<p>The set $\{\mathbb{p}:\exists \alpha \in (0,1), \mathbb{p}_1=\alpha b_i, \mathbb{p}_2=(1-\alpha)b_i\}$ coincides with the hyperplane $\{\mathbb{p}:p_1+p_2=b_i\}$, therefore you are not finding extra hyperplanes by doing this and varying $\alpha$.</p>

<p>More generally, if you fix $x=(x_1,x_2)$ and $b_i$, you obtain one hyperplane characterized by the equation $\{\mathbb{p}:p_1 x_1+p_2 x_2=b_i\}$. Since there are finitely many such $x$ (and $b_i$ is given), there are finitely many such hyperplanes.</p>

CC BY-SA 3.0
2015-06-04T06:18:09.283Z
2015-06-04T06:18:09.283Z
Oliv
<p>This question relates to a <a href="http://faculty.chicagobooth.edu/eric.budish/research/budish-approxceei-jpe-2011.pdf" rel="nofollow">specific paper by Eric Budish, published in 2011 in JPE</a>, but I've tried to put all relevant information in this question. On page 1072, he defines <em>budget constraint hyperplanes</em> as follows:</p>

<blockquote>
  <p>Let $H(i,x) = \{\mathbf{p}:\mathbf{p} \cdot x = b_i \}$ denote the
  hyperplane in $M$-dimensional price space along which agent $i$ can
  exactly afford bundle $x$. As prices cross $H(i,x)$ from below, bundle
  $x$ goes from being affordable for $i$ to being unaffordable for $i$.</p>
</blockquote>

<p>$\textbf{p}$ represents a price vector for the $M$ goods, which, importantly, are indivisible. $b_i$ is agent $i$'s budget. I think the rest is sufficiently self-explanatory. My confusion is with this subsequent statement:</p>

<blockquote>
  <p>Importantly, the number of such hyperplanes is finite because the
  number of agents and the number of bundles are finite. This is an
  advantage of having only indivisible goods.</p>
</blockquote>

<p>This I don't see. For example, suppose $M=2$. Then aren't all $\textbf{p}=(\alpha b_i, (1-\alpha)b_i)$ such that $\alpha \in [0,1]$ hyperplanes meeting that definition, and thus I have infinitely many?</p>

<p>Having said that, a hyperplane is the set of those price vectors, not each of the vectors, so maybe the full set of price vectors I've just described together define just one hyperplane? I suppose my issue is understanding what a hyperplane is in this setting, and how the indivisibility gives us just a finite set of hyperplanes to work with. Any guidance would be very much appreciated.</p>

CC BY-SA 3.0
2015-06-03T21:59:50.373Z
I think the key here is that he is fixing the bundle $x$. Notice the hyperplane is a function of which agent it is $i$ and what bundle he wants $x$. Might be helpful to stick to your $\mathcal{R}^2$ example and use $x = (1, 1)$ and $i$ gives us $b_i = B$. This should give you something like $H(i, x) = \{ p | p_1 = B - p_2 \}$ (where notationally I'm writing $p = (p_1, p_2) \in \mathcal{R}^2$). Does that help?
cc7768
CC BY-SA 3.0
2015-06-03T22:06:16.987Z
That does help a little - thanks. So in $\mathcal{R}^2$, a hyperplane is a line, though, right? And then for it to exactly equal the budget that line must go through $x$. But then, why can't I pivot that line around $x$ by a few degrees and call it a new hyperplane? What he describes as a hyperplane, $H(i,x)$, seems more like a set of hyperplanes to me. What am I missing?
Shane
CC BY-SA 3.0
2015-06-03T22:09:30.603Z
Exactly, a hyperplane in $\mathcal{R}^2$ is a line. Are you thinking about the hyperplane being in price space or goods space? $x$ isn't in the same space as the hyperplane, so I don't know what you mean pivot the line around $x$.
cc7768
CC BY-SA 3.0
2015-06-03T22:12:41.713Z
I'm thinking of a space with two goods where $x$ is the $(1,1)$ coordinate. I'm thinking of the budget constraint hyperplane/line as being any line that goes through that point. Basically, I'm thinking of the standard intro micro plot with apples, oranges, and a budget constraint. I realize the budget set is just the vertices, not the hull, because the goods are indivisible.
Shane
CC BY-SA 3.0
2015-06-03T22:15:09.323Z
This hyperplane is in the price space which means the axis that it would be plotted on are $p_1$ and $p_2$ so you can't really plot $x$ with this hyperplane
cc7768
CC BY-SA 3.0
2015-06-03T22:18:09.963Z
Indeed. OK - progress. Let me ponder this now with that added insight. OK, so then suppose $b_i=B$ and $x=(1,1)$. Then $H(i,x)$ is just a line going from $(0,B)$ to $(B,0)$ in the price space, yes?
Shane
CC BY-SA 3.0
2015-06-03T22:25:24.273Z
And then the reason we may have infinitely many of these with divisible goods is just that there are infinitely many $x$, which is not true in the finite indivisible goods case?
Shane
<microeconomics><mathematical-economics>
Budget constraint hyperplanes
Shane
# Budget constraint hyperplanes

