5
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ 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 $$\sum_{n\geq 0} \mathbb{P}\left(\left| \bar{X}_{n} - \mu\right|\geq \epsilon\right) < \infty.$$ $\endgroup$
    – Michh
    Nov 29, 2023 at 22:48
  • $\begingroup$ @Michh That's a good idea. This actually gives an easy counterexample (see my answer) but I would still be interested in positive results. $\endgroup$ Dec 1, 2023 at 13:36

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .