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Weak Law of Large Numbers:
let $X_1, X_2, ...$ be a sequence of independent random variables with the same probability distribution, expected value $E(X_k) = m < \infty$ and $Var(X_k) = \sigma_k < \infty$. Then $\forall_{\epsilon > 0}$: $$\lim_{n \to \infty} P \big( \big|\frac{X_1 + ... + X_n}{n} - m\big| > \epsilon \big) = 0. \tag{1}$$ Strong Law of Large Numbers:
let $X_1, X_2, ...$ be a sequence of independent random variables with the same probability distribution, expected value $E(X_k) = m < \infty$. Then: $$P \big(\lim_{n \to \infty} \frac{X_1 + ... + X_n}{n} = m\big) = 1. \tag{2}$$


My problem

Let $X_n$ be a sequence of independent random variables. We know that:

  1. $P(X_n = \pm n) = 2^{-n},$
  2. $P(X_n = 0) = 1 - 2^{-n + 1}.$

Our task is check whether the sequence $X_n$ satisfies $(1)$ or $(2)$.

I started to calculate the expected value of $E(X_n):$ $$E(X_n) = n2^{-n} + (-n)2^{-n} + 0(1 - 2^{-n + 1}) = 0.$$ Unfortunately I don't really know where to go from here. I would appreciate any hints or tips.

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  • $\begingroup$ What are you trying to show? $\endgroup$
    – Titus
    Commented Jun 2, 2018 at 18:25
  • $\begingroup$ @Titus I'm sorry. It was written only in the topic. Now I fixed that issue. $\endgroup$
    – Hendrra
    Commented Jun 2, 2018 at 18:38

1 Answer 1

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The series $\sum_{n\geqslant 1}\Pr\left\{X_n\neq 0\right\}$ is convergent hence by the Borel-Cantelli lemma, there exists $\Omega'\subset \Omega$ of probability one such that for all $\omega\in \Omega'$, there is an integer $N\left(\omega\right)$ such that for all $n\geqslant N\left(\omega\right)$, $X_n\left(\omega\right)=0$.

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  • $\begingroup$ It is because from the Borel-Cantelli lemma $\limsup \text{Pr} \{ X_n \neq 0 \} = 0$? How can I apply this result to the Laws of Large Numbers? $\endgroup$
    – Hendrra
    Commented Jun 2, 2018 at 21:39
  • $\begingroup$ Yes for your first question. For the second only the first $N(\omega)$ terms may be non-zero hence $S_n(\omega)/n:=\sum_{j=1}^nX_j(\omega)/n=\sum_{j=1}^{N(\omega)}X_j(\omega)/n$. $\endgroup$ Commented Jun 2, 2018 at 21:45
  • $\begingroup$ I did some research on Borel-Cantelli lemma. I did the proofs and tried to understand it as best as I can. However I still don't see how can I get from the conclusion of the lemma to the laws of large numbers. Can I ask for a deeper answer? Thank you! $\endgroup$
    – Hendrra
    Commented Jun 4, 2018 at 18:20
  • $\begingroup$ Once you have $S_n(\omega)/n:=\sum_{j=1}^nX_j(\omega)/n=\sum_{j=1}^{N(\omega)}X_j(\omega)/n$, we can derive that $S_n(\omega)/n\to 0$. $\endgroup$ Commented Jun 13, 2018 at 13:51

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