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Let $(\Omega, F, P)$ be probability theory.

Suppose that $X_1, X_2, X_3,...$ be sequence of random variable and $E(X_i)=0$ for all $i\in \mathbb{N}$.

Let $Y_n=\frac{X_1+X_2+... +X_n}{n}$.

claim . $Y_n$ converges to 0 in probability.

Let $A_n=\{w : |Y_n(w)| > \epsilon >0\}$

$$\int _{A_n} |Y_n| dP \geq \int_{A_n} \epsilon dP \geq P(A_n) \epsilon$$

But $|Y_n| \leq |X_1|/n + |X_2|/n +... |X_n|/n$ so $P(A_n)=0$.

is it right?

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  • $\begingroup$ What if $X_1=X_2=\ldots =X_n$? $\endgroup$ – Robert Israel Sep 14 '16 at 17:16
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If you assume the $X_i$ are iid, this is the Weak Law of Large Numbers. Without that assumption (or some slight generalizations of it), the statement is false.

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The result does not hold. Robert has a fine counterexample.

You are stating that

$|Y_n| \leq |X_1|/n + ... + |X_n|/n$

and concluding that therefore, $P(A_n)=0$.

How do you arrive at this conclusion? I suppose you evaluate

$\int_{A_n} |Y_n| \, \mathrm{d}P \leq \frac{1}{n} \sum_{i=1}^n \int_{A_n} |X_i| \, \mathrm{d}P$,

but how do you argue hereafter?

Maybe useful: $EX_i=0$ does not imply $E|X_i|=0$. Actually $E|X_i|=0$ iff $X_i=0$ almost surely, in which case your statement is trivial.

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