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Suppose $\{X_1,X_2,.....\}$ sequence of independent and identical random variable.

Let $\mathbb{E}(X_1^{+})<\infty$, i.e. expectation of positive part of the random variable $X_1$ is finite. Instead of saying $\mathbb{E}(X_1)<\infty$

From here, can I conclude that $$ \frac{1}{n}\sum_{i=1}^{n}X_i \xrightarrow{a.s.} \mathbb{E}(X_1) $$ (a.s. = almost surely)

Thanks in advance

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I suppose you can! If $E|X_i| <\infty$, then the regular SLLN is in force. If not, then $X_i=X_i^{+}-X_i^{-}$, and $E[X_1^{-}] = \infty$, with $EX_1^+ < \infty$. Let $X_{i,M}^{-} = X_i^{-}1_{\{X_i^{-}\le M\}} \le X_1^{-}$. Then $EX_{i,M}^{-} \to \infty$, as $M\to \infty$. Hence

$\frac{1}{n}\sum_{i=1}^n X_i^{+}-X_i^{-} \le \frac{1}{n}\sum_{i=1}^n X_i^{+}-X_{i,M}^{-} \stackrel{a.s.}{\to} EX_1^+ - EX_{i,M}^{-} $.

Since $EX_1^+ < \infty$, the right hand side converges down to $-\infty$ as $M\to \infty$, so $\frac{1}{n}\sum_{i=1}^n X_i \to -\infty$, a.s.

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  • $\begingroup$ Good point @AaronMontgomery! I guess I was thinking $EX_1$ was supposed to be finite in the statement. I adjusted the answer accordingly. $\endgroup$ Jun 17, 2020 at 14:09

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