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Given a utility function $u(\cdot)$ and two bundles $x$ and $y$.

Assuming $u(x)=u(y)$.

I am to prove or disprove that $x \succcurlyeq y$.

Now I'm confused by this. We say $x$ is strictly preferred to $y$, or $x \succ y$, if $x \succcurlyeq y$ but not $y \succcurlyeq x$. We say the agent is indifferent between $x$ and $y$, or $x \sim y$, if $x \succcurlyeq y$ and $y \succcurlyeq x$. Thus if $x \sim y$, we should be able to say $y \sim x$.

So we know that if $x \sim y$, $y \sim x $ and $x \succcurlyeq y$ and $y \succcurlyeq x$.

Now what does $\succcurlyeq$ say at all? It's either $\succ$ or $\sim$.

So can we say $\succcurlyeq$ is just a superset of $\succ$ and $\sim$ when we're unsure? Because if $x \succcurlyeq y$, $x$ can either be strictly preferred or indifferent to $y$. But if $x \sim y$ (which could be, because $\succcurlyeq$ means either indifferent or strict preference), $x$ can not be strictly preferred to y anymore. This is confusing to me. Aren't these two statements contradictory? It can be strictly preferred first, but under the second assumption, it can't?

And for the example above, this would mean because $u(x)=u(y)$, we know $x \sim y$ and $y \sim x$ and thus $x \succcurlyeq y$ (and $y \succcurlyeq x$) are correct?

I know this is a very basic question. I just got confused because I think I have a slight disagreement with a fellow student who says that for the example above, we cannot say that $x \sim y$ (which, obviously, is false).

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    $\begingroup$ You have $x \sim y$ when $x \succcurlyeq y$ and $y \succcurlyeq x$. $\endgroup$
    – Bayesian
    Jun 7 at 12:46
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    $\begingroup$ @Bayesian Right. So if $u(x) = u(y)$, this means $x \sim y$ and therefore $x \succcurlyeq y$ as well as $y \succcurlyeq x$, correct? Thanks! $\endgroup$ Jun 7 at 12:48
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Assuming $u(x)=u(y)$.

I am to prove or disprove that $x \succcurlyeq y$.

Proof:

The utility function $u$ represents your preferences $\succcurlyeq$ over consumption bundles. By definition of the term "represents", this means that $u(x)\ge u(y)\Leftrightarrow x\succcurlyeq y$ for all bundles $x$ and $y$.

If we have $u(x)=u(y)$, then we have $u(x)\ge u(y)$. Thus we get $x\succcurlyeq y$. QED

Remark: From $u(x)=u(y)$, we also get $u(y)\ge u(x)$ and therefore $y\succcurlyeq x$. Together with $x\succcurlyeq y$ then, by definition of the relation $\sim$, we have $x\sim y$. However, this is not really needed if you are only asked to prove that $x\succcurlyeq y$.

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  • $\begingroup$ Cheers. I'm still unclear about this: Why do we have $x \succcurlyeq y$ if $u(x) = u(y)$ with no further information on $x$ nor $y$? Why do we not know right away that $x \sim y$ or $y \sim x$? Like why is it you're mentioning $x \succcurlyeq y$ first, not $y \succcurlyeq x$ $\endgroup$ Jun 7 at 14:19
  • $\begingroup$ @j3141592653589793238, that's because of the way these symbols are defined. We know everything "right away", but for a proof you have to use the exact definition, and the indifference relation is defined via the at-least-as-good-as relation, so that's the sequence of steps you have to follow. $\endgroup$
    – VARulle
    Jun 8 at 15:33

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