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Mean, mean-square and covariance bounds for second-order linearization error

Statement

Let $m \geq 1$ , let $U \subseteq R^{m}$ be open and convex, let $h : U \to R$ be of class $C^{2}$ , and let $B \in [0, + \infty)$ satisfy $∥ H_{h} (x) ∥ \leq B$ for every $x \in U$ (A uniform Hessian bound controls the second-order Taylor remainder). Let $X : Ω \to R^{m}$ be a random element with $X (ω) \in U$ for every $ω$ , with $E [X] = μ \in U$ and $E ∣ X - μ ∣^{2} < + \infty$ , and put $R : = h (X) - h (μ) - \nabla h (μ) \cdot (X - μ) .$ Then $R$ is integrable and $∣ E [R] ∣ \leq E ∣ R ∣ \leq \frac{B}{2} E ∣ X - μ ∣^{2} .$ If moreover $E ∣ X - μ ∣^{4} < + \infty$ , then $R^{2}$ is integrable and $E [R^{2}] \leq \frac{B^{2}}{4} E ∣ X - μ ∣^{4} .$

Vector covariance error. Let $f = (f_{1}, \dots, f_{d}) : U \to R^{d}$ , $d \geq 1$ , have $C^{2}$ components satisfying $∥ H_{f_{i}} (x) ∥ \leq B_{i}$ on $U$ , $B_{i} \geq 0$ . Keep the same input hypotheses and assume $M_{4} : = E ∣ X - μ ∣^{4} < \infty$ . Put $A = D f (μ)$ , $L = f (μ) + A (X - μ)$ , $S = A Cov (X) A^{T}$ and $q_{i} = B_{i}^{2} M_{4} / 4$ . Then $f (X)$ is square-integrable and for every $1 \leq i, j \leq d$ , $∣ Cov (f (X))_{i j} - S_{i j} ∣ \leq \sqrt{S_{i i} q_{j}} + \sqrt{S_{j j} q_{i}} + \sqrt{q_{i} q_{j}} .$ In particular, for the scalar function $h$ above, writing $s^{2} = Var (h (μ) + \nabla h (μ) \cdot (X - μ))$ and $q = B^{2} M_{4} / 4$ gives $∣ Var (h (X)) - s^{2} ∣ \leq 2 \sqrt{s^{2} q} + q = B \sqrt{s^{2} M_{4}} + \frac{B^{2}}{4} M_{4} .$ No independence of input coordinates or remainders is assumed. These covariance bounds require the fourth moment; the second-moment mean bound alone does not control a covariance difference.

Facts & Assumptions

Given: Objects and hypotheses as in the Statement.

[L1]

Under the stated convexity and Hessian bound, every $y \in U$ satisfies $∣ h (y) - h (μ) - \nabla h (μ) \cdot (y - μ) ∣ \leq \frac{B}{2} ∣ y - μ ∣^{2}$ . (A uniform Hessian bound controls the second-order Taylor remainder)

[L2]

A continuous real-valued map on an open subset of $R^{m}$ is Borel measurable, composition of a Borel map with a random element is a random variable, and sums, differences and real multiples of measurable real-valued functions are measurable. (A continuous map has Borel preimages of Borel sets, A measurable function between measurable spaces, Random elements and real random variables, Closure properties of measurable functions used by the integral)

[L3]

A measurable real-valued function $g$ with $∣ g ∣ \leq c f$ pointwise, where $c \geq 0$ and $f$ is integrable with $f \geq 0$ , is integrable: monotonicity and homogeneity of the nonnegative integral give $\int ∣ g ∣ \leq c \int f < + \infty$ . (Integrable real and complex functions, and their integrals, Monotonicity and nonnegative homogeneity of the nonnegative integral)

[L4]

For integrable random variables, $∣ E [G] ∣ \leq E [∣ G ∣]$ , and $0 \leq G \leq H$ almost surely with $G, H$ integrable gives $E [G] \leq E [H]$ . (Linearity, monotonicity, and the modulus bound for expectation)

[L5]

Square-integrable covariance is symmetric and bilinear, with $Var (Z) = E [Z^{2}] - E [Z]^{2} \leq E [Z^{2}]$ . The affine linearization has covariance $A Cov (X) A^{T}$ . (Variance and covariance identities for random variables, The covariance of the linearization is $A Σ A^{T}$ , Mean vector and covariance matrix of a square-integrable random vector)

[L6]

Coordinatewise measurability is equivalent to measurability into $R^{d}$ . (A map into $R^{n}$ is measurable exactly when its coordinates are measurable)

Proof

technique · direct

1.1givenL2

The function $h$ is continuous because it is $C^{2}$ , hence Borel measurable, so $h \circ X$ is a random variable; the function $x \mapsto \nabla h (μ) \cdot (x - μ)$ is an affine map, hence Borel measurable, so its composition with $X$ is a random variable too. Therefore $R$ , their difference, is a random variable.

