I'm struggling with this exercise:

For each $k\in \mathbb{N}$, let $(X^{(k)}_n)_{n\in\mathbb{N}}$ be a sequence of real random variables converging to $0$ in probabilty as $n\to\infty$. Define for $k,n\in\mathbb{N}$, $$Y_n^{(k)}:=\sum\limits_{i=1}^k X_n^{(i)}$$ and for $\epsilon>0$ arbitrary $$f_n(k):=P(|Y_n^{(k)}|>\epsilon)$$ Let $k\in \mathbb{N}$. Show that $f_n(k)\to 0\quad (Y_n^{(k)} \xrightarrow{P}0)$, as $n\to \infty$

So the $k$ is the amount of sequences and the $n$ is at which position the sequence begins. So we have to show $\lim\limits_{n\to\infty}P(|Y_n^{(k)}-0|>\epsilon)=0$ This is the same as $$\lim\limits_{n\to\infty}P(|Y_n^{(1)}-0|>\epsilon,\cdots,|Y_n^{(k)}-0|>\epsilon)=0$$ $$=\lim\limits_{n\to\infty}P(|\sum\limits_{i=1}^1 X_n^{(i)}-0|>\epsilon,\cdots,|\sum\limits_{i=1}^k X_n^{(i)}-0|>\epsilon)=0 $$

So as far as I understand, this means that we have from left to right an increasing sequence of random variables where each random variable converges to $0$.

$$=\lim\limits_{n\to\infty}P(|\underbrace{X_n^{(1)}-0}_{= converges}|>\epsilon,\cdots,|\sum\limits_{i=1}^k X_n^{(i)}-0|>\epsilon)=0 $$

So can I some how argue, that each therefore the whole expression must also converge to $0$ in probabilty? Also we do not know whether $(X_n)_{n\in \mathbb{N}}$ are independent, but are the $X^{(i)}_k$ independent? Since I not know anything about independence I can't use something like the law of large numbers.

Thanks for help

  • $\begingroup$ "So we have to show $\lim\limits_{n\to\infty}P(|Y_n^{(k)}-0|>\epsilon)=0$ This is the same as $\lim\limits_{n\to\infty}P(|Y_n^{(1)}-0|>\epsilon,\cdots,|Y_n^{(k)}-0|>\epsilon)=0$" Well, no this is not the same. What makes you think this is? $\endgroup$
    – Did
    May 5, 2016 at 20:56

1 Answer 1


If $Y_n^k=\sum_{i=1}^k X_n^i>\epsilon$, one of those summands must be greater than $\epsilon/k$. So $P\{Y_n^k>\epsilon\}\le P\{\cup_{i=1}^k \{X_n^i>\epsilon/k\}\}\le \sum_{i=1}^k P\{X_n^i>\epsilon/k\} \to 0$ as $n\to\infty$.


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