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For two goods, the MRS is defined as the amount of one good you would trade off for the other. Mathematically, $$\text{MRS} = \frac{dy}{dx}$$ where the amount of goods $X$ and $Y$ are denoted by $x$ and $y$ respectively.

How do we do this for three or more goods? The way (as per my understanding) is to find the tangent (hyper)plane (call it $T$) first. Now consider a point $q \in B(p,\epsilon) \cap T$ for an infinitesimally small $\epsilon > 0$. This tells us that we can get/give $q_1$ units of good $X_1$, $q_2$ units of good $X_2$, $\cdots$, $q_{n-1}$ units of good $X_{n-1}$ to give/get (trade) an unit of $x_n$.

How do I find the tangent $T$ and these elements $q$? For this, I need to have a general formula to compute all the $(x_1, \cdots, x_n)$ satisfying the equation. How do I do that?

The goal is to find the amount of units of each good we need to get/give for an unit of $x_n$. This does not have to be $x_n$, but we are looking at this particular case for brevity. Change of variable is not the main concern here.

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Pairwise MRS

The Marginal Rate of Substitution is usually defined as a pairwise thing. E.g., in case of three goods $x,y$ and $z$, you would have $\text{MRS}_{xy}, \text{MRS}_{xz}$ and $\text{MRS}_{yz}$. (And you could also flip any of these, i.e. $\text{MRS}_{yx}$.)

These are defined as the slope of the tangent line of the two dimensional indifference curve while holding the quantities of all other goods constant. Mathematically they are the same as the "usual" MRS for two goods, e.g., $$ \text{MRS}_{yz}(x,y,z) = -\frac{\frac{\partial U(x,y,z)}{\partial y}}{\frac{\partial U(x,y,z)}{\partial z}}. $$

Tangential hyperplanes

If I understand correctly, you are trying to find the hyperplane that is tangential to an indifference curve. When there are more than two but finitely many goods, this can be found using the Jacobian matrix of the utility function.

The Fréchet derivative of a function $f: \mathbb{R}^n \to \mathbb{R}$ is the vector $A$ when $$f(x + h) = f(x) + Ah +o(h)$$ where $\lim_{||h|| \to 0} \frac{o(h)}{||h||} = 0$.

If the vectorspace we are considering is finite dimensional and $f$ is Fréchet differentiable everywhere then the Fréchet derivative is given by the Jacobian matrix of $f$, denoted by $J_f$.

Consider a differentiable utility function $U$ and let $\underline{x}$ be a basket of goods. This also defines an indifference curve. Consider some nearby baskets $\underline{x} + \underline{h}$ which are on the same indifference curve, so $$ U(\underline{x} + \underline{h}) = U(\underline{x}). $$ According to the Fréchet derivative $$ U(\underline{x} + \underline{h}) = U(\underline{x}) + J_U(\underline{x}) \underline{h} +o(\underline{h}), $$ which means $$ J_U(\underline{x})\underline{h} = - o(\underline{h}). $$ Since multiplying by $J_U(\underline{x})$ is a linear operator this implies $$ J_U(\underline{x})\underline{h} = 0. $$ Thus the Jacobian defines the tangential hyperplane.

Example :
Let $U(x,y) = xy$. Then the Jacobian is $ J_U(x,y) = \begin{bmatrix} y & x \end{bmatrix} $, and $\text{MRS}(x,y)$ is the solution to $$J_U(x,y)\begin{bmatrix} 1 \\ \text{MRS}(x,y) \end{bmatrix} = 0.$$ In this case $$ \begin{bmatrix} y & x \end{bmatrix} \begin{bmatrix} 1 \\ -y/x \end{bmatrix} = 0.$$

Relation of the hyperplane and the pairwise MRSs, an example

The pairwise MRSs can be gleaned in a similar manner. Going back to the example with three goods $x,y$ and $z$, the pairwise MRSs are solutions to the following matrix equations: $$ J_U(x,y,z)\begin{bmatrix} 1 \\ \text{MRS}_{xy}(x,y,z) \\ 0 \end{bmatrix} = 0, $$ $$ J_U(x,y,z)\begin{bmatrix} 1 \\ 0 \\ \text{MRS}_{xz}(x,y,z) \end{bmatrix} = 0, $$ $$ J_U(x,y,z)\begin{bmatrix} 0 \\ 1 \\ \text{MRS}_{yz}(x,y,z) \end{bmatrix} = 0. $$

The hyperplane and multigood-trades

The Jacobian also shows us how one can trade goods of multiple sorts and stay "near" the same indifference curve. If $$ J_U(x,y,z)\begin{bmatrix} 1 \\ \Delta y \\ \Delta z \end{bmatrix} = 0 $$ holds then trading small quantities at the ratios defined by this vector will keep us "near" the indifference curve. In my opinion this is as close as one can get to a multidimensional MRS.

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  • $\begingroup$ Hi, thanks for the clarification. I think there was a misunderstanding with the notation, $\nabla U$ denotes the Jacobian matrix of $U$. This notation is also written in the wiki article you shared about Jacobian Matrix. $\endgroup$ Jul 26 at 6:30
  • $\begingroup$ While I did not explicitly mention that is the MRS (and I just realized that I forgot to do so), I did mean that. I have edited the equation of the tangent hyperplane to fix the error with dimension and write $x_i - x_i'$ instead of just $x_i'$ for the scalar. Is the tangent correct in my post (question) now? $\endgroup$ Jul 26 at 6:32
  • $\begingroup$ @Gang'sBigBoss Hi! Unfortunately, I do not understand the indexing in your notation. What is $n$? And how are there $n-1$ things in $\langle \rangle$? This operator is only defined for two inputs? $\endgroup$
    – Giskard
    Jul 26 at 7:36
  • $\begingroup$ Oh well then, the expression on the RHS is the summation $\displaystyle \sum_{i=1}^{n-1} \frac{\partial U}{\partial x_i} \cdot (x_i - x_i')$. Here $n$ is the dimensions, and in this particular sense, we are finding $\text{MRS}_{{x_1}, \cdots, x_{n-1}} (x_1, \cdots, x_n)$. There are $n-1$ elements because we the Utility function is of dim $n$. In general, we can change the variables to find $\text{MRS}_{x_{i_1}, \cdots, x_{i_{n-1}}} (x_1, \cdots, x_n)$ where each $i_j \in [1,n] \cap \mathbb{N}$ is distinct. Does this make sense? $\endgroup$ Jul 26 at 8:43
  • $\begingroup$ @Gang'sBigBoss To me it does not. You should probably edit this information into the body of your question though. $\endgroup$
    – Giskard
    Jul 26 at 16:35

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