# Cantelli's Law of Large Numbers

In Probability, Shiryaev Chapter IV, Section 3 there's the following theorem:

Let $$X_1, X_2,...$$ be indipendent random variables with finite fourth moments and let $$\mathbb{E}[|X_n - \mathbb{E}[X_n]|^4] \leq C$$ for $$n \geq 1$$ and $$C$$ some constant. Then, as $$n \rightarrow \infty$$,

$$\frac{S_n - \mathbb{E}[S_n]}{n} \rightarrow 0$$ almost surely (where $$S_n = X_1 + ... + X_n)$$.

The proof is the following:

WLOG, $$\mathbb{E}[X_n] = 0 \,\forall n$$, so thesis is now $$\frac{S_n}{n} \rightarrow 0$$ almost surely. This follows if $$\sum \mathbb{P}(|\frac{S_n}{n}| \geq \varepsilon) < \infty, \,\forall \varepsilon > 0$$.

Until this bit, it's all good. Then it says:

"by Chebyshev's inequality, this will follow from $$\sum \mathbb{E}[|\frac{S_n}{n}|^4] \ < \infty$$", and the proof goes on.

I can't see where that "4" comes from; I mean, Cheb's inequality says $$\mathbb{P}(|\frac{S_n}{n}| \geq \varepsilon) \leq \frac{Var(|\frac{S_n}{n}|)}{\varepsilon^2} = \frac{\mathbb{E}(|\frac{S_n}{n}|^2)}{\varepsilon^2}$$ (assuming $$\mathbb{E}[X_n] = 0$$), so I sould show that $$\sum \mathbb{E}[|\frac{S_n}{n}|^2] \ < \infty$$.

Somebody can tell me where I'm wrong? Thanks

• To many authors, the Chebyshev's or Markov's inequality is also $\mathbb P(|X|>t) \le \frac{\mathbb E(|X|^p)}{t^p}$ for any $p,t > 0$. The 4'th power is important in this proof, it won't work with $2$ without additional work. Commented Sep 4 at 18:40
• Durrett's "Probability: Theory and examples", 1991 Theorem 6.3 has a terse proof of this. Commented Sep 5 at 17:37

$$\sum \frac {E(S_n^{2})} {{n^{2}}}=\sum \frac {n Var X_1} {n^{2}}=\infty$$ so you have to use the $$4-$$th power.
[$$P(|Y| >c)\le \frac {E|Y|^{p}} {c^{p}}$$ for any $$p \in [1,\infty)$$. Take $$p=4$$].