From what I understand, $MRS$ is calculated as

$$dU = U_x dx + U_y dy =0$$ which by rearrangement yields $$\frac{dy}{dx}= -\frac{U_x}{U_y}$$

So suppose I have $$U(x,y) = \ln x +\ln y$$ Then $$ \frac{dy}{dx}= -\frac{1/x}{1/y} = -\frac{y}{x}$$

Okay. So I have a function $y$ in terms of $x$.


Now consider my $U(x,y)$ again. Let $$\mathbf{a} = \begin{bmatrix} 1\\ 1 \end{bmatrix}$$ and $$U(\mathbf{a})=0$$ We have $$DU(x,y) = \begin{bmatrix} \frac{1}{x} & \frac{1}{y} \end{bmatrix} $$ and $$\frac{\partial U}{\partial y} (\mathbf{a}) = \begin{bmatrix}\frac{1}{y}\end{bmatrix}= 1$$ which is nonsingular since $\det(1) = 1$ and so by the Implicit Function Theorem, $$U = 0$$ defines $y$ implicitly as a function of $x$ in a neighborhood of $\mathbf{a}$.

My Question:

How are these two trains of thought connected? The first is stated in terms of differentials. But the second is not. So I am confused why the definition of $MRS$ follows from the implicit function theorem.


It's actually pretty straight forward. The implicit function theorem for two variables is given as follows (as long as some regularity conditions hold):

For $F(x, y) = 0$,

$ \frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y} $

In the case of MRS, we want the marginal change in $x$ associated with a marginal change in $y$ required to maintain a certain level of utility, $c$, such as (conveniently) $c=0$. So, starting with

$U = U(x, y) = 0$,

we have

$\frac{dy}{dx} = -\frac{U_x}{U_y}$

Note that $c=0$ is just a simplification for exposition. For a general $c$, you can just subtract it from either side of the equation and you get the same result since $c$ disappears in the derivative.

| improve this answer | |
  • $\begingroup$ And utility satisfies such regularity conditions? $\endgroup$ – BCLC Feb 16 '16 at 15:19
  • 1
    $\begingroup$ @BCLC Usually. The two main conditions are 1) $U(.,.)$ needs to be continuously differentiable and 2) the partial of F wrt y needs to be nonzero. The second is generally innocuous. The first is usually assumed either as a general primitive or by functional form, with the primary exception being behavioral econ. studies that break the assumption intentionally (eg. Prospect Theory) $\endgroup$ – philE Feb 16 '16 at 15:30

This is intended to be a partial answer. I hope more knowledgeable people will answer.

Apart from asserting that $\exists \phi(\mathbf{x}) =\mathbf{y}$ the implicit function theorem also asserts

\begin{equation} D \phi(\mathbf{x}) = - \left(\frac{\partial U}{\partial \mathbf{y}} \left( \begin{array}{c} x\\ \phi(\mathbf{x}) \\\end{array} \right) \right)^{-1}\left( \frac{\partial U}{\partial \mathbf{x}} \left( \begin{array}{c} x\\ \phi(\mathbf{x}) \\\end{array} \right) \right) \end{equation} in this case that corresponds to

$$\frac{d \phi(x)}{dx} = - \left(\frac{1}{y}\right)^{-1}\left(\frac{1}{x}\right) = -\frac{y}{x}$$ so thus $$ \frac{dy}{dx}= -\frac{y}{x}$$

This shows that the definition of $MRS$ follows from the Implicit Function Theorem.


Consider $$ dy = - (U_y)^{-1} U_x dx$$ which we agree follows from the implicit function theorem. Then by multiplying both sides by $U_y$ and rearranging we have

\begin{align*} U_y dy &= -U_x dx \\ U_x dx + U_y dy &= 0 \end{align*} and thus by the definition of a differential,

$$ U_x dx + U_y dy = 0 = dU$$

so this is also a consequence of the implicit function theorem. This can also be seen trivially since $U = 0$, then by the definition of a differential, $dU =0$.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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