Owings paradox

Gérard Letac^{∗}

4 octobre 2001

Résumé

In Problem 10687 of the American Mathematical Monthly, October 1998, page 769, J. Owings observes that if X is a real random variable with mean μ and standard deviation σ, then although Chebyshev’s Inequality P r(|X −μ| ≥ ασ) ≤ 1/α^{2} implies that

P r(|X − μ| ≥ 2σ) + P r(|X − μ| ≥ 3σ) ≤ 1/4 + 1/9,

actually the better following result

P r(|X − μ| ≥ 2σ) + P r(|X − μ| ≥ 3σ) ≤ 1/4

holds. This note provides an explanation of this phenomena by a proper generalization.

Theorem. Let Y be a non negative random variable with mean 1. Let m be a positive measure on [1, +∞) such that

c = sup_{b≥1} \frac{1}{b}m([1, b]) < +∞.

Then ∫_{1}^{∞} P r[Y ≥ y]m(dy) ≤ c, and c is the smallest number with this property.

For instance if m = δ_{4} + δ_{9}, where δ_{a} is the Dirac mass on a, the constant c is 1/4, since

f(b) = \frac{1}{b}m([1, b]) here is decreasing on [1, 4), [4, 9) and [9, ∞) and since f(1) = 0, f(4) =

1/4 and f(9) = 2/9. Applying the theorem to this m and to Y = (X − μ)^{2}/σ^{2} explains the Owings paradox. Proof. Denote by C the convex set of probabilities ν on [0, +∞) such that ∫_{0}^{∞} yν(dy) = 1 and consider the function I on C defined by

I(ν) = ∫_{1}^{∞} ν([y, +∞))m(dy).

∗Laboratoire de Statistique et Probabilités, Université Paul Sabatier, 31062 Toulouse.

By Fubini, I(ν) = ∫_{1}^{∞} m([1, x])ν(dx). To find sup{I(ν) ; ν ∈ C}, we are essentially looking for the set of extreme points of C, and finding the sup of the linear form I on this set. Denote by S_{0} the set {(a, b) ; 0 ≤ a < 1 < b} and write S = S_{0} ∪ {(1, 1)}. For (a, b) in S_{0} consider the element ν_{a,b} of C defined by

ν_{a,b} = \frac{b − 1}{b − a}δ_{a} + \frac{1 − a}{b − a}δ_{b}

and write ν_{1,1} = δ_{1}. The following lemma shows that (ν_{a,b} ; (a, b) ∈ S) is the set of extreme points of C : Lemma : Given ν in C there exists at least one probability γ(da, db) on S such that for all positive measurable functions f : [0, +∞) → [0, +∞] one has

∫_{0}^{∞} f(x)ν(dx) = ∫_{S} (∫_{0}^{∞} f(x)ν_{a,b}(dx)) γ(da, db).

Accepting the lemma for a while, we finish the proof of the theorem by applying the lemma to f(x) = m([1, x]). We obtain

I(ν) = ∫_{S_{0}} \frac{1 − a}{b − a}m([1, b])γ(da, db) + m(\{1\})γ(\{1, 1\}) ≤ ∫_{S} \frac{1}{b}m([1, b])γ(da, db) ≤ c.

To see that c is the best constant, observe that I(ν_{0,b}) = \frac{1}{b}m([1, b]), thus sup\{I(ν) ; ν ∈ C} = c. Proof of the Lemma : We prove it first when ν({1}) = 0. Denote v = ∫_{1}^{∞}(b−1)ν(db) =

∫_{0}^{1}(1 − a)ν(da). Let γ_{1}(da, db) a probability on S_{0} whose margins are respectively

α(da) = v^{−1}(1 − a)1_{[0,1)}(a)ν(da), β(db) = v^{−1}(b − 1)1_{(1,+∞)}[b)ν(db).

For instance, γ_{1} can be taken as α(da)β(db). Then

γ(da, db) = \frac{v(b − a)}{(1 − a)(b − 1)}γ_{1}(da, db)

has the required property. To see this, one writes

∫_{S_{0}} (∫_{0}^{∞} f(x)ν_{a,b}(dx)) γ(da, db)
= ∫_{S_{0}} (\frac{b − 1}{b − a}f(a) + \frac{1 − a}{b − a}f(b)) \frac{v(b − a)}{(1 − a)(1 − b)}γ_{1}(da, db)
= v ∫_{S_{0}} f(a)(1 − a)^{−1}γ_{1}(da, db) + v ∫_{S_{0}} f(b)(b − 1)^{−1}γ_{1}(da, db)
= v ∫_{0}^{1} f(a)(1 − α)^{−1}α(da) + v ∫_{1}^{∞} f(b)(b − 1)^{−1}β(db) = ∫_{0}^{∞} f(x)ν(dx).

If f = 1, this proves also that γ is a probability. If ν({1}) = p ∈ (0, 1), we apply the above proof to ν_{0} = (1 − p)^{−1}1_{x≠1}ν(dx), obtaining a γ_{0}. The corresponding γ is now (1 − p)γ_{0} + pδ_{(1,1)}. The case p = 1 is trivial.