(Asked by: username_0 on 2015-06-03)

This question relates to a [specific paper by Eric Budish, published in 2011 in JPE](http://faculty.chicagobooth.edu/eric.budish/research/budish-approxceei-jpe-2011.pdf), but I've tried to put all relevant information in this question. On page 1072, he defines *budget constraint hyperplanes* as follows:

> Let $H(i,x) = \{\mathbf{p}:\mathbf{p} \cdot x = b\_i \}$ denote the
> hyperplane in $M$-dimensional price space along which agent $i$ can
> exactly afford bundle $x$. As prices cross $H(i,x)$ from below, bundle
> $x$ goes from being affordable for $i$ to being unaffordable for $i$.

$\textbf{p}$ represents a price vector for the $M$ goods, which, importantly, are indivisible. $b\_i$ is agent $i$'s budget. I think the rest is sufficiently self-explanatory. My confusion is with this subsequent statement:

> Importantly, the number of such hyperplanes is finite because the
> number of agents and the number of bundles are finite. This is an
> advantage of having only indivisible goods.

This I don't see. For example, suppose $M=2$. Then aren't all $\textbf{p}=(\alpha b\_i, (1-\alpha)b\_i)$ such that $\alpha \in [0,1]$ hyperplanes meeting that definition, and thus I have infinitely many?

Having said that, a hyperplane is the set of those price vectors, not each of the vectors, so maybe the full set of price vectors I've just described together define just one hyperplane? I suppose my issue is understanding what a hyperplane is in this setting, and how the indivisibility gives us just a finite set of hyperplanes to work with. Any guidance would be very much appreciated.

**Tags:** microeconomics, mathematical-economics
> **Comments:**
> **username_1**: I think the key here is that he is fixing the bundle $x$. Notice the hyperplane is a function of which agent it is $i$ and what bundle he wants $x$. Might be helpful to stick to your $\mathcal{R}^2$ example and use $x = (1, 1)$ and $i$ gives us $b_i = B$. This should give you something like $H(i, x) = \{ p | p_1 = B - p_2 \}$ (where notationally I'm writing $p = (p_1, p_2) \in \mathcal{R}^2$). Does that help?  
> **username_0**: That does help a little - thanks. So in $\mathcal{R}^2$, a hyperplane is a line, though, right? And then for it to exactly equal the budget that line must go through $x$. But then, why can't I pivot that line around $x$ by a few degrees and call it a new hyperplane? What he describes as a hyperplane, $H(i,x)$, seems more like a set of hyperplanes to me. What am I missing?  
> **username_1**: Exactly, a hyperplane in $\mathcal{R}^2$ is a line. Are you thinking about the hyperplane being in price space or goods space? $x$ isn't in the same space as the hyperplane, so I don't know what you mean pivot the line around $x$.  
> **username_0**: I'm thinking of a space with two goods where $x$ is the $(1,1)$ coordinate. I'm thinking of the budget constraint hyperplane/line as being any line that goes through that point. Basically, I'm thinking of the standard intro micro plot with apples, oranges, and a budget constraint. I realize the budget set is just the vertices, not the hull, because the goods are indivisible.  
> **username_1**: This hyperplane is in the price space which means the axis that it would be plotted on are $p_1$ and $p_2$ so you can't really plot $x$ with this hyperplane  
> **username_0**: Indeed. OK - progress. Let me ponder this now with that added insight. OK, so then suppose $b_i=B$ and $x=(1,1)$. Then $H(i,x)$ is just a line going from $(0,B)$ to $(B,0)$ in the price space, yes?  
> **username_0**: And then the reason we may have infinitely many of these with divisible goods is just that there are infinitely many $x$, which is not true in the finite indivisible goods case?  

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## 1 answers

### Answer by username_2 on 2015-06-04. Score: 1 | Accepted answer

The set $\{\mathbb{p}:\exists \alpha \in (0,1), \mathbb{p}\_1=\alpha b\_i, \mathbb{p}\_2=(1-\alpha)b\_i\}$ coincides with the hyperplane $\{\mathbb{p}:p\_1+p\_2=b\_i\}$, therefore you are not finding extra hyperplanes by doing this and varying $\alpha$.

More generally, if you fix $x=(x\_1,x\_2)$ and $b\_i$, you obtain one hyperplane characterized by the equation $\{\mathbb{p}:p\_1 x\_1+p\_2 x\_2=b\_i\}$. Since there are finitely many such $x$ (and $b\_i$ is given), there are finitely many such hyperplanes.


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