Adeimantus
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This is going to take a while to set up, so I apologize for that. This came up in the course of thinking about the Strong Law of Large Numbers. It's not homework.
Suppose you have a doubly infinite sequence of random variables X_{i,n} that obey the following almost sure convergence relations. For each i = 1,2,3,...,
X_{i,n} \xrightarrow{a.s} a_i \quad \mbox{ as } \quad n\xrightarrow{} \infty.
Further, we have that \sum_{i=1}^\infty a_i = \mu < \infty. Since this series converges, for any \delta > 0, there is some smallest I such that \left| \sum_{i=1}^m a_i - \mu \right | < \delta for all m \geq I. Consider a sequence of deltas decreasing to zero, and the increasing sequence of their corresponding I's.
\delta_1 > \delta_2 > ... \xrightarrow{} 0 \quad \mbox{and} \quad I_1 < I_2 < ...
Consider some particular pair (\delta_j, I_j). Since almost sure convergence is linear,
\sum_{i=1}^{I_j}X_{i,n} \xrightarrow{a.s} \sum_{i=1}^{I_j} a_i \quad \mbox{ as } \quad n\xrightarrow{} \infty
This is the same thing as saying the set
\{ \omega: \left| \sum_{i=1}^{I_j}X_{i,n}(\omega) - \sum_{i=1}^{I_j} a_i \right| > \epsilon \quad i.o. \quad n\xrightarrow{} \infty \}
has probability zero for any choice of \epsilon > 0. From the definition of the deltas and I's, the set
A_j = \{ \omega: \left| \sum_{i=1}^{I_j}X_{i,n}(\omega) - \mu \right| > \epsilon + \delta_j \quad i.o. \quad n\xrightarrow{} \infty \}
also probability zero. My question is, does the sequence of sets A_j have a limit of
A = \{ \omega: \left| \sum_{i=1}^{\infty}X_{i,n}(\omega) - \mu \right| > \epsilon \quad i.o. \quad n\xrightarrow{} \infty \}
Thanks for wading through that!
Suppose you have a doubly infinite sequence of random variables X_{i,n} that obey the following almost sure convergence relations. For each i = 1,2,3,...,
X_{i,n} \xrightarrow{a.s} a_i \quad \mbox{ as } \quad n\xrightarrow{} \infty.
Further, we have that \sum_{i=1}^\infty a_i = \mu < \infty. Since this series converges, for any \delta > 0, there is some smallest I such that \left| \sum_{i=1}^m a_i - \mu \right | < \delta for all m \geq I. Consider a sequence of deltas decreasing to zero, and the increasing sequence of their corresponding I's.
\delta_1 > \delta_2 > ... \xrightarrow{} 0 \quad \mbox{and} \quad I_1 < I_2 < ...
Consider some particular pair (\delta_j, I_j). Since almost sure convergence is linear,
\sum_{i=1}^{I_j}X_{i,n} \xrightarrow{a.s} \sum_{i=1}^{I_j} a_i \quad \mbox{ as } \quad n\xrightarrow{} \infty
This is the same thing as saying the set
\{ \omega: \left| \sum_{i=1}^{I_j}X_{i,n}(\omega) - \sum_{i=1}^{I_j} a_i \right| > \epsilon \quad i.o. \quad n\xrightarrow{} \infty \}
has probability zero for any choice of \epsilon > 0. From the definition of the deltas and I's, the set
A_j = \{ \omega: \left| \sum_{i=1}^{I_j}X_{i,n}(\omega) - \mu \right| > \epsilon + \delta_j \quad i.o. \quad n\xrightarrow{} \infty \}
also probability zero. My question is, does the sequence of sets A_j have a limit of
A = \{ \omega: \left| \sum_{i=1}^{\infty}X_{i,n}(\omega) - \mu \right| > \epsilon \quad i.o. \quad n\xrightarrow{} \infty \}
Thanks for wading through that!