Physical relative scores without a density cap

First prove finiteness rather than assuming the integration by parts is legitimate. Choose even compact smooth cutoffs $0\leq\chi_R\leq1$, nonincreasing in $|\xi_i|$, increasing to one, with uniformly bounded $\xi_i\chi_R'$. Set $M_R=\mathbb E(\xi_i^2\chi_R)$. Integration by parts against the compact test $\xi_i\chi_R$ gives

Spectator cutoffs can be exhausted: the active test is bounded and its score term is integrable by Cauchy. Solving the quadratic gives $M_R\leq[(\sqrt{J_i}+\sqrt{J_i+4})/2]^2$; monotone convergence proves $M_i<\infty$. Now $\xi_i s_i$ is integrable. Removing the cutoff, its bounded derivative term tends to zero by dominated convergence, giving $\mathbb E\xi_i s_i=M_i-1$. The ordinary score is $s_i-\xi_i$, so

This proves (5.2) without presupposing the moment.

For almost every $x$, the one-coordinate marginal $p_{i,x}$ has weak derivative equal to the integrated joint derivative. Conditional expectation and Cauchy imply $\|p_{i,x}'\|_1\leq\sqrt{I_{i,x}^{\rm abs}}$. An integrable nonnegative $W^{1,1}(\mathbb R)$ density tends to zero at both infinities, so $2\sup p_{i,x}\leq\|p_{i,x}'\|_1$. On the box interior, differentiating in $u_i$ introduces $1/\phi(\xi_i)\leq D_{R_i}$; score Cauchy bounds the integrated interior variation by $D_{R_i}\sqrt{J_i}$. At each of the two quantile faces the zero extension contributes the trace of the density. Integrating all other coordinates bounds their sum by

BV traces or a limiting regular face justify the same assertion for weak densities. This proves (5.3). Finally apply weighted Cauchy to the coefficients $(2,1,\ldots,1)$ in $S_n$. Their squared sum is $n+4$; including unused coordinates in $J$ and $d$ only enlarges the upper bound. This proves (5.4).

The elementary exponential enclosure gives $D_{10}<1.31\times10^{22}$. Using $\sqrt{102}<10.1$, $\sqrt{104}<10.2$ and $\sqrt{304}<17.44$ in Lemma 5.1 gives

The last inequality retains the positive mass term $100$ in $H_{100}^*$. Choose $c=3$, $\eta=10^{-110}$, $k=24$. Gaussian upper tails and exponential bounds give

Thus $\delta<2\times10^{-130}$ and $\beta<3\times10^{-110}$. In particular $\sqrt{\beta+4N\delta}<3\times10^{-45}$ and $\sqrt\beta<2\times10^{-55}$. Therefore

The ideal boxed writer fee is bounded by $2.6331\times10^{23}/(4\,2^{100})$ $<5.192874123844\times10^{-8}$. The actual complete-box tail is at most $10^{15}/10^{20}=10^{-5}$. Substitution in Theorem 4.2 gives the stated rational decimal ceilings; the copy ceiling includes the additional $4\times10^{-31}$ bad-set term.

For reproducible directed inequalities one may use $\sqrt{2\pi}<2.51$ and bound $e^{50}$ by its positive Taylor sum through degree $200$ plus the next term divided by $1-50/202$. For the lower exponential bounds used in the tails, $e>\sum_{j=0}^5 1/j!=163/60$ suffices, together with $\Phi(-x)\leq e^{-x^2/2}/(2x)$ for $x>0$. All row decisions consequently reduce to rational inequalities.

Choose $R_w^2\geq8(J_*+2)/\varepsilon$. The writer tail is at most $\varepsilon/8$ and

independently of $n$. Choose finite $n\geq1$ so that $V_*/(4\,2^n)\leq\varepsilon/4$. For this chosen full prefix, choose each of the $n+1$ source cutoffs to satisfy $R_i^2\geq8(n+1)(J_i(n)+2)/\varepsilon$. Their aggregate tail is at most $\varepsilon/8$; hence $\tau\leq\varepsilon/4$. All boxed source variations are finite by Lemma 5.1. Set $S=S_n$ and $H=n+S/2$ and choose

The two terms in $E_n$ are at most $\varepsilon/(16n)$ each, so $E_n\leq\varepsilon/(8n)$. Thus freshness is at most $\varepsilon/4+\varepsilon/4+\varepsilon/(4n)\leq3\varepsilon/4$, and the all-record bound is at most

Here $\sqrt{512}>89/4$. The factor $n$ in (5.10) prices all own-event records; a schedule controlling freshness alone does not automatically do so.

