<p>There is <span class="math-container">$C_t$</span> the number and <span class="math-container">$C_t()$</span> the function. This is already so in your first equation. Your second equation says that the function is constant over <span class="math-container">$[0,1]$</span>. In which case it is true that <span class="math-container">$C_t$</span> the number equals <span class="math-container">$C_t()$</span> the function's value times 1. </p>

<p>The same thing with a different notation:
<span class="math-container">$$ C_t \triangleq \int_0^1 f_{C,t}(i) di.$$</span>
If there exists a constant <span class="math-container">$c_t$</span>, where for all <span class="math-container">$i$</span>
<span class="math-container">$$f_{C,t}(i) = c_t.$$</span>
then
<span class="math-container">$$ C_t \triangleq \int_0^1 f_{C,t}(i) di = \int_0^1 c_t di = c_t. $$</span></p>

CC BY-SA 4.0
2018-12-09T18:52:09.197Z
2018-12-09T18:52:09.197Z
Giskard
<p>Both the other two answers are of course correct, but they do not provide a way to understand (or assess the suitability of) this correct but counter intuitive result.  </p>

<p>To complete the mathematical discussion, indeed, it holds, if <span class="math-container">$c(i) = c$</span>, that </p>

<p><span class="math-container">$$\int_0^1 c di = c \int_0^1  di = c \tag{1}$$</span></p>

<p>This is one of those useful cases where we can see clearly that treating an integral as "more or less like a sum", doesn't always work. By the integral we do <em>not</em> "sum" distinct consumed quantities for the fundamental reason that the index <span class="math-container">$i \in \mathbb R$</span> and not <span class="math-container">$i \in \mathbb N$</span> nor <span class="math-container">$i \in \mathbb Q$</span>. So we <em>cannot even write</em> a sum expression, which by definition requires an index that is at most countably infinite.  </p>

<p>Then the issue arises: is the specific mathematical tool appropriate in modeling the economic phenomenon that it attempts to model here? The OP's puzzlement is all too real and valid, since indeed, the expression says "I consume a continuum of goods each at the same quantity <span class="math-container">$c$</span> and the result is that my total consumption is just <span class="math-container">$c$</span>". </p>

<p>In the real world we say "I consume <span class="math-container">$n \in \mathbb N$</span> goods each at quantity <span class="math-container">$c$</span>, and my total consumption is then <span class="math-container">$n \cdot c$</span>." </p>

<p>Note that the discrepancy would have evaporated if the author of the book has used "mass" <span class="math-container">$n$</span> of goods rather then <span class="math-container">$1$</span> (for dummy variable of integration <span class="math-container">$v$</span> and consumption per good related to <span class="math-container">$[0,n]$</span> goods): </p>

<p><span class="math-container">$$\int_0^n c_n dv = c_n \int_0^n  dv = n\cdot c_n \tag{2}$$</span></p>

<p>This also helps us to understand why the <span class="math-container">$[0,1]$</span> formulation is used after all in many economic models that use such a depiction of cunsumption: it is just a re-scaling issue.</p>

<p>Assume that we start with eq. <span class="math-container">$(2)$</span>, which accords better with intuition. Define the variable </p>

<p><span class="math-container">$$i \equiv v/ n \implies di = \frac {dv}{n}\implies ndi =dv,\;\;\; v=0\implies i=0,\;\;\; v=n \implies i=1 $$</span></p>

<p>Applying the change of variables we have </p>

<p><span class="math-container">$$\int_0^n c_n dv = \int_0^1 c_n ndi $$</span></p>

<p>Then set <span class="math-container">$c \equiv n\cdot c_n$</span> to get </p>

<p><span class="math-container">$$\int_0^n c_n dv = \int_0^1 cdi = c\int_0^1 di = c  $$</span></p>

<p>In other words, when we have a continuum of goods <span class="math-container">$[0,1]$</span> as the authors did, and we say that we consume the same <span class="math-container">$c$</span> "per good", it is equivalent to say that we have a continuum of goods <span class="math-container">$[0,n]$</span> and we consume <span class="math-container">$c_n=c/n$</span> from each.</p>

