I am asked to construct a sequence of independent random variables $X_n$ with $E(X_n)=0$ for any $n$, but the sample mean of the $X_n$’s diverges almost surely to minus infinity.
I understand that the only condition not mentioned here to fall into Strong Law of Large Numbers is the « identically distributed » condition. So I know I need to build a sequence of independent but not identically distributed random variables.
But I get stuck…. Would appreciate any leads
Suppose $X_n$ is supported on $\{-n, a_n\}$ and that $\sum_{n=1}^\infty P(X_n = a_n) < \infty$. By the Borel-Cantelli lemma, we immediately have $$P\left(X_n = a_n \quad \text{infinitely often}\right) = 0,$$ which in particular means $$P(X_n=-n \quad \text{ for all but finitely many } n) = 1,$$ so that $P(\lim_n X_n = -\infty) = 1$. Finally, by the elementary real analysis of sequences, if $X_n \to a$, then $\frac1n(X_1+X_2+\ldots+X_n)\to a$, so the sample mean tends to $-\infty$ almost surely.
The above skipped over the details of whether it's possible to define $X_n$ in this manner, so I'll briefly deal with how to construct it.
First, choose some positive sequence $p_n := P(X_n=a_n)$ such that $\sum_n p_n < \infty$. This can be done arbitrarily, but for this example, I'll choose $p_n = 2^{-n}$. Since $P(X_n = -n) + P(X_n=a_n) = 1,$ we must have $P(X_n = -n) = 1-p_n = 1-2^{-n}.$
Then we'll solve for $a_n$: Since $E[X_n] = 0$, we have $$(-n)(1-p_n) + a_np_n = 0 \\ a_n = \frac{n(1-p_n)}{p_n} = n2^n(1-2^{-n})$$ and that's all. Specifically, if we let $X_n$ satisfy $$P(X_n = -n) = 1-2^{-n}, \qquad P(X_n = n2^n(1-2^{-n})) = 2^{-n},$$ then we may conclude, by the above argument, that the sample mean tends to $-\infty$ almost surely.
Note that we didn't need to use the independence of $X_n$ anywhere, as the result holds no matter the dependence/independence of the sequence $X_n$.