<p>As I've written in the edit I believe this equality does indeed hold. We can think of x as a function of i. In a symmetric steady-state the function <span class="math-container">$x(i)$</span> is simply the constant function which returns the constant <span class="math-container">$x_{ss}$</span>, thus <span class="math-container">$$\int_0^1f(x_i)di=\int_0^1f(x(i))di=\int_0^1f(x_{ss})di=f(x_{ss})\int_0^1di=f(x_{ss})$$</span></p>

CC BY-SA 4.0
2023-07-18T13:06:30.630Z
2023-07-18T13:06:30.630Z
user43994
<p>Suppose I have a model with a continuum of agents denoted by <span class="math-container">$i \in [0;1]$</span>. In a symmetric equilibrium, i.e. in a equilibrium where <span class="math-container">$x_i = x_j$</span> for all <span class="math-container">$x,i$</span>, would I be able to conclude that: and some expression which I know holds:
<span class="math-container">$$ \int^1_0 f(x_i) di = f \left( \int^1_0 x_i di \right) $$</span>
The reasoning being that <span class="math-container">$\int^1_0 f(x_i) di$</span> can heuristically be thought of as a normalized sum over agents, i.e. as:
<span class="math-container">$$ \sum_i^n \frac{1}{n} f(x_i) $$</span>
Which in a symmetric equilibrium equals
<span class="math-container">$$ \frac{n}{n} f(x_i) = f(x_i) = f\left(\sum_i^n \frac{1}{n} x_i\right) $$</span>
Which can again heuristically be thought of as <span class="math-container">$f \left( \int^1_0 x_i di \right) $</span>.</p>
<p>Are my heuristics wrong? Also if someone has a more rigorous approach to understanding these types of integrals please tell me. I have been unable to find a better answer than &quot;think of the integral as a sum&quot; while searching this website and others.</p>
<p>EDIT: I believe this equality does indeed hold. We can think of <span class="math-container">$x$</span> as a function of <span class="math-container">$i$</span>. In a symmetric steady-state the function <span class="math-container">$x(i)$</span> is simply the constant function which returns the constant <span class="math-container">$x_{ss}$</span>, thus <span class="math-container">$\int_0^1 f(x_i) di = \int^1_0 f(x_{ss}) di = f(x_{ss}) \int^1_0 di = f(x_{ss})$</span></p>

CC BY-SA 4.0
2023-07-17T21:02:16.600Z
What is $F$? This formulas seem related to some theorem of passage of the limit or of a derivative under the integral sign. In general, these theorems have to do with measure theory, which is used in probability, and the Lebesgue's integral.
BakerStreet
CC BY-SA 4.0
2023-07-18T09:56:04.723Z
$F$ is any arbitrary function. I have replaced $F$ with $f$ in the question.
user43994
<macroeconomics><mathematical-economics><dsge>
What rules apply integrals over a continuum of agents in a symmetric equilibrium?
user43994
# What rules apply integrals over a continuum of agents in a symmetric equilibrium?

(Asked by: username_0 on 2023-07-17)

Suppose I have a model with a continuum of agents denoted by $i \in [0;1]$. In a symmetric equilibrium, i.e. in a equilibrium where $x\_i = x\_j$ for all $x,i$, would I be able to conclude that: and some expression which I know holds:
$$ \int^1\_0 f(x\_i) di = f \left( \int^1\_0 x\_i di \right) $$
The reasoning being that $\int^1\_0 f(x\_i) di$ can heuristically be thought of as a normalized sum over agents, i.e. as:
$$ \sum\_i^n \frac{1}{n} f(x\_i) $$
Which in a symmetric equilibrium equals
$$ \frac{n}{n} f(x\_i) = f(x\_i) = f\left(\sum\_i^n \frac{1}{n} x\_i\right) $$
Which can again heuristically be thought of as $f \left( \int^1\_0 x\_i di \right) $.

Are my heuristics wrong? Also if someone has a more rigorous approach to understanding these types of integrals please tell me. I have been unable to find a better answer than "think of the integral as a sum" while searching this website and others.

EDIT: I believe this equality does indeed hold. We can think of $x$ as a function of $i$. In a symmetric steady-state the function $x(i)$ is simply the constant function which returns the constant $x\_{ss}$, thus $\int\_0^1 f(x\_i) di = \int^1\_0 f(x\_{ss}) di = f(x\_{ss}) \int^1\_0 di = f(x\_{ss})$

**Tags:** macroeconomics, mathematical-economics, dsge
> **Comments:**
> **username_1**: What is $F$? This formulas seem related to some theorem of passage of the limit or of a derivative under the integral sign. In general, these theorems have to do with measure theory, which is used in probability, and the Lebesgue's integral.  
> **username_0**: $F$ is any arbitrary function. I have replaced $F$ with $f$ in the question.  

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## 1 answers

### Answer by username_0 on 2023-07-18. Score: 1

As I've written in the edit I believe this equality does indeed hold. We can think of x as a function of i. In a symmetric steady-state the function $x(i)$ is simply the constant function which returns the constant $x\_{ss}$, thus $$\int\_0^1f(x\_i)di=\int\_0^1f(x(i))di=\int\_0^1f(x\_{ss})di=f(x\_{ss})\int\_0^1di=f(x\_{ss})$$


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