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?
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3$\begingroup$ 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? $\endgroup$– VCGCommented Sep 11, 2016 at 22:39
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4$\begingroup$ @VCG Why are you hating on non-continuous monotonically increasing transformations bro? $\endgroup$– GiskardCommented Sep 11, 2016 at 22:47
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1$\begingroup$ @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. $\endgroup$– VCGCommented Sep 11, 2016 at 22:53
2 Answers
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 $\succeq$
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.
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.