> [!quote] Remark > > While the convergence [[Convergence in Measure|in probability]] from the [[Weak Law of Large Numbers|weak law]] guarantees the existence of an [[Almost Everywhere|a.s.]] convergent subsequence, the following theorem guarantees a.s. convergence for an entire class of sequences. > [!definition] > > Let $\seq{n_k}$ be a [[Sequence|sequence]] of natural numbers. The sequence is **lacunary** if there exists $c > 1$ such that $n_{k + 1} > c_{k}$ eventually. > [!theorem] > > Let $\seq{X_n} \subset L^1$ be a family of [[Probabilistic Independence|independent]] identically [[Probability Distribution|distributed]] [[Random Variable|random variables]] and $S_n = \sum_{k = 1}^nX_k$. Then for any *lacunary* sequence of natural numbers > $ > \frac{S_{n_k}}{n_k} \to \ev(X_1) > $ > [[Almost Everywhere|a.e.]] > > *Proof*. Let $\seq{n_k}$ be a lacunary sequence, $c > 1$ such that $n_{k + 1} \ge c_{n_k}$ eventually (after some $k_0$). > > ### Notation > > Let $X_n^{\le N} = \one_{\bracs{\abs{X_n} \le N}}$ and $X_n^{>N} = \one_{\bracs{\abs{X_n} > N}}$. Denote > $ > S_n^{\le N} = \sum_{k = 1}^nX_n^{\le N} \quad S_n^{> N} = \sum_{k = 1}^nX_n^{> N} > $ > and $\ol{S}_n^{\le N} = S_n^{\le N}/n$, $\ol{S}_n^{>N} = S_n^{>N}/n$. > > ### The Body > > For any $\varepsilon > 0$, $N > 0$ and $n \ge 1$, the truncation allows the [[Variance|variance]] to exists, independence allows adding it up: > $ > \begin{align*} > \bp\bracs{\abs{\ol{S}_n^{\le N} - \ev\paren{\ol{S}_n^{\le N}}} > \varepsilon/2} &\le\frac{\var{X_1^{\le N}}}{(\varepsilon/2)^2n} \\ > &\le\frac{\ev\braks{(X_1^{\le N})^2}}{(\varepsilon/2)^2n} \\ > &\le \frac{4N^2}{\varepsilon^2n} > \end{align*} > $ > Plugging in the lacunary sequence places a finite bound on the sum of all the probabilities, which allows invoking the [[Borel-Cantelli Lemmas|First Borel-Cantelli Lemma]]: > $ > \begin{align*} > &\sum_{k \ge 1}\bp\bracs{\abs{\ol{S}_{n_k}^{\le N} - \ev\paren{\ol{S}_{n_k}^{\le N}}} > \varepsilon/2} \\ > &\le k_0 + \sum_{k > k_0}\bp\bracs{\abs{\ol{S}_{n_k}^{\le N} - \ev\paren{\ol{S}_{n_k}^{\le N}}} > \varepsilon/2} \\ > &\le k_0 + \sum_{k > k_0}\frac{4N^2}{(\varepsilon/2)^2n_k} \\ > &\le k_0 + \frac{4N^2}{\varepsilon/2 n_{k_0}} \sum_{k > k_0}\frac{1}{c^{k - k_0}} \\ > &< \infty > \end{align*} > $ > therefore > $ > \bp\bracs{\limsup_{k \to \infty}\bracs{\abs{S_{n_k}^{\le N} - \ev\paren{\ol{S}_{n_k}^{\le N}}} > \varepsilon/2}} = 0 > $ > and the body converges to $0$ a.e. for all $N \in \nat$. > > > ### Index Shift > > Instead of working with a fixed bound, let > $ > J(\omega) = \min\bracs{k \ge k_0: n_k \ge \abs{X_1}} > $ > then for fixed $\omega$, $X_{n_k}^{\le n_k}(\omega) = 0$ for all $k \in [k_0, J(\omega))$ from the definition of this parameter. Replace by the lacunary sequence, then we can trim off the head as irrelevant: > $ > \begin{align*} > &\sum_{k \ge 1}\bp\bracs{\abs{\ol{S}_{n_k}^{\le n_k} - \ev\paren{\ol{S}_{n_k}^{\le n_k}}} > \varepsilon/2} \\ > &\le k_0 + \sum_{k > k_0}\bp\bracs{\abs{\ol{S}_{n_k}^{\le n_k} - \ev\paren{\ol{S}_{n_k}^{\le n_k}}} > \varepsilon/2} \\ > &\le k_0 + \sum_{k > J}\frac{\ev\braks{(X_1^{\le n_k})^2}}{(\varepsilon/2)^2n_k} \\ > &\le k_0 + \ev\braks{\sum_{k > J}\frac{\abs{X_1^{\le n_k}} \cdot \abs{X_1^{\le n_k}}}{(\varepsilon/2)^2n_k}}\\ > &\le k_0 + \ev\braks{\frac{\abs{X_1^{\le n_k}}}{(\varepsilon/2)^2} \cdot \sum_{k > J}\frac{\abs{X_1^{\le n_k}}}{n_k}}\\ > &\le k_0 + \ev\braks{\frac{\abs{X_1^{\le n_k}}}{(\varepsilon/2)^2} \cdot \sum_{k > J}\frac{1}{c^{k - J(\omega)}}}\\ > &< \infty > \end{align*} > $ > > ### The Tail > > On the other end of things > $ > \abs{\ol{S}_{n_k}^{>n_k} - \ev\paren{\ol{S}_{n_k}^{>n_k}}} \le \frac{1}{n}\abs{X_{n_k}^{>n_k} - \ev\paren{E_{n_k}^{>n_k}}} > $ > If the above sum is greater than $\varepsilon/2$, then there exists $i \in [n_k]$ such that $\abs{X_{i}^{>i} - \ev\paren{X_{i}^{>i}}} > \varepsilon/2$. Therefore > $ > \begin{align*} > &\bp\bracs{\abs{\ol{S}_{n_k}^{>n_k} - \ev\paren{\ol{S}_{n_k}^{>n_k}}} > \varepsilon/2} \\ > &\le \bp\bracs{\exists i \in [n_k]: \abs{X_{i}^{>i} - \ev\paren{X_{i}^{>i}}} > \varepsilon/2} \\ > &\le n_k\bp\bracs{\abs{X_{1}^{>n_k} - \ev\paren{X_{1}^{>n_k}}} > \varepsilon/2} > \end{align*} > $ > where the last inequality comes from independence. > > Let $k_1$ such that $\ev\abs{X_1^{>n_k}} < \varepsilon/2$, and let $k_2 = \max(k, k_1)$. For any $k \ge k_2$, if $\abs{X_1^{>n_k} - \ev(X_1^{>n_k})} > \varepsilon/2$, then $X_{1}^{>n_k} \ne 0$ and $\abs{X_1} > n_k$. This simplifies the bound to > $ > \bp\bracs{\abs{\ol{S}_{n_k}^{>n_k} - \ev\paren{\ol{S}_{n_k}^{>n_k}}} > \varepsilon/2} \le n_k\bp\bracs{\abs{X_1} > n_k} > $ > From here, since $\abs{X_1} > n_k$ if and only if $k < J$, > $ > \begin{align*} > \sum_{k > k_2}n_k\bp\bracs{\abs{X_1} > n_k} &= \ev\braks{\sum_{k = k_1}^{J - 1}n_k \one_{\bracs{X_1 > n_k}}} \\ > &= \ev\braks{\sum_{k = k_1}^{J - 1}n_k} \\ > &\le \ev\braks{\sum_{k = k_1}^{J - 1}\frac{X_1}{c^{-(J-1-k)}}}\\ > &< \infty > \end{align*} > $ > By the First Borel-Cantelli lemma again, > $ > \bp\bracs{\abs{\ol{S}_{n_k}^{>n_k} - \ev\paren{\ol{S}_{n_k}^{>n_k}}} > \varepsilon/2 \ \text{i.o.}} = 0 > $ > Combining the two results, we get the desired a.e. convergence.