Let $\{X_n\}_{n=1}^{\infty}$ be a sequence of independent random variables such that $$P(X_n=n+1)=P(X_n=-(n+1))=\frac{1}{2(n+1)\log(n+1)}$$
$$P(X_n=0)=1-\frac{1}{(n+1)\log(n+1)}$$
Prove that $X_n$ satisfies weak law of large numbers but doesn't satisfy strong law of large numbers.
I thought that I could show that $\mathbb EX_n=0$ and then I tried to show that $\frac{ \mathrm{Var}X_n}{n^2} \rightarrow 0$. But I don't think this is the right way, I think I need to use $S_n$ somehowe. How should it be done?
Best Answer
To show convergence in probability, $$ \mathbb E[X_n] = \frac{n+1}{2(n+1)\log(n+1)} - \frac{n+1}{2(n+1)\log(n+1)} = 0$$ and $$ \mathrm{Var}(X_n)=\mathbb E[X_n^2] = \frac{2(n+1)^2}{2(n+1)\log(n+1)} = \frac{n+1}{\log(n+1)}. $$ Hence $\mathbb E[S_n]=0$, and $$\begin{align*} \frac1{\varepsilon^2}\mathbb E\left[\left(\frac{S_n}n\right)^2\right] &= \frac1{n^2\varepsilon^2} \mathbb E[S_n^2]\\ &= \frac1{n^2\varepsilon^2} \mathrm{Var}\left(\sum_{i=1}^n X_i\right)\\ &= \frac1{n^2\varepsilon^2} \sum_{i=1}^n \mathrm{Var}(X_i)\\ &= \frac1{n^2\varepsilon^2} \sum_{i=1}^n \frac{i+1}{\log(i+1)}\\ &\leqslant \frac1{n^2\varepsilon^2}\left(\frac{n(n+1)}{\log(n+1)}\right)\\ &= \frac{n+1}{n\log(n+1)\varepsilon^2}\\ &= \frac1{\log(n+1)\varepsilon^2} + \frac1{n\log(n+1)\varepsilon^2}\stackrel{n\to\infty}{\longrightarrow}0. \end{align*}$$ By Markov's inequality, $$ \mathbb P\left(\frac{S_n}n \geqslant\varepsilon \right)\leqslant \frac{\mathbb E\left[\left(\frac{S_n}n\right)^2 \right]}{\varepsilon^2}\stackrel{n\to\infty}{\longrightarrow}0.$$
To show that the convergence is not almost sure, for each $n$ we have, as pointed out by @Frank $$ \{X_n=n+1\} \subset \left\{|S_n|\geqslant \frac n2\right\} \cup \left\{|S_{n-1}|\geqslant \frac n2\right\}.$$ Since $$\sum_{n=1}^\infty \mathbb P(X_n=n+1) = \sum_{n=1}^\infty\frac1{2(n+1)\log(n+1)}=+\infty,$$ by Borel-Cantelli we have $$\limsup_{n\to\infty} \mathbb P\left(\frac{|S_n|}n\geqslant \frac12\right)=1,$$ and hence $$\mathbb P\left(\lim_{n\to\infty} \frac{S_n}n = 0\right)<1.$$