3
$\begingroup$

Is it known when a marginal-rate-of-substitution function can be rationalized by some utility function?

More precisely, and focusing on the case of two goods, what conditions are required on $M: (\mathbb R_{\geq 0})^2 \to \mathbb R$ in order for there to exist $u: (\mathbb R_{\geq 0})^2 \to \mathbb R$ such that for all $x,y \geq 0$, $$ M(x,y)=\frac{u_x(x,y)}{u_y(x,y)} $$

$\endgroup$
1
  • $\begingroup$ Here is one thought: One way to look at this condition is as a linear, homogeneous, partial differential equation $$ u_x(x,y) = M(x,y) u_y(x,y) $$ The question then boils down to what conditions on $M(x,y)$ guarantee that such a PDE has a solution. I am unfortunately not knowledgeable on this topic. $\endgroup$
    – John Sturm
    Commented Apr 5 at 16:57

1 Answer 1

2
$\begingroup$

This is a rather indirect way.

For $\omega, z \in \mathbb{R}_{++}$, define the (demand) correspondence:

$$ D(\omega, z) = \left\{(x,y) \in \mathbb{R}^2_+| MRS(x,y) = \omega \text{ and } \omega x + y = z.\right\} $$ The idea is that $D(\omega, z)$ gives the demand correspondence for (relative prices $\omega = MRS(x,y)$ and (normalized) income $z = \omega x + y$.

If $D(\omega, z)$ would turn out to be a (smooth) function, one could then check if it satisfies the usual integrability conditions: Slutsky symmetry and negative definiteness. (Alternatively, one could check if $D(\omega, z)$ satisfies the SARP (strong axiom of revealed preference).)

$\endgroup$
1
  • $\begingroup$ Thanks so much. That was the trick I needed for my application. $\endgroup$
    – John Sturm
    Commented Apr 8 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.