Limit derivation of generalized mean, $-\infty$ case yields contradiction with ones as input

Question

Consider the following function (a CES utility function or weighted power mean):

$$u_\rho(\vec x) = \left(\sum_{i=1}^n a_i x_i^\rho\right)^{1/\rho}$$

for parameters $a_i \geq 0$, $\sum a_i =1$, and $\rho\leq1$.

It can be shown (see the Wikipedia link above or this Economics SE question) that taking the limit as $\rho\to -\infty$ yields

$$u_{-\infty}(\vec x) = \min_i a_i x_i$$

Consider the case where $x = \vec 1$. Clearly

$$u_\rho(\vec 1) = \left(\sum_{i=1}^n a_i \right)^{1/\rho} = 1^{1/\rho} = 1, \quad\forall \rho$$

But the limit above says that $$u_{-\infty}(\vec 1) = \min_i a_i \neq1$$ (unless $n=1$).

What explains the contradiction?

The formulas above can also be found in the following reference at p. 139:

• Nisan, Noam, Tim Roughgarden, Èva Tardos, and Vijay V. Vazirani, eds. 2007. Algorithmic Game Theory. Cambridge University Press.

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Here is a graph that may shed some light (or perhaps just add confusion):

Due to numerical errors, the function evaluates to zero for $\rho < -900$ or so, but it seems from the graph that the limit (blue) is approaching a value between $0.45$ and $0.50$, not the lower value of $0.11$ predicted (red).

import numpy as np
import matplotlib.pyplot as plt
np.random.seed(987)
def ces(a, x, rho):
    return sum(a * xrho)(1/rho)
def leontief(a, x):       # Limit of CES as rho -> -inf
    return min(a * x)
a = np.random.rand(3)
a /= sum(a)
x = np.random.rand(3)
rho = list(range(-1000, 2))
rho.remove(0)
plt.figure(figsize=(10, 7), facecolor="white")
plt.plot(rho, [ces(a, x, r) for r in rho], label="CES")
plt.xlabel(r"$ρ$"), plt.ylabel(r"$u_ρ(x)$")
plt.axhline(leontief(a, x), color="crimson", label="Leontief")
plt.legend()
print(a*x)
[0.11114134 0.19854696 0.48172427]

Answer 1

Supplementing the previous answers, a sketch of the general case:

Omitting the terms with $a_i=0$, summing terms with equal $x_i$, and renumbering the $x_i$, without loss of generality, we can assume $a_i>0$ and $0<x_i<x_{i+1}$ ($1\le i <n'\le n$). Then, in the limit $\rho\to-\infty$, the term $a_1x_1^\rho$ dominates all other terms in the sum, and we have $u_\rho(\vec x)\sim(a_1x_1^\rho)^{1/\rho}=a_1^{1/\rho}x_1\to x_1=\min_{\{i|a_i\neq0\}} x_i$.

If needed, this can be worked out in more detail similar to Ricardo Cavalcanti's answer.

Answer 2

As the example shows the limit expression is not correct. The true one should be

$$u_{-\infty}(\vec x) = \min_i x_i$$

(this requires that all $a_i>0$; but $a_i=0$ doesn't really make sense because then $u_\rho(\vec x)$ wouldn't depend on $x_i$ at all).

As the name mean suggests, $u_\rho(\vec x)$ lies always inside the interval $[\min_i x_i, \max_i x_i]$.

The Wikipedia link states that the ordinary (non-weighted) power mean converges to $\min_i x_i$ as $\rho \to -\infty$: $$\lim_{\rho\to-\infty} \left(\frac{x_1^\rho+\dots+x_n^\rho}{n}\right)^{1/\rho} = \min_i x_i$$

How does this relate to the weighted power mean?

An interesting observation is that if the weights $a_i$ are rational numbers, then the weighted power mean can be written as an ordinary power mean. Example: Suppose $a_1 = \frac{1}{6}$, $a_2 = \frac{2}{6}$, $a_3 = \frac{3}{6}$. Then $$(a_1x_1^\rho + a_2x_2^\rho + a_3x_3^\rho)^{1/\rho} = \left(\frac{x_1^\rho + 2x_2^\rho + 3x_3^\rho}{6}\right)^{1/\rho} = \left(\frac{x_1^\rho + x_2^\rho + x_2^\rho + x_3^\rho + x_3^\rho + x_3^\rho}{6}\right)^{1/\rho} $$ so the weighted power mean of $x_1,x_2,x_3$ with weights $a_1,a_2,a_3$ is the same as the ordinary power mean of $x_1,x_2,x_2,x_3,x_3,x_3$. (Essentially: giving more weight to variables is simulated by repeating them.)

Using this it follows that the weighted power mean also converges to $\min_i x_i$ as $\rho \to -\infty$.

Note: This method only works for rational weights $a_i$ but since real numbers can be approximated arbitrarily well by rational numbers it should be intutive that the result also holds for real weights.