CC BY-SA 4.0
2018-12-10T02:44:32.413Z
2018-12-10T10:51:22.047Z
2018-12-10T10:51:22.047Z
Alecos Papadopoulos
<p>I guess that you would have no problem in the general case of <span class="math-container">$n \ (&gt;0)$</span> goods, instead of <span class="math-container">$1$</span>. In such a case, the <em>identical-consumption</em> hypothesis becomes <span class="math-container">$C_t(i) = C_t(j) = C_t/n$</span>, which leads to</p>

<p><span class="math-container">$$C_t = \int_0^n C_t(i) di = \int_0^n \frac{C_t}{n} di = \frac{C_t}{n} \int_0^n di = \frac{C_t}{n} [ i ]^n_0 = \frac{C_t}{n} (n- 0) = C_t$$</span></p>

<p>Simply set <span class="math-container">$n=1$</span>, and you will get your answer.</p>

<hr>

<blockquote>
  <p>That's like saying <span class="math-container">$4 = 2 + 2$</span>, and then say <span class="math-container">$4 = 2$</span>.</p>
</blockquote>

<p>Actually, it is more like saying that, e.g. <span class="math-container">$4 = 1 + 3$</span> and then say <span class="math-container">$4 = 2 \times \frac{4}{2}$</span></p>

CC BY-SA 4.0
2018-12-09T22:02:35.700Z
Any question @Mashim ?
keepAlive
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2019-10-17T20:37:19.610Z
[Last seen Dec 9 '18 at 19:40](https://economics.stackexchange.com/users/20432/mashim)
Giskard
CC BY-SA 4.0
2019-10-17T21:01:40.783Z
Indeed, I regularly remove my old chatty comments. Thank you for showing me this math notation. Seems weird to set a bot for this, but that is just my opinion.
Giskard
CC BY-SA 4.0
2019-10-17T21:08:39.383Z
Actually @Giskard I just wanted to test the generalization-capability of an algorithm of web scraping. It did pretty well. Thx for the upvote.
keepAlive
CC BY-SA 4.0
2018-12-09T21:36:22.380Z
2018-12-13T20:29:18.777Z
2018-12-13T20:29:18.777Z
keepAlive
<p>I'm reading a book on monetary policy, and it defines the representative household's consumption as <span class="math-container">$$C_t = \int_0^1 C_t(i) di$$</span></p>

<p>so, basically a sum over the consumption of good <span class="math-container">$i$</span>. </p>

<p>Then later on, it says that the consumption of each good is identical <span class="math-container">$$C_t(i) = C_t(j) = C_t$$</span></p>

<p>This equation makes no logical sense. How can <span class="math-container">$C_t(i) = C_t$</span> for all <span class="math-container">$i$</span>? <span class="math-container">$C_t$</span> is DEFINED as an integral sum over all those <span class="math-container">$C_t(i)$</span>. They cannot be identical.</p>

<p>That's like saying 4 = 2 + 2, and then say 4 = 2. </p>

<macroeconomics>
Logically inconsistent optimality condition
Mashim
# Logically inconsistent optimality condition

(Asked by: username_0 on 2018-12-09)

I'm reading a book on monetary policy, and it defines the representative household's consumption as $$C\_t = \int\_0^1 C\_t(i) di$$

so, basically a sum over the consumption of good $i$.

Then later on, it says that the consumption of each good is identical $$C\_t(i) = C\_t(j) = C\_t$$

This equation makes no logical sense. How can $C\_t(i) = C\_t$ for all $i$? $C\_t$ is DEFINED as an integral sum over all those $C\_t(i)$. They cannot be identical.

That's like saying 4 = 2 + 2, and then say 4 = 2.

**Tags:** macroeconomics


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## 3 answers

### Answer by username_1 on 2018-12-09. Score: 2

There is $C\_t$ the number and $C\_t()$ the function. This is already so in your first equation. Your second equation says that the function is constant over $[0,1]$. In which case it is true that $C\_t$ the number equals $C\_t()$ the function's value times 1.

