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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

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    $\begingroup$ 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. $\endgroup$ Commented Sep 4 at 18:40
  • $\begingroup$ Durrett's "Probability: Theory and examples", 1991 Theorem 6.3 has a terse proof of this. $\endgroup$
    – copper.hat
    Commented Sep 5 at 17:37

1 Answer 1

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$\sum \frac {E(S_n^{2})} {{n^{2}}}=\sum \frac {n Var X_1} {n^{2}}=\infty$ 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 $p=4$].

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  • $\begingroup$ Thanks, have a nice day :-) $\endgroup$ Commented Sep 5 at 8:21

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