# Thread: Preference relation that is represented by a utility function problem

1. ## Preference relation that is represented by a utility function problem

I have a problem and i'm stuck already...
Given a preference relation $\displaystyle \preceq$ that is represented by the utility function u. If $\displaystyle \preceq$ is monotonic, show that $\displaystyle \preceq$ is represtented by $\displaystyle \bar{u}$, where $\displaystyle \bar{u}(0)=0$ and $\displaystyle \bar{u}(x) \geq 0$, for every $\displaystyle x \in \mathbb{R}^n$+ (n-dimension positive real numbers)
Can anybody help?

2. Its not clear if you want to show that any utility function with $\displaystyle \bar{u}(0)=o$ that represents the preferences must have the specified properties, or you just want to show that one exists.

Ill assume you only want to show that one exists. The simplest way to do this is to just identify a valid $\displaystyle \bar{u}$, and show that it represents the preference relation. There are many possibilities, but ill use $\displaystyle \bar{u}(x)= u(x) - u(0)$.

its clear that $\displaystyle \bar{u}(0) = u(0) - u(0) =0$ as required.

Now you only have to show that $\displaystyle \bar{u}$ represents your preference relation.

You're told that u represents the preferences, so it must be true that:
$\displaystyle u(x) \leq u(y) \iff x \preceq y$

You need to show that
$\displaystyle \bar{u}(x) \leq \bar{u}(y) \iff x \preceq y$

But you already know u() represents the preferences, so its sufficient to show that $\displaystyle \bar{u}$ is the same ordering as u().

ie
$\displaystyle \bar{u}(x) \leq \bar{u}(y) \iff u(x) \leq u(y)$
$\displaystyle u(x) - u(0) \leq u(y) - u(0) \iff u(x) \leq u(y)$

This is clearly true as u(0) is constant.

If you want to show rigorously that the above condition is sufficient to show the preferences are represented by $\displaystyle \bar{u}$ then;

$\displaystyle u(x) - u(0) \leq u(y) - u(0) \iff u(x) \leq u(y) \iff x \preceq y$
$\displaystyle u(x) - u(0) \leq u(y) - u(0) \iff x \preceq y$

$\displaystyle \bar{u}(x) \leq \bar{u}(y) \iff u(x) \leq u(y) \iff x \preceq y$

The final property that you must show is that $\displaystyle \bar{u}(x) > 0$ if $\displaystyle x > 0$. This is straightforward from the definition of $\displaystyle \bar(u)$ and the monotonicity of u(x).

u() monotinic:
$\displaystyle u(x) \geq u(y) \iff x \geq y$

substitute y=0
$\displaystyle u(x) \geq u(0) \iff x \geq 0$
$\displaystyle u(x) - u(0) \geq 0 \iff x \geq 0$
$\displaystyle \bar{u}(0) \geq 0 \iff x \geq y$