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Kolmogorov's strong law for square integrable random variables states that if $X_1,X_2,\ldots$ is a sequence of independent random variables with $\sum_n\operatorname{Var}[X_n]/n^2<\infty$ then $n^{-1}\sum_{k=1}^n(X_k-\mathbb{E}[X_k])$ converges to zero with probability one as $n\rightarrow\infty$ (see
martingale proof of Kolmogorov's strong law for square integrable variables). We show that the following version of the strong law for IID random variables follows from this.
Theorem (Kolmogorov) Let $X_1,X_2,\ldots$ be independent and identically distributed random variables with $\mathbb{E}[|X_n|]<\infty$ . Then, $n^{-1}\sum_{k=1}^n(X_k-\mathbb{E}[X_k])\rightarrow 0$ as $n\rightarrow\infty$ , with probability one.
Note that here, the random variables $X_n$ are not necessarily square integrable. Let us set $\tilde X_n=X_n-\mathbb{E}[X_n]$ , so that $\tilde X_n$ are IID random variables with $\mathbb{E}[\tilde X_n]=0$ . Then, set
Using the fact that $X_n$ has the same distribution as $X_1$ gives \begin{equation}\label{eq:1}\begin{split} \sum_n\mathbb{E}[Y_n^2]/n^2 &= \sum_n\mathbb{E}\left[1_{\{|\tilde X_n|<n\}}n^{-2}\tilde X_n^2\right]\\ &= \sum_n\mathbb{E}\left[1_{\{|\tilde X_1|<n\}}n^{-2}\tilde X_1^2\right]\\ &= \mathbb{E}\left[\sum_n1_{\{|\tilde X_1|<n\}}n^{-2}\tilde X_1^2\right]. \end{split}\end{equation}Letting $N$ be the smallest integer greater than $|\tilde X_1|$ , \begin{equation*}\begin{split} \sum_n1_{\{|\tilde X_1|<n\}}n^{-2}&\le\sum_{n=N}^\infty\frac{4}{4n^2-1}=\sum_{n=N}^\infty\left(\frac{2}{2n-1}-\frac{2}{2n+1}\right)\\
&=\frac{2}{2N-1}\le\frac{2}{N}\le\frac{2}{|\tilde X_1|}. \end{split}\end{equation*}So, putting this into equation (![[*]](http://images.planetmath.org:8080/cache/objects/11289/js//usr/share/latex2html/icons/crossref.png "[*]") ), \begin{equation*} \sum_n\operatorname{Var}[Y_n]/n^2\le\sum_n\mathbb{E}[Y_n^2]/n^2\le\mathbb{E}[2|\tilde X_1|]<\infty. \end{equation*}Therefore, $Y_n$ satisfies the required properties to apply the strong law for square integrable random variables, \begin{equation}\label{eq:2} n^{-1}\sum_{k=1}^n(Y_k-\mathbb{E}[Y_k])\rightarrow 0 \end{equation}as $n\rightarrow\infty$ , with probability one. Also, \begin{equation*}
\mathbb{E}[Y_n]=\mathbb{E}[Y_n-\tilde X_n]=-\mathbb{E}[1_{\{|\tilde X_n|\ge n\}}\tilde X_n]=-\mathbb{E}[1_{\{|\tilde X_1|\ge n\}}\tilde X_1] \end{equation*}converges to $0$ as $n\rightarrow\infty$ (by the dominated convergence theorem). So, the $\mathbb{E}[Y_k]$ terms in () vanish in the limit, giving \begin{equation}\label{eq:3} n^{-1}\sum_{k=1}^nY_k\rightarrow 0
\end{equation}as $n\rightarrow\infty$ with probability one.
We finally note that \begin{equation*} \mathbb{E}\left[\sum_n1_{\{\tilde X_n\not=Y_n\}}\right]=\mathbb{E}\left[\sum_n1_{\{|\tilde X_1|\ge n\}}\right]\le\mathbb{E}[|\tilde X_1|]<\infty, \end{equation*}so $\sum_n1_{\{\tilde X_n\not=Y_n\}}<\infty$ , and $\tilde X_n=Y_n$ for large $n$ (with probability one). So, $Y_k$ can be replaced by $\tilde X_k$ in (), giving the result.
- 1
- David Williams, Probability with martingales, Cambridge Mathematical Textbooks, Cambridge University Press, 1991.
- 2
- Olav Kallenberg, Foundations of modern probability, Second edition. Probability and its Applications. Springer-Verlag, 2002.
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