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Is this possible? I've been trying so many times, without success, yet if I plot a two good CES on Wolfram, this seems true... I can't find a single numerical example where this does not hold, but I can't prove it either differentiating...

EDIT: The question is, can someone prove that the CES is increasing in the elasticity of substitution?

EDIT: I simply want to show whether the function $U(x_1,x_2;\sigma)$ increases as the parameter $\sigma$ does, which seems to be the case.

$U(\bullet)$ is a standard two good CES.

$$U(x_1,x_2;\sigma) = \left[\alpha x_1^{\frac{\sigma -1}{\sigma}} +(1-\alpha)x_2^{\frac{\sigma -1}{\sigma}} \right]^{\frac{\sigma}{\sigma -1}} $$

enter image description here

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  • $\begingroup$ Can you please edit your post so that it shows some of these plots & examples? I am not quite sure what you are trying to do as the elasticity of substitution in a CES function should be, by default, constant. You are probably changing some parameter, but I am not sure which one. $\endgroup$
    – Giskard
    Oct 7, 2022 at 9:26
  • $\begingroup$ He seems to be showing how output/utility is increasing in the EOS.. $\endgroup$ Oct 7, 2022 at 13:17

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Ok, thanks to Giskard, I found the proof: I think it is so elegant it deserves to be shared.

Let's take $$ U(x_1, ..., x_L) = \left( \sum_{l=1}^{L}\alpha_l x_l^{\rho} \right)^{1/\rho} $$

With $\rho = \frac{\sigma-1}{\sigma}$.

As $x_l \geq 0$, our goal is simply to prove that, for any $\rho_2 > \rho_1 > 0$,

$$ \left( \sum_{l=1}^{L}\alpha_l x_l^{\rho_2} \right)^{1/\rho_2} \geq \left( \sum_{l=1}^{L}\alpha_l x_l^{\rho_1} \right)^{1/\rho_1} $$

Take a scalar $k>1$, then, for any $y\in \mathbb{R}_{+} \cup \{0\}$, $f(y) = y^{k}$ is convex. Thus, by convexity and the Jensen's inequality, it holds that:

$$ \sum_{l=1}^{L} \alpha_l y_l^k \geq \left(\sum_{l=1}^{L} \alpha_l y_l \right)^k $$

Substitute $y_l = x_l^{\rho_1}$ and $k = \rho_2/\rho_1$, and take the $\rho_2$th root, and the statement is proved!

$$ \left( \sum_{l=1}^{L}\alpha_l x_l^{\rho_2} \right)^{1/\rho_2} \geq \left( \sum_{l=1}^{L}\alpha_l x_l^{\rho_1} \right)^{1/\rho_1} $$

As I commented below, I think that this feature of CES preferences makes them quite questionable at best, for most of the applications in Microeconomics. However, for trade theory and inter-temporal decisions, I think this can make sense.

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The statement is not generally true. For a counterexample check the function at $x_1=x_2=1$. This is $$ (\alpha \cdot 1 + (1- \alpha) \cdot 1)^{\frac{\sigma}{\sigma-1}} = 1 $$ which is constant in $\sigma$.

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  • $\begingroup$ Thanks, but this is not the point. Let's say non-decreasing, then. I updated. $\endgroup$ Oct 8, 2022 at 23:27
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    $\begingroup$ Alright; then I recommend looking into weighted power means. planetmath.org/weightedpowermean Final property seems to be what you want. $\endgroup$
    – Giskard
    Oct 9, 2022 at 2:09
  • $\begingroup$ Thanks a lot, Giskard! This is indeed what I wanted, but in my view, this is startling: it means that for the same allocation, a consumer with Leontief preferences has lower utility than a consumer with Cobb-Douglas preferences, hence that consumers dislike the fact that goods become more complementary. This, in my view, doesn't make any sense (it makes sense for a CES production function, not for preferences). You're basically claiming that for the same allocation of, say, oranges and lemons and bike and wheels, consumer prefer the first, in gross surplus terms. $\endgroup$ Oct 9, 2022 at 8:56
  • $\begingroup$ It does not really make sense to compare utility from two different utility functions. Utility is not a real unit of measurement, it serves as a tool for ordinal comparisons given a utility function. $\endgroup$
    – Giskard
    Oct 9, 2022 at 14:00
  • $\begingroup$ Well, it's the same utility function! CES preferences converge to a Leontief as sigma goes to 0, to a Cobb-Douglas as sigma goes to 1, and to linear preferences as sigma goes to infinity $\endgroup$ Oct 10, 2022 at 11:42

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