<p>Let <span class="math-container">$N=\max\{N_1,\ldots, N_J\}$</span> and <span class="math-container">$\sigma&lt;0$</span>.</p>
<p><span class="math-container">$$N=\left[N^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}\leq\left[\sum_{j=1}^{J} N_{j}^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}\leq \left[J N^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}=J^{\sigma /(\sigma-1)}  N.$$</span>
Since <span class="math-container">$\lim_{\sigma\uparrow 0} J^{\sigma /(\sigma-1)}=1,$</span> the result follows. The argument is adopted from <a href="https://math.stackexchange.com/a/807768/21674">here</a>.</p>
<p>Similarly, let <span class="math-container">$N'=\min\{N_1,\ldots, N_J\}$</span>, <span class="math-container">$K&gt;0$</span> be such that <span class="math-container">$KN'&gt;N_j$</span> for all <span class="math-container">$j$</span>,  and <span class="math-container">$0&lt;\sigma&lt;1$</span>.</p>
<p><span class="math-container">$$N'=\left[N'^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}\geq\left[\sum_{j=1}^{J} N_{j}^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}\geq \left[KJ N'^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}=[KJ]^{\sigma /(\sigma-1)}  N'.$$</span></p>
<p>Since <span class="math-container">$\lim_{\sigma\downarrow 0} [KJ]^{\sigma /(\sigma-1)}=1,$</span> the result follows.</p>
<p>The difference between the two results should not be too surprising, since you work with positive and negative exponents, respectively, and they are clearly diametrically opposed.</p>

CC BY-SA 4.0
2022-03-26T14:10:06.237Z
2022-03-26T14:10:06.237Z
Michael Greinecker
<p>I just find it seems that a CES aggregator e.g. <span class="math-container">$\left[\sum_{j=1}^{J} N_{j}^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}$</span> with <span class="math-container">$\sigma&lt;0$</span> is called a general convex aggregator and its limit as <span class="math-container">$\sigma \uparrow 0$</span> is <span class="math-container">$\max \left\{N_{1}, \ldots, N_{J}\right\}$</span>, which is exactly the inverse of the result that Leontief function is a special case of the CES aggregator when <span class="math-container">$\sigma \downarrow 0$</span>.</p>
<p>Is there any easy way to prove this and more importantly what is the intuition behind these two distinguished results?</p>

<mathematical-economics><ces-function>
Convex CES Aggregator
Alalalalaki
# Convex CES Aggregator

(Asked by: username_0 on 2022-03-26)

I just find it seems that a CES aggregator e.g. $\left[\sum\_{j=1}^{J} N\_{j}^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}$ with $\sigma<0$ is called a general convex aggregator and its limit as $\sigma \uparrow 0$ is $\max \left\{N\_{1}, \ldots, N\_{J}\right\}$, which is exactly the inverse of the result that Leontief function is a special case of the CES aggregator when $\sigma \downarrow 0$.

Is there any easy way to prove this and more importantly what is the intuition behind these two distinguished results?

**Tags:** mathematical-economics, ces-function

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## 1 answers

### Answer by username_1 on 2022-03-26. Score: 5 | Accepted answer

Let $N=\max\{N\_1,\ldots, N\_J\}$ and $\sigma<0$.

$$N=\left[N^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}\leq\left[\sum\_{j=1}^{J} N\_{j}^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}\leq \left[J N^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}=J^{\sigma /(\sigma-1)} N.$$
Since $\lim\_{\sigma\uparrow 0} J^{\sigma /(\sigma-1)}=1,$ the result follows. The argument is adopted from [here](https://math.stackexchange.com/a/807768/21674).

Similarly, let $N'=\min\{N\_1,\ldots, N\_J\}$, $K>0$ be such that $KN'>N\_j$ for all $j$, and $0<\sigma<1$.

$$N'=\left[N'^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}\geq\left[\sum\_{j=1}^{J} N\_{j}^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}\geq \left[KJ N'^{(\sigma-1) / \sigma}\right]^{\sigma /(\sigma-1)}=[KJ]^{\sigma /(\sigma-1)} N'.$$

Since $\lim\_{\sigma\downarrow 0} [KJ]^{\sigma /(\sigma-1)}=1,$ the result follows.

The difference between the two results should not be too surprising, since you work with positive and negative exponents, respectively, and they are clearly diametrically opposed.


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