# A What is the limit of this (complicated) set?

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

$$A = \{ \omega: \left| \sum_{i=1}^{\infty}X_{i,n}(\omega) - \mu \right| > \epsilon \quad i.o. \quad n\xrightarrow{} \infty \}$$

Related Set Theory, Logic, Probability, Statistics News on Phys.org

#### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

#### andrewkirk

Homework Helper
Gold Member
I waded some of the way, then got stuck here:
$$\left\{ \omega: \left| \sum_{i=1}^{I_j}X_{i,n}(\omega) - \sum_{i=1}^{I_j} a_i \right| > \epsilon \quad i.o. \quad n\xrightarrow{} \infty \right\}$$
I have not come across the initials $i.o.$ before. What do they mean?

I feel that perhaps the set is
$$\left\{ \omega: \forall M\in\mathbb N\ \exists n\ge M\ :\ \left| \sum_{i=1}^{I_j}X_{i,n}(\omega) - \sum_{i=1}^{I_j} a_i \right| > \epsilon \right\}$$
in which case the statement is that the set of $\omega$ for which the $n$-indexed sequence of sums of the first $I_j$ RVs does not converge to the sum of the first $I_j$ $a_i$s, has probability measure zero.

Is that what you meant?

"What is the limit of this (complicated) set?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving