# Constructive van Dantzig pairs

## Problem, conditions and results

Let $U$ be symmetric, nondegenerate, with $M_U(t)=\mathbb E e^{tU}<\infty$ for every real $t$ (hypothesis H). Seek an independent real $V$ such that
$$
\mathbb E e^{t(U+iV)}=1\quad(t\in\mathbb R),\qquad
\varphi_V(t)=1/M_U(t).
$$
All constructed pairs use independent representatives. The manuscript establishes existence/recovery criteria (Theorems 2.2–2.3), a polynomial realization theorem (3.1), explicit families and samplers, and a mean-value identity with an integrability condition (2.5). Fourier inversion, Stieltjes representation and Gaussian quadrature are classical tools. Priority of the proposed constructions is not established by this summary.

### Existence tests (Section 2; Appendices A–B)

Choose $r>0$ with $\rho=\Pr(|U|\ge r)>0$. Symmetry gives
$$
M_U(t)\ge\rho\cosh(rt),\qquad 0<M_U(t)^{-1}\le2\rho^{-1}e^{-r|t|}.
$$
The reciprocal is automatically integrable. Therefore
$$
f_U^*(x)=\frac1\pi\int_0^\infty\frac{\cos(tx)}{M_U(t)}\,dt,\qquad
V\text{ exists}\ \Longleftrightarrow\ f_U^*(x)\ge0\ \text{for every }x\in\mathbb R.
$$
If it exists, this is the unique conjugate density. Truncating at $T$ has uniform error at most $2e^{-rT}/(\pi\rho r)$; also $\sup_x|(f_U^*)'(x)|\le2/(\pi\rho r^2)$. A negative value with certified integration error excludes existence. Finite grid positivity does not certify global positivity.

For $u_n=\mathbb EU^{2n}$ define
$$
v_0=1,\qquad v_n=\sum_{k=1}^n(-1)^{k+1}\binom{2n}{2k}u_kv_{n-k}.
$$
Existence is equivalent to
$$
H_N=(v_{i+j})_{i,j=0}^N\succeq0,\qquad
\widetilde H_N=(v_{i+j+1})_{i,j=0}^N\succeq0\quad\text{for every }N\ge0.
$$
Under H, these are also equivalent to strict positive definiteness at every size. In particular $v_1=u_1$, $v_2=6u_1^2-u_2$, so $\det H_1>0$ implies $\mathbb EU^4<5(\mathbb EU^2)^2$. Passing finitely many Hankel tests is necessary evidence, not the full criterion.

## Core mechanisms and derivations

### 1. Recover normalization and a global identity from local information

For the Fourier test, nonnegativity alone must still imply mass one. Gaussian regularization gives
$$
\int f_U^*(x)e^{-\varepsilon x^2/2}\,dx
=\frac1{\sqrt{2\pi\varepsilon}}\int \frac{e^{-t^2/(2\varepsilon)}}{M_U(t)}\,dt\longrightarrow1.
$$
Monotone convergence on the left and the approximate identity on the right establish normalization. Applying the same device after a Fourier modulation recovers the reciprocal transform.

For the Hankel test, the Stieltjes theorem yields $X\ge0$ with moments $v_n$; an independent fair sign attached to $\sqrt X$ gives symmetric $V$. Since $q=1/M_U$ is analytic near zero, Cauchy estimates give $v_n\le C(2n)!r_0^{-2n}$. Thus $\mathbb E e^{a|V|}<\infty$ for $0<a<r_0$. On a connected horizontal strip the product
$$
M_U(z)\varphi_V(z)-1
$$
is analytic and vanishes near zero, hence everywhere on that strip. Using this product avoids extending a local Taylor series beyond its disk. Exponential moments also yield moment determinacy.

To recover discrete approximations, use the full symmetric sequence $m_{2n}=v_n$, $m_{2n+1}=0$. The orthogonal-polynomial Jacobi matrices have zero diagonal and positive off-diagonals. Their first-coordinate spectral measures match moments through degree $2N-1$ and converge weakly to $\mathcal L(V)$ by tightness, higher-moment uniform integrability and determinacy. Applying this zero-diagonal construction to $V^2$ would solve a different moment problem.

### 2. Separate two positivity problems (Theorem 3.1; Appendix C)

For every nonconstant real $P$ with $P(0)=1$ and $P(s)>0$ for all $s\ge0$, there exists $b_0>0$ such that for each $b\ge b_0$, all sufficiently large $\tau\ge\tau_0(b)$ realize
$$
M_U(t)=e^{\tau t^2/2}(1+bt^2)P(t^2),\qquad
\varphi_V(t)=\frac{e^{-\tau t^2/2}}{(1+bt^2)P(t^2)}.
$$
Both laws have strictly positive symmetric densities. The quantifier order is $b$ first, then $\tau$.