1.2L4L5algebra

For any square-integrable real variables $V, W$ , center them as $V_{0} = V - E [V]$ , $W_{0} = W - E [W]$ , and write $a = E [V_{0}^{2}]$ , $b = E [W_{0}^{2}]$ , $c = E [V_{0} W_{0}]$ . Products are integrable by $2 ∣ V_{0} W_{0} ∣ \leq V_{0}^{2} + W_{0}^{2}$ . Monotonicity and linearity give $0 \leq E [(V_{0} + t W_{0})^{2}] = a + 2 t c + t^{2} b$ for every real $t$ . If $b > 0$ , take $t = - c / b$ to obtain $c^{2} \leq a b$ ; if $b = 0$ , the affine expression $a + 2 t c$ is nonnegative for all $t$ , forcing $c = 0$ . Consequently $∣ Cov (V, W) ∣ \leq \sqrt{Var (V) Var (W)}$ , including zero variances.

2.1givenstep 1.1L1algebra

Applying [L1] at $y = X (ω)$ for every $ω$ gives the pointwise bound $∣ R ∣ \leq \frac{B}{2} ∣ X - μ ∣^{2}$ , and squaring gives $R^{2} \leq \frac{B^{2}}{4} ∣ X - μ ∣^{4}$ .

3.1step 1.1step 2.1L3L4algebra

The function $∣ X - μ ∣^{2}$ is integrable by hypothesis, so step 2.1 with $g = R$ , $c = B / 2$ and $f = ∣ X - μ ∣^{2}$ makes $R$ integrable by [L3], and [L4] applied to $R$ and to the pair $(0, ∣ R ∣)$ gives $∣ E [R] ∣ \leq E ∣ R ∣ \leq \frac{B}{2} E ∣ X - μ ∣^{2}$ .

3.2step 1.1step 2.1givenL3L4algebra

If $E ∣ X - μ ∣^{4} < + \infty$ , then step 2.1 with $g = R^{2}$ , $c = B^{2} / 4$ and $f = ∣ X - μ ∣^{4}$ makes $R^{2}$ integrable by [L3], and [L4] gives $E [R^{2}] \leq \frac{B^{2}}{4} E ∣ X - μ ∣^{4}$ .

4.1givenstep 1.1step 3.2L5L6algebra

In the vector case apply steps 1.1–3.2 componentwise to $f_{i}$ and put $R_{i} = f_{i} (X) - L_{i}$ . Each $R_{i}$ is measurable and $E [R_{i}^{2}] \leq q_{i}$ . Each $L_{i}$ is square-integrable as a finite linear combination of the square-integrable coordinates of $X$ : $(\sum_{k = 1}^{m} a_{k} v_{k})^{2} \leq (\sum_{k = 1}^{m} a_{k}^{2}) ∣ v ∣^{2}$ . Thus $f_{i} (X) = L_{i} + R_{i}$ is square-integrable by $(u + v)^{2} \leq 2 u^{2} + 2 v^{2}$ . Coordinatewise measurability gives a random vector $f (X)$ , and [L5] gives $Cov (L) = S$ and $Var (R_{i}) \leq q_{i}$ .

5.1step 4.1step 1.2L5algebra

By covariance bilinearity, $Cov (f_{i} (X), f_{j} (X)) - S_{i j} = Cov (L_{i}, R_{j}) + Cov (R_{i}, L_{j}) + Cov (R_{i}, R_{j})$ . Apply step 1.2 to these three terms and use $Var (L_{i}) = S_{i i}$ and $Var (R_{i}) \leq q_{i}$ from step 4.1. The triangle inequality yields $∣ Cov (f (X))_{i j} - S_{i j} ∣ \leq \sqrt{S_{i i} q_{j}} + \sqrt{S_{j j} q_{i}} + \sqrt{q_{i} q_{j}}$ .

6.1step 3.1step 3.2step 5.1algebra∎

Steps 3.1 and 3.2 establish the scalar mean and mean-square conclusions; step 5.1 proves the full vector covariance conclusion, and setting $d = 1$ with $B_{1} = B$ gives $2 \sqrt{s^{2} q} + q = B \sqrt{s^{2} M_{4}} + B^{2} M_{4} / 4$ .

Mean, mean-square and covariance bounds for second-order linearization error

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources