Show the following series convergent to $EX_1$ almost surely.

Question

Suppose random variables $X_{1},\ X_{2},\cdots$ are integrable independent indentical distributed

I want to show that $\sum_{k=1}^{n}\frac{X_{k}}{k} \sim EX_{1} \log n $ a.s.

Answer 1

Just follow the same way as the proof of Kolomogorov's SLLN. Firstly, \begin{align} \sum_{k=1}^\infty&\mathsf{P}(|X_k|> k)=\sum_{k=1}^\infty\mathsf{P}(|X_1|> k)\le \mathsf{E}|X_1|<\infty\\ \text{by Borel-Cantelli Lemma}\quad&\implies \mathsf{P}(\{|X_k|>k\}\; \text{i.o.})=0.\tag{1} \end{align} Secondly, \begin{align} &\mathsf{E}[X_k1_{\{|X_k|\le k\}}]=\mathsf{E}[X_11_{\{|X_1|\le k\}}]\to \mathsf{E}[X_1]\qquad \text{as $k\to\infty$}\\ &\qquad \implies\frac1{\log n}\sum_{k=1}^n\frac{\mathsf{E}[X_k1_{\{|X_k|\le k\}}]}{k}\to \mathsf{E}[X_1]\qquad \text{as $n\to\infty$} \tag{2} \end{align} At last, \begin{align} \sum_{k=2}^\infty&\frac{\mathsf{Var}[X_k1_{\{|X_k|\le k\}}]}{k^2\log^2 k} \le \sum_{k=2}^\infty\frac{\mathsf{E}[X_k^21_{\{|X_k|\le k\}}]}{k^2}<\infty\\ &\implies \sum_{k=2}^\infty\frac{X_k1_{\{|X_k|\le k\}}-\mathsf{E}[X_k1_{\{|X_k|\le k\}}]}{k\log k}\qquad\text{convergence a.s.}\\ \text{by Kronecker Lemma}&\implies \lim_{n\to\infty}\frac1{\log n} \sum_{k=1}^n\frac{X_k1_{\{|X_k|\le k\}}-\mathsf{E}[X_k1_{\{|X_k|\le k\}}]}{k}= 0 \;\text{a.s.}\\ \text{by (2)}\quad&\implies \lim_{n\to\infty}\frac1{\log n} \sum_{k=1}^n\frac{X_k1_{\{|X_k|\le k\}}}{k}=\mathsf{E}[X_1]\qquad\text{a.s.}\tag{3}\\ \text{by (1)}\quad&\implies \lim_{n\to\infty}\frac1{\log n} \sum_{k=1}^n\frac{X_k}{k}=\mathsf{E}[X_1]\qquad\text{a.s.}\tag{4} \end{align} Complement: (3): Using following fact: If $\lim\limits_{n\to\infty}a_n=a$, then $\lim\limits_{n\to\infty}\dfrac{1}{\log n}\sum\limits_{k=1}^n\dfrac{a_k}{k}=a$, (3) could be proved directly or be deduced from Stolz-Cesàro theorem.
(4): \begin{align} \{\omega:(|X_k(\omega)|>k)\;\text{i.o.}\}^c&=\{\omega:(X_k(\omega)\ne X_k(\omega)1_{\{|X_k(\omega)|\le k\}}) \;\text{except a finite number of $k$}\}\\ &\subset\Bigl\{\omega: \lim_{n\to\infty}\Bigl[\frac1{\log n} \sum_{k=1}^n\frac{X_k(\omega)}{k}-\frac1{\log n} \sum_{k=1}^n\frac{X_k(\omega)1_{\{|X_k(\omega)|\le k\}}}{k}\Bigr]=0\Bigr\}. \end{align} $\sum\limits_{k=1}^\infty\frac{\mathsf{E}[X_k^21_{\{|X_k|\le k\}}]}{k^2}<\infty$: \begin{align} &\sum_{k=1}^\infty\frac{\mathsf{E}[X_k^21_{\{|X_k|\le k\}}]}{k^2} =\sum_{k=1}^\infty\frac{\mathsf{E}[X_1^21_{\{|X_1|\le k\}}]}{k^2} \le \sum_{k=1}^\infty\int_{k}^{k+1}\frac4{t^2}\mathsf{E}[|X_1|^21_{|X_1|\le t}] \,dt\\ &\quad=4\int_0^\infty\frac1{t^2}\mathsf{E}[|X_1|^21_{|X_1|\le t}]\,dt \qquad\text{(Exchange the order of integration)}\\ &\quad=4\int_0^\infty y^2\Bigl(\int_y^\infty\frac1{t^2}\,dt\Bigr)dF_{|X_1|}(y) =4\int_0^\infty ydF_{|X_1|}(y)=4\mathsf{E}[|X_1|]<\infty. \end{align}