# Doubt on strong law of large number theorem

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)

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