<p><strong>Pairwise MRS</strong></p>
<p>The Marginal Rate of Substitution is usually defined as a pairwise thing. E.g., in case of three goods <span class="math-container">$x,y$</span> and <span class="math-container">$z$</span>, you would have
<span class="math-container">$\text{MRS}_{xy}, \text{MRS}_{xz}$</span> and <span class="math-container">$\text{MRS}_{yz}$</span>. (And you could also flip any of these, i.e. <span class="math-container">$\text{MRS}_{yx}$</span>.)</p>
<p>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 &quot;usual&quot; MRS for two goods, e.g.,
<span class="math-container">$$
\text{MRS}_{yz}(x,y,z) = -\frac{\frac{\partial U(x,y,z)}{\partial y}}{\frac{\partial U(x,y,z)}{\partial z}}.
$$</span></p>
<p><strong>Tangential hyperplanes</strong></p>
<p>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.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative" rel="nofollow noreferrer">Fréchet derivative</a> of a function <span class="math-container">$f: \mathbb{R}^n \to \mathbb{R}$</span> is the vector <span class="math-container">$A$</span> when
<span class="math-container">$$f(x + h) = f(x) + Ah +o(h)$$</span>
where <span class="math-container">$\lim_{||h|| \to 0} \frac{o(h)}{||h||} = 0$</span>.</p>
<p>If the vectorspace we are considering is finite dimensional and <span class="math-container">$f$</span> is Fréchet differentiable everywhere <a href="https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative#Finite_dimensions" rel="nofollow noreferrer">then</a> the Fréchet derivative is given by the <a href="https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant" rel="nofollow noreferrer">Jacobian matrix</a> of <span class="math-container">$f$</span>, denoted by <span class="math-container">$J_f$</span>.</p>
<p>Consider a differentiable utility function <span class="math-container">$U$</span> and let <span class="math-container">$\underline{x}$</span> be a basket of goods. This also defines an indifference curve. Consider some nearby baskets <span class="math-container">$\underline{x} + \underline{h}$</span> which are on the same indifference curve, so
<span class="math-container">$$
U(\underline{x} + \underline{h}) = U(\underline{x}).
$$</span>
According to the Fréchet derivative
<span class="math-container">$$
U(\underline{x} + \underline{h}) = U(\underline{x}) + J_U(\underline{x}) \underline{h} +o(\underline{h}),
$$</span>
which means
<span class="math-container">$$
J_U(\underline{x})\underline{h} = - o(\underline{h}).
$$</span>
Since multiplying by <span class="math-container">$J_U(\underline{x})$</span> is a linear operator this implies
<span class="math-container">$$
J_U(\underline{x})\underline{h} = 0.
$$</span>
Thus the Jacobian defines the tangential hyperplane.</p>
<p><em>Example :</em> <br>
Let <span class="math-container">$U(x,y) = xy$</span>. Then the Jacobian is
<span class="math-container">$
J_U(x,y) =
\begin{bmatrix}
y &amp; x
\end{bmatrix}
$</span>,
and <span class="math-container">$\text{MRS}(x,y)$</span> is the solution to
<span class="math-container">$$J_U(x,y)\begin{bmatrix}
1 \\
\text{MRS}(x,y)
\end{bmatrix} = 0.$$</span>
In this case
<span class="math-container">$$
\begin{bmatrix}
y &amp; x
\end{bmatrix}
\begin{bmatrix}
1 \\
-y/x
\end{bmatrix} = 0.$$</span></p>
<p><strong>Relation of the hyperplane and the pairwise MRSs, an example</strong></p>
<p>The pairwise MRSs can be gleaned in a similar manner. Going back to the example with three goods <span class="math-container">$x,y$</span> and <span class="math-container">$z$</span>, the pairwise MRSs are solutions to the following matrix equations:
<span class="math-container">$$
J_U(x,y,z)\begin{bmatrix}
1 \\
\text{MRS}_{xy}(x,y,z) \\
0
\end{bmatrix} = 0,
$$</span>
<span class="math-container">$$
J_U(x,y,z)\begin{bmatrix}
1 \\
0 \\
\text{MRS}_{xz}(x,y,z)
\end{bmatrix} = 0,
$$</span>
<span class="math-container">$$
J_U(x,y,z)\begin{bmatrix}
0 \\
1 \\
\text{MRS}_{yz}(x,y,z)
\end{bmatrix} = 0.
$$</span></p>
<p><strong>The hyperplane and multigood-trades</strong></p>
<p>The Jacobian also shows us how one can trade goods of multiple sorts and stay &quot;near&quot; the same indifference curve. If
<span class="math-container">$$
J_U(x,y,z)\begin{bmatrix}
1 \\
\Delta y \\
\Delta z
\end{bmatrix} = 0
$$</span>
holds then trading small quantities at the ratios defined by this vector will keep us &quot;near&quot; the indifference curve. In my opinion this is as close as one can get to a multidimensional MRS.</p>

