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Suppose that ${X_n}$ are independent, $S_n = X_1 + ... X_n \rightarrow S$ in probability for some random variable S.

If $Y_n$ are identically distributed with $E|Y_1|<\infty$,

show that $T_n = \sum_{i=1}^{n} X_i 1(|Y_i|\leq i) \rightarrow T$ almost surely for some random variable T.

Is there any good way to solve this? Usually, the form can be solved because the variance of infinite sum is finite, so we can use 3-series theorem. I think it requires similar way, but the existence of unrelated variable $Y_i$ makes it hard.

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  • $\begingroup$ Are $Y_n$'s independent? $\endgroup$ – d.k.o. Jun 7 '17 at 16:45
  • $\begingroup$ @d.k.o. In the problem, there is no such condition. Maybe the problem omits independence of Yn, but I'm not sure. $\endgroup$ – artes75 Jun 7 '17 at 17:40
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First, $S_n\xrightarrow{p}S \Leftrightarrow S_n\xrightarrow{a.s.}S$. Since $\sum_{i\ge 1}X_i(\omega)$ converges (for almost all $\omega$), it suffices to show that $\sum_{i\ge 1}|Z_i-Z_{i+1}|<\infty$ a.s., where $Z_i:=1\{|Y_i|\le i\}$. But $$ \mathsf{E}\sum_{i\ge 1}|Z_i-Z_{i+1}|\le 2\sum_{i\ge 1}\mathsf{P}(|Y_1|>i)< \infty $$ because $\mathsf{E}|Y_1|<\infty$ so that $\sum_{i}|Z_i-Z_{i+1}|<\infty$ a.s.

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