martingales, almost sure convergence

Question

I am given a sequence of independent random variables $(X_n)$ with respective laws given by $P(X_n=-n^2)=\dfrac{1}{n^2}$ and $P(X_n=\dfrac{n^2}{n^2-1})=1-\dfrac{1}{n^2}$, and letting $S_n=X_1+...+X_n$ I am asked to show $\dfrac{S_n}{n}$ tends to 1 as $n$ tends to $\infty$. Furthermore I am asked to deduce $(S_n)$ is a martingale which converges to infinity.

For the first part, I tried using the strong law of large numbers but I keep getting $0=1$ which is definitely wrong. Could somebody give me a hint? Thanks. Harry

Answer 1

You can use the Borel-Cantelli lemma to deduce the first part, i.e. $$\sum_{n=1}^\infty\mathbb{P}(X_n=-n^2)<\infty.$$

Answer 2

You can not use SLLN because $(X_n)_{n \geq 2}$ in your case is a sequence of independent but not identically distributed random variables. You can review the conditions of SLLN here. According to the hint of Morris, I will complete the proof:

$$ \because \sum_{n=2}^\infty\mathbb{P}(X_n=-n^2)<\infty\,, $$

By first Borel-Cantelli lemma, we arrive that

$$ \mathbb{P}(X_n=-n^2 \quad \text{infinitely often}) = 0\,, $$

thus

$$ X_n = \frac{n^2}{n^2-1} \text{ eventually, a.s} \,, $$

that is

$$ X_n \to 1 \text{ a.s}\,, $$

which implies

$$ \frac{S_n}{n} \to 1 \text{ a.s}. $$

We also arrive that the martingale $(S_n)_{n \geq 2}$ converges to (positive) infinity since $X_n \to 1$ almost surely.