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Dominated exponential normalization and covariance

Statement

Let $(Ω, F, m)$ be a sigma-finite measure space, let $w \geq 0$ be measurable with $m ({w > 0}) > 0$ , let $r \geq 1$ and let $X = (X_{1}, \dots, X_{r}) : Ω \to R^{r}$ be measurable and finite, and let $Θ \subseteq R^{r}$ be open. For each compact $K \subset Θ$ , assume an integrable $G_{K}$ satisfies $w (x) e^{θ \cdot X (x)} (1 + \sum_{a} ∣ X_{a} (x) ∣^{2}) \leq G_{K} (x) (θ \in K) .$ Then $Z (θ) = \int w e^{θ \cdot X} d m$ is finite, strictly positive and $C^{2}$ , and the probability density $ρ_{θ} = Z (θ)^{- 1} w e^{θ \cdot X}$ defines $d P_{θ} = ρ_{θ} d m$ as in The measure with density $f$ relative to $μ$ . For any integrable real $A$ , write $⟨ A ⟩_{θ} : = \int A d P_{θ}$ ; for square-integrable real $A, B$ , define ${Cov}_{θ} (A, B) : = \int (A - ⟨ A ⟩_{θ}) (B - ⟨ B ⟩_{θ}) d P_{θ}$ . These extend the phase-space conventions of Statistical states, densities, observables and expectations to the present measure space. Then $\partial_{a} \log Z = ⟨ X_{a} ⟩_{θ}, \partial_{a} \partial_{b} \log Z = {Cov}_{θ} (X_{a}, X_{b}) .$ In particular, the Hessian of $\log Z$ is positive semidefinite.

Facts & Assumptions

Given: The measurable model above and one common integrable envelope on each compact parameter neighborhood.

[A1]

For every compact $K \subset Θ$ , the stated $G_{K}$ is integrable and dominates the zeroth, first and second exponential moments.

[F1]

Dominated convergence passes pointwise limits through an integral when the functions have a common integrable majorant (Dominated convergence).

[F2]

The real exponential is differentiable with derivative itself; the chain rule and mean-value theorem apply to its parameterized composition (The real exponential function and the number $e$ by a power series, The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F3]

The exponential is positive, and the natural logarithm is defined on positive values with derivative $1 / x$ (The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , The natural logarithm as the inverse of the exponential function, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

[F4]

The measure with density is defined on an arbitrary measure space (The measure with density $f$ relative to $μ$ ). Integration against it equals integration of the density-weighted nonnegative function (Integrating against a density agrees with integrating the product); applying this to positive and negative parts gives the same identity for integrable real functions. Expectations and covariances here have the explicit integral definitions in the Statement, agreeing with Statistical states, densities, observables and expectations when the reference is a phase-reference space.

Proof

Given: [A1] and the integration, exponential and state interfaces [F1]–[F4].

1.1A1F3algebra

For each fixed $θ \in Θ$ , choose a compact neighborhood $K$ of $θ$ . The bound in [A1] implies $\int w e^{θ \cdot X} d m < \infty$ . On ${w > 0}$ the integrand is strictly positive by [F3]; since this set is the countable union of the sets ${w e^{θ \cdot X} \geq 1 / k}$ , at least one has positive measure. Hence $Z (θ) > 0$ as well as finite.

2.1A1F1F2step 1.1algebra

For coordinate $a$ , choose a compact neighborhood $K$ of $θ$ containing every $θ + s e_{a}$ with $∣ s ∣ \leq ϵ$ . The mean-value theorem and chain rule [F2] bound the difference quotient by $w ∣ X_{a} ∣ \sup_{∣ s ∣ \leq ϵ} e^{(θ + s e_{a}) \cdot X}$ . Since $∣ X_{a} ∣ \leq 1 + \sum_{c} ∣ X_{c} ∣^{2}$ , [A1] bounds this entire supremum pointwise by the single integrable $G_{K}$ . Thus [F1] gives $\partial_{a} Z = \int w X_{a} e^{θ \cdot X} d m$ . Applying the same argument to the first derivative, with $∣ X_{a} X_{b} ∣ \leq (∣ X_{a} ∣^{2} + ∣ X_{b} ∣^{2}) / 2$ , gives $\partial_{b} \partial_{a} Z = \int w X_{a} X_{b} e^{θ \cdot X} d m$ . The same local envelopes and [F1] give continuity of these derivatives, hence $Z \in C^{2}$ .

3.1F3F4step 2.1algebra∎

By [F3], dividing the first-derivative formula by $Z > 0$ gives $\partial_{a} \log Z = ⟨ X_{a} ⟩_{θ}$ . Differentiating once more yields $\partial_{b} \partial_{a} \log Z = \frac{\int X_{a} X_{b} w e^{θ \cdot X} d m}{Z} - \frac{\int X_{a} w e^{θ \cdot X} d m \int X_{b} w e^{θ \cdot X} d m}{Z^{2}} = ⟨ X_{a} X_{b} ⟩ - ⟨ X_{a} ⟩ ⟨ X_{b} ⟩ .$ The envelope makes $\int X_{a}^{2} d P_{θ}$ finite; $∣ X_{a} ∣ \leq 1 + X_{a}^{2}$ and $2 ∣ X_{a} X_{b} ∣ \leq X_{a}^{2} + X_{b}^{2}$ give finite means and mixed moments. Expanding the centered product in the Statement's covariance definition, and using $P_{θ} (Ω) = 1$ , identifies the displayed difference with that covariance. For $v \in R^{r}$ its quadratic form is $⟨ (v \cdot (X - ⟨ X ⟩))^{2} ⟩ \geq 0$ , proving positive semidefiniteness.

Dominated exponential normalization and covariance

Statement

Facts & Assumptions

Proof

Depends on

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