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Tail integration and a median-to-mean bridge

Statement

Assume countable choice. Let $(Ω, F, P)$ be a probability space, let $Z$ be an integrable real random variable, and let $m$ be a median: $P (Z \leq m) \geq 1 / 2$ and $P (Z \geq m) \geq 1 / 2$ . Suppose that $L > 0$ and, for every $t > 0$ , both $P (Z > m + t)$ and $P (Z < m - t)$ are at most $2 e^{- t^{2} / (4 L^{2})}$ . Then

$∣ E Z - m ∣ \leq \int_{0}^{\infty} \min {1, 2 e^{- t^{2} / (4 L^{2})}} d t \leq 2 \sqrt{π} L < 4 L .$

The two one-sided expectations $E (Z - m)^{+}$ and $E (Z - m)^{-}$ equal their respective tail integrals. No symmetry of $Z$ is required. The first inequality in fact holds under the displayed tail bounds for any real $m$ , whether or not it is a median.

Facts & Assumptions

Given: The probability space, $Z$ , $m$ , $L$ and countable choice as in the statement. For a real $x$ , write $x^{+} = \max {x, 0}$ and $x^{-} = \max {- x, 0}$ .

[F1]

Expectation is integration against the probability measure; the expectation of an integrable real variable is the difference of the expectations of its positive and negative parts (Expectation of a nonnegative or integrable random variable). The nonnegative integral is defined as the supremum over simple nonnegative minorants (The nonnegative Lebesgue integral).

[F2]

The layer-cake formula gives $E X = \int_{0}^{\infty} P (X > t) d t$ for nonnegative measurable $X$ , allowing infinite values (Layer-cake formulas for random variables). The integral is linear on integrable functions (The Lebesgue integral is linear on L¹(μ)), and the nonnegative integral is monotone and positively homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F3]

The improper Gaussian integral is $\int_{- \infty}^{\infty} e^{- u^{2}} d u = \sqrt{π}$ (The Gaussian integral ∫_-∞^∞e^-x² dx=√π), and $π < 3.185 < 4$ (Bartle–Sherbert bounds for pi).

[F4]

Continuous functions on compact intervals are Riemann integrable, and substitution applies there (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion, Substitution: if φ is differentiable on [c,d] with φ' integrable and f is continuous on an interval containing φ([c,d]), then ∫_φ(c)^φ(d) f = ∫_cᵈ (f∘φ) φ'). Under countable choice, bounded Riemann integrable functions have the same Lebesgue integral (The Axiom of Countable Choice (AC_ω), A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral). Monotone convergence passes increasing nonnegative integrals to their limit (Monotone convergence for the integral).

[F5]

A probability space has $P (Ω) = 1$ , and its events have probabilities in $[0, 1]$ (Probability measures and probability spaces). A real random variable is a measurable real-valued map (Random elements and real random variables). An integrable real function has finite integrals of both its positive and negative parts (Integrable real and complex functions, and their integrals). Sums, real scalar multiples, and positive and negative parts of measurable real functions are measurable (Closure properties of measurable functions used by the integral).

[F6]

For a nonnegative simple function $s = \sum_{j} c_{j} χ_{E_{j}}$ on disjoint measurable sets, its simple integral is $\sum_{j} c_{j} P (E_{j})$ (The integral of a nonnegative simple function), and its nonnegative Lebesgue integral agrees with this value (The nonnegative integral agrees with the simple integral on simple functions).

Proof

Given: The data and assumptions in the statement.

Goal: Prove the one-sided tail identities and the displayed median-to-mean estimates.

1.1F1F2F5F6givenalgebra

Constants are measurable because inverse images of Borel sets are either $Ω$ or $\emptyset$ . The nonnegative constants $m^{+}$ and $m^{-}$ are simple, so [F5] and [F6] give $E m^{+} = m^{+} P (Ω) = m^{+}$ and $E m^{-} = m^{-} P (Ω) = m^{-}$ . Thus the constant $m$ is integrable by [F5] and has expectation $m^{+} - m^{-} = m$ by [F1]. Put $Y = Z - m$ . By the closure properties in [F5], $Y$ and its nonnegative parts $Y^{+}$ and $Y^{-}$ are measurable. Since $Z$ and the constant $m$ are integrable, the vector-space and linearity assertions in [F2] give integrability of $Y$ and $E Y = E Z - m$ . By [F5], both $E Y^{+}$ and $E Y^{-}$ are finite.

2.1F1F2step 1.1algebra

By [F1], $E Y = E Y^{+} - E Y^{-}$ . For $t > 0$ , the strict superlevel events of the nonnegative measurable variables $Y^{+}$ and $Y^{-}$ from step 1.1 are respectively ${Z > m + t}$ and ${Z < m - t}$ . Applying [F2] to $Y^{+}$ and $Y^{-}$ gives $E Y^{+} = \int_{0}^{\infty} P (Z > m + t) d t$ and $E Y^{-} = \int_{0}^{\infty} P (Z < m - t) d t$ . Both expectations are finite by step 1.1.

3.1F2F5step 1.1step 2.1givenalgebra

Each probability in step 2.1 is at most one by [F5] and at most the stated Gaussian bound. By monotonicity in [F2], each of the two nonnegative expectations is at most $I = \int_{0}^{\infty} \min {1, 2 e^{- t^{2} / (4 L^{2})}} d t$ . For nonnegative $a, b \leq I$ , $∣ a - b ∣ \leq I$ ; apply this to the two expectations in step 2.1 and use the expectation difference from steps 1.1 and 2.1. This argument does not use the median property or symmetry.

4.1F2F3F4step 3.1algebra∎

On every finite interval, substitute $u = t / (2 L)$ using [F4]. The integrand is continuous, so [F4] also identifies its compact-interval Riemann and Lebesgue integrals. Applying monotone convergence to increasing compact intervals and using the evenness of $e^{- u^{2}}$ and [F3] yields $\int_{0}^{\infty} 2 e^{- t^{2} / (4 L^{2})} d t = 2 \sqrt{π} L$ . Hence $I \leq 2 \sqrt{π} L$ by [F2]. Finally [F3] gives $\sqrt{π} < 2$ , so $2 \sqrt{π} L < 4 L$ . Together with step 3.1 this proves the claim.

Depends on

Used by · 1

Median–mean gap for convex Lipschitz concentration

Dependency tree · 2 levels

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