The same thing with a different notation:
$$ C\_t \triangleq \int\_0^1 f\_{C,t}(i) di.$$
If there exists a constant $c\_t$, where for all $i$
$$f\_{C,t}(i) = c\_t.$$
then
$$ C\_t \triangleq \int\_0^1 f\_{C,t}(i) di = \int\_0^1 c\_t di = c\_t. $$


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### Answer by username_2 on 2018-12-10. Score: 1

Both the other two answers are of course correct, but they do not provide a way to understand (or assess the suitability of) this correct but counter intuitive result.

To complete the mathematical discussion, indeed, it holds, if $c(i) = c$, that

$$\int\_0^1 c di = c \int\_0^1 di = c \tag{1}$$

This is one of those useful cases where we can see clearly that treating an integral as "more or less like a sum", doesn't always work. By the integral we do *not* "sum" distinct consumed quantities for the fundamental reason that the index $i \in \mathbb R$ and not $i \in \mathbb N$ nor $i \in \mathbb Q$. So we *cannot even write* a sum expression, which by definition requires an index that is at most countably infinite.

Then the issue arises: is the specific mathematical tool appropriate in modeling the economic phenomenon that it attempts to model here? The OP's puzzlement is all too real and valid, since indeed, the expression says "I consume a continuum of goods each at the same quantity $c$ and the result is that my total consumption is just $c$".

In the real world we say "I consume $n \in \mathbb N$ goods each at quantity $c$, and my total consumption is then $n \cdot c$."

Note that the discrepancy would have evaporated if the author of the book has used "mass" $n$ of goods rather then $1$ (for dummy variable of integration $v$ and consumption per good related to $[0,n]$ goods):

$$\int\_0^n c\_n dv = c\_n \int\_0^n dv = n\cdot c\_n \tag{2}$$

This also helps us to understand why the $[0,1]$ formulation is used after all in many economic models that use such a depiction of cunsumption: it is just a re-scaling issue.

Assume that we start with eq. $(2)$, which accords better with intuition. Define the variable

$$i \equiv v/ n \implies di = \frac {dv}{n}\implies ndi =dv,\;\;\; v=0\implies i=0,\;\;\; v=n \implies i=1 $$

Applying the change of variables we have

$$\int\_0^n c\_n dv = \int\_0^1 c\_n ndi $$

Then set $c \equiv n\cdot c\_n$ to get

$$\int\_0^n c\_n dv = \int\_0^1 cdi = c\int\_0^1 di = c $$

In other words, when we have a continuum of goods $[0,1]$ as the authors did, and we say that we consume the same $c$ "per good", it is equivalent to say that we have a continuum of goods $[0,n]$ and we consume $c\_n=c/n$ from each.


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### Answer by username_3 on 2018-12-09. Score: 1

I guess that you would have no problem in the general case of $n \ (>0)$ goods, instead of $1$. In such a case, the *identical-consumption* hypothesis becomes $C\_t(i) = C\_t(j) = C\_t/n$, which leads to

$$C\_t = \int\_0^n C\_t(i) di = \int\_0^n \frac{C\_t}{n} di = \frac{C\_t}{n} \int\_0^n di = \frac{C\_t}{n} [ i ]^n\_0 = \frac{C\_t}{n} (n- 0) = C\_t$$

Simply set $n=1$, and you will get your answer.

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> That's like saying $4 = 2 + 2$, and then say $4 = 2$.

Actually, it is more like saying that, e.g. $4 = 1 + 3$ and then say $4 = 2 \times \frac{4}{2}$

> **Comments:**
> **username_3**: Any question @username_0 ?  
> **username_1**: [Last seen Dec 9 '18 at 19:40](https://economics.stackexchange.com/users/20432/mashim)  
> **username_1**: Indeed, I regularly remove my old chatty comments. Thank you for showing me this math notation. Seems weird to set a bot for this, but that is just my opinion.  
> **username_3**: Actually @username_1 I just wanted to test the generalization-capability of an algorithm of web scraping. It did pretty well. Thx for the upvote.  

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