<p>Many economic models consider a continuum of agents, <span class="math-container">$i \in [0,1]$</span>. Suppose these agents have characteristics <span class="math-container">$(x_i, y_i)$</span>, which are independently distributed. Are all possible values of <span class="math-container">$(x_i,y_i) \in \mathbb{R} \times \mathbb{R}$</span> held by some agent <span class="math-container">$i$</span>? If I pin down a specific value <span class="math-container">$\bar{x}$</span>, is there still a continuum of individuals with <span class="math-container">$x_i = \bar{x}$</span> and holding all possible values of <span class="math-container">$y_i$</span>? Some models appear to rely on this being the case (see e.g. the 'island-economy' proof in Appendix B.1 of <a href="http://www.jonathanheathcote.com/hsv_taxation_final.pdf" rel="nofollow noreferrer">http://www.jonathanheathcote.com/hsv_taxation_final.pdf</a> - it seems like on each 'island' there is a continuum of agents with different values for <span class="math-container">$\epsilon_i$</span>).</p>
<p>Is there a more mathematically formal way of setting this up which could clarify it? Does my confusion have something to do with the agent space being one-dimensional, whereas the characteristic space is two-dimensional?</p>
<p>Is the case different if each agent has characteristics <span class="math-container">$(x_i, y_i(j))$</span>, where <span class="math-container">$y_i(j)$</span> are a set of values for each <span class="math-container">$j \in [0,1]$</span>? In this sense, the characteristic vector exists in an infinite-dimensional space. In the previous example, the characteristic vectors have values in <span class="math-container">$\mathbb{R}^2$</span>, which has the same cardinality as the agent space <span class="math-container">$[0,1]$</span>. But in this example, the characteristic space <span class="math-container">$\mathbb{R} \times \mathbb{R}^{[0,1]}$</span> has greater cardinality than the agent space. I'm not sure if this is relevant though.</p>
<p>Thanks!</p>

CC BY-SA 4.0
2020-09-09T05:28:28.953Z
Seems like several of these would depend on the parameter space of the model used. Are you asking about the specific model you linked or models in general? (In the latter case the answer is "it depends".)
Giskard
CC BY-SA 4.0
2020-09-09T05:29:26.920Z
"*If I pin down a specific value $\bar{x}$, is there still a continuum of individuals with $x_i = \bar{x}$ and holding all possible values of $y_i$?*" Seems like this is an immediate consequence of the independence of $x_i$ and $y_i$.
Giskard
CC BY-SA 4.0
2020-09-09T05:31:40.073Z
"*where $y_i(j)$ are a set of values for each $j \in [0,1]$*" What is this undefined $j$ here?
Giskard
CC BY-SA 4.0
2020-09-09T06:40:31.030Z
I suppose my question is basically - what does it depend on? Is there some implicit assumption about the underlying probability space being made which determines whether or not this is the case? The undefined $j$ could be, for instance, some set of options over which individuals have random utility (e.g. products, firms, residential locations, etc).
John
CC BY-SA 4.0
2020-09-09T06:46:30.983Z
Maybe to put it more concretely - on the one hand, it seems intuitive that given because there is a continuum of agents, the entire bivariate distribution will be replicated in the population. On the other hand, if you imagine drawing $x_i$ for every point on the continuum, and then drawing $y_i$ again for each point, it does not seem intuitive (to me) to say you'll draw the whole bivariate distribution. The intuition breaks down even more for me in the infinite-dimensional case. I assume there's an often skipped-over measure-theoretic set-up which formalises this, but I could be wrong.
John
CC BY-SA 4.0
2020-09-09T07:41:59.060Z
One can make all these arguments rigorous using what is an "often skipped-over measure-theoretic set-up which formalises this" but this set up is quite subtle. See [here](https://www.sciencedirect.com/science/article/abs/pii/S0022053104002224) for the appropriate set-up and [here](https://link.springer.com/article/10.1007/s00199-009-0458-9) for results that show what exactly you have to assume in terms of the space of agents.
Michael Greinecker
CC BY-SA 4.0
2020-09-10T02:20:34.153Z
Thanks @MichaelGreinecker ! Those papers are very helpful (if a little beyond my mathematical ability right now). Two quick follow-up questions. (1) I can see how a rich Fubini extension can exist when the characteristic space in $\mathbb{R}^2$, as in my first example, but can it still exist when the characteristic space is $\mathbb{R} \times \mathbb{R}^{[0,1]}$, as in my second example. I'm not familiar enough with the maths to see whether this space (with a suitable $\sigma$-algebra and measure) would be super-atomless (Def 5 and Corollary 2 of Podczeck), for example.
John
CC BY-SA 4.0
2020-09-10T02:24:01.113Z
(2) @MichaelGreinecker - in cases where agents only draw their characteristics once, why not just model the agents as the measure space of characteristics directly? Why construct an agent space (e.g. $[0,1]$) and use it as an index for a continuum of random variables, only to then have to do all this work to show that the sample distribution equals the population distribution? It seems like a very roundabout way of getting the distribution of characteristics you wanted in the first place. Thanks again!
John
CC BY-SA 4.0
2020-09-10T11:20:47.213Z
@John "...can it still exist when the characteristic space is ...$\mathbb{R}^{[0,1]}$?"---According to Theorem 5 of Podczeck, it's enough that the original factor spaces in the product are Polish with atomless measures. So perhaps $\mathbb{R}^{[0,1]}$ is too much to ask for but, e.g. $C[0,1]$ or the Skorohod space $D[0,1]$ with an atom-less measure would be covered by the construction.
Michael
CC BY-SA 4.0
2020-09-10T13:23:21.197Z
@Michael that makes sense. I suppose that is somewhat intuitive, given Polish spaces all have the same cardinality as $[0,1]$, so you can have a surjective mapping from agents to the characteristic space, whereas $\mathbb{R}^[0,1]$ does not (I think). Thanks!
John
<mathematical-economics><theory>
Conditional distributions in model with continuum of agents
John
# Conditional distributions in model with continuum of agents

