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I've been trying to prove that given a sequence of independent random variables with identical distribution $\{X_n\}_{n \in \mathbb{N}}$ such that $P(X_1 \neq 0)>0$, so also $P(X_i \neq 0) >0 \ \ \forall i$, the series $$\sum_{n \in \mathbb{N}} X_n \ \ \text{is diverent almost surely}.$$

That is, I need to prove that $P(\omega| \ \exists \varepsilon>0 : \forall N \in \mathbb{N} \exists m, n \ge N : |S_m(\omega) - S_n(\omega)| > \varepsilon)$.

We are dealing with a series of random variables, so I thought I could use Borel-Cantelli lemma. It implies that the series $\sum_{n \in \mathbb{N}} P(X_m \neq 0) $ is divergent.

This is because $\forall n \in \mathbb{N} : P(X_n \neq 0) = c >0$, because the variables have identical distributions.

But I don't think it is helpful, because by Markov's inequality we also have for $a \ge 0$:

$$\mathbb P (|X| \ge a) \le \frac{\mathbb E(|X|)}{a}.$$

So it would seem that the series is divergent in $L^1.$ But why is it divergent almost surely?

Could I use the law of large numbers? If so, how? The variables satisfy all necessary conditions for the following to hold:

$$\frac{S_n}{n} = \frac{X_1 + ... + X_n}{n} \to \mathbb{E}X_1 \ \ \text{a.s.}$$

But does that imply that the series is divergent?

Could you help me finish that?

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    $\begingroup$ Your LLN approach is good if you assume $E[X_1] \neq 0$. Then $\frac{1}{n} \sum_{i=1}^n X_i \rightarrow E[X_1]$ almost surely, so for any $\epsilon$ you can (with prob 1) find an $N$ such that $E[X_1] + \epsilon \geq \frac{1}{n}\sum_{i=1}^n X_i \geq E[X_1] - \epsilon$ for all $n \geq N$. $\endgroup$ – Michael May 1 '15 at 11:40
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    $\begingroup$ So if $E[X_1]$ is a real number such that $E[X_1] \neq 0$, you can in fact show that $\sum_{i=1}^n X_i$ diverges to either $\infty$ or $-\infty$ (depending on the sign of $E[X_1]$). If you just want to show the limit of $\sum_{i=1}^n X_i$ does not exist, just consider that the difference between successive terms is $X_{n+1}$, which is not converging to 0. $\endgroup$ – Michael May 1 '15 at 11:49
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    $\begingroup$ @Michael Thank you. So $E[X_1] + \epsilon \geq \frac{1}{n}\sum_{i=1}^n X_i \geq E[X_1] - \epsilon$ means that the arithmetic mean is from a certain moment between $\mathbb{E}X_1 - \varepsilon $ and $\mathbb{E}X_1 + \varepsilon$. But how can I extract the sum itself from that? How can I conclude that the series is divergent? $\endgroup$ – Bruce May 1 '15 at 12:12
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    $\begingroup$ If $E[X_1]=c>0$ then choose $\epsilon =c/2$ to conclude there is an $N>0$ such that $\frac{1}{n}\sum_{i=1}^n X_i \geq c/2$ for all $n \geq N$. So $\sum_{i=1}^n X_i \geq cn/2$ for all $n \geq N$. Take limit as $n\rightarrow\infty$. $\endgroup$ – Michael May 1 '15 at 22:02
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Hints:

  1. Since $\mathbb{P}(X_1 \neq 0)>0$, there exists $\epsilon>0$ such that $\mathbb{P}(|X_1|>\epsilon)>0$.
  2. Conclude from $$\sum_{n \geq 1} \mathbb{P}(|X_n|>\epsilon)= \sum_{n \geq 1} \mathbb{P}(|X_1|>\epsilon) =\infty$$ and the Borel-Cantelli lemma that $|X_n(\omega)|>\epsilon$ happens infinitely often for almost all $\omega \in \Omega$.
  3. Deduce that $$\sum_{n \geq 1} X_n $$ is divergent almost surely.
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