I'm trying to show the following:

Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of i.i.d random variables with $\mathbb{E}[|X_1|]<\infty$ and $\mathbb{E}[X_1]=\mu$. Consider $$S_n:=X_1+\cdots+X_n,\quad n\in \mathbb{N}$$ Show that if $\mu>0$ or $\mu<0$ then $S_n\to\infty$ resp. $S_n\to -\infty$ almost surely

Now I know that I somehow must argue with the law of large numbers. Here is what I thought. Consider $$W_n=nS_n=n(X_1+\cdots+X_n)$$ Then the law of large numbers tells us that $$p(\lim\limits_{n\to\infty}\frac{1}{n}W_n=\mathbb{E}[nX_1])=1$$ $$p(\lim\limits_{n\to\infty}\frac{1}{n}nS_n=\mathbb{E}[nX_1])=1$$ $$p(\lim\limits_{n\to\infty}S_n=n\mathbb{E}[X_1])=1$$ $$p(\lim\limits_{n\to\infty}S_n=n\mathbb{E}[X_1])=1$$

But here I'm stucked since the limit is only on the left side.

  • 1
    $\begingroup$ There should not be $n$ on the right side, it should be just $E(X_1)$. Also, $S_n=X_1+\dots + X_n$, not $n(X_1+\dots + X_n)$. $\endgroup$ – Augustin May 5 '16 at 11:44
  • $\begingroup$ @Augustin Hi, I'm trying to show the statement for $nS_n$ (I forgot the $n$) so that I can somehow remove the factor $\frac{1}{n}$, such that it goes to infinity $\endgroup$ – Matriz May 5 '16 at 12:06

$S_{n}=n\bar{X}_{n}$ where $\bar{X}_{n}:=\frac{1}{n}\left(X_{1}+\cdots+X_{n}\right)$ and $P\left(\bar{X}_{n}\to\mu\right)=1$ according to the strong law of large numbers.

If $\bar{X}_{n}\left(\omega\right)\to\mu>0$ then $S_n(\omega)=n\bar{X}_{n}\left(\omega\right)\rightarrow+\infty$.

So: $$\left\{ \bar{X}_{n}\to\mu\right\} \subseteq\left\{ S_{n}\to+\infty\right\}$$ and consequently $$P(S_n\to+\infty)\geq P(\bar{X}_{n}\to\mu)=1$$

  • $\begingroup$ Great answer! Any guidance on the case where $\mu=0$? $\endgroup$ – mbiron May 20 at 14:42
  • $\begingroup$ @mbiron Thank you. If $\mu=0$ then I am speechless on this ;-). $\endgroup$ – drhab May 20 at 17:29

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