Checking whether a sequence $X_n$ satisfies the laws of large numbers

275 Views Asked by At

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.

1

There are 1 best solutions below

4
On BEST ANSWER

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 $n\geqslant N\left(\omega\right)$, $X_n\left(\omega\right)=0$.