Uniform row partial sums of a nested finite-variance array

If $C\le 0$ the claim is vacuous. Suppose $C>0$, set $v=E{Z}_{11}^2$ and fix $\epsilon >0$. For a dyadic integer $m=2^{\mathrm{\ell }}$ put $N=2m$, $I=\lceil 2Cm\rceil$, $x=\epsilon m$, $r=4$ and $y=x\mathrm{/}r$. For one row write $Z_j=Z_{ij}$ and $W_j=Z_j{\mathbf{1}}_{\{Z_j\le y\}}$. These variables are independent with $E{W}_j=-E[Z_j{\mathbf{1}}_{\{Z_j>y\}}]\le 0$ and $E{W}_j^2\le v$. For $t>0$ and $w\le y$, $e^{tw}\le 1+tw+(e^{ty}-1-ty)w^2\mathrm{/}{y}^2$: the function $(e^u-1-u)\mathrm{/}{u}^2={\int }_0^1(1-s)e^{su}ds$, extended continuously at zero, is increasing in $u$. Consequently $E{e}^{t{W}_j}\le \exp \{v(e^{ty}-1-ty)\mathrm{/}{y}^2\}$.

Set $\psi =v(e^{ty}-1-ty)\mathrm{/}{y}^2\ge 0$. Independence shows that $M_k=\exp (t{\sum }_{j\le k}{W}_j-k\psi )$ has conditional expected next value at most $M_k$. If $\tau$ is the first $k\le N$ with ${\sum }_{j\le k}{W}_j\ge x$, stopping this finite sequence at $\tau$ or $N$ preserves the expectation bound $1$: condition each increment and observe that the indicator of not having stopped is determined by the preceding variables. On a crossing its stopped value is at least $e^{tx-N\psi }$. Nonnegativity therefore gives $P({\max }_{k\le N}{\sum }_{j\le k}{W}_j\ge x)\le e^{-tx+N\psi }$. This proves the needed maximal bound directly.

If $v=0$, every $Z_j=0$ almost surely and the conclusion follows. Otherwise choose $t>0$ by $e^{ty}=1+xy\mathrm{/}(Nv)$. Then $N\psi \le x\mathrm{/}y=r$, so the preceding probability is at most $e^r(Nv\mathrm{/}(xy){)}^r\le K{m}^{-r}$, with $K$ depending only on $v,\epsilon$. On the event that all $Z_j\le y$, the original and truncated partial sums agree; the complement has probability at most $NP(Z_{11}>y)$. Thus the positive original maximal tail is at most $NP(Z_{11}>y)+K{m}^{-r}$. Apply the same argument to $-Z_j$ for the negative tail and use a union bound over the $I$ rows. The result is $P({\max }_{i\le I,k\le N}\mid {\sum }_{j\le k}{Z}_{ij}\mid \ge \epsilon m)\le K_C{m}^{2}P(\mathrm{\mid }{Z}_{11}\mathrm{\mid }>\epsilon m\mathrm{/}4)+K_C^{\mathrm{\prime }}{m}^{1-r}\mathrm{.}$

The right-hand side is summable over dyadic $m$: for every $a>0$, ${\sum }_{\mathrm{\ell }}{2}^{2\mathrm{\ell }}P(\mathrm{\mid }{Z}_{11}\mathrm{\mid }>a{2}^{\mathrm{\ell }})\le K_{a}E\mathrm{\mid }{Z}_{11}{\mathrm{\mid }}^2<\mathrm{\infty }$, by summing the finite geometric progression ${\sum }_{\mathrm{\ell }:a{2}^{\mathrm{\ell }}<\mathrm{\mid }{Z}_{11}\mathrm{\mid }}{2}^{2\mathrm{\ell }}$ under expectation. Also ${\sum }_{\mathrm{\ell }}{2}^{(1-r)\mathrm{\ell }}<\mathrm{\infty }$. First Borel–Cantelli gives the displayed maximal estimate eventually on each dyadic block. Whenever $m\le n\le 2m$, every $i\le Cn$ and $k\le n$ occurs in the corresponding block; division by $n\ge m$ bounds the desired ratio by $\epsilon$ eventually. Intersecting over positive rational $\epsilon$ proves convergence to zero. Exchanging the two coordinates preserves the iid law and gives the column version.