(Asked by: username_0 on 2020-09-09)

Many economic models consider a continuum of agents, $i \in [0,1]$. Suppose these agents have characteristics $(x\_i, y\_i)$, which are independently distributed. Are all possible values of $(x\_i,y\_i) \in \mathbb{R} \times \mathbb{R}$ held by some agent $i$? If I pin down a specific value $\bar{x}$, is there still a continuum of individuals with $x\_i = \bar{x}$ and holding all possible values of $y\_i$? Some models appear to rely on this being the case (see e.g. the 'island-economy' proof in Appendix B.1 of <http://www.jonathanheathcote.com/hsv_taxation_final.pdf> - it seems like on each 'island' there is a continuum of agents with different values for $\epsilon\_i$).

Is there a more mathematically formal way of setting this up which could clarify it? Does my confusion have something to do with the agent space being one-dimensional, whereas the characteristic space is two-dimensional?

Is the case different if each agent has characteristics $(x\_i, y\_i(j))$, where $y\_i(j)$ are a set of values for each $j \in [0,1]$? In this sense, the characteristic vector exists in an infinite-dimensional space. In the previous example, the characteristic vectors have values in $\mathbb{R}^2$, which has the same cardinality as the agent space $[0,1]$. But in this example, the characteristic space $\mathbb{R} \times \mathbb{R}^{[0,1]}$ has greater cardinality than the agent space. I'm not sure if this is relevant though.

Thanks!

**Tags:** mathematical-economics, theory
> **Comments:**
> **username_1**: Seems like several of these would depend on the parameter space of the model used. Are you asking about the specific model you linked or models in general? (In the latter case the answer is "it depends".)  
> **username_1**: "*If I pin down a specific value $\bar{x}$, is there still a continuum of individuals with $x_i = \bar{x}$ and holding all possible values of $y_i$?*" Seems like this is an immediate consequence of the independence of $x_i$ and $y_i$.  
> **username_1**: "*where $y_i(j)$ are a set of values for each $j \in [0,1]$*" What is this undefined $j$ here?  
> **username_0**: I suppose my question is basically - what does it depend on? Is there some implicit assumption about the underlying probability space being made which determines whether or not this is the case? The undefined $j$ could be, for instance, some set of options over which individuals have random utility (e.g. products, firms, residential locations, etc).  
> **username_0**: Maybe to put it more concretely - on the one hand, it seems intuitive that given because there is a continuum of agents, the entire bivariate distribution will be replicated in the population. On the other hand, if you imagine drawing $x_i$ for every point on the continuum, and then drawing $y_i$ again for each point, it does not seem intuitive (to me) to say you'll draw the whole bivariate distribution. The intuition breaks down even more for me in the infinite-dimensional case. I assume there's an often skipped-over measure-theoretic set-up which formalises this, but I could be wrong.  
> **username_2**: One can make all these arguments rigorous using what is an "often skipped-over measure-theoretic set-up which formalises this" but this set up is quite subtle. See [here](https://www.sciencedirect.com/science/article/abs/pii/S0022053104002224) for the appropriate set-up and [here](https://link.springer.com/article/10.1007/s00199-009-0458-9) for results that show what exactly you have to assume in terms of the space of agents.  
> **username_0**: Thanks @MichaelGreinecker ! Those papers are very helpful (if a little beyond my mathematical ability right now). Two quick follow-up questions. (1) I can see how a rich Fubini extension can exist when the characteristic space in $\mathbb{R}^2$, as in my first example, but can it still exist when the characteristic space is $\mathbb{R} \times \mathbb{R}^{[0,1]}$, as in my second example. I'm not familiar enough with the maths to see whether this space (with a suitable $\sigma$-algebra and measure) would be super-atomless (Def 5 and Corollary 2 of Podczeck), for example.  
> **username_0**: (2) @MichaelGreinecker - in cases where agents only draw their characteristics once, why not just model the agents as the measure space of characteristics directly? Why construct an agent space (e.g. $[0,1]$) and use it as an index for a continuum of random variables, only to then have to do all this work to show that the sample distribution equals the population distribution? It seems like a very roundabout way of getting the distribution of characteristics you wanted in the first place. Thanks again!  
> **username_3**: @John "...can it still exist when the characteristic space is ...$\mathbb{R}^{[0,1]}$?"---According to Theorem 5 of Podczeck, it's enough that the original factor spaces in the product are Polish with atomless measures. So perhaps $\mathbb{R}^{[0,1]}$ is too much to ask for but, e.g. $C[0,1]$ or the Skorohod space $D[0,1]$ with an atom-less measure would be covered by the construction.  
> **username_0**: @Michael that makes sense. I suppose that is somewhat intuitive, given Polish spaces all have the same cardinality as $[0,1]$, so you can have a surjective mapping from agents to the characteristic space, whereas $\mathbb{R}^[0,1]$ does not (I think). Thanks!  

---

## 0 answers