The inverse $k=\mathcal F^{-1}[1/P(t^2)]$ can be signed. Partial fractions show it is real, even, of mass one and has an exponential absolute moment. For $\ell_\lambda(x)=\lambda e^{-\lambda|x|}/2$,
$$
\left|\frac{\ell_\lambda(x-y)}{\ell_\lambda(x)}-1\right|\le e^{\lambda|y|}-1,
$$
so, uniformly in $x$,
$$
\left|\frac{(\ell_\lambda*k)(x)}{\ell_\lambda(x)}-1\right|
\le\delta_k(\lambda):=\int |k(y)|(e^{\lambda|y|}-1)\,dy\longrightarrow0.
$$
Choose $\delta_k(\lambda)<1$ and $b=\lambda^{-2}$. Then $\ell_\lambda*k$ is a positive probability density for $Y_b$ with transform $[(1+bt^2)P(t^2)]^{-1}$. This relative estimate controls the whole line, including tails; a bounded absolute error would not suffice.

For the original variable, set $Q(s)=(1+bs)P(s)$, $R(y)=Q(y^2)$, $D=\deg Q$. Gaussian differentiation yields the finite exact formula
$$
f_U(x)=g_\tau(x)S_\tau(x/\tau),\qquad
S_\tau(y)=\sum_{j=0}^D\frac{(-1)^jR^{(2j)}(y)}{(2\tau)^j j!}.
$$
Since $R>0$ and derivative degrees decrease, $C_j=\sup_y|R^{(2j)}(y)|/R(y)<\infty$. Thus
$$
S_\tau\ge R(1-A_Q(\tau)),\qquad
A_Q(\tau)=\sum_{j=1}^D C_j/[(2\tau)^j j!]\longrightarrow0.
$$
Positive-order Gaussian derivatives integrate to zero, giving normalization. Integration by parts recovers $M_U$. Finally $V=\sqrt\tau Z+Y_b$ with independent standard normal $Z$ completes the pair. Positivity of $P$ is necessary because $M_U$ is positive.

Section 3.3 gives sufficient parameter certificates: exponential partial-fraction bounds control $\delta_k$, and univariate inequalities $\overline C_jR\pm R^{(2j)}\ge0$ certify $C_j\le\overline C_j$. Taking $\tau\ge\max\{1,\sum_{j=1}^D\overline C_j/j!\}$ ensures $A_Q(\tau)\le1/2$. Exact admissible parameter regions remain open.

### 3. An explicit positive example and a sharp rational threshold (Section 4; Appendix D)

For $P(s)=1+s^2$, $b=2$, $\tau=4$,
$$
M_U(t)=e^{2t^2}(1+2t^2)(1+t^4),\qquad V=2Z+Y,
$$
$$
f_Y(x)=\frac{e^{-|x|/\sqrt2}}{10\sqrt2}
\left(4-\cos\frac{|x|}{\sqrt2}+3\sin\frac{|x|}{\sqrt2}\right),
$$
$$
S_4(y)=2y^2(y^2-13/8)^2+(27/32)y^2+7/32>0.
$$
The trigonometric bracket is at least $4-\sqrt{10}>0$; its integral gives mass one. Both variances are $8$. The rational reciprocal $[(1+bt^2)(1+t^4)]^{-1}$ is a characteristic function exactly when $b\ge2$ for $b>0$: for $b\ge2$, factor through the $b=2$ law and a point-mass/Laplace mixture; for $b<2$, a more slowly decaying oscillatory inverse term forces negative tail values. This threshold alone does not settle positivity of the original density for a given $\tau$.

Choosing $P$ with specified complex roots gives families outside the even Laguerre–Pólya class. The additional exclusion of the KPS class uses its specific Bernstein representation as an external premise: a probability Laplace transform is analytic and bounded in the right half-plane, incompatible with the constructed poles. This is not a criterion for arbitrary even characteristic functions, and that external theorem was not reproved in this upload check.

### 4. Mean-value identities require integrability (Theorem 2.5)

For $W=U+iV$, some $a_0>0$ satisfies $\mathbb E e^{a_0|W|}<\infty$, and transform cancellation yields $\mathbb E W^n=0$ for $n\ge1$. Every polynomial is preserved. For an entire $F$, require
$$
\sum_{n\ge0}\frac{|F^{(n)}(x)|}{n!}\mathbb E|W|^n<\infty.
$$
Then termwise expectation gives $\mathbb E F(x+W)=F(x)$. A sufficient envelope is $|F(z)|\le C e^{a|z|}$ with $a<a_0$: choose $c>1$ with $ac<a_0$ and apply Cauchy estimates on radius $c|W|$ to obtain an integrable geometric majorant. For the Rademacher conjugate and $F(z)=e^{-z^2}$, both positive and negative parts can have infinite expectation. Entire analyticity alone does not justify interchange.

Known-class recovery and samplers are given in Sections 5 and Appendices E–F. The original remains authoritative; the supplementary historical report has missing prompt/date provenance and is not treated as a new full review.