, y) = (u(x,y))^9 = x^3y^3$, your desired result.</p>

CC BY-SA 3.0
2016-09-12T15:26:54.397Z
2016-09-12T15:26:54.397Z
Kitsune Cavalry
<p>Another trick is to compare their marginal rate of substitution. If for a similar marginal rate of technical substitution (i.e. relative prices) they yield the same relative inputs, then they represent the same preferences (basically one is a monotonic transformation of the other; see below).</p>

<p>For the first preferences you give, this is:</p>

<p>$$ \frac{\dfrac{\partial u}{\partial x}}{\dfrac{\partial u}{\partial y}} = \frac{y}{x} $$</p>

<p>It is trivial to show that the second function yields the same MRS. Hence, they represent the same preferences.</p>

<hr>

<p>Another way to look at this is to notice that optimising a function $f(x)$ is the same than optimising a monotonically transformation of such function. Thus, apply the natural logarithm to the two functions and you get, respectively:</p>

<p>$$ \ln u = \frac{1}{3}\left(\ln x + \ln y \right) $$</p>

<p>$$ \ln v = 3\left(\ln x + \ln y \right) $$</p>

<p>Notice that $\ln v = 9\ln u$. Hence, they are indeed the same.</p>

CC BY-SA 3.0
2017-09-21T17:03:10.447Z
2017-09-21T17:08:24.503Z
2017-09-21T17:08:24.503Z
luchonacho
<p>I need to verify that $u(x,y)=x^{1/3}y^{1/3}$ represents the same preferences as $v(x,y)=x^3y^3$. Obviously these are completely different functions with different derivatives, so what am I comparing? What makes 2 utility functions represent the same preferences?</p>

CC BY-SA 3.0
2016-09-11T22:39:45.570Z
If you can find a monotonically increasing continuous transformation that turns one into the other, then they have the same preferences. Can you find that?
VCG
CC BY-SA 3.0
2016-09-11T22:47:03.017Z
@VCG Why are you hating on non-continuous monotonically increasing transformations bro?
Giskard
CC BY-SA 3.0
2016-09-11T22:53:20.623Z
@denesp Oh jeez ya oops. No idea why I wrote that. Too late to edit. Oh I guess I was thinking of needing the utility function to be continuous for sufficiency.
VCG
<utility><preferences>
How can I tell if 2 different utility functions represent the same preferences?
Henry
# How can I tell if 2 different utility functions represent the same preferences?

(Asked by: username_0 on 2016-09-11)

I need to verify that $u(x,y)=x^{1/3}y^{1/3}$ represents the same preferences as $v(x,y)=x^3y^3$. Obviously these are completely different functions with different derivatives, so what am I comparing? What makes 2 utility functions represent the same preferences?

**Tags:** utility, preferences

> **Comments:**
> **username_1**: If you can find a monotonically increasing continuous transformation that turns one into the other, then they have the same preferences. Can you find that?  
> **username_2**: @username_1 Why are you hating on non-continuous monotonically increasing transformations bro?  
> **username_1**: @denesp Oh jeez ya oops. No idea why I wrote that. Too late to edit. Oh I guess I was thinking of needing the utility function to be continuous for sufficiency.  

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## 2 answers

### Answer by username_3 on 2016-09-12. Score: 7

Recall the definition:

> The function $u: X \rightarrow \mathbb{R}$ represents $\succeq$ on $X$ if for any $x,y \in X$, then $x \succeq y \iff u(x) \geq u(y)$

We can show that if $u: X \rightarrow \mathbb{R}$ represents $\succeq$ on $X$, then for any strictly increasing function, $f: \mathbb{R} \rightarrow \mathbb{R}$, the function $v(x) = f(u(x))$ also represents $\succeq$

(Short) **Proof:** For any $x,y \in X$, if $u$ represents $\succeq$, then $x \succeq y \iff u(x) \geq u(y)$ by definition.

Because $v$ is strictly increasing, $f(u(x)) \geq f(u(y)), \implies v(x) \geq v(y)$.

Thus $x \succeq y \iff v(x) \geq v(y), \implies v$ represents $\succeq$

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In your case, you have $u(x,y) = x^\frac{1}{3}y^\frac{1}{3}$, and can set $v(x, y) = (u(x,y))^9 = x^3y^3$, your desired result.


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### Answer by username_4 on 2017-09-21. Score: 1

Another trick is to compare their marginal rate of substitution. If for a similar marginal rate of technical substitution (i.e. relative prices) they yield the same relative inputs, then they represent the same preferences (basically one is a monotonic transformation of the other; see below).

For the first preferences you give, this is:

$$ \frac{\dfrac{\partial u}{\partial x}}{\dfrac{\partial u}{\partial y}} = \frac{y}{x} $$

It is trivial to show that the second function yields the same MRS. Hence, they represent the same preferences.

---

Another way to look at this is to notice that optimising a function $f(x)$ is the same than optimising a monotonically transformation of such function. Thus, apply the natural logarithm to the two functions and you get, respectively:

$$ \ln u = \frac{1}{3}\left(\ln x + \ln y \right) $$

$$ \ln v = 3\left(\ln x + \ln y \right) $$

Notice that $\ln v = 9\ln u$. Hence, they are indeed the same.


---

