# Strong Law of Large Numbers for increasing index sets

Let $$(X_n)_{n \in \mathbb{N}}$$ be a sequence of integrable i.i.d random variables. Let $$I_n \subset \mathbb{N}$$ satisfy $$\lim_{n \to \infty} |I_n| = \infty$$, where $$|I_n|$$ is the cardinality of $$I_n$$. Is it true without further hypotheses that $$\lim_{n \to \infty} \frac{1}{|I_n|} \sum_{k \in I_n} X_k = \mathbb{E}[X_1] \quad \text{a.s.}$$

You can assume that the $$I_n$$ are disjoint and the cardinalities are strictly increasing for simplicity.

• Some thoughts: set $Y_n = \frac{1}{|I_n|} \sum_{k\in I_n} X_k$. Then assuming that the $I_n$ are disjoint, it is clear that the $Y_n$ are independent. Thus, by the Borel-Cantelli lemma, $Y_n$ converges to $\mu = \mathbb{E}[X_1]$ a.s. if and only if $\sum_{n\geq 0} \mathbb{P}(|Y_n-\mu|\geq \epsilon) < \infty$ for every $\epsilon >0$. Letting $k_n = |I_n|$, notice that $Y_n$ is distributed as $\bar{X}_{k_n}$. Thus, the question is pretty much equivalent to showing $$\sum_{n\geq 0} \mathbb{P}\left(\left| \bar{X}_{n} - \mu\right|\geq \epsilon\right) < \infty.$$ Nov 29, 2023 at 22:48
• @Michh That's a good idea. This actually gives an easy counterexample (see my answer) but I would still be interested in positive results. Dec 1, 2023 at 13:36

Using @Michh's suggestion, we can provide an easy counterexample. Fix $$\varepsilon > 0$$ and let $$(X_n)_{n \in \mathbb{N}}$$ be an arbitrary i.i.d. sequence. Define

$$\begin{gather} S_n = \sum_{k = 1}^n X_k \\ b_n = \mathbb{P}\left(\left|\frac{S_n}{n} - \mathbb{E}[X_1]\right| \ge \varepsilon \right) \end{gather}$$

Take a sequence $$(m_k)_{k \in \mathbb{N}}$$ of natural numbers such that $$m_k \cdot b_k > 1$$. Set $$M_k = \sum_{j = 1}^k m_j$$. Define $$I_n$$ inductively as follows: Choose $$I_1, \dots, I_{m_1}$$ to be disjoint sets with cardinality $$1$$ each. Then, having defined $$I_1, \dots, I_{M_k}$$, choose $$I_{M_k + 1}, \dots, I_{M_{k + 1}}$$ to be sets with cardinality $$k + 1$$ each such that the family $$\{I_n\}_{n = 1}^{M_{k + 1}}$$ is pairwise disjoint. Then, for each $$k$$ there are exactly $$m_k$$ index sets with cardinality $$k$$. Clearly, $$\lim_{n \to \infty} |I_n| = \infty$$ as well.

Set $$Y_n = \frac{1}{|I_n|}\sum_{k \in I_n} X_k$$. Finally, we get

\begin{align} \sum_{n = 1}^\infty \mathbb{P}(|Y_n - \mathbb{E}[X_1]| \ge \varepsilon) &= \sum_{n = 1}^\infty \mathbb{P}\left(\left|\frac{S_{|I_n|}}{|I_n|} - \mathbb{E}[X_1]\right| \ge \varepsilon \right) \\ &= \sum_{n = 1}^\infty m_n \cdot \mathbb{P}\left(\left|\frac{S_n}{n} - \mathbb{E}[X_1]\right| \ge \varepsilon \right) = \sum_{n = 1}^\infty m_n \cdot b_n. \end{align} But the last sum clearly diverges.

There is also an elementary positive result. If $$I_n$$ is a set of consecutive integers such that $$\frac{|I_n|}{\max I_n}$$ stays bounded, then SLLN holds for $$(I_n)$$. To see this, simply write

$$Y_n = \frac{1}{|I_n|} \sum_{k = 1}^{\max I_n} X_k - \frac{1}{|I_n|} \sum_{k = 1}^{\min I_n} X_k$$

By an easy argument and SLLN, the right-hand side converges to $$\mathbb{E}[X_1]$$. I would be interested in having more general conditions under which we can apply SLLN.