Nested iid fourth-moment truncation and uniform diagonal control

Fix a finite $D$ with $R_n,C_n\le Dn$ and $N_n\ge n\mathrm{/}D$ eventually. For every $\epsilon >0$, the union bound on a square of side $D{2}^{\mathrm{\ell }+1}$ gives a constant times $2^{2\mathrm{\ell }}P(\mathrm{\mid }{\xi }_{11}\mathrm{\mid }>\epsilon 2^{\mathrm{\ell }\mathrm{/}2})$ as a bound for the probability of any such large entry in that square. Its series over $\mathrm{\ell }$ is finite: for every $t\ge 0$ the geometric sum ${\sum }_{\mathrm{\ell }:\epsilon 2^{\mathrm{\ell }\mathrm{/}2}<t}{2}^{2\mathrm{\ell }}$ is at most a constant times ${\epsilon }^{-4}{t}^4$, so expectation proves the assertion. F1 on dyadic blocks, followed by intersection over rational $\epsilon >0$, proves ${\max }_{i\le R_n,j\le C_n}\mathrm{\mid }{\xi }_{ij}\mathrm{\mid }\mathrm{/}\sqrt{N_n}\to 0$ almost surely.

The dyadic probability series in step 1.1 converges separately for every threshold $\epsilon =1\mathrm{/}a$. Choose deterministic integers $L_a$ increasing so that its tail from $L_a$ onward is at most $2^{-a}$; impose also $L_a\ge a$, and every prescribed slow-rate constraint by increasing $L_a$ further. On the blocks $L_a\le \mathrm{\ell }<L_{a+1}$ choose the truncation threshold as a suitable fixed multiple of $2^{\mathrm{\ell }\mathrm{/}2}\mathrm{/}a$, large enough to cover the comparable dimensions and the adjacent dyadic interval. The sum of the resulting block probabilities is at most ${\sum }_a{2}^{-a}$. First Borel–Cantelli proves eventual equality to these truncations. A stepwise deterministic enlargement gives ${\delta }_n\to 0$ as slowly as required, with ${\delta }_n\ge 1\mathrm{/}\log (n+2)$; enlarging a threshold preserves eventual equality. One may absorb every fixed dimensional comparison constant into ${\delta }_n$. Since ${\delta }_n\sqrt{N_n}\to \mathrm{\infty }$, enlarge finitely many initial values further so that ${\delta }_n\sqrt{N_n}\ge \max (1,E\mathrm{\mid }{\xi }_{11}\mathrm{\mid })$ for every $n$; this preserves all asymptotic and eventual-equality conclusions.

Since $E\xi =0$, $\mathrm{\mid }{\mu }_n\mathrm{\mid }\le E[\mathrm{\mid }\xi \mathrm{\mid }{\mathbf{1}}_{\mathrm{\mid }\xi \mathrm{\mid }>{\delta }_n\sqrt{N_n}}]\le E\mathrm{\mid }\xi {\mathrm{\mid }}^4\mathrm{/}({\delta }_n^3{N}_n^{3\mathrm{/}2})$. Therefore $\sqrt{R_n{C}_n\mathrm{/}{N}_n}\mathrm{\mid }{\mu }_n\mathrm{\mid }\le K{\delta }_n^{-3}\mathrm{/}{N}_n\to 0$, by the selected slow rate. Dominated convergence gives $E\mathrm{\mid }{x}_{11}^{(n)}{\mathrm{\mid }}^2=E[{\xi }^2{\mathbf{1}}_{\mathrm{\mid }\xi \mathrm{\mid }\le {\delta }_n\sqrt{N_n}}]-{\mu }_n^2\to 1$, and this variance is at most $1$. Moreover $\mathrm{\mid }{\mu }_n\mathrm{\mid }\le E\mathrm{\mid }\xi \mathrm{\mid }\le {\delta }_n\sqrt{N_n}$ for every $n$, so $\mathrm{\mid }{x}_{ij}^{(n)}\mathrm{\mid }\le L_n:=2{\delta }_n\sqrt{N_n}$ for every $n$. The third absolute moments are bounded by $4(E\mathrm{\mid }\xi {\mathrm{\mid }}^3+\mathrm{\mid }{\mu }_n{\mathrm{\mid }}^3)\le 4(E\mathrm{\mid }\xi {\mathrm{\mid }}^3+(E\mathrm{\mid }\xi \mathrm{\mid }{)}^3)$, uniformly in $n$. Thus for $q\ge 3$, $E\mathrm{\mid }x{\mathrm{\mid }}^q\le L_n^{q-3}E\mathrm{\mid }x{\mathrm{\mid }}^3\le K{L}_n^{q-3}$.

Apply F2 to the infinite centered finite-variance array $Z_{ij}={\xi }_{ij}^2-1$. Using a fixed enlargement of the row bound and of the maximal partial-sum index covers every $R_n,C_n\le Dn$; division by $N_n\ge n\mathrm{/}D$ proves the stated uniform row and column second-moment laws for the original entries. On the probability-one event of eventual truncation equality, $x_{ij}^{(n)}={\xi }_{ij}-{\mu }_n$ throughout the rectangle. The difference of a normalized row squared sum from the original one is bounded by $2\mathrm{\mid }{\mu }_n\mathrm{\mid }\sqrt{(C_n\mathrm{/}{N}_n)(N_n^{-1}{\sum }_j{\xi }_{ij}^2)}+(C_n\mathrm{/}{N}_n){\mu }_n^2$, by Cauchy–Schwarz; its maximum tends to zero because the original row laws bound these sums and ${\mu }_n\to 0$. The column calculation is identical. This proves both centered-truncation diagonal laws explicitly.

Enumerate the cells of the infinite quadrant by successive square shells $\max (i,j)=k$, in fixed lexicographic order on each finite shell. The resulting sequence of $\mathrm{\mid }{\xi }_{ij}{\mathrm{\mid }}^4$ is iid and integrable; F3 at the square cardinalities $k^2$ gives $k^{-2}{\sum }_{i,j\le k}\mathrm{\mid }{\xi }_{ij}{\mathrm{\mid }}^4\to E\mathrm{\mid }\xi {\mathrm{\mid }}^4$. Enclosing each rectangle in a square of side $\lceil Dn\rceil$ gives $N_n^{-2}{\sum }_{i\le R_n,j\le C_n}\mathrm{\mid }{\xi }_{ij}{\mathrm{\mid }}^4=O(1)$. The maximum-entry conclusion for the centered entries follows from step 1.1 and $\mathrm{\mid }{\mu }_n\mathrm{\mid }\mathrm{/}\sqrt{N_n}\to 0$. Also $\mathrm{\mid }\xi -{\mu }_n{\mathrm{\mid }}^4\le 8(\mathrm{\mid }\xi {\mathrm{\mid }}^4+\mathrm{\mid }{\mu }_n{\mathrm{\mid }}^4)$, so eventual equality to the truncations and $R_n{C}_n\mathrm{/}{N}_n^2=O(1)$ give the fourth-power bound for the centered entries.

For every fixed $q>2$, the square of the Frobenius norm of the indicated power matrix is bounded by $(\max \mathrm{\mid }{x}_{ij}^{(n)}\mathrm{\mid }\mathrm{/}\sqrt{N_n}{)}^{2q-4}{N}_n^{-2}\sum \mathrm{\mid }{x}_{ij}^{(n)}{\mathrm{\mid }}^4$, which tends to zero by step 5.1. Multiplying its entries by signs does not change the Frobenius norm. The original-entry assertion follows from the same calculation. All conclusions hold on the intersection of the countably many probability-one events already specified.