<p>Yes it is a function, just like you have $f(x)$, read as $f$ of $x$, $c(w)$ is $c$ of $w$ (in layman's words). Simply $c$ is the function of $w$ goods. Depending on the shape of the consumption function, the result is the area under the $c$ curve evaluated at each infinitesimally small intervals on $w$. </p>

CC BY-SA 3.0
2015-11-16T15:50:04.077Z
2015-11-16T15:50:04.077Z
london
<p>It is my impression that the correct expression for "Dixit-Stiglitz" preferences is </p>

<p>$$U=\left(\int_{0}^{1}c(\omega)^{\rho}d\omega\right)^{\frac{1}{\rho}}$$</p>

<p>which then can be seen as a continuous incarnation (in [0,1]) of, say</p>

<p>$$\left (\sum_{i=1}^na_i\omega_i^{\rho}\right)^{\frac{1}{\rho}}$$</p>

<p>with $c(\omega_i) = a_i^{1/\rho}\omega_i$.</p>

<p>In other words, a definite (Riemann) integral is indeed conceived as a sum of infinitesimally small rectangles, but it can also be seen as the continuous incarnation of a sum.</p>

<p>A formal link between an integral and a sum is provided by the <strong><a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula">Euler-MacLaurin formula</a></strong>.</p>

CC BY-SA 4.0
2018-06-10T12:09:06.487Z
Could you elaborate a bit more on this? I'm having a similar conceptual problem. It's not clear to me that this is a Riemann integral, given that $c(\omega)$ is not necessarily continuous and hence the Riemann integral not necessarily exists..
FooBar
CC BY-SA 3.0
2015-12-05T20:12:47.250Z
2015-12-05T20:12:47.250Z
Alecos Papadopoulos
<p>I asked this question on math.stackexchange but deleted it from there and brought it here.</p>

<p>I had a question about Constant Elasticity of substitution type preferences of the form: $$U=\int_{0}^{1}(c(\omega)^{\rho}d\omega)^{\frac{1}{\rho}}$$
 where the paramter $\rho$
  governs the degree of substitutability between goods. Here, $c(\omega)$
  represents consumption of good $\omega$
  which exists on the unit interval. As such, this type of preference specification aggregates over consumption of different goods.</p>

<p>My question is as follows. I have always thought of integrals of the form: $$I=\int f(x)dx$$
  as approximating sums of areas of infinitesimally small rectangles (in terms of their base) and heights being determined by $f(x)$
 . In the case of the example above, what really is $c(\omega)$
 ? Is it a function?</p>

<mathematical-economics><consumer-theory>
CES preferences intuition
ChinG
# CES preferences intuition

(Asked by: username_0 on 2015-11-16)

I asked this question on math.stackexchange but deleted it from there and brought it here.

I had a question about Constant Elasticity of substitution type preferences of the form: $$U=\int\_{0}^{1}(c(\omega)^{\rho}d\omega)^{\frac{1}{\rho}}$$
where the paramter $\rho$
governs the degree of substitutability between goods. Here, $c(\omega)$
represents consumption of good $\omega$
which exists on the unit interval. As such, this type of preference specification aggregates over consumption of different goods.

My question is as follows. I have always thought of integrals of the form: $$I=\int f(x)dx$$
as approximating sums of areas of infinitesimally small rectangles (in terms of their base) and heights being determined by $f(x)$
. In the case of the example above, what really is $c(\omega)$
? Is it a function?

**Tags:** mathematical-economics, consumer-theory

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## 2 answers

### Answer by username_2 on 2015-12-05. Score: 5 | Accepted answer

It is my impression that the correct expression for "Dixit-Stiglitz" preferences is

$$U=\left(\int\_{0}^{1}c(\omega)^{\rho}d\omega\right)^{\frac{1}{\rho}}$$

which then can be seen as a continuous incarnation (in [0,1]) of, say

$$\left (\sum\_{i=1}^na\_i\omega\_i^{\rho}\right)^{\frac{1}{\rho}}$$

with $c(\omega\_i) = a\_i^{1/\rho}\omega\_i$.

In other words, a definite (Riemann) integral is indeed conceived as a sum of infinitesimally small rectangles, but it can also be seen as the continuous incarnation of a sum.

A formal link between an integral and a sum is provided by the **[Euler-MacLaurin formula](https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula)**.

> **Comments:**
> **username_3**: Could you elaborate a bit more on this? I'm having a similar conceptual problem. It's not clear to me that this is a Riemann integral, given that $c(\omega)$ is not necessarily continuous and hence the Riemann integral not necessarily exists..  

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### Answer by username_1 on 2015-11-16. Score: 0

Yes it is a function, just like you have $f(x)$, read as $f$ of $x$, $c(w)$ is $c$ of $w$ (in layman's words). Simply $c$ is the function of $w$ goods. Depending on the shape of the consumption function, the result is the area under the $c$ curve evaluated at each infinitesimally small intervals on $w$.


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