Possible for $X_{n} \to - \infty$ when $E[\xi_{i}]=0$ and $X_{n}=\sum\limits_{i=1}^{n}\xi_{i}$

Question

I am attempting to construct an example where: $X_{n} \to - \infty-$a.s. and $E[\xi_{i}]=0$ and $X_{n}=\sum\limits_{i=1}^{n}\xi_{i}$

My idea: we would need a process that has greater weighting to the negative side, e.g.

$1-\frac{1}{i}=P(\xi_{i}=-\frac{1}{i})=1-P(\xi_{i}=\frac{1}{i})\Rightarrow P(\xi_{i}=1-\frac{1}{i})=\frac{1}{i}$

This is the example I had in mind, however, I am not sure this assures that $X_{n} \to \infty$ a.s. since I always have positive weighting (albeit with dwindling probability)

Answer 1

Consider the random variable $\zeta_i$ with $P(\zeta_i=2^i)=2^{-i}$ and $P\left(\zeta_i=\frac{-2^i}{2^i-1}\right)=\frac{2^i-1}{2^i}$ for $i\geq 1$. Then it can be verified that $E\zeta_i=0$ and furthermore $P(\zeta_i>0\quad \text{i.o})=0$ by the Borel cantelli lemma. Hence eventually $\zeta_i<0$ with probability one so $X_n=\sum_{i=1}^n \zeta_i\to -\infty$ a.s. as $n\to \infty$.