Convergence of sum of triangular array of random variables

Solution 1:

The following will give a proof of what you want. Using your notations, let \begin{align*} M_{k,n}&=\sum_{j=1}^{k}(\xi_{j,n}- \mathsf{E}_{j-1}[\xi_{j,n}]), \quad k\le l =\Big[\frac{t}{\Delta_n}\Big], \\ \langle M\rangle_{k,n}&=\sum_{j=1}^{k}\mathsf{E}_{j-1}[ (\xi_{j,n}- \mathsf{E}_{j-1}[\xi_{j,n}])^2 ], \qquad k\le l. \end{align*} Then it is easy to verify that $\{M_{k,n}^2-\langle M\rangle_{k,n},\; 1\le k \le l\} $ is a martingale for each $n\ge 1$ and following inequality is true for $\epsilon>0, \delta>0$, \begin{equation*} \mathsf{P}(\sup_{k\le l}|M_{k,n}|\ge \delta)\le \frac{\epsilon}{\delta^2} +\mathsf{P}(\langle M\rangle_{l,n}\ge \epsilon). \tag{3} \end{equation*} (The proof of (3) and its more general cases, please refer to J. Jacod, and A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003. p.35, Lemma 1.3.30. ).

Meanwhile, \begin{gather*} (2): \sum_{j=1}^{l}\mathsf{E}_{j-1}[\xi_{j,n}^2]\overset{p}{\longrightarrow}0\\ \text{$\;$}\hspace{3cm}\Downarrow (\text{by } \mathsf{E}_{j-1}[(\xi_{j,n} - \mathsf{E}_{j-1}[\xi_{j,n}])^2] \le \mathsf{E}_{j-1}[\xi_{j,n}^2]) \\ \langle M\rangle_{l,n} =\sum_{j=1}^{l}\mathsf{E}_{j-1}[(\xi_{j,n} - \mathsf{E}_{j-1}[\xi_{j,n}])^2] \overset{p}{\longrightarrow}0\\ \Downarrow (\text{by } (3))\\ \sup_{k\le l}\Big|\sum_{j=1}^{k}(\xi_{j,n}- \mathsf{E}_{j-1}[\xi_{j,n}])\Big| = \sup_{k\le l}|M_{k,n}| \overset{p}{\longrightarrow}0\\ \Downarrow (\text{by } (1))\\ \sup_{k\le l}\Big|\sum_{j=1}^{k}\xi_{j,n}\Big|\overset{p}{\longrightarrow}0. \end{gather*}

Solution 2:

In your first line, plus-and-subtract before using triangle inequality, $$\biggl|\sum_{j=1}^{n}\xi_{j,n}\biggr|=\biggl|\sum_{j=1}^{n}(\xi_{j,n}-\mathbb{E}_{j-1}[\xi_{j,n}])+\sum_{j=1}^{n}\mathbb{E}_{j-1}[\xi_{j,n}]\biggr|\leq\biggl|\sum_{j=1}^{n}(\xi_{j,n}-\mathbb{E}_{j-1}[\xi_{j,n}])\biggr|+\biggl|\sum_{j=1}^{n}\mathbb{E}_{j-1}[\xi_{j,n}]\biggr|.$$ It gives a stronger upper bound: above line only uses the triangle inequality once (counting of the basic form $|a+b|\leq|a|+|b|$), while your attempt uses $2(n-1)$ times.