# Weak LLN holds but strong LLN fails

Show there exist independent random variables $$\{X_n\}$$ with $$X_n\in\{-n, n, 0\}$$, $$\mathbb{E}(X_n)=0$$, and $$Y_n=\tfrac{1}{n}\sum_{k=1}^nX_k$$ for all $$n$$, $$\mathbb{P}(|Y_n|\ge\epsilon)\to 0$$ for all $$\epsilon>0$$, and $$\mathbb{P}(Y_n\to 0)<1$$.

$$\mathbb{E}(X_n)=0$$, so $$p_n=\mathbb{P}(X_n=-n)=\mathbb{P}(X_n=n)$$. By Kolmogorov's criterion in http://mathworld.wolfram.com/StrongLawofLargeNumbers.html, if $$S=\sum_n\tfrac{\mathbb{V}(X_n)}{n^2}$$ converges, then $$\mathbb{P}(Y_n\to 0)=1$$, so we definitely need $$S$$ to diverge if we want $$\{X_n\}$$ to work. $$\mathbb{V}(X_n)=\mathbb{E}(X_n^2)=2p_nn^2,$$ so $$S=\sum_n2p_n$$ diverges.

If $$\mathbb{V}(Y_n)\to 0$$, we can use Chebyshev's inequality as in the proof of the weak LLN to get $$\mathbb{P}(|Y_n|\ge\epsilon)\to 0$$ for all $$\epsilon>0$$. If $$\{p_nn^2\}$$ is increasing, $$\mathbb{V}(Y_n)=\dfrac{1}{n^2}\sum_{k=1}^n\mathbb{V}(X_k)\le\dfrac{1}{n^2}\sum_{k=1}^n2p_nn^2=2np_n,$$ so the weak LLN will hold if $$np_n\to 0$$.

For all positive integers $$k$$, $$p_n=\tfrac{1}{n\log^k(n)}$$ satisfies $$np_n\to 0$$ and $$\sum_np_n\to\infty$$, where $$\log^k(n)=\log(...\log(n)...)$$, since $$\sum_np_n$$ grows like $$\log^{k+1}(x)$$, the integral of $$p_x$$. I know how to adapt the proof of the weak LLN to show that $$\{X_n\}$$ satisfies the weak LLN, but I don't know how to show that $$\mathbb{P}(Y_n\to 0)< 1$$. It might not even be true, since I don't know if the converse of Kolmogorov's criterion holds. At any rate, I can't think of a good way to show $$\mathbb{P}(Y_n\to 0)<1$$. You might be able to write a complicated expression for $$\mathbb{P}(Y_n\to 0)$$ and bound it, but that seems hard.

If $$p_n=\frac 1 {n log \, n}$$ for $$n >1$$ then $$\{Y_n\}$$ does not tend to $$0$$ almost surely. This is because $$\frac {X_n} n =\frac {nY_n-(n-1)Y_{n-1}} n$$ so if $$Y_n$$ tends to $$0$$ then $$\frac {X_n} n \to 0$$. By Borel Cantelli Lemma this implies $$\sum P\{|\frac {X_n} n| >\frac 1 2\}<\infty$$ which becomes $$\sum 2p_n <\infty$$ and this is not true.