It remains to realize these positive $\beta,\delta$ thresholds with finite Gaussian parameters. Fix $c=3$ and choose $\eta$ sufficiently small for the source-end part of $\beta$. Choose finite $k$ with $\Phi(-k)<\eta/2$. Increase $D$ until $e^{-2Dc}$ is below both the desired displacement allocation and $\eta/2$, and until $\Phi(-(D-c))$ and $\Phi(-(D+c))$ meet their respective allocations. Increase $A>k$ until $e^{-2A(A-k)}$ and $\Phi(-(2A-k))$ meet the remaining allocations. Lemma 3.3 proves the required map and copy bounds. All choices are finite after this finite prefix has been selected. The translations and conjugated holding swaps described above implement the corresponding smooth prescribed parent.

Take independent standard normals $Y,G_1,G_2,\ldots$ and set $Z_j^{\rm in}=Y+2^{-6j}G_j$; use writer $u=\Phi(Y)$ and sources $v_j=\Phi(Z_j^{\rm in})$. Every finite covariance is nonsingular. For a prefix containing the writer and $m$ sources, differentiation at fixed source coordinates gives

A bank with $n$ warmup archives and an untouched receiver has $m=n+1$. Thus this family fails the uniform full conditional writer hypothesis, although its writer marginal is exactly standard normal. The final archive $b_n$ reveals $v_n=2b_n-\lfloor2b_n\rfloor$, hence $Z_n^{\rm in}$. Define the archive prediction $\widehat r=\{2^n\Phi(Z_n^{\rm in})\}$. Circle distance obeys

Since $\sqrt{2\pi}>2.5$, the event $|G_n|\leq0.8$ implies this distance is at most $0.01$ for all $n\geq1$. Under an independent uniform remainder, the same archive-dependent circle interval has probability $0.02$. Meanwhile

Subtracting $0.02$ gives $5189/12550$. The strict inequality follows from $e^{-0.32}>1-0.32$ and $\sqrt{2\pi}<2.51$.

For fixed $u<1/2$ and $v\in(0,1)$, $F_D^{-1}(u)=-D+\Phi^{-1}(2u)+o(1)$ and its minority posterior tends to zero. The copy position is $-A+\Phi^{-1}(v)+o(1)$, and the return and archive outputs tend to $(2u,v/2)$. Reflection gives the other branch. Consequently $B^{-1}T_{D,A}\to I$ almost everywhere. Finite iteration excludes only the finite family of dyadic cuts and their preimages. The corresponding full difference maps preserve Lebesgue measure. For bounded continuous $h$ on the closed cube, dominated convergence gives $\|h\circ K-h\|_1\to0$. Approximate an arbitrary $f\in L^1$ by such $h$ and use

This proves TV convergence on each $X$ fibre. Dominated convergence in the actual $X$ law completes the argument.

Take actual writer $Q_0\sim N(10,1)$ and independent standard-normal actual archives. The entrance ratio is $e^{10Q_0-50}$, with physical relative score $10$ in the writer direction and zero in every archive direction. Its relative density is unbounded. The Gaussian upper-tail inequality implies $\Phi(-8.1)<0.005/2^{44}$ and $\Pr(Q_0>8.1)>0.96$. On this event every ideal digit is right and every ideal remainder is greater than $0.995$. For each actual archive, exclude its initial quantiles outside $[10^{-12},1-10^{-12}]$; the aggregate exclusion probability is at most $88\times10^{-12}$. With $c=3,k=8$, Lemma 3.3 gives, on the right good branch, a ready-coordinate error per cycle at most $10^{-48}+10^{-41}$. Induction bounds the accumulated error by $(2^{44}-1)(10^{-48}+10^{-41})<10^{-20}$. Every actual remainder therefore stays in the right good branch and the final one exceeds $0.99$. Thus

whereas uniform assigns $0.01$ to this event. The difference exceeds $0.94$. Complete archive-conditioned distance is at least this marginal distance by projection. This is a finite-resource counterexample, not a failure of an upper estimate.