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Two-sided Mills bounds for the standard normal tail

Statement

Let $φ (x) = (2 π)^{- 1 / 2} e^{- x^{2} / 2}$ and $\overline{Φ} (x) = \int_{x}^{\infty} φ (y) d y$ for $x \in R$ . Then for every $x > 0$ $\frac{x φ (x)}{1 + x^{2}} \leq \overline{Φ} (x) \leq \frac{φ (x)}{x},$ and consequently, for every $x > 1$ , $(\frac{1}{x} - \frac{1}{x^{3}}) φ (x) \leq \overline{Φ} (x) \leq \frac{φ (x)}{x} .$ Both bounds are sharp as $x \to \infty$ in the sense that the ratio of each side to $\overline{Φ} (x)$ tends to $1$ .

Facts & Assumptions

Given: AC, AC $_{ω}$ , DC, and a real $x > 0$ .

[F1]

$φ$ is the strictly positive, Borel measurable standard normal density, with $\int_{R} φ = 1$ , and $N (0, 1)$ is the probability measure $φ d y$ . Standard normal and normal laws The standard normal density has total mass one

[F2]

On a compact interval every $C^{1}$ function is Lipschitz, hence absolutely continuous and of bounded variation; in particular $φ$ and $y \mapsto 1 / y$ are absolutely continuous on $[x, R]$ for $0 < x < R$ . $C^{1}$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation

[F3]

Integration by parts for absolutely continuous functions: $\int_{a}^{b} F G^{'} + \int_{a}^{b} F^{'} G = F (b) G (b) - F (a) G (a)$ , under AC $_{ω}$ and DC. Integration by parts for absolutely continuous functions The Axiom of Countable Choice ( ${A C}_{ω}$ ) The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain

[F4]

Substitution computes $\int_{x}^{R} y e^{- y^{2} / 2} d y = e^{- x^{2} / 2} - e^{- R^{2} / 2}$ , and monotone convergence justifies passing to the limit $R \to \infty$ in the integrals of the nonnegative functions $φ$ and $φ (y) / y^{2}$ over $[x, R]$ . Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ Monotone convergence for the integral

[F5]

AC is the ambient assumption; AC $_{ω}$ and DC are the hypotheses of the integration-by-parts interface used in [F3]. The Axiom of Choice The Axiom of Countable Choice ( ${A C}_{ω}$ ) The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain

Proof

technique · direct

1.1

For $y \geq x > 0$ one has $φ (y) \leq (y / x) φ (y)$ ; integrating over $[x, R]$ and letting $R \to \infty$ with [F4] gives $\overline{Φ} (x) \leq \frac{1}{x} \int_{x}^{\infty} y φ (y) d y$ , and [F4] computes $\int_{x}^{R} y φ (y) d y = (2 π)^{- 1 / 2} (e^{- x^{2} / 2} - e^{- R^{2} / 2}) \to φ (x)$ , so $\overline{Φ} (x) \leq φ (x) / x$ .

givenF1F4

1.2

For $0 < x < R$ , [F2] and [F3] applied to $F (y) = 1 / y$ and $G = φ$ on $[x, R]$ , together with $φ^{'} (y) = - y φ (y)$ , give $\int_{x}^{R} φ (y) d y = φ (x) / x - φ (R) / R - \int_{x}^{R} φ (y) y^{- 2} d y$ .

F2F3

2.1

Letting $R \to \infty$ in [step 1.2] with [F4] gives $\overline{Φ} (x) = φ (x) / x - \int_{x}^{\infty} φ (y) y^{- 2} d y$ , and since $y^{- 2} \leq x^{- 2}$ on $[x, \infty)$ one gets $\overline{Φ} (x) \geq φ (x) / x - \overline{Φ} (x) / x^{2}$ , that is, $\overline{Φ} (x) (1 + x^{- 2}) \geq φ (x) / x$ and hence $\overline{Φ} (x) \geq x φ (x) / (1 + x^{2})$ .

step 1.2F4

3.1

The algebraic comparison $x / (1 + x^{2}) \geq (1 / x - 1 / x^{3})$ for $x > 0$ is equivalent to $x^{4} \geq (x^{2} - 1) (1 + x^{2}) = x^{4} - 1$ , which holds; combining it with [step 2.1] gives the displayed form for $x > 1$ , and the ratio claim follows because $(1 / x - 1 / x^{3}) / (1 / x) = 1 - x^{- 2} \to 1$ and $(x / (1 + x^{2})) / (1 / x) = 1 / (1 + x^{- 2}) \to 1$ while the upper bound already is $φ (x) / x$ .

step 1.1step 2.1

4.1

The endpoint and degenerate cases are covered: $x > 0$ is required so that $1 / x$ and the integration interval $[x, R]$ are meaningful and $F = 1 / y$ is $C^{1}$ on it; $x = 0$ is excluded by the statement because the upper bound would divide by zero, while $\overline{Φ} (0) = \frac{1}{2}$ is finite; the limit $R \to \infty$ is handled by monotone convergence over the increasing family $[x, R]$ ; the constants AC $_{ω}$ and DC are those declared for [F3] and are used nowhere else; and AC enters only through [F5].

step 1.2step 2.1F3F5given∎

Source notes

Durrett, Lemma 1.2.6 and the estimates (8.5.2) in the proof of Theorem 8.5.1, states the two-sided bound $(x^{- 1} - x^{- 3}) e^{- x^{2} / 2} \leq \int_{x}^{\infty} e^{- y^{2} / 2} d y \leq x^{- 1} e^{- x^{2} / 2}$ for $x > 0$ (up to the normalization constant), together with the asymptotic ratio $1$ used in the law of the iterated logarithm. The proof above derives the stronger lower bound $x φ (x) / (1 + x^{2})$ from the identity obtained by integrating by parts, which is the form consumed by the LIL item.

Depends on

Used by

Brownian law of the iterated logarithm at infinity Theorem

Dependency tree · two levels

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Sources

Rick Durrett, Probability: Theory and Examples, fifth edition, Lemma 1.2.6 and the Gaussian tail estimates (8.5.2) in the proof of Theorem 8.5.1 (standard reference, not scraped)