# Arithmetic mean converges almost surely

I've been asked to show that, for a sequence of independent random variables $(X_n)$, if

$$\mathbb{P}[\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n} X_k = 1] > 0$$

then $$\frac{1}{n}\sum_{k=1}^{n} X_k \rightarrow 1$$ almost surely.

I'm really not sure how to begin. I thought it might be possible with the Borel Cantelli lemma, but the sequence of sums is not an independent sequence, which is required in the statement. Any help would be appreciated.

• There's nothing to show here! This is just the definition of "almost surely". – David C. Ullrich Nov 21 '17 at 20:01
• Not quite,the first line is saying that the probability is greater than zero, not that it's one. I need to show that the fact that its greater than zero implies that it is one, and therefore that its almost sure convergence – Fahrenheit997 Nov 21 '17 at 20:06
• Show that the event in the first display is in the tail $\sigma$-field for the sequence $\{X_n\}_{n \geq 1}$, and then use the Kolmogorov 0-1 law. – gogurt Nov 21 '17 at 20:12
• You only have a few "zero-one laws" in your quiver, and that definitely reeks of one of them. Kolmogorov's is the most relevant -- try it. – Clement C. Nov 21 '17 at 20:13
• @Fahrenheit997 Not really. See e.g. this remark from the link above: "For example, the event that the sequence converges, and the event that its sum converges are both tail events. In an infinite sequence of coin-tosses, a sequence of 100 consecutive heads occurring infinitely many times is a tail event." You basically are looking at Cesaro convergence of the sequence. – Clement C. Nov 21 '17 at 20:17