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A Note on a Moment Inequality

  • Seungwon (Eugene) Jeong EMAIL logo
Published/Copyright: September 21, 2023

Abstract

Using Chebyshev’s sum inequality, we generalize E[X 2] ≥ E 2[X] into a moment inequality with a general combination of moments that compares E[X r+s ] and E[X r ]E[X s ]. We also provide its economic application.

JEL Classification: C02; D44; D81; H57

1 Introduction

While Chebyshev’s inequality is one of the most widely used inequalities in many fields, some other inequalities of Chebyshev are less commonly known or used. Chebyshev’s algebraic inequality is introduced in Simonovits (1995), yet Chebyshev’s sum inequality may not have received enough attention, especially in social sciences.[1] For instance, as also quoted in Marinescu and Monea (2018), Fink (2000) says,

The most obviously important named inequalities are those of Hölder and Minkowski; but the watershed paper, in my estimation, is the paper of Chebyshev.

From the fact that variances are nonnegative, we have an inequality: Var(X) = E[X 2] − E[X]E[X] ≥ 0. While it is a natural step to generalize this inequality into the one with higher order moments or general combinations of moments, to the best of our knowledge, the generalized version has not received attention, which may be because potential applications are not clear.

We generalize E[X 2] ≥ E[X]E[X] by comparing E[X r+s ] and E[X r ]E[X s ] by using Chebyshev’s sum inequality, and provide its economic application. We first recall Chebyshev’s sum inequality (or Chebyshev’s order inequality (Chebyshev 1882)).[2]

1.1 Chebyshev’s Sum Inequality

Let f 1 and f 2 be increasing functions, and X be a real random variable with a probability density function f. Then,

(1) E [ f 1 ( X ) f 2 ( X ) ] E [ f 1 ( X ) ] E [ f 2 ( X ) ] .

where (1) holds with equality only when f 1 or f 2 is constant on the support of f. It can easily be shown that the inequality is reversed if one of f 1 and f 2 is increasing and the other is decreasing.

Note that (1) can also be stated as Cov(f 1(X), f 2(X)) = E[f 1(X)f 2(X)] − E[f 1(X)]E[f 2(X)] ≥ 0, which provides an intuitive interpretation: if f 1 and f 2 are both increasing, the covariance of f 1(X) and f 2(X) is nonnegative, but if one of f 1 and f 2 is increasing and the other is decreasing, the covariance is nonpositive.

2 Results

We state the result with continuous random variables, but the result also holds for discrete random variables by the discrete version of Chebyshev’s sum inequality.

Proposition 1

Let X be a nondegenerate nonnegative random variable with a probability density function f. For any r , s R and f except for f(0) > 0 with r + s = 0 or rs = 0,[3]

(2) E [ X r + s ] > E [ X r ] E [ X s ] , r s > 0 E [ X r + s ] = E [ X r ] E [ X s ] , r s = 0 E [ X r + s ] < E [ X r ] E [ X s ] , r s < 0 .

Proof

Note that x k is increasing (decreasing, resp.) in x R + if k > 0 (k < 0, resp.), and x 0 = 1 unless x = 0. Thus, Chebyshev’s sum inequality (1) implies (2). Note that (2) trivially holds when rs = 0, but it can also be derived as the result from

E [ X r + s ] E [ X r ] E [ X s ] , r s 0 E [ X r + s ] E [ X r ] E [ X s ] , r s 0 .

where each (weak) inequality holds by Chebyshev’s sum inequality.□

Proposition 1 can be understood more intuitively with the covariance interpretation. When rs > 0, r and s have the same signs; thus, Cov(X r , X s ) > 0.

We state a corollary below, as the case of s = 1 often appears in applications, e.g. Proposition 5.

Corollary 1

Let X be a nondegenerate nonnegative random variable with a probability density function f. For any α R and f except for f(0) > 0 with α = 0 or α = 1,

(3) E [ X α ] > E [ X α 1 ] E [ X ] , α > 1 E [ X α ] = E [ X α 1 ] E [ X ] , α = 1 E [ X α ] < E [ X α 1 ] E [ X ] , α < 1 .

When X is not a nonnegative random variable, a moment inequality (2) may not hold. This is because x k may not be monotone any more, e.g. x 2. For instance, in Corollary 1, either E[X α ] > E[X α−1]E[X] or E[X α ] < E[X α−1]E[X] is possible as below.

Example 1

Consider α = 0 so to compare E[X −1 X] = 1 and E[X −1]E[X]. If X is 1 or −1 with equal probability, then E[X] = 0. Thus, E[X −1 X] = 1 > 0 = E[X −1]E[X]. More generally, if X is a or b with equal probability, then E[X −1 · X] > E[X −1]E[X] if ab < 0, and E[X −1 · X] < E[X −1]E[X] if ab > 0.□

Nevertheless, there are some useful cases when the moment inequality holds for real-valued (i.e. not necessarily nonnegative) random variables. For instance, when r and s are both even or odd natural numbers, i.e. when r + s is an even number, we have an inequality below.

Proposition 2

Let X be a real random variable with a probability density function f. For any r , s N { 1,2 , } such that X r and X s are nondegenerate[4] and r + s is even,

(4) E [ X r + s ] > E [ X r ] E [ X s ] .

