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.


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 $$ C_t \triangleq \int_0^1 f_{C,t}(i) di = \int_0^1 c_t di = c_t. $$


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.


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.

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 $4 = 2 \times \frac{4}{2}$

  • $\begingroup$ @denesp This at least adds non-ambiguity to your answer. And...indeed $di = 1di$.. What is your point? $\endgroup$ – Kanak Dec 9 '18 at 21:48
  • $\begingroup$ Any question @Mashim ? $\endgroup$ – Kanak Dec 9 '18 at 22:02
  • $\begingroup$ Sorry, I was unfamiliar with this $\int di$ notation. I see Alecos uses it as well. I thought it was a typo and a "1" was missing. $\endgroup$ – denesp Dec 10 '18 at 5:52

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