> [!theoremb] Theorem > > Let $\seq{X_n} \subset L^1$ be a family of [[Probabilistic Independence|independent]] identically [[Probability Distribution|distributed]] [[Random Variable|random variables]]. Let $S_n = \sum_{i = 1}^nX_i$, then > $ > \limv{n}\frac{S_n}{n} = \ev(X_1) \quad \text{a.e.} > $ > > ### Non-Negative Case > > Assume that $\seq{X_n} \subset L^+ \cap L^1$, then $S_n$ is non-decreasing [[Almost Everywhere|a.s.]]. For each $\omega$, this is a non-decreasing sequence, so by the result on the [[Deviation on Lacunary Subsequences]], > $ > \bracs{\omega: \limsup\abs{\frac{S_n(\omega)}{n\ev(E_1)} - 1} > 3 \varepsilon} > $ > is contained in > $ > \bracs{\omega: \limsup\abs{\frac{S_{n_k}(\omega)}{n_k\ev(X_1)} - 1} > \varepsilon} > $ > for some lacunary sequence $\seq{n_k}$. By the [[Lacunary Strong Law of Large Numbers]], the above set has zero probability. Therefore the original set also has zero probability. Since the above holds for all $\varepsilon > 0$, $S_n/n \to \ev(X_1)$ almost surely. > > ### Arbitrary $L^1$ > > Let $S_n^+ = \sum_{i = 1}^nX_i^+$ and $S_n^- = \sum_{i = 1}^n X_i^-$, then by the above result, > $ > \frac{S_n^+}{n} \to \ev(X_1^+) \quad \frac{S_n^-}{n} \to \ev(X_1^-) > $ > almost surely. Therefore > $ > \frac{S_n}{n} = \frac{S_n^+ - S_n^-}{n} \to \ev(X_1^+) - \ev(X_1^-) = \ev(X_1) > $ > [!theorem] > > If $\seq{X_n} \subset L^2$ is independent, identically distributed with $\ev(X_n) = 0$ for all $n$, and $\seq{b_n} \subset \real^+$ is increasing with $\sum_{n \in \nat}\frac{1}{b_n^2} < \infty$, then $S_n/b_n \to 0$ almost surely as well. > > *Idea.* By the given assumptions, > $ > \sum_{n \in \nat}\ev\braks{\frac{X_n}{b_n}}^2 = \ev(X_n^2)\sum_{n \in \nat}\frac{1}{b_n^2} < \infty > $ > which according to Kolmogorov's Maximal Inequality, means that $\sum_{n = 1}^NX_n/b_n$ converges almost surely. By Kronecker's lemma, this is converges to zero.