Convergence in probability of series of random variables

Question

From the standard literature it is well known that for sequences of random variables $X_{1, n} \stackrel{P}{\rightarrow} X_1$ and $X_{2, n} \stackrel{P}{\rightarrow} X_2$ as $n \rightarrow \infty$ it holds that $(X_{1, n}, X_{2, n}) \stackrel{P}{\rightarrow} (X_1, X_2)$ for $n \rightarrow \infty$. Using a continuous mapping theorem argument this can be used to establish that $X_{1, n} + X_{2, n} \stackrel{P}{\rightarrow} X_1 + X_2$ for $n \rightarrow \infty$.

Question in general case

Under which conditions can the arguments of the standard literature be extended from a simple sum of two sequences to a sum of sequences whose number of summands also increases to infinity as $n$ goes to infinity. Specifically, given a sequences $X_{l, n}$, $l \in \mathbb{N}$, with $X_{l, n} \stackrel{P}{\rightarrow} X_l$ $(n \rightarrow \infty)$ for all $l\in \mathbb{N}$ under which conditions does it holds that $$ \lim_{n \rightarrow \infty} \sum_{l = 1}^n X_{l, n} \stackrel{P}{\rightarrow} \sum_{l = 1}^\infty X_{l} $$ as $n \rightarrow \infty$?

Question in special case

In my particular use case something more specific would also suffice. Though, I think the more general question is also interesting. If we know that $X_{l, n}$, $l \in \mathbb{N}$, are nonnegative random sequences, i.e. each sequence element is a nonnegative random variable, and $X_{l, n} \stackrel{P}{\rightarrow} 0$ $(n \rightarrow \infty)$ for all $l \in \mathbb{N}$ under which conditions does it holds that $$ \lim_{n \rightarrow \infty} \sum_{l = 1}^n X_{l, n} \stackrel{P}{\rightarrow} 0 $$ as $n \rightarrow \infty$? Possibly, what would also be interesting and might be easier to answer is when the sample mean converges to zero, i.e., $$ \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{l = 1}^n X_{l, n} \stackrel{P}{\rightarrow} 0 $$ as $n \rightarrow \infty$

Answer 1

$\newcommand\ep\varepsilon\newcommand\de\delta\newcommand{\P}[1]{\overset P{\underset{#1}\longrightarrow}}$What you need is the uniform summability (in probability).

Here are details: Let $Y_{l,n}:=X_{l,n}-X_l$, so that $$Y_{l,n}\P{n\to\infty}0 \tag{0}$$ for each $l$. We want to have $$S_{n,n}\P{n\to\infty}0,\tag{1}$$ where $$S_{m,n}:=\sum_{l=1}^m Y_{l,n}.$$ The mentioned sufficient uniform summability condition is that $$S_{n,n}-S_{m,n}\P{n\ge m\to\infty}0. \tag{2}$$

Indeed, take any real $\de>0$ and $\ep>0$. Then, by (2), for some natural $m_1$ we have the following implication: $$n\ge m\ge m_1\implies P(|S_{n,n}-S_{m,n}|>\ep/2)\le\de/2. \tag{3}$$ Also, (0) implies $S_{m_1,n}\P{n\to\infty}0$, so that for some natural $m_2$ we have the following implication: $$n\ge m_2\implies P(|S_{m_1,n}|>\ep/2)\le\de/2. \tag{4}$$ Letting now $m_3:=\max(m_1,m_2)$, by (3) and (4) we have $$n\ge m_3\implies P(|S_{n,n}|>\ep)\le P(|S_{n,n}-S_{m_1,n}|>\ep/2)+P(|S_{m_1,n}|>\ep/2)\le\de. $$ So, (1) holds.

Answer 2

To see a deterministic example where even the averages do not tend to zero, consider $X_{\ell,n}=0$ if $\ell^2 \le n$ and $X_{\ell,n}=1$ if $\ell^2 >n$. For each $\ell$ we have $X_{\ell,n} \to 0$ as $ n \to \infty$ but the averages considered $\frac{1}{n} \sum_{\ell = 1}^n X_{\ell, n}$ tend to 1.