CC BY-SA 4.0
2021-07-26T06:30:30.577Z
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](https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant#:~:text=common%20notations%20include).
not tdm's twin
CC BY-SA 4.0
2021-07-26T06:32:33.133Z
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?
not tdm's twin
CC BY-SA 4.0
2021-07-26T07:36:13.780Z
@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?
Giskard
CC BY-SA 4.0
2021-07-26T08:43:18.743Z
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?
not tdm's twin
CC BY-SA 4.0
2021-07-26T16:35:50.660Z
@Gang'sBigBoss To me it does not. You should probably edit this information into the body of your question though.
Giskard
CC BY-SA 4.0
2021-07-26T16:37:58.963Z
Is it your intention to get the quantities of all other goods one is willing to give up for an additional (marginal) unit of good $n$? I think my answer clarified that there are multiple vectors that solve this problem, specifically see the "Relation of the hyperplane and the pairwise MRSs, an example" section, consider the vectors containing $\text{MRS}_{xy}$ and $\text{MRS}_{xz}$.
Giskard
CC BY-SA 4.0
2021-07-26T21:31:03.753Z
My intention was to calculate the amount of units of each (of the $n−1$) goods I get/give when I trade (give/get) one unit of xn. Like for $2$-goods case, we trade an unit of $y$ for MRS units of $x$. I have edited my original post.
not tdm's twin
CC BY-SA 4.0
2021-07-26T23:42:57.567Z
This is the exact question I have answered above.
Giskard
CC BY-SA 4.0
2021-07-27T03:12:42.493Z
I am not able to see a way to explicitly write the elements $q$. For example in the 2D case, $x→x+ϵ$ gives a $y$ always (for as long $x$ lies in the domain). But in higher dimensions, how do we know there'll always be an element if we set the value of $\epsilon_i$ for a *proper* subset of $\{x_1, \cdots, x_n\}$? What if the IC is discontinuous? I have understood the rest, thanks!
not tdm's twin
CC BY-SA 4.0
2021-07-27T03:46:31.537Z
If the IC is discontinuous that means $U$ is not continuous and as such it is not continuously differentiable. I don't know how you would define MRS in this case - seems like the concept just does not apply.
Giskard
CC BY-SA 4.0
2021-07-27T14:33:12.163Z
Exactly, it does not apply there. Not even in 2D, my bad. It does not apply in that case.  think this clears everything, thanks @Giskard.
not tdm's twin
CC BY-SA 4.0
2021-07-25T05:25:56.873Z
2021-07-27T00:25:49.263Z
2021-07-27T00:25:49.263Z
Giskard
<p>For two goods, the MRS is defined as the amount of one good you would trade off for the other. Mathematically, <span class="math-container">$$\text{MRS} = \frac{dy}{dx}$$</span> where the amount of goods <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are denoted by <span class="math-container">$x$</span> and <span class="math-container">$y$</span> respectively.</p>
<p>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 <span class="math-container">$T$</span>) first. Now consider a point <span class="math-container">$q \in B(p,\epsilon) \cap T$</span> for an infinitesimally small <span class="math-container">$\epsilon &gt; 0$</span>. This tells us that we can get/give <span class="math-container">$q_1$</span> units of good <span class="math-container">$X_1$</span>, <span class="math-container">$q_2$</span> units of good <span class="math-container">$X_2$</span>, <span class="math-container">$\cdots$</span>, <span class="math-container">$q_{n-1}$</span> units of good <span class="math-container">$X_{n-1}$</span> to give/get (trade) an unit of <span class="math-container">$x_n$</span>.</p>
<p>How do I find the tangent <span class="math-container">$T$</span> and these elements <span class="math-container">$q$</span>? For this, I need to have a general formula to compute all the <span class="math-container">$(x_1, \cdots, x_n)$</span> satisfying the equation. How do I do that?</p>
<p>The goal is to find the amount of units of each good we need to get/give for an unit of <span class="math-container">$x_n$</span>. This does not have to be <span class="math-container">$x_n$</span>, but we are looking at this particular case for brevity. Change of variable is not the main concern here.</p>

<utility>
How do you measure the MRS for more than two goods?
not tdm's twin
# How do you measure the MRS for more than two goods?

(Asked by: username_0 on 2021-07-24)

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.

**Tags:** utility

---

## 1 answers

### Answer by username_1 on 2021-07-25. Score: 3 | Accepted answer

**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](https://en.wikipedia.org/wiki/Fr%C3%A9chet_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](https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative#Finite_dimensions) the Fréchet derivative is given by the [Jacobian matrix](https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant) 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.

> **Comments:**
> **username_0**: 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](https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant#:~:text=common%20notations%20include).  
> **username_0**: 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?  
> **username_1**: @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?  
> **username_0**: 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?  
> **username_1**: @Gang'sBigBoss To me it does not. You should probably edit this information into the body of your question though.  
> **username_1**: Is it your intention to get the quantities of all other goods one is willing to give up for an additional (marginal) unit of good $n$? I think my answer clarified that there are multiple vectors that solve this problem, specifically see the "Relation of the hyperplane and the pairwise MRSs, an example" section, consider the vectors containing $\text{MRS}_{xy}$ and $\text{MRS}_{xz}$.  
> **username_0**: My intention was to calculate the amount of units of each (of the $n−1$) goods I get/give when I trade (give/get) one unit of xn. Like for $2$-goods case, we trade an unit of $y$ for MRS units of $x$. I have edited my original post.  
> **username_1**: This is the exact question I have answered above.  
> **username_0**: I am not able to see a way to explicitly write the elements $q$. For example in the 2D case, $x→x+ϵ$ gives a $y$ always (for as long $x$ lies in the domain). But in higher dimensions, how do we know there'll always be an element if we set the value of $\epsilon_i$ for a *proper* subset of $\{x_1, \cdots, x_n\}$? What if the IC is discontinuous? I have understood the rest, thanks!  
> **username_1**: If the IC is discontinuous that means $U$ is not continuous and as such it is not continuously differentiable. I don't know how you would define MRS in this case - seems like the concept just does not apply.  
> **username_0**: Exactly, it does not apply there. Not even in 2D, my bad. It does not apply in that case.  think this clears everything, thanks @Giskard.  

---