Proof

If k N is an odd number, x k is increasing in x R . Thus, when both r and s are odd, by Chebyshev’s sum inequality (1), the desired inequality (4) holds. On the other hand, when both r and s are even, applying Chebyshev’s sum inequality to the nondecreasing rearrangements of X r and X s concludes the proof as follows: let Y 1 and Y 2 be the nondecreasing rearrangements of X r and X s . That is, the pdfs of Y 1 and Y 2 are f Y 1 ( y ) = inf { t > 0 : m ( { x : f ( x r ) > t } ) y } and f Y 2 ( y ) = inf { t > 0 : m ( { x : f ( x s ) > t } ) y } where m(⋅) is a measure. Since f Y 1 ( x r ) = f Y 2 ( x s ) for all x in the support of f, applying Chebyshev’s sum inequality (1) to Y 1 and Y 2 concludes the proof.□

Note that r and s should be natural numbers rather than integers. For instance, x −1 is not monotone, as illustrated in Example 1. Proposition 2 can also be understood more intuitively with the covariance interpretation: x k is increasing for odd k; thus, when both r and s are odd, Cov(X r , X s ) > 0. When both r and s are even, for intuition, assume that X is discrete. Then, X r and X s can be rearranged in the nondecreasing order with the same permutation of the indices. Thus, the covariance of the two nondecreasing rearranged random variables is positive.

3 Application

Jeong and Kim (2021) study a procurement, where the total cost of the procurement is proportional to the number of bidders. In particular, they compare bidding strategies in two cases depending on the presence of uncertainty—whether bidders know the exact number of bidders n (no uncertainty); or bidders do not know n but they believe E[N] = n (uncertainty) where N is the random variable that denotes the number of bidders. They show that bidders bid more aggressively without uncertainty than with uncertainty, under a certain condition (which is (5) in Proposition 3).

More specifically, bidder i’s unit cost c i is independently drawn from an identical distribution F, with density f that is continuously differentiable and positive on ( c ̲ , c ̄ ) R + . Bidder i’s total cost is ρ(n, α)c i , where ρ(n, α) denote an impact function with an impact factor α, and ρ(n, α) is increasing in both n and α. Let β(c) and β ̃ ( c ) denote the bidding strategies without uncertainty and with uncertainty, respectively. Then, the following Propositions 3 and 4 hold.

Proposition 3

(Theorem 2 of Jeong and Kim (2021)) If

(5) β ( c ̲ ) β ̃ ( c ̲ ) ,

then β ( c ) < β ̃ ( c ) for all c ( c ̲ , c ̄ ) .

That is, if (5) holds, the desired result—bidders bid more aggressively without uncertainty than with uncertainty—holds. Note that, in procurements, c ̲ is an end point (not a start point), so showing that (5) holds is not trivial. However, it can be shown that (5) is equivalent to (6) in the following proposition.

Proposition 4

(Proposition 2 of Jeong and Kim (2021)) Condition (5), β ( c ̲ ) β ̃ ( c ̲ ) , holds if and only if

(6) E [ ρ ( N , α ) ] β ( c ̲ ; n ) E [ ρ ( N , α ) β ( c ̲ ; N ) ] .

There are some special cases where (6) can be shown easily. For instance, when cU[0, 1] and ρ(n, α) = n α , by using the main result, Proposition 1, or more specifically, Corollary 1, we can further characterize the relationship between (5) and (6).

Proposition 5

If c is distributed according to U[0, 1] and ρ(n, α) = n α , then the condition (5) holds if and only if α ≤ 1. More specifically,

β ( c ̲ ) < β ̃ ( c ̲ ) , α < 1 β ( c ̲ ) = β ̃ ( c ̲ ) , α = 1 β ( c ̲ ) > β ̃ ( c ̲ ) , α > 1 .

Proof

Note that β ( c ̲ ; n ) = β ( 0 ; n ) = 1 / n (Jeong and Kim 2021). Thus, (6) with ρ(n, α) = n α becomes E[N α ]/nE[N α−1] or, equivalently, E[N α ] ≤ E[N α−1]E[N]. By Corollary 1,

E [ N α ] < E [ N α 1 ] E [ N ] , α < 1 E [ N α ] = E [ N α 1 ] E [ N ] , α = 1 E [ N α ] > E [ N α 1 ] E [ N ] , α > 1 .


Corresponding author: Seungwon (Eugene) Jeong, College of Business, KAIST, 291 Daehak-ro, Yuseong-gu, Daejeon, 34141, Republic of Korea, E-mail:

Funding source: British Academy/Leverhulme Small Research Grants

Award Identifier / Grant number: SRG21\211503

Funding source: National Research Foundation of Korea (NRF), Korea Government (MSIT)

Award Identifier / Grant number: RS-2022-00166647

Acknowledgment

I thank Ken Binmore and Dong-Hyuk Kim for their valuable comments.

  1. Research funding: This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. RS-2022-00166647). This work was also supported by the British Academy/Leverhulme Small Research Grants SRG21/211503 when I was at the University of Bristol.

References

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Received: 2023-05-14
Accepted: 2023-07-23
Published Online: 2023-09-21

© 2023 the author(s), published by De Gruyter, Berlin/Boston

This work is licensed under the Creative Commons Attribution 4.0